I. A. Shelykh and V. K. Kozin
Introduction to Wave and
Quantum Mechanics
August 18, 2022
Contents
1
Particles and fields in classical physics . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1 Classical mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Classical fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Plane waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4 Superpositions of plane waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5 Interference and diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6 Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1
6
8
12
19
22
23
2
Particle-wave dualism and wavefunction of quantum particles . . . . . . .
2.1 Photoelectric effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Compton scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Fabry-Perot resonator and Heisenberg uncertainty relation for
photons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4 Signatures of a wave behavior of matter . . . . . . . . . . . . . . . . . . . . . . . .
2.5 Dispersion of the de Broglie waves . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6 Quantization of the energy of a quantum particle in a rigid box . . . .
2.7 Bohr-Sommerfeld quantization rule . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.8 When does a gas becomes quantum? . . . . . . . . . . . . . . . . . . . . . . . . . .
2.9 Wave function of a quantum particle and the Schrodinger equation .
2.10 Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25
25
27
3
Physical observables in quantum mechanics . . . . . . . . . . . . . . . . . . . . . . .
3.1 Operators of physical observables . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Stationary Schrodinger equation, eigenvalues and eigenfunctions
of the operators of physical observables . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Supplementary material and tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.1 Differential operators in different coordinate systems . . . . . .
30
31
32
33
34
36
37
41
43
43
49
53
53
v
vi
Contents
4
Free particles and quantum tunneling . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1 States of a propagating quantum particle . . . . . . . . . . . . . . . . . . . . . . .
4.2 Scattering of 1D quantum particles, tunneling effect . . . . . . . . . . . . . .
4.3 Supplementary material and tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3.1 Evolution of a wave function from the initial condition . . . . .
4.3.2 Quantum tunneling through a rectangular potential . . . . . . . .
4.3.3 Quantum tunneling through a 𝛿-barrier . . . . . . . . . . . . . . . . . .
4.3.4 Task: reflection from a two dimensional step . . . . . . . . . . . . .
57
57
63
73
73
74
75
75
5
Bound states of quantum 1D and 2D particles . . . . . . . . . . . . . . . . . . . . . 77
5.1 Bound states in 1D square potential well . . . . . . . . . . . . . . . . . . . . . . . 77
5.2 Quantum harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
5.3 Quantum particle in 1D periodic potential, Kronig-Penny model . . . 95
5.4 Quantum particle in a magnetic field and Landau quantization . . . . . 101
5.5 Supplementary material and tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
5.5.1 Infinite Square Well . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
5.5.2 Dirac-Kronig-Penney model. Bound states in a 𝛿-potential . . 110
5.5.3 Classical and Semi-classical Electrodynamics . . . . . . . . . . . . 112
5.5.4 Quantum harmonic oscillator in an electric field . . . . . . . . . . 113
5.5.5 2D quantum harmonic oscillator in a magnetic field . . . . . . . 113
6
Quantum particle in a central potential . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
6.1 Stationary Schrodinger equation in a central potential, separation
of variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
6.2 Operators of angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
6.3 Spherical harmonics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
6.4 Hydrogen atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
6.5 Hydrogen atom in the external magnetic field and Zeeman effect . . . 136
6.6 Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
6.6.1 Bound states in the 3D two-body problem . . . . . . . . . . . . . . . . 142
6.6.2 Spherical quantum well in 3D . . . . . . . . . . . . . . . . . . . . . . . . . . 143
7
Spin of electron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
7.1 Mathematical description of spin of electron . . . . . . . . . . . . . . . . . . . . 145
7.2 Spin precession in the external magnetic field and magnetic
resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
7.3 Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
Chapter 1
Particles and fields in classical physics
Abstract In this chapter we describe some basic aspects from classical mechanics
of particles and fields.
1.1 Classical mechanics
In classical physics there are two types of the objects: particles and fields. From the
classical point of view, an elementary particle is a material point, which has a mass,
but does not really have any finite spatial dimension. The position of a particle in
three dimensional space is thus characterized by a set of 3 coordinates (𝑞 1 ,𝑞 2 ,𝑞 3 ).
Their particular choice is dictated mostly by the symmetry of the problem. The
simplest case is Cartesian coordinates, which are constructed as projections of a
radius vector of the particle r on three orthogonal axes 𝑥, 𝑦, 𝑧
r = e 𝑥 𝑥 + e 𝑦 𝑦 + e𝑧 𝑧 =
3
Õ
(1.1)
e𝑗𝑥 𝑗.
𝑗=1
Here e 𝑥 = e1 , e 𝑦 = e2 , e𝑧 = e3 are unit vectors in the directions of x, y, z. They obey
the following condition:
(
0, 𝑖 ≠ 𝑗
e𝑖 · e 𝑗 = 𝛿𝑖 𝑗 =
,
(1.2)
1, 𝑖 = 𝑗
where e𝑖 · e 𝑗 - scalar product, 𝛿𝑖 𝑗 - Kronecker delta. If a particle moves, its position
changes with time: r = r (𝑡). We can then introduce the vector of the velocity:
3
V=
𝑑r ·
𝑑𝑥
𝑑𝑦
𝑑𝑧 Õ
≡ r = e𝑥
+ e𝑦
+ e𝑧
=
e 𝑗𝑉𝑗 ;
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑗=1
(1.3)
1
2
1 Particles and fields in classical physics
Fig. 1.1 Cartesian coordinates
𝑑𝑥 𝑗
.
(1.4)
𝑑𝑡
Knowing the velocity, we can introduce the vector of the kinetic momentum:
𝑉𝑗 =
(1.5)
p = 𝑚V.
As well, we can introduce the vector of the acceleration:
3
W=
𝑑V 𝑑 2 r ·· Õ
= 2 ≡r=
e𝑗𝑤 𝑗;
𝑑𝑡
𝑑𝑡
𝑗=1
(1.6)
𝑑𝑉 𝑗
𝑑2𝑥 𝑗
.
(1.7)
=
𝑑𝑡
𝑑𝑡 2
Three vectors r, V, W in classical mechanics are not independent, but connected by
the 2𝑛𝑑 law of Newton:
𝑤𝑗 =
𝑑p
= F (r, V, 𝑡) ,
(1.8)
𝑑𝑡
where the vector F describes the force acting on the particle, which can be a function
of particles position, velocity and time, but never depends on acceleration or any
higher order time derivatives of r. From the mathematical point of view, 2𝑛𝑑 law of
Newton represents a system of three coupled second order differential equations, for
3 components of the radius vector Fig. 1.1
𝑑𝑥 𝑑𝑦 𝑑𝑧
𝑑2𝑥
𝑚 2 = 𝐹𝑥 𝑥, 𝑦, 𝑧, , , , 𝑡 ,
𝑑𝑡 𝑑𝑡 𝑑𝑡
𝑑𝑡
𝑚W =
1.1 Classical mechanics
3
𝑑𝑥 𝑑𝑦 𝑑𝑧
𝑑2 𝑦
=
𝐹
𝑥,
𝑦,
𝑧,
,
,
,
𝑡
,
𝑦
𝑑𝑡 𝑑𝑡 𝑑𝑡
𝑑𝑡 2
𝑑2 𝑧
𝑑𝑥 𝑑𝑦 𝑑𝑧
𝑚 2 = 𝐹𝑧 𝑥, 𝑦, 𝑧, , , , 𝑡 .
𝑑𝑡 𝑑𝑡 𝑑𝑡
𝑑𝑡
𝑚
(1.9)
According to the mathematical theorem, the general solution of this set of equations depends on 6 constants. To specify them, one needs to know 6 initial conditions,
namely initial values of 3 coordinates and 3 velocities. From all type of forces, one
can pick a particular important example: potential forces, for which the vector of
the force can be expressed as a gradient of the scalar function 𝑈 (r, 𝑡), known as
potential energy or simply potential:
F (r, 𝑡) = −∇𝑈 (r, 𝑡) ≡ −grad𝑈 (r, 𝑡) .
(1.10)
For potential forces one can introduce a classical Hamiltonian:
𝐻 (r, p, 𝑡) =
p2
+ 𝑈 (r, 𝑡) = 𝐸,
2𝑚
(1.11)
where 𝐸 is the total energy of a particle. If potential 𝑈 does not depend on time, the
value of the 𝐻 is constant, and the energy is conserved:
𝑑 p2
𝑑𝐻 𝑑𝐸
=
=
+ 𝑈 (r) =
𝑑𝑡
𝑑𝑡
𝑑𝑡 2𝑚
3
3 𝑑 ©Õ 𝑝 𝑗 2
ª Õ 𝑝 𝑗 𝑑𝑝 𝑗 𝜕𝑈 𝑑𝑥 𝑗
+ 𝑈 (r) ® =
+
=
­
𝑑𝑡 𝑗=1 2𝑚
𝑚 𝑑𝑡
𝜕𝑥 𝑗 𝑑𝑡
«
¬ 𝑗=1
Õ
3 3 Õ
𝜕𝑈
𝜕𝑈
𝑉𝑗 𝐹𝑗 + 𝑉𝑗
=
𝑉 𝑗 = 0,
𝐹𝑗 +
𝜕𝑥 𝑗
𝜕𝑥 𝑗
𝑗=1
𝑗=1
(1.12)
𝜕𝑈
. The latter follows from the definition of the gradient
as 𝐹 𝑗 = − 𝜕𝑥
𝑗
𝜕𝑈
𝜕𝑈
𝜕𝑈
F = −∇𝑈 = − e 𝑥
+ e𝑦
+ e𝑧
.
𝜕𝑥
𝜕𝑦
𝜕𝑧
(1.13)
Note, that in classical physics not all of the forces are conservative. A good example
is the force of friction, which leads to the dissipation of mechanical energy and its
transformation to heat. In quantum physics, consideration of dissipative systems is
extremely complicated, and we will not consider it in this course, focusing on the
conservative systems only.
Other systems of the coordinates. Cartesian coordinates are not always the smart
choice. For example, consider the case, when the potential depends on the distance
to the origin of the coordinates:
4
1 Particles and fields in classical physics
𝑈 = 𝑈 (𝑟) = 𝑈
q
𝑥 2 + 𝑦2 + 𝑧2 .
(1.14)
Oone important example of central forces in physics is the coulomb force,
𝑈 (𝑟) =
𝑞1 𝑞2
,
4𝜋𝜖0 𝑟
(1.15)
where 𝑞 1 , 𝑞 2 are charges. Another example is the gravitational force
𝑈 (𝑟) = −𝐺
𝑀1 𝑀2
,
𝑟
(1.16)
where 𝑀1 , 𝑀2 are masses. In Cartesian coordinates, the components of the central
force read:
v
u
t 3
𝑥 𝑗 𝜕𝑈
𝜕𝑈 𝜕𝑟
𝜕𝑈
𝜕
© Õ 2ª
𝑥 𝑗® = −
𝐹𝑗 = −
=−
𝑈­
=−
,
(1.17)
𝜕𝑥 𝑗
𝜕𝑥 𝑗
𝜕𝑟 𝜕𝑥 𝑗
𝑟 𝜕𝑟
𝑗=1
«
¬
v
u
tÕ
3
𝑟=
𝑥 2𝑗 .
𝑗=1
If one writes the system of differential equations corresponding to the 2𝑛𝑑 law of
Newton
𝑑2𝑥 𝑗
𝑥 𝑗 𝜕𝑈
𝑚 2 = 𝐹𝑗 = −
.
(1.18)
𝑟 𝜕𝑟
𝑑𝑡
One immediately sees that there is no way to separate the variable using Cartesian
coordinates, as all 3 equations are are related to each other, and to separate the
variables, i.e. split the system of 3 equations into 3 independent equations, one needs
to use spherical coordinates, introduced in the picture below (Fig. 1.2):
𝑧 = 𝑟 cos 𝜃,
𝑦 = 𝑟 sin 𝜃 sin 𝜙,
𝑥 = 𝑟 sin 𝜃 cos 𝜙,
and the inverse transformations
𝑟=
q
𝑥 2 + 𝑦2 + 𝑧2 ,
𝑥2 + 𝑦2
,
𝑧
𝑦
𝜙 = arctan ,
𝑥
p
𝜃 = arctan
(1.19)
1.1 Classical mechanics
5
Fig. 1.2 Spherical coordinates
where 𝑟 is the length of the radius vector r; 𝜃 is an angle between z axis and the
radius vector r; 𝜙 is an angle between x axis and the projection of r into x-y plane.
Another system of coordinates, which is widely used corresponds to cylindrical
coordinates. It naturally appears, if the potential depends not on the distance to a
certain point, as in the case of spherical coordinates, but to a certain line, which
can be chosen to coincide with the z-axis. Cylindrical coordinates are introduced as
shown in the figure below (Fig. 1.3):
Fig. 1.3 Cylindrical coordinates
6
1 Particles and fields in classical physics
(1.20)
𝑧 = 𝑧,
𝑦 = 𝜌 sin 𝜙,
𝑥 = 𝜌 cos 𝜙.
The inverse transformations read
(1.21)
𝑧 = 𝑧,
q
𝜌 = 𝑥2 + 𝑦2 ,
𝑦
𝜙 = arctan .
𝑥
i.e. z is the same z projection as in the cartesian coordinates, 𝜌 is the length of the
projection of the radius vector into the x-y plane, and 𝜙 is an angle between this
projection and the x-axis, 2D version of cylindrical coordinates are known as polar
coordinates.
1.2 Classical fields
In classical physics, fields are very different from particles. We have seen that
dynamics of the particle is characterized by its trajectory, defined by the dependence
of the radius vector of the particle on time, r = r(𝑡), which can be found by solving
the set of three coupled 2𝑛𝑑 order differential equations, corresponding to the second
law of Newton.
The field on the other hand, is a delocalized object, described by the distribution of
some field function in real space and time [1]. For an electromagnetic field, the functions, which characterize it, are values of electric and magnetic fields E(r, 𝑡), B(r, 𝑡),
which are related to scalar and vector potentials 𝜙(r, 𝑡), A(r, 𝑡):
E(r, 𝑡) = −∇𝜙(r, 𝑡) −
𝜕A(r, 𝑡)
,
𝜕𝑡
(1.22)
B(r, 𝑡) = ∇ × A(r, 𝑡),
where in the last equation we meet a vector operator curl, which can be formally
considered as a cross product of the vector operator "nabla"
3
∇ = e𝑥
and the vector A, one gets:
Õ
𝜕
𝜕
𝜕
𝜕
+ e𝑦
+ e𝑧
=
e𝑗
𝜕𝑥
𝜕𝑦
𝜕𝑧 𝑗=1 𝜕𝑥 𝑗
(1.23)
1.2 Classical fields
7
e 𝑥 e 𝑦 e𝑧
∇×A=
𝜕 𝜕 𝜕
𝜕𝑥 𝜕𝑦 𝜕𝑧
=
𝐴 𝑥 𝐴 𝑦 𝐴𝑧
𝜕 𝐴𝑦 𝜕 𝐴𝑥
𝜕 𝐴𝑧 𝜕 𝐴 𝑦
𝜕 𝐴 𝑥 𝜕 𝐴𝑧
e𝑥 (
−
) + e𝑦 (
−
) + e𝑧 (
−
).
𝜕𝑦
𝜕𝑧
𝜕𝑧
𝜕𝑥
𝜕𝑥
𝜕𝑦
(1.24)
Particles and fields are connected to each other, as fields act on particles producing
the forces; for example, a charged particle, being placed in an electromagnetic field
experiences the Lorentz force
(1.25)
F 𝐿 = 𝑞E + 𝑞 [v × B],
where 𝑞 is the charge of the particle. On the other hand, particles create fields; charged
particles, for example, create an electromagnetic field. This process is described by
the set of Maxwell’s equations:
∇ · E ≡ div E =
𝜌
,
𝜖0
∇ × E ≡ curl E = −
(1.26)
𝜕B
,
𝜕𝑡
(1.27)
div B ≡ ∇ · B = 0,
(1.28)
1 𝜕E
,
(1.29)
𝑐2 𝜕𝑡
where, in addition to the operation of the curl (∇ × E, ∇ × B), one meets the operation
of the divergency, which can be formally considered as an inner (or dot) product of
the vector operator ∇ and the given vector:
∇ × B ≡ curl B = 𝜇0 j +
div E = ∇ · E =
3
𝜕𝐸 𝑥 𝜕𝐸 𝑦 𝜕𝐸 𝑧 Õ 𝜕𝐸 𝑗
+
+
=
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑥 𝑗
𝑗=1
(1.30)
In the Maxwell equations 𝜌 = 𝜌(r, 𝑡) stands for the density of charge, and j = j(r, 𝑡)
- for the density of current j = 𝜌V, where V is velocity.
In addition to the operators of gradient, divergency, and curl, one can construct
the Laplacian operator, as divergency of the gradient, or dot product of ∇ with itself:
∇ · ∇ = ∇2 =
3
Õ 𝜕2
𝜕2
𝜕2
𝜕2
+
+
=
𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 𝑗=1 𝜕𝑥 2𝑗
(1.31)
Importantly, vector differential operators look simple in Carthesian coordinates, but
in other systems of the coordinates they look more cumbersome, as we will see later
on.
8
1 Particles and fields in classical physics
1.3 Plane waves
Let us now consider a solution of the Maxwell equations if charges and currents are
absent, 𝜌 = 0, j = 0. Then, one has
div E = 0,
∇×E=−
𝜕B
,
𝜕𝑡
div B = 0,
(1.32)
(1.33)
(1.34)
1 𝜕E
.
(1.35)
𝑐2 𝜕𝑡
This system, of course, has a trivial solution corresponding to E = B = 0, but
non-trivial solutions are also possible. Indeed, take the time derivative of the equation (4.86):
∇×B=
1 𝜕2E
𝜕
𝜕B
(∇ × 𝐵) = ∇ ×
=
= −∇ × ∇ × E.
2
2
𝜕𝑡
𝜕𝑡
𝑐 𝜕𝑡
(1.36)
Then, one can use a well known identity ∇ × ∇× = grad · div −∇2 , and then get:
1 𝜕2E
= −∇ × ∇ × E = −∇ (∇ · E) + ∇2 E = ∇2 E,
(1.37)
𝑐2 𝜕𝑡 2
where we used that in the absence of the charges ∇ · E ≡ div E = 0.
Dynamics of the free electromagnetic field is thus described by the well known
wave equation:
1 𝜕2E
− ∇2 E ≡ 2 E = 0,
𝑐2 𝜕𝑡 2
1 𝜕2
2 = 2 2 − ∇ 2 ,
𝑐 𝜕𝑡
(1.38)
(1.39)
where - D’Alambert operator. The equation for the magnetic field B is similar
(derive it yourselves):
1 𝜕2B
− ∇2 B = 0.
(1.40)
𝑐2 𝜕𝑡 2
The wave equations for E and B allow for the solution in the form of a plane
wave, which can be cast in the following form (convince yourself by plugging into
the wave equation):
(
E(r, 𝑡) = E0 cos(k · r − 𝜔𝑡),
(1.41)
B(r, 𝑡) = B0 cos(k · r − 𝜔𝑡),
1.3 Plane waves
9
where 𝜔 is the frequency of a wave, k = n𝑘 - wave vector (n is a unity vector in the
direction of the propagation of a wave), E0 and B0 are amplitudes of the electric and
magnetic field in the wave.
The quantities 𝜔, k, E0 , B0 are not independent from one another), but are connected by the following relations:
𝜔 = 𝑐|k|
(1.42)
To show this, let us choose the wave to propagate along the x-axis, k = 𝑘 · e 𝑥 n = e 𝑥 .
Then:
E = E0 cos(𝑘𝑥 − 𝜔𝑡) =⇒
𝜕2E
(1.43)
𝜕 𝜕
· E0 cos(𝑘𝑥 − 𝜔𝑡) = −E0 𝜔2 cos(𝑘𝑥 − 𝜔𝑡),
(1.44)
𝜕𝑡 𝜕𝑡
2
𝜕
𝜕2
𝜕2
𝜕2
+
+
E
cos(𝑘𝑥
−
𝜔𝑡)
=
E0 cos(𝑘𝑥 − 𝜔𝑡) =
∇2 E =
0
𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧2
𝜕𝑥 2
𝜕𝑡 2
=
− 𝑘 2 cos(𝑘𝑥 − 𝜔𝑡).
(1.45)
Plugging this into the wave equation
1 𝜕2 E
𝑐 2 𝜕𝑡 2
− ∇2 E = 0 one immediately gets:
𝜔2
2
−
𝑘
cos(𝑘𝑥 − 𝜔𝑡) = 0 =⇒
𝑐2
𝜔 = 𝑐𝑘 ≡ 𝑐 |k| .
E0
(1.46)
(1.47)
The wave number k is connected to the wavelength; this can be easily seen if we
plot the absolute value of E at a given time (as an example 𝑡 = 0) as function of x:
2𝜋
𝑘
2𝜋
=⇒ 𝑘 =
.
𝜆
=⇒ 𝜆 =
(1.48)
(1.49)
In a similar way, plotting 𝐸 as function of time for a given x, one immediately sees,
that 𝜔 is related to the oscillation period, 𝜔 = 2𝑇𝜋 . The argument of the cosine in the
plane wave is known as its phase:
𝜙(𝑡, 𝑥) = 𝑘𝑥 − 𝜔𝑡
(1.50)
(we take n = e 𝑥 ).
Let us plot the profile of the wave as a function of 𝑥 at 𝑡 = 0, and 𝑡 = 𝛿𝑡 > 0. Let us
monitor the displacement of the point of a constant phase (say, 𝜙 = 0 : 𝑘𝑥 − 𝜔𝑡 = 0).
From the plot above it is clear that the position of this point displaces for 𝛿𝑥 = 𝜔𝑘 𝛿𝑡
for the time period 𝛿𝑡, so one can introduce the velocity of this displacement:
10
1 Particles and fields in classical physics
Fig. 1.4 Plane wave solution.
Fig. 1.5 Evolution of a plane wave in time
𝛿𝑥 𝜔(𝑘)
=
(1.51)
𝛿𝑡
𝑘
this is known as phase velocity of a wave. For electromagnetic waves, 𝜔 = 𝑐𝑘, so
that 𝑣 𝑝ℎ = 𝑐, meaning that phase velocity is equal to the speed of light, and doesn’t
depend on 𝑘.
Other types of waves can have different dispersion relations 𝜔(𝑘). In these cases
𝑣 𝑝ℎ (𝑘) = 𝜔 𝑘(𝑘) becomes 𝑘 - dependent, and waves become dispersive (we will see
what this means when we consider wave packets).
𝑣 𝑝ℎ =
1.3 Plane waves
11
Electromagnetic waves are [transverse waves, which means that the vectors of
electric and magnetic fields and the wavevector are orthogonal to each other,
E0 ⊥ 𝐵0 ⊥ k
(1.52)
(⊥ means perpendicular) =⇒ electromagnetic waves are transverse waves.
Let us first show that E0 ⊥ k. According to the first Maxwell equation
∇ · E ≡ div E = 0.
(1.53)
Let us put here
E = E0 cos(kr − 𝜔𝑡) = E0 cos(𝑘 𝑥 𝑥 + 𝑘 𝑦 𝑦 + 𝑘 𝑧 𝑧 − 𝜔𝑡),
𝜕
𝜕𝐸 𝑥 𝜕𝐸 𝑦 𝜕𝐸 𝑧
+
+
= 𝐸 0𝑥
cos(𝑘 𝑥 𝑥 + 𝑘 𝑦 𝑦 + 𝑘 𝑧 𝑧 − 𝜔𝑡)+
∇·E=
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑥
(1.54)
(1.55)
𝜕
𝜕
cos(𝑘 𝑥 𝑥 + 𝑘 𝑦 𝑦 + 𝑘 𝑧 𝑧 − 𝜔𝑡) + 𝐸 0𝑧
cos(𝑘 𝑥 𝑥 + 𝑘 𝑦 𝑦 + 𝑘 𝑧 𝑧 − 𝜔𝑡)
𝜕𝑦
𝜕𝑧
= −(𝐸 0𝑥 𝑘 𝑥 + 𝐸 0𝑦 𝑘 𝑦 + 𝐸 0𝑧 𝑘 𝑧 ) cos(kr − 𝜔𝑡) = −E0 k cos(kr − 𝜔𝑡) = 0
𝐸 0𝑦
=⇒ E0 · k = 0 =⇒ E0 ⊥ k.
Using another Maxwell equation ∇ · B = 0, one can demonstrate that B0 ⊥ k.
Now, we demonstrate that B0 ⊥ E0 ; knowing that E0 ⊥ k, B0 ⊥ k, let us choose:
k = e𝑥 𝑘
(1.56)
(make the x-axis coincide with direction of the propagation of the wave). We can
also always choose the y-axis to coincide with E0 , E0 = 𝐸 0 e 𝑦 . As for B0 , we only
know that B0 ⊥ k, and thus lies in the z-y plane, so, most generally, can be written
as:
B0 = 𝐵0𝑦 e 𝑦 + 𝐵0𝑧 e𝑧 .
(1.57)
Let us now put E = 𝐸 0 e 𝑦 cos(𝑘𝑥 − 𝜔𝑡) and B = 𝐵0𝑦 e 𝑦 + 𝐵0𝑧 e𝑧 cos(𝑘𝑥 − 𝜔𝑡) into
the following Maxwell equation:
∇×E=−
and one has:
𝜕B
,
𝜕𝑡
(1.58)
12
1 Particles and fields in classical physics
𝜕
𝜕B
= 𝐵0𝑦 e 𝑦 + 𝐵0𝑧 e𝑧
cos(𝑘𝑥 − 𝜔𝑡) = 𝐵0𝑦 e 𝑦 + 𝐵0𝑧 e𝑧 𝜔 sin(𝑘𝑥 − 𝜔𝑡),
𝜕𝑡
𝜕𝑡
(1.59)
∇×E=
e𝑥
e𝑦
e𝑧
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
0 𝐸 0 cos(𝑘𝑥 − 𝜔𝑡) 0
= e𝑧
𝜕
𝐸 0 cos(𝑘𝑥 − 𝜔𝑡) = −𝑘 𝐸 0 e𝑧 sin(𝑘𝑥 − 𝜔𝑡).
𝜕𝑥
(1.60)
Thus, one gets:
𝜔 e 𝑦 𝐵0𝑦 + e𝑧 𝐵0𝑧 sin(𝑘𝑥 − 𝜔𝑡) = −𝑘 𝐸 0 e𝑧 sin(𝑘𝑥 − 𝜔𝑡)
(1.61)
=⇒ 𝐵0𝑦 = 0; 𝜔𝐵0𝑧 = −𝑘 𝐸 0 , 𝐵0𝑧 = − 𝐸𝑐0 =⇒ B0 = − 𝐸𝑐0 e𝑧 , E0 = 𝐸 0 e 𝑦 =⇒
E0 ⊥ B0 .
As well, we see that the absolute values of the oscillations of electric and magnetic
fields are related as:
|E0 |
|B0 | =
.
(1.62)
𝑐
1.4 Superpositions of plane waves
In the previous section we have seen that the system of Maxwell equations for
electromagnetic fields in vacuum can be reduced to the wave equation for electric
and magnetic fields:
1
𝑐2
1
𝑐2
𝜕2E
− ∇2 E = 0,
𝜕𝑡 2
𝜕2B
− ∇2 B = 0.
𝜕𝑡 2
(1.63)
(1.64)
These equations allow a solution in the form of linearly polarized plane waves:
E = E0 cos(kr − 𝜔𝑡),
B = B0 cos(kr − 𝜔𝑡),
(1.65)
where E0 ⊥ B0 ⊥ k, 𝜔 = 𝑐|k|, |B0 | = |E0 |/𝑐. Let us now analyze solutions of the
wave equation further.
First, let us note that solution in the form of a linearly polarized wave can also be
represented in the following form:
E = E0 sin(kr − 𝜔𝑡),
B = B0 sin(kr − 𝜔𝑡),
(1.66)
1.4 Superpositions of plane waves
13
After that, one can note, that the wave equation is linear, and thus, any linear
superposition of its solutions is in turn a solution as well. Consider, for example,
superposition corresponding to solutions (1.65) and (1.66) with the same k, with
linear polarizations orthogonal to each other. Let us choose the coordinate axis z
to coincide with a direction of the propagation of the wave, k = 𝑘e𝑧 . Then, we can
write (± - sign of rotation):
E = e 𝑥 𝐸 0𝑥 cos(𝑘 𝑧 − 𝜔𝑡) ± e 𝑦 𝐸 0𝑦 sin(𝑘 𝑧 − 𝜔𝑡).
(1.67)
This corresponds to a general case of elliptically polarized waves if one looks how
the electric field evolves as a function of time in a certain point, one will see an
ellipse
Fig. 1.6 Elliptically polarized light
Direction of rotation is determined by the sign of linear superposition. If 𝐸 0𝑥 = 0
(or 𝐸 0𝑦 = 0) we deal with linear polarization. If 𝐸 0𝑥 = 𝐸 0𝑦 , we deal with circular
polarization
Fig. 1.7 Circularly polarized light
Now, let us consider possible superpositions of the plane waves, corresponding
to different k. For simplicity, let us focus on the case of 1D propagation (say, along
14
1 Particles and fields in classical physics
the x-axis) for linear polarization only (i.e. direction of E is constant). In this case,
the wave equation for a single component of the electric field has the form:
1 𝜕2 𝐸 𝜕2 𝐸
−
= 0.
𝑐2 𝜕𝑡 2
𝜕𝑥 2
(1.68)
Let us show, that solution of this equation can be written as 𝐸 (𝑥, 𝑡) = 𝐸 (𝑥 ± 𝑐𝑡).
Indeed, introduce 𝜉 = 𝑥 ± 𝑐𝑡 so that
𝜕𝐸 𝑑𝐸
=
𝜕𝑡
𝑑𝜉
𝜕𝐸 𝑑𝐸
=
𝜕𝑥
𝑑𝜉
2
1 𝜕 𝐸
−
𝑐2 𝜕𝑡 2
𝑑2 𝐸
𝜕𝜉
𝑑𝐸 𝜕 2 𝐸
= ±𝑐 , 2 = 𝑐2 2 ,
𝜕𝑡
𝑑𝜉 𝜕𝑡
𝑑𝜉
2
2
𝜕𝜉 𝑑𝐸 𝜕 𝐸 𝑑 𝐸
=
=⇒
=
,
𝜕𝑥
𝑑𝜉 𝜕𝑥 2
𝑑𝜉 2
𝜕2 𝐸 𝑑2 𝐸 𝑑2 𝐸
=
−
= 0.
𝜕𝑥 2
𝑑𝜉 2
𝑑𝜉 2
(1.69)
(1.70)
(1.71)
Solutions in the form 𝐸 = 𝐸 (𝑥 ± 𝑐𝑡) correspond to wavepackets, propagating
from left to right (sign "+") or right to left (sign "-").
Fig. 1.8 Propagation of wave packets.
Each wavepacket can be represented as a linear combination of the plane waves.
This can be shown using the tools of Fourier analysis. Indeed, take 𝑡 = 0. Then,
according to the fundamental theorem of Fourier analysis, if 𝐸 (𝑥)|𝑡=0 = 𝐸 0 (𝑥) goes
to zero at 𝑥 −→ ±∞, then it can be represented in the following form:
∫ +∞
1
𝐸 (𝑥) = √
𝐸 (𝑘)𝑒 −𝑖𝑘 𝑥 𝑑𝑘,
(1.72)
2𝜋 −∞
where 𝐸 (𝑘) is Fourier image (or Fourier representation) of 𝐸 (𝑥), which can be
calculated as:
∫ +∞
1
𝐸 (𝑘) = √
𝐸 (𝑥)𝑒 𝑖𝑘 𝑥 𝑑𝑥.
(1.73)
2𝜋 −∞
1.4 Superpositions of plane waves
15
Note, that although the electric field 𝐸 (𝑥) is real, its Fourier image 𝐸 (𝑘) can be
complex. What is the meaning of this expression? 𝑒 𝑖𝑘 𝑥 = cos 𝑘𝑥+𝑖 sin 𝑘𝑥 corresponds
to a plane wave (at 𝑡 = 0). The integral is equivalent to the summation by different
values of wavevectors =⇒ wavepacket at 𝑡 = 0 corresponds to linear superposition
of plane waves with different 𝑘.
Now, how do these wavepackets evolve with time? Before detailed analysis of the
corresponding mathematical solution, let us note the following:
1) For fixed 𝑘, we can have waves propagating in both positive and negative
direction of 𝑥, i.e. when 𝑡 ≠ 0 we should make replacement 𝑒 −𝑖𝑘 𝑥 −→ 𝑒 −𝑖 (𝑘 𝑥±𝜔𝑡) ,
where "±" corresponds to the direction of the propagation.
2) At 𝑡 ≠ 0 each solution should account for the plane waves propagating in both
directions.
3)The wave equation is second order in time. Thus, to know 𝐸 (𝑥, 𝑡) at any moment
in time, we need to know not only
𝐸 (𝑥, 𝑡)|𝑡=0 = 𝐸 0 (𝑥),
(1.74)
but also its time derivative:
𝜕𝐸 (𝑥, 𝑡)
𝜕𝑡
= 𝐸¤ 0 (𝑥).
(1.75)
𝑡=0
Suppose both of these function are known. Then, according to the main theorem of
Fourier analysis, we can write for 𝑡 = 0:
∫ +∞
1
𝐸 0 (𝑥) = √
𝐸 0 (𝑘)𝑒 −𝑖𝑘 𝑥 𝑑𝑘,
(1.76)
2𝜋 −∞
∫ +∞
1
𝐸 0 (𝑥)𝑒 𝑖𝑘 𝑥 𝑑𝑥,
(1.77)
𝐸 0 (𝑘) = √
2𝜋 −∞
∫ +∞
1
𝐸¤ 0 (𝑥) = √
𝐸¤ 0 (𝑘)𝑒 −𝑖𝑘 𝑥 𝑑𝑘,
(1.78)
2𝜋 −∞
∫ +∞
1
𝐸¤ 0 (𝑘) = √
𝐸¤ 0 (𝑥)𝑒 𝑖𝑘 𝑥 𝑑𝑥.
(1.79)
2𝜋 −∞
Now using the linear superposition principle we can write at 𝑡 ≠ 0:
∫ +∞ h
i
1
𝐸 (𝑥, 𝑡) = √
𝐸 + (𝑘)𝑒 −𝑖 (𝑘 𝑥+𝜔𝑡) + 𝐸 − (𝑘)𝑒 −𝑖 (𝑘 𝑥−𝜔𝑡) 𝑑𝑘,
2𝜋 −∞
where 𝜔 = 𝜔(𝑘) = 𝑐𝑘.
Its time derivative is:
∫ +∞ h
i
𝜕𝐸
1
=√
𝑖𝜔 𝐸 − (𝑘)𝑒 −𝑖 (𝑘 𝑥−𝜔𝑡) − 𝐸 + (𝑘)𝑒 −𝑖 (𝑘 𝑥+𝜔𝑡) 𝑑𝑘.
𝜕𝑡
2𝜋 −∞
(1.80)
(1.81)
16
1 Particles and fields in classical physics
Compare these expressions with those for 𝑡 = 0 by placing 𝑡 = 0 into them, one
gets:
(
𝐸 (𝑘) = 𝐸 + (𝑘) + 𝐸 − (𝑘)
(1.82)
𝐸¤ (𝑘) = 𝑖𝜔 [𝐸 − (𝑘) − 𝐸 + (𝑘)]
𝐸¤ (𝑘)
1
𝐸 (𝑘) ± 𝑖
,
(1.83)
=⇒ 𝐸 ± (𝑘) =
2
𝜔(𝑘)
This can be placed into Eq. (1.80), and thus the problem is solved. Now, suppose,
that we construct a wavepacket from the waves, propagating from left to right only:
∫ ∞
1
𝐸 (𝑥, 𝑡) = √
𝐸 (𝑘)𝑒 −𝑖 (𝑘 𝑥−𝜔 (𝑘)𝑡) 𝑑𝑘,
(1.84)
2𝜋 0
Suppose as well, that the main contribution is given by the wave with 𝑘 = 𝑘 0 , and
the width of the wavepacket in 𝑘 space 𝛿𝑘 is much less than 𝑘 0 , and, moreover, 𝜔(𝑘)
does not change much at the scale of 𝛿𝑘:
Fig. 1.9 Fourier-image of the electric field.
Then, we can write 𝜔(𝑘) in the form of Taylor series developed near 𝑘 = 𝑘 0 :
𝜔(𝑘) = 𝜔(𝑘 0 ) +
Then, one gets:
𝜕𝜔
𝜕𝑘
(𝑘 − 𝑘 0 ) = 𝜔(𝑘 0 ) + 𝜔 0 (𝑘 0 ) (𝑘 − 𝑘 0 ).
𝑘=𝑘0
(1.85)
1.4 Superpositions of plane waves
17
∫ +∞
0
1
𝐸 (𝑥, 𝑡) = √
𝐸 (𝑘)𝑒 −𝑖 (𝑘 𝑥−𝜔 (𝑘0 )𝑡−𝜔 (𝑘0 ) (𝑘−𝑘0 )𝑡) 𝑑𝑘 =
2𝜋 −∞
∫ +∞
0
1
𝐸 (𝑘)𝑒 −𝑖 ( (𝑘−𝑘0 ) 𝑥+𝑘0 𝑥−𝜔 (𝑘0 )𝑡−𝜔 (𝑘0 ) (𝑘−𝑘0 )𝑡) 𝑑𝑘 =
√
2𝜋 −∞
∫ +∞
0
1
𝐸 (𝑘)𝑒 −𝑖 (𝑘−𝑘0 ) [𝑥−𝜔 (𝑘0 )𝑡 ] 𝑑𝑘.
√ 𝑒 −𝑖 (𝑘0 𝑥−𝜔 (𝑘0 )𝑡)
−∞
2𝜋
(1.86)
The profile of the field can thus be represented as the product of the plane wave
𝑒 −𝑖 (𝑘0 𝑥−𝜔 (𝑘0 )𝑡) with being the value of the wavevector, which gives the main contribution to the wavepacket, and envelope function
∫ +∞
1
𝐸 (𝑘)𝑒 −𝑖 (𝑘−𝑘0 ) [𝑥−𝑉𝑔𝑟 𝑡 ] 𝑑𝑘,
(1.87)
𝑓 (𝑥 − 𝑉𝑔𝑟 𝑡) = √
2𝜋 −∞
where 𝑉𝑔𝑟 = 𝜕𝜔
𝜕𝑘 - is known as group velocity of a wave, being the velocity of the
displacement of a wavepacket. For electromagnetic waves: phase velocity 𝑉 𝑝ℎ =
𝜔 (𝑘)
𝜕𝜔 (𝑘)
𝜕
= 𝑐𝑘
= 𝜕𝑘
(𝑐𝑘) =
𝑘
𝑘 = 𝑐 (already considered by us), group velocity 𝑉𝑔𝑟 = 𝜕𝑘
𝑐 =⇒ 𝑉 𝑝ℎ = 𝑉𝑔𝑟 = 𝑐.
The equality of phase and group velocity for light means, that wavepackets propagate while conserving their shape:
∫ +∞
1
𝐸 (𝑘)𝑒 −𝑖 (𝑘 𝑥−𝜔𝑡) 𝑑𝑘 =
𝐸 (𝑥, 𝑡) = √
2𝜋 −∞
∫ +∞
1
𝐸 (𝑘)𝑒 −𝑖𝑘 ( 𝑥−𝑐𝑡) 𝑑𝑘 = 𝐸 (𝑥 − 𝑐𝑡).
(1.88)
√
2𝜋 −∞
However, if 𝜔 ≠ 𝑉 𝑝ℎ 𝑘, the shape of the wavepacket will necessary be distorted.
Indeed, in this case
∫ +∞
1
𝐸 (𝑥, 𝑡) = √
𝐸 (𝑘)𝑒 −𝑖 (𝑘 𝑥−𝜔 (𝑘)𝑡) 𝑑𝑘 ≠ 𝐸 (𝑥 − 𝑉 𝑝ℎ 𝑡).
(1.89)
2𝜋 −∞
This phenomenon is known as dispersion of the waves. If takes place, for example
when light propagates not in vacuum, but in dielectric media, for which frequency
dependent refractive index is introduced.
Let us now consider how the size of a wavepacket in the real space (i.e. effective
extension of the function 𝐸 (𝑥)) correlates with its size in k-space (i.e. effective
extension of 𝐸 (𝑘)). For that, consider a Gaussian wavepacket:
𝐸 (𝑥) = 𝐸 0 𝑒
−𝑥 2
𝑎2
0
.
Now, let us calculate the Fourier image for a Gaussian wavepacket:
(1.90)
18
1 Particles and fields in classical physics
Fig. 1.10 Wave packet and its size.
∫ +∞
∫ +∞
2
− 𝑥2
− 12 [ 𝑥 2 −𝑖𝑘 𝑥𝑎02 ]
𝐸0
𝐸0
𝑎
𝑖𝑘 𝑥
0
𝑒 𝑒
𝑒 𝑎0
𝐸 (𝑘) = √
𝑑𝑥 = √
𝑑𝑥 =
2𝜋 −∞
2𝜋 −∞
"
#
2
∫ +∞ − 1 𝑥− 𝑖𝑘𝑎02 + 𝑘 2 𝑎04
∫ +∞ 𝑘 2 𝑎2 ( 𝑥−e𝑥 ) 2
2
4
𝐸0
𝐸0
0
2
𝑎2
0
𝑒
𝑒 − 4 𝑒 𝑎0 𝑑𝑥 =
𝑑𝑥 = √
√
2𝜋 −∞
2𝜋 −∞
∫ +∞ ( 𝑥−e𝑥 ) 2
∫ +∞ 𝑥2
𝑘 2 𝑎2
𝑘 2 𝑎2
− 2
𝐸0 − 0
𝐸0 − 0
𝑎2
4
4
0
𝑒
𝑑𝑥 = √ 𝑒
𝑒 𝑎0 𝑑𝑥 =
√ 𝑒
−∞
−∞
2𝜋
2𝜋
𝐸 0 𝑎 0 − 𝑘 2 𝑎02
(1.91)
√ 𝑒 4 ,
2
2
∫ +∞
− 𝑥2
𝑑𝑥 = 𝑎 0 𝜋. From here, one
where we used, that the Gaussian integral is −∞ 𝑒
obviously sees, that the Fourier image of a Gaussian wavepacket is also Gaussian:
2
𝐸 (𝑥) = 𝐸 0 𝑒
− 𝑥2
𝑎
0
𝐸0 𝑎0
𝐸 (𝑘) = √ 𝑒
2
= 𝐸0 𝑒
𝑘 2 𝑎2
− 40
−
𝑥2
Δ 𝑥2
𝑎
0
(1.92)
, Δ𝑥 = 𝑎 0 ,
𝐸0 𝑎0
= √ 𝑒
2
2
− 𝑘2
Δ𝑘
, Δ𝑘 =
2
,
𝑎0
(1.93)
Δ 𝑘 gives the size of the Fourier image, Δ 𝑘 = Δ2𝑥 - one sees, that the narrower the
wavepacket is in x-space, the wider its Fourier image becomes,
Δ𝑥Δ 𝑘 ' 2 ∼ 1
(1.94)
(the sign ∼ denotes the order of magnitude). This is classical analog of the famous
Heisenberg uncertainty relation,
Δ𝑥Δ 𝑘 ≥ ℏ/2
(1.95)
1.5 Interference and diffraction
19
as we will see in the next chapter.
1.5 Interference and diffraction
To conclude with phenomena of the superposition of the waves, let us briefly consider
the phenomena of their interference and diffraction. Their essence is, that when we
have two (or more) sources of electromagnetic waves, we need to summate their
amplitudes, not intensities
Fig. 1.11 Interfence of the waves from several sources.
(1.96)
E = E1 + E2 + E3 .
The fact, that amplitudes should be summated, follows from linearity of the wave
equation
1 𝜕2E
− ∇2 E = 0.
(1.97)
𝑐2 𝜕𝑡 2
The fact, that summation of amplitudes is not equivalent to summation of intensities, is due to quadratic dependence of the intensity of an electromagnetic wave
E = E0 cos(kr − 𝜔𝑡) on its amplitude E0 . Indeed, from school electrodynamics, we
know, that for an electromagnetic wave the intensity 𝐼, defined as energy of the plane
wave passing through a unit of square per unit of time is
𝐼=
𝑐𝜖0 𝐸 02
2
(1.98)
,
and if we have several sources, then
Õ
E=
E𝑛 ,
(1.99)
𝑛
2
𝐼∼E =
!2
Õ
𝑛
E𝑛
=
Õ
𝑛,𝑚
E𝑛 E𝑚 ≠
Õ
E2𝑛 ,
(1.100)
𝑛
Í
due to the presence of so called interference terms in the sum, 𝑛≠𝑚 E𝑛 ·E𝑚 . Consider
two important examples of interference phenomena.
1)Double slit experiment. Consider that coherent light from the source falls to
20
1 Particles and fields in classical physics
Fig. 1.12 Double slit experiment
a screen with two slits, separated by a distance 𝑑. At distance 𝐿 from it there is a
detecting screen. Let us find, how the intensity on the screen will depend on the
distance x (see Fig. 1.12) slits, using the Huygens principle, according to which the
slits serve serve as sources of secondary waves. The distances from these sources to
the detection point can be calculated as:
s
2 r
𝑑
𝑑2
2
𝑥1 = 𝐿 +
− 𝑥 = 𝐿2 +
− 𝑑𝑥 + 𝑥 2 =
2
4
s
r
r
r
s
𝑑2
𝑑𝑥 − 𝑥 2
𝑑2
𝑑𝑥
𝑑 2 𝛼𝑥
2
2
= 𝐿 +
1−
' 𝐿 +
1−
' 𝐿2 +
−
,
2
2
𝑑
𝑑
4
4
4
2
𝐿2 +
𝐿2 +
4
4
(1.101)
q 𝑑
, and we supposed that 𝑥 𝑑, 𝐿, thus neglecting the term 𝑥 2 ,
2
𝐿 2 + 𝑑4
and using 1𝑠𝑡 order Taylor expansion by small parameter 2𝑑 𝑥𝑑2 = 𝛼𝑥. Similarly, one
𝐿 + 4
where 𝛼 =
gets:
s
𝑥2 =
𝐿2
𝑑
+ 𝑥+
2
2
r
'
𝐿2 +
𝑑 2 𝛼𝑥
+
.
4
2
(1.102)
Now, summate the two waves:
h i
𝐸 = 𝐸 0 [cos(𝑘𝑥 1 − 𝜔𝑡) + cos(𝑘𝑥2 − 𝜔𝑡)] = 𝑅𝑒 𝐸 0 𝑒 𝑖 (𝑘 𝑥1 −𝜔𝑡) + 𝑒 𝑖 (𝑘 𝑥2 −𝜔𝑡) =
#
q
"
𝑖 𝐿 2 + 𝑑2 −𝜔𝑡
2
=
= 𝑅𝑒 𝐸 0 𝑒 𝑖𝑘 𝛼𝑥/2 + 𝑒 −𝑖𝑘 𝛼𝑥/2 𝑒
r
2𝐸 0 cos(𝑘𝛼𝑥) cos 𝑘
!
!
r
2
2
𝑑
𝑑
e0 cos 𝑘 𝐿 2 +
𝐿2 +
− 𝜔𝑡 = 𝐸
− 𝜔𝑡 ,
4
4
(1.103)
1.5 Interference and diffraction
21
e0 cos(𝑘𝛼𝑥) - is a periodic function of
where the amplitude of the sum of the waves 𝐸
x. Its absolute value has maxima at 𝑘𝛼𝑥/2 = 𝜋𝑛, where 𝑛 = 0, 1, . . . , and goes to
zero at 𝑘𝛼𝑥 = 𝜋2 + 𝜋𝑛 - i.e. on the detection screen one sees the system of interference
fringes when measuring the intensity
e2 = 4𝐸 0 cos2 (𝑘𝛼𝑥/2) = 2𝐸 0 [1 + cos(𝑘𝛼𝑥)]
𝐼∼𝐸
0
(1.104)
2) Diffraction grating. Consider the reflection of the light, which falls perpendicularly onto the diffraction grating, i.e. the surface with parallel scratches, each
of them presenting the source of the scattered wave, propagating in all directions.
Suppose, that scratches are very thin, and separated by a distance 𝑑.
Fig. 1.13 Diffraction grating
Suppose that we detect the Intensity of the reflected wave at angle 𝜃 (note, that
due to the presence of the scratches the wave can be scattered at any angle, not just
at the angle of incidence). If the detection screen is far, then summation of the fields
produced by two neighbouring scratches 1 and 2 gives:
𝐸 = 𝐸 0 (cos(𝑘𝑥1 − 𝜔𝑡) + cos(𝑘𝑥 2 − 𝜔𝑡)) ,
(1.105)
where 𝑥1 = 𝑥 2 + 𝑑 sin 𝜃. One thus gets:
𝑥 +𝑥
h
i
𝑥1 −𝑥2
𝑥1 −𝑥2 𝑖 𝑘 1 2 2 −𝜔𝑡
𝑒 𝑖𝑘 2 + 𝑒 −𝑖𝑘 2
𝐸 = 𝐸 0 𝑅𝑒 𝑒 𝑖 (𝑘 𝑥1 −𝜔𝑡) + 𝑒 𝑖 (𝑘 𝑥2 −𝜔𝑡) = 𝐸 0 𝑅𝑒 𝑒
=
𝑥 +𝑥
𝑥 +𝑥
𝑥 − 𝑥 sin 𝜃
1
2
1
2
1
2
2𝐸 0 cos
𝑘 cos 𝑘
− 𝜔𝑡 = 2𝐸 0 cos 𝑘 𝑑
cos 𝑘
− 𝜔𝑡 ,
2
2
2
2
(1.106)
22
1 Particles and fields in classical physics
e0 the amplitude of the diffracted wave, we get
Denoting by 𝐸
sin 𝜃
2
2
2
e
𝐼 ∼ 𝐸 0 = 4𝐸 0 cos 𝑘 𝑑
= 2𝐸 0 [1+ cos(𝑘 𝑑 sin 𝜃)].
2
(1.107)
From this equation, one immediately sees, that diffraction is maximal for certain
angles, given by the so called Bragg condition:
𝑑𝜋 sin 𝜃
sin 𝜃
=
𝑘𝑑
= 𝜋𝑛,
(1.108)
2
𝜆
where 𝜆 = 2𝑘𝜋 - wavelength of the diffracted light =⇒
𝑑 sin 𝜃 = 𝑛𝜆, 𝑛 =
0, 1, , 2, . . . =⇒ sin 𝜃 = 𝑛𝜆
one
sees,
that
if
𝜆
>
𝑑
the
only
diffraction
maximum
𝑑
is at 𝜃 = 0. To see more diffraction maxima, one needs to have wavelength being
smaller than the distance between the scratches.
On the other hand, if 𝜆 𝑑, there will be so many diffraction maxima lying
so close to each other, that in experiment it will not be possible to resolve them
=⇒ from here we have an important qualitative conclusion that interference and
diffraction phenomena are most pronounced, when wavelength of light is comparable
with the characteristic size of the object, at which the diffraction occurs.
1.6 Tasks
• Exercise 1: Prove that
∇ · (A × B) = B · (∇ × A) − A · (∇ × B)
• Exercise 2: Compute the Jacobian of the coversion from Cartesian coordinates to
spherical coordinates.
• Exercise 3: Consider a 1D space divided into 3 regions with the following speeds
of light in each part: 𝑣 1 = 𝑐 (vacuum), 𝑣 2 = 𝑐/𝑛 (glass plate), 𝑣 3 = 𝑣 1 = 𝑐 (vacuum). The incident wave is coming from the left. The 1st region is 𝑥 ∈ (−∞, 0),
the 2nd is 𝑥 ∈ [0, 𝐿] and the 3d is 𝑥 ∈ (𝐿, +∞). Calculate the transmission 𝑇 (𝜔)
(and the reflection 𝑅(𝜔)) coefficient, defined as the ratio between the intensity of
the transmitted (reflected) light and the intensity of the incident light of frequency
𝜔. Show that 𝑅(𝜔) + 𝑇 (𝜔) = 1. Explain the physical meaning of this equality.
Plot 𝑇 (𝜔) for a glass plate of width 1 𝜇m for the frequencies of the visible light
range.
Find analytically the frequencies at which 𝑇 (𝜔) (transmission coefficient) reaches
its maximum and minimum. How are they linked to the frequencies of the FabryPerot resonator?
References
23
References
1. L.D. Landau and E.M. Lifschits. The Classical Theory of Fields, volume Volume 2 of Course
of Theoretical Physics. Pergamon Press, Oxford, 1975.
Chapter 2
Particle-wave dualism and wavefunction of
quantum particles
In the beginning of the 20𝑡 ℎ century experiments started to appear demonstrating,
that light, which is from the classical point of view an electromagnetic wave, in
some cases behaves as a flux of individual particles, know as photons. On the other
hand, in some cases, fluxes of elementary particles, such as electrons and neutrons,
demonstrate clear signatures of wave behavior, such as interference and diffraction.
Let us consider some examples of these effects.
2.1 Photoelectric effect
This effect occurs, when a surface of a metal is illuminated by light, it is absorbed
by electrons, which, in consequence, gain some energy, and thus can leave a metal
producing the photocurrent. To be able to leave a metal, each electron needs to gain
certain minimal amount of energy 𝐴, which depends on the material and is known
as work function
Fig. 2.1 𝐴 - potential barrier, that electrons need to surpass to participate in the photocurrent.
25
26
2 Particle-wave dualism and wavefunction of quantum particles
The dependence of photoelectric current on frequency and intensity of light was
investigated, and the following rules were established:
1) Photoelectric current appears only if frequency of the illumination exceeds
some characteristic threshold frequency 𝜔0 , which depends on the type of material,
but does not depend on the intensity of light.
2) If the frequency surpasses the threshold frequency, 𝜔 > 𝜔0 , the maximal value
of the velocity of photoelectrons is determined from the following equation:
𝑚 𝑒𝑉 2
= ℏ𝜔 − 𝐴 = ℎ𝜈 − 𝐴,
2
where 𝐴 is a work function of the material, 𝜈 = 2𝜔𝜋 , and ℏ =
constant, known as the Plank constant, which in SI units is:
(2.1)
ℎ
2𝜋
is a fundamental
ℏ ≈ 1, 054 · 10−34 𝐽 · 𝑠−1 = 6, 582 · 10−16 𝑒𝑉 · 𝑠−1
(2.2)
(we give values in Joules/second and electron-volts/second, 1 𝑒𝑉 ≈ 1, 6 · 10−19 𝐽).
The Planck constant, as we will see, is ubiquitous in quantum mechanic. For the first
time it was proposed by Max Planck, when he buil the theory of radiation spectrum
of a black body, not considered in this course.
These experimental results are inconsistent with classical theory of light as electromagnetic wave. Indeed, in this case the energy transferred to the electrons by the
wave depends on the intensity of the latter, not its frequency, and the existence of
the threshold frequency 𝜔0 , as well as relation defining dependence of velocity of
photoelectrons on 𝜔 are puzzling.
However, experimental data can be explained, if one suggests that in photoelectric
effect light behaves as a flux of individual particles, with energy
𝐸 = ℏ𝜔 = ℎ𝜈.
(2.3)
And each electron can absorb single light quantum, known as photon.
Then, existence of threshold 𝜔0 is fully understandable: if frequency is less than
𝜔0 = 𝐴ℏ there is simply not enough energy for one photon to give to an electron
to surpass the potential barrier 𝐴. The relation 𝑚 𝑒𝑉 2 /2 = ℏ𝜔 − 𝐴 is also fully
understandable, and is nothing more than energy conversation law for absorption of
a single photon.
Note, that from the connection between energy and frequency
𝐸 = ℏ𝜔.
(2.4)
It also leads to the connection between the momentum and the wavelength of an
individual photon. Indeed, we know that for light the following dispersion relation
holds:
𝜔 = 𝑐𝑘,
(2.5)
where 𝑘 is the wavenumber, 𝑘 =
2𝜋
𝜆 .
Multiplying this by ℏ, one gets:
2.2 Compton scattering
27
ℏ𝜔 = 𝐸 = 𝑐ℏ𝑘.
(2.6)
Now, take the relativistic expression for the dependence of kinetic energy on momentum:
𝑚𝑐2
𝑚V
𝐸=q
,p = q
,
2
2
1 − 𝑉𝑐2
1 − 𝑉𝑐2
q
𝐸 2 − 𝑝 2 𝑐2 = 𝑚 2 𝑐4 =⇒ 𝐸 = 𝑚 2 𝑐4 + 𝑝 2 𝑐2 ,
(2.7)
(2.8)
where 𝑚 is the mass of the particle. Photons are massless and therefore for them
𝑚 = 0, and 𝐸 = 𝑝𝑐. Comparing this with Eq. (2.6), one immediately gets:
𝑝 = ℏ𝑘 =
2𝜋ℏ
2𝜋ℏ
=⇒ 𝜆 =
.
𝜆
𝑝
(2.9)
2.2 Compton scattering
Another important example of particle-like behavior of light is the effect of the
Compton scattering, appearing when x-rays or 𝛾-rays (i.e. light with very high
frequency, invisible to the naked eye) scatter on free electrons. Experimentally, it
was shown, that in the process of such a scattering the wavelength changes, depending
on the scattering angle. From the point of view of classical light scattering theory,
the change of wavelength (=change of frequency) is not understandable. Indeed,
a classical electromagnetic wave will make a free electron to oscillate with the
frequency, corresponding to the frequency of the wave. As the electron will now
have non-zero acceleration, it will itself emit electromagnetic waves, but again with
the same frequency.
Fig. 2.2 Classical picture of the scattering of an electromagnetic wave by a free electron.
28
2 Particle-wave dualism and wavefunction of quantum particles
To understand Compton scattering, one should represent it as the process of the
scattering of individual photons with energy 𝐸 = ℏ𝜔 and momentum p = ℏk = ℏ𝜔
𝑐 ·n
(n corresponds to direction of propagation, k = n · 𝑘).
Consider the scattering process in the case, when an electron is initially not
moving:
Fig. 2.3 Quantum picture of the scattering of an electromagnetic wave by a free electron.
Let 𝜖 𝛾 , p𝛾 denote energy and momentum of an incident photon, 𝜖 𝛾0 , p𝛾0 - energy
and momentum of scattered photon, 𝜖 𝑒 , p𝑒 energy and momentum for scattered
electron. Suppose, that the incident photon propagates along the x-axis, 𝜃 demonstrates the scattering angle of a photon, 𝜙 - scattering angle of an electron. Energy
and momentum are conserved in the scattering act, thus:
(2.10)
p𝛾 = p𝛾0 + p𝑒 ,
2
𝜖𝛾 + 𝑚𝑒 𝑐 =
𝜖 𝛾0
+ 𝜖𝑒
(2.11)
(we describe electronp relativistically, so before scattering it has rest energy 𝑚 𝑒 𝑐2 ,
after scattering 𝜖 𝑒 = 𝑚 2𝑒 𝑐4 + 𝑝 2𝑒 𝑐2 ).
Let us now project the momentum conservation law on the x and y axes, and use
ℏ𝜔0
0
𝑝 𝛾 = ℏ𝑘 = ℏ𝜔
𝑐 , 𝑝 𝛾 = ℏ𝑘 = 𝑐 . One gets:
- x projection,
- y projection, so that
ℏ𝜔 ℏ𝜔 0
=
cos 𝜃 + 𝑝 𝑒 cos 𝜙
𝑐
𝑐
(2.12)
ℏ𝜔 0
sin 𝜃 = 𝑝 𝑒 sin 𝜙
𝑐
(2.13)
2.2 Compton scattering
29
𝑐 𝑝 𝑒 cos 𝜙 = ℏ𝜔 − ℏ𝜔 0 cos 𝜃,
(2.14)
𝑐 𝑝 𝑒 sin 𝜙 = ℏ𝜔 sin 𝜃.
(2.15)
0
Squaring these two equations and adding them, one gets:
𝑝 2𝑒 𝑐2 = ℏ𝜔2 + ℏ2 𝜔 02 cos2 3𝜃 − 2ℏ2 𝜔𝜔 0 cos 𝜃 + ℏ2 𝜔 02 sin2 𝜃 =
ℏ2 𝜔2 + ℏ2 𝜔 02 − 2ℏ2 𝜔𝜔 0 cos 𝜃.
(2.16)
Now, use an energy conversation law:
q
ℏ𝜔 + 𝑚 𝑒 𝑐2 = ℏ𝜔 0 + 𝑝 2𝑒 𝑐2 + 𝑚 2𝑒 𝑐4 =⇒
q
𝑝 2𝑒 𝑐2 + 𝑚 2𝑒 𝑐4 = ℏ(𝜔 − 𝜔 0) + 𝑚 𝑒 𝑐2 =⇒
2
𝑝 2𝑒 𝑐2 + 𝑚 2𝑒 𝑐4 = ℏ(𝜔 − 𝜔 0) + 𝑚 𝑒 𝑐2 =
ℏ2 (𝜔 − 𝜔 0) 2 + 2ℏ(𝜔 − 𝜔 0)𝑚 𝑒 𝑐2 + 𝑚 2𝑒 𝑐4 =⇒
𝑝 2𝑒 𝑐2 = ℏ2 (𝜔 − 𝜔 0) 2 + 2ℏ(𝜔 − 𝜔 0)𝑚 𝑒 𝑐2 .
(2.17)
Combining this with the previous equation which we got from momentum conservation, we get:
ℏ2 𝜔2 + ℏ2 𝜔 02 − 2ℏ2 𝜔𝜔 0 cos 𝜃 = ℏ2 (𝜔 − 𝜔 0) 2 + 2ℏ(𝜔 − 𝜔 0)𝑚 𝑒 𝑐2 =
ℏ2 𝜔2 + ℏ2 𝜔 02 − 2ℏ2 𝜔𝜔 02 + 2ℏ(𝜔 − 𝜔 0)𝑚 𝑒 𝑐2 =⇒
ℏ𝜔 · 𝑚 𝑒 𝑐2 = ℏ𝜔 0 𝑚 𝑒 𝑐2 + ℏ𝜔(1 − cos 𝜃) .
(2.18)
From here we can get the frequency of a scattered photon:
𝜔0 =
𝑚 𝑒 𝑐2
𝜔=
𝑚 𝑒 𝑐2 + ℏ𝜔(1 − cos 𝜃)
1+
𝜔
ℏ𝜔
𝑚𝑒 𝑐 2
(1 − cos 𝜃)
.
(2.19)
For visible light ℏ𝜔 𝑚 𝑒 𝑐2 , thus 𝜔 0 ' 𝜔, so no frequency change should be observed. However, for photons with very high frequencies, with energies comparable
to the rest energy of an electron 𝑚 𝑒 𝑐2 , one clearly sees angle-dependent frequency
change (Compton effect).
Let us obtain an elegant equation for the shift of the wavelength Using 𝜔 =
2 𝜋𝑐
2 𝜋𝑐
0
𝜆 , 𝜔 = 𝜆0 we get
1
1
= ·
0
𝜆
𝜆 1+
Multiplying this by 𝜆𝜆 0, one gets:
1
2 𝜋ℏ𝑐
(1
𝑚𝑒 𝑐 2 𝜆
− cos 𝜃)
.
(2.20)
30
2 Particle-wave dualism and wavefunction of quantum particles
𝜆=
𝜆0
=⇒
− cos 𝜃)
2𝜋ℏ𝑐
2𝜋ℏ𝑐
(1 − cos 𝜃) = 𝜆 +
(1 − cos 𝜃) = 𝜆 0 =⇒
𝜆 1+
2
𝑚𝑒 𝑐 𝜆
𝑚 𝑒 𝑐2𝜆
2𝜋ℏ𝑐
𝜆 0 − 𝜆 = 4𝜆 =
(1 − cos 𝜃) = 𝜆 𝑐 (1 − cos 𝜃),
𝑚 𝑒 𝑐2𝜆
1+
2 𝜋ℏ𝑐
(1
𝑚𝑒 𝑐 2 𝜆
(2.21)
where we introduced the parameter 𝜆 𝑐 = 𝑚2 𝜋ℏ𝑐
= 𝑚 ℎ𝑐2 𝜆 known as the Compton
2
𝑒𝑐 𝜆
𝑒
wavelength of an electron.
The obtained expression for 4𝜆 is in perfect agreement with experimental data. Let
us now discuss further some important consequences of the relations 𝐸 = ℏ𝜔, p = ℏk
for photons.
2.3 Fabry-Perot resonator and Heisenberg uncertainty relation
for photons
Consider a one dimensional Fabry-Perot resonator, in which an electromagnetic field
is confined between two ideal parallel mirrors. In this case, a standing wave is formed.
The electric field should go to zero at boundaries of the resonator (at 𝑥 = 0, 𝑥 = 𝐿),
Fig. 2.4 Modes of a resonator corresponding to standing waves.
and a standing wave is thus formed if the length of the resonator equals an integer
number of half waves
𝜆𝑛
𝑛 = 𝐿 =⇒
2
2𝐿
𝜆𝑛 =
=⇒
𝑛
2𝜋 𝜋𝑛
𝑘𝑛 =
=
=⇒
𝜆𝑛
𝜆
𝜋𝑐
𝜔 𝑛 = 𝑐𝑘 𝑛 =
𝑛.
𝐿
(2.22)
(2.23)
(2.24)
(2.25)
2.4 Signatures of a wave behavior of matter
31
This discrete set of frequencies corresponds to the modes of the resonator. But now
we know, that frequency of an electromagnetic wave corresponds to the energy of
the counterpart photons, 𝐸 = ℏ𝜔. Thus, discretization of frequencies of the standing
waves automatically gives us quantization of the energy of the photons, confined in
the resonator:
ℏ𝜋𝑐
𝑛.
(2.26)
𝐸 𝑛 = ℏ𝜔 𝑛 =
𝐿
This conclusion hold for any type of the waves: confinement of a wave + relations
𝐸 = ℏ𝜔, p = ℏk =⇒ quantization of energy.
Now, consider a wavepacket of light. We have demonstrated, that the size of a
wavepacket in real space is connected to its size in k-space (Fourier image) as:
4𝑥4𝑘 ∼ 1
(2.27)
Now, we know that 𝑝 = ℏ𝑘, and so
4𝑝 = ℏ4𝑘 =⇒ 4𝑥4𝑝 ∼ ℏ
(2.28)
- the product of uncertainties of position and momentum of photon is about the Planck
constant. Again, this conclusion is general, and holds for any object demonstrating
the properties of particle-wave dualism.
2.4 Signatures of a wave behavior of matter
After having considered the examples of particle-like behavior of light, we should
consider as well the experiments showing that fluxes of elementary particles behave
as waves. Those are famous experiments on diffraction of electrons and neutrons
(later referred as quantum particles) on the surface of crystals. Consider the flux of
the quantum particles hitting the surface of a solid. The latter consists of atoms or
molecules, packed in a regular crystalline lattice. The flux of quantum particles will
scatter somehow, and if one measures the intensity of the scattered beam as function
of the scattering angle, one should see some smooth distribution, defined by the
properties of individual scatters. This is not the case though - scattering intensity
demonstrates clear maxima for certain scattering directions.
This phenomenon can be only explained, if one suggests, that quantum particles
demonstrate wave like behavior. Indeed, surface of a solid creates periodic scattering
potential, which will form an analog of a diffraction grating.
We have seen already, that interference of the waves, scattered by individual nodes
of a diffraction grating leads to the appearance of diffraction maxima, if the Bragg
condition is satisfied:
𝑑 sin 𝜃 = 𝑛𝜆, 𝑛 = 0, 1, 2 . . .
(2.29)
- and this is exactly what happens when quantum particles are scattered by the surface
of a solid. Now, two questions naturally appear:
32
2 Particle-wave dualism and wavefunction of quantum particles
Fig. 2.5 Scattering of particles by a solid crystal.
1. What is the physical nature of the waves, associated with quantum particles,
known as the de Broglie waves?
2. What defines the frequency and wavevector of these waves?
Keeping the answer for the first question for later, we can immediately answer
the second one: for any quantum particle frequency and wavelength of a de Broglie
wave are determined by its energy and momentum with the same formulae as for
light:
(2.30)
𝐸 = ℏ𝜔,
𝑝 = ℏ𝑘 =
2𝜋ℏ
.
𝜆
(2.31)
The wavelength 𝜆 = 2 𝜋ℏ
𝑝 is known as de Broglie wavelength. Let us consider some
properties of the de Broglie waves.
2.5 Dispersion of the de Broglie waves
For light, 𝜔 = 𝑐𝑘, 𝐸 = 𝑝𝑐 and phase velocity is equal to group velocity,
𝜔 𝐸
=
= 𝑐,
𝑘
𝑝
𝜕𝜔 𝜕𝐸
=
=
= 𝑐 = 𝑉 𝑝ℎ .
𝜕𝑘
𝜕𝑝
𝑉 𝑝ℎ =
(2.32)
𝑉𝑔𝑟
(2.33)
we have seen, that this leads to non-dispersive propagation of wavepackets of light
in vacuum (i.e. wavepackets preserve their shape). This is not the case of massive
quantum particles, for which (in the non-relativistic limit)
2.6 Quantization of the energy of a quantum particle in a rigid box
𝑚𝑉 2
𝑝2
ℏ𝑘 2
=
=⇒ 𝜔 =
,
2
2𝑚
2𝑚
𝐸 𝜔
𝑝
𝑉
𝑉 𝑝ℎ =
=
=
= ,
𝑝
𝑘
2𝑚
2
𝜕𝐸
= 𝑉 = 2𝑉 𝑝ℎ .
𝑉𝑔𝑟 =
𝜕𝑝
𝐸=
33
(2.34)
(2.35)
(2.36)
From the relation above one can make couple of important conclusions:
1) 𝑉𝑔𝑟 ≠ 𝑉 𝑝ℎ , which means that wavepackets corresponding to massive particles
will be distorted during the propagation:
Fig. 2.6 Comparison between an electromagnetic wave packet, and a matter wave packet.
2) What velocity will one measure in an experiment? Usually it is group velocity,
which corresponds to the velocity of a classical particle, and not phase velocity,
which is half of it. In general, group velocity is more "real". For example, only group
velocity is limited by the speed of light, while phase velocity can easily be bigger
than 𝑐.
Indeed, take a relativistic massive particle, for which
q
𝐸 = 𝑝 2 𝑐2 + 𝑚 2 𝑐4 ,
(2.37)
𝜕𝐸
𝑝𝑐2
𝑝
=p
𝑐 < 𝑐,
=p
2
2
2
4
2
𝜕𝑝
𝑝 𝑐 +𝑚 𝑐
𝑝 + 𝑚 2 𝑐2
s p
𝑝 2 𝑐2 + 𝑚 2 𝑐4
𝐸
𝑚𝑐
=
=
= 𝑐2 1 +
,
𝑝
𝑝
𝑝
𝑉𝑔𝑟 =
(2.38)
𝑉 𝑝ℎ
(2.39)
so that 𝑉𝑔𝑟 −→ ∞ when 𝑝 −→ 0.
2.6 Quantization of the energy of a quantum particle in a rigid
box
Consider a quantum analog of a Febry-Perot resonator, i.e. an object, that confines
the motion of a quantum particle in the region 𝑥 ∈ [0, 𝐿].
34
2 Particle-wave dualism and wavefunction of quantum particles
Fig. 2.7
In full analogy with light, we expect formation of standing waves with
𝜆𝑛
𝑛 = 𝐿 =⇒ 𝜆 𝑛 =
2
2𝜋 𝜋𝑛
𝑘𝑛 =
=
,
𝜆𝑛
𝐿
𝑝2
ℏ2 𝑘 𝑛2
𝐸𝑛 = 𝑛 =
=
2𝑚
2𝑚
2𝐿
,
𝑛
(2.40)
(2.41)
𝜋 2 ℏ2 𝑛 2
2𝑚𝐿 2
(2.42)
- we got an expression for the quantization of energies of quantum particle in a rigid
box. As for photons, this is a direct consequence of particle wave dualism.
2.7 Bohr-Sommerfeld quantization rule
In reality, motion of quantum particles is restricted by smooth confining potentials,
and rigid box approximation can often be too rough. Let us formulate a simple rule
for estimation of the quantization of energies in an arbitrary potential 𝑈 (𝑥).
For this, let us note that condition of the formation of a standing wave is equivalent
to a condition, that a running wave propagating from left mirror and back gains
phaseshift 4𝜙 = 2𝜋𝑛
Fig. 2.8
2.7 Bohr-Sommerfeld quantization rule
35
𝑒 𝑖 ( 4 𝜙1 +4 𝜙2 ) = 𝑒 𝑖 4 𝜙 = 1,
(2.43)
4𝜙 = 4𝜙1 + 4𝜙2 = 2𝑘 𝐿 = 2𝜋𝑛 =⇒
𝜋𝑛
2𝜋
𝑘𝑛 =
, 𝜆𝑛 =
= 2𝐿𝑛
𝐿
𝑘𝑛
(2.44)
(2.45)
- same condition of equivalency of the resonator length to an integer number of half
waves.
Now, consider a potential 𝑈 (𝑥) in the form of a potential well.
Fig. 2.9
For a classical particle with energy 𝐸, placed inside a well 𝑈 (𝑥) there exist
two turning points 𝑥1 and 𝑥2 , beyond which the particle can not penetrate, as there
2
𝑈 (𝑥) > 𝐸 = 𝑈 + 𝑇, and 𝑇 = 𝑚𝑉
2 should become formally negative.
Turning points are defined as solutions of the equation 𝐸 = 𝑈 (𝑥). Now, consider
the phase shift, which the de Broglie wave gains propagating from 𝑥1 = 𝑥1 (𝐸) to
𝑥2 = 𝑥2 (𝐸). We can not calculate it as 4𝜙1 = 𝑘 𝐿 = 𝑘 (𝑥1 − 𝑥 2 ) = 𝑝ℏ (𝑥1 − 𝑥2 ), as
value of 𝑝 changes during the propagation. Indeed, from the conservation of energy
we have
𝑝2
= 𝑐𝑜𝑛𝑠𝑡 =⇒
2𝑚
p
𝑝 = 𝑝(𝑥) = 2𝑚 [𝐸 − 𝑈 (𝑥)].
𝐸 = 𝑈 (𝑥) +
(2.46)
(2.47)
Let us cut the potential 𝑈 (𝑥) in the interval 𝑥 ∈ [𝑥 1 , 𝑥2 ] to very thin slices
of thepthickness 4𝑥𝑖 each, so thin, that in each interval the value of momentum
𝑝 𝑖 = 2𝑚 [𝐸 − 𝑈 (𝑥𝑖 )] is almost constant:
4𝜙1 =
Õ
𝑖
4𝜙𝑖 =
Õ
𝑖
𝑘 𝑖 4𝑥 𝑖 =
1Õ
1 Õp
𝑝 𝑖 4𝑥𝑖 =
2𝑚 [𝐸 − 𝑈 (𝑥𝑖 )]4𝑥 𝑖
ℏ 𝑖
ℏ 𝑖
(2.48)
Now, take the limit when each 4𝑥 𝑖 −→ 0. In this case, the sum above transforms
into the integral:
36
2 Particle-wave dualism and wavefunction of quantum particles
Fig. 2.10
4𝜙1 =
1
ℏ
∫
𝑥2 (𝐸)
𝑝(𝑥)𝑑𝑥 =
𝑥1 (𝐸)
4𝜙 = 24𝜙1 =
∫
𝑥2 (𝐸)
2
ℏ
∫
𝑥2 (𝐸)
1
ℏ
∫
𝑥2 (𝐸)
p
2𝑚 [𝐸 − 𝑈 (𝑥𝑖 )]𝑑𝑥,
(2.49)
𝑥1 (𝐸)
𝑝(𝑥, 𝐸)𝑑𝑥 = 2𝜋𝑛 =⇒
(2.50)
𝑥1 (𝐸)
𝑝(𝑥, 𝐸)𝑑𝑥 = 𝜋ℏ𝑛, 𝑛 = 0, 1, 2 . . . .
(2.51)
𝑥1 (𝐸)
This equation, which is a transcendential equation for a single unknown parameter p
𝐸 (as 𝑥1 , 𝑥 2 are defined from 𝐸 = 𝑈 (𝑥) as functions of 𝐸, and
𝑝(𝑥) = 2𝑚 [𝐸 − 𝑈 (𝑥𝑖 )]𝑑𝑥) is known as Bohr-Sommerfeld quantization rule. It
is sometimes written in the form
∮
𝑝(𝑥, 𝐸)𝑑𝑥 = 2𝜋ℏ𝑛,
(2.52)
∮
where stands for the integral along closed trajectory (i.e. from 𝑥 1 to 𝑥2 and back
for 1D case).
2.8 When does a gas becomes quantum?
Consider the gas of quantum particles with concentration 𝑛 and temperature 𝑇. We
√
1
can estimate the mean value between the particles as 𝑙 ∼ 3 𝑛 = 𝑛 3 , and mean energy
per particle, which is related to the temperature 𝑇 : 𝐸 = 23 𝑘 𝐵 𝑇, where 𝑘 𝐵 is
the Boltzmann constant. We know, that at high temperatures particles will behave
as classical ones. While at low temperatures deviations from classical behavior
can occur. Can we estimate characteristic temperatures of quantum to classical
transitions?
We can expect, that quantum effects start to play a role, when de Broglie wavelength 𝜆 becomes comparable with the mean value of the distance between particles:
2.9 Wave function of a quantum particle and the Schrodinger equation
1
𝜆 ≈ 𝑙 ≈ 𝑛3 .
37
(2.53)
To estimate 𝜆 we have:
p
𝑝2
3
= 𝑘𝑇 =⇒ 𝑝 = 3𝑘 𝐵 𝑇 − 𝑚,
2𝑚 2
1
2𝜋ℏ
2𝜋ℏ
=√
𝜆=
≈ 𝑛 3 =⇒
𝑝
3𝑘 𝐵 𝑇 − 𝑚
𝐸=
𝑘 𝐵𝑇 ≈
4𝜋 2 ℏ2
2
3𝑚𝑛 3
.
(2.54)
(2.55)
(2.56)
From this expression, one can easily see, that the smaller the mass of the particles 𝑚
is, the lower is the temperature at which transitions to a quantum state occurs.
This is the general rule: the smaller the mass is, the more pronounced the quantum
effects are.
2.9 Wave function of a quantum particle and the Schrodinger
equation
After having understood, that quantum particles can demonstrate clear signatures
of wave - like behavior, we should introduce the corresponding wave function,
characterizing the state of the particle, explain its physical meaning, and propose
corresponding wave equation, which will allow us to predict the value of the wave
function 𝜓(r, 𝑡) at any moment of time, if we know it at 𝑡 = 0, 𝜓0 (r) = 𝜓(r, 𝑡)|𝑡=0 .
Differently from the case of electromagnetic field, the wave function of a quantum
particle, if we do not account for such important property as spin (which we will
introduce later in the course) is a complex scalar 𝜓(r, 𝑡) = 𝜓1 (r, 𝑡) + 𝑖𝜓2 (r, 𝑡).
Introduction of the complexity of the wave function is absolutely necessary and has
physical meaning, as we will see shortly.
The physical meaning of the wave function is that it gives the amplitude of the
probability to find a quantum particle in the point r at moment 𝑡. This means, that
corresponding probability density 𝑝(r, 𝑡) = |𝜓(r, 𝑡)| 2 = 𝜓 ∗ (r, 𝑡)𝜓(r, 𝑡).
It is extremely important, that according to most widely accepted interpretation
of quantum mechanics, proposed by Niels Bohr, and known as the Copenhagen
interpretation, the wave function gives most possible complete description of any
quantum system, which makes quantum mechanics essentially probabilistic science:
knowing the wave function, we can’t say where exactly the particle is, but can only
compute the probability of this or that location.
Naturally, in this picture the concept of classical trajectory totally loses its meaning.
If wavefunction is known, we can calculate the probability to find a particle at
any volume Ω at a given time as:
38
2 Particle-wave dualism and wavefunction of quantum particles
(𝑓 )
Fig. 2.11 𝑃Ω
=
∫
Ω
𝜌(r, 𝑡) 𝑑 3 r =
∫
Ω
𝜓 ∗ (r, 𝑡) 𝜓 (r, 𝑡) 𝑑 3 r.
Naturally, if we integrate over all space, the probability should be equal to 1, as
particle should exist somewhere, and thus
∫
∫
𝜓 ∗ (r, 𝑡)𝜓(r, 𝑡)𝑑 3 r =
|𝜓(r, 𝑡)| 2 𝑑 3 r = 1.
(2.57)
𝐴𝑙𝑙 𝑠 𝑝𝑎𝑐𝑒
𝐴𝑙𝑙 𝑠 𝑝𝑎𝑐𝑒
This condition is known as the normalization condition, and should always be
satisfied. This means, that a correct dynamic equation for the wave function should
give
∫
𝑑
|𝜙(r, 𝑡)| 2 𝑑 3 r = 0.
(2.58)
𝑑𝑡
What are other condition required from the equation for the wave function? We
would like to have it first order in time, in order to be able to define 𝜓(r, 𝑡) if
only the wavefunction itself at initial time 𝜓(r, 𝑡)|𝑡=0 = 𝜓0 (r) is known. Indeed,
the introduction of higher order time derivatives into the wave equation will indeed
automatically require the increase of the number of necessary initial condition in
terms of values of time derivatives of 𝜓 at 𝑡 = 0, which does not have any physical
meaning.
As well, solutions of the wave equation in terms of the plane waves should give
correct dispersion relation
𝐸=
p2
ℏk2
,𝜔 =
.
2𝑚
2𝑚
(2.59)
The equation which satisfied all these condition is the time dependent Schrodinger
equation:
𝜕𝜓(r, 𝑡)
ℏ2 2
𝑖ℏ
=−
∇ 𝜓(r, 𝑡) + 𝑈 (r, 𝑡)𝜓(r, 𝑡),
(2.60)
𝜕𝑡
2𝑚
2.9 Wave function of a quantum particle and the Schrodinger equation
2
2
39
2
𝜕
𝜕
𝜕
where ∇2 = 𝜕𝑥
2 + 𝜕𝑦 2 + 𝜕𝑧 2 - the Laplacian, and 𝑈 (r, 𝑡) is an external potential
acting on the quantum particle.
In our course we will often restrict ourselves to the consideration of one dimensional (1D) motion, and thus use the 1D form of Schrodinger equation
𝑖ℏ
𝜕𝜓(𝑥, 𝑡)
ℏ2 𝜕 2 𝜓(𝑥, 𝑡)
+ 𝑈 (𝑥, 𝑡)𝜓(𝑥, 𝑡).
=−
𝜕𝑡
2𝑚 𝜕𝑥 2
(2.61)
Let us search its solution in the form of a plane wave. In electrodynamics plane
waves were written as sines or cosines of an argument kr − 𝜔𝑡 (𝑘𝑥 − 𝜔𝑡 in 1D case).
Equation for 𝜓(𝑥, 𝑡) is complex, so instead of trigonometric functions, we can write
complex exponents,
𝜓(𝑥, 𝑡) = 𝐴𝑒 𝑖 (±𝑘 𝑥−𝜔𝑡) ,
(2.62)
where signs ” ± ” correspond to the direction of the propagation.
Naturally, one can only have plane wave solutions if a particle is free, i.e. for
𝑈 (r, 𝑡) = 0. In this case for a 1D particle one has:
𝑖ℏ
ℏ2 𝜕 2 𝜓
𝜕𝜓
=−
.
𝜕𝑡
2𝑚 𝜕𝑥 2
(2.63)
Putting here Eq. (2.62) one gets:
𝜕𝜓
𝜕
=
𝐴𝑒 𝑖 (±𝑘 𝑥−𝜔𝑡) = −𝑖𝜔𝐴𝑒 𝑖 (±𝑘 𝑥−𝜔𝑡) = −𝑖𝜔𝜓,
𝜕𝑡
𝜕𝑡
𝜕2
𝜕2𝜓
=
𝐴𝑒 𝑖 (±𝑘 𝑥−𝜔𝑡) = −𝑘 2 𝐴𝑒 𝑖 (±𝑘 𝑥−𝜔𝑡) = −𝑘𝜓.
𝜕𝑥 2
𝜕𝑥 2
(2.64)
(2.65)
And thus:
ℏ2 𝑘 2
ℏ𝜔 −
𝜓 = 0 =⇒
2𝑚
ℏ𝜔 = 𝐸 =
ℏ2 𝑘 2
𝑝2
=
2𝑚
2𝑚
(2.66)
(2.67)
- we have the correct dispersion relation.
Now, let us show, that the 1D Schrodinger equation conserves normalization, i.e.
∫ +∞
𝑑
|𝜓(𝑥, 𝑡)| 2 𝑑𝑥 = 0.
(2.68)
𝑑𝑡 −∞
∫ +∞
First, let us note, that in order to have a converging integral −∞ |𝜓(𝑥, 𝑡)| 2 𝑑𝑥 = 1,
we need the wave function to decay at infinities, i.e. 𝜓(𝑥, 𝑡)| 𝑥−→±∞ = 0 ∀𝑡. We then
have:
∫ +∞
∫ +∞ ∗
𝑑
𝜕𝜓
𝜕𝜓 ∗
𝜓 ∗ (𝑥, 𝑡)𝜓(𝑥, 𝑡)𝑑𝑥 =
𝜓+
𝜓 𝑑𝑥.
(2.69)
𝑑𝑡 −∞
𝜕𝑡
𝜕𝑡
−∞
Let us put here:
40
2 Particle-wave dualism and wavefunction of quantum particles
𝜕𝜓
1 ℏ2 𝜕 2 𝜓 1
𝑖ℏ 𝜕 2 𝜓 𝑖
+ 𝑈𝜓 =
− 𝑈𝜓,
=−
2
𝜕𝑡
𝑖ℏ 2𝑚 𝜕𝑥
𝑖ℏ
2𝑚 𝜕𝑥 2 ℏ
𝜕𝜓 ∗
𝑖ℏ 𝜕 2 𝜓 ∗ 𝑖
+ 𝑈𝜓 ∗ (𝑈 ∗ = 𝑈).
=−
𝜕𝑡
2𝑚 𝜕𝑥 2
ℏ
(2.70)
(2.71)
And:
𝑖ℏ 𝜕 2 𝜓 𝑖
𝑖ℏ 𝜕 2 𝜓 ∗ 𝑖
∗
∗
|𝜓| 𝑑𝑥 =
− 𝑈𝜓 𝜓 −
− 𝑈𝜓 𝜓 𝑑𝑥 =
2𝑚 𝜕𝑥 2 ℏ
2𝑚 𝜕𝑥 2
ℏ
−∞
∫ +∞
∫ +∞
2𝜓
2𝜓∗
𝑖ℏ
𝜕
𝜕
𝜕𝜓
𝜕
𝑖ℏ
𝜕𝜓 ∗
∗
∗
=
𝜓
𝜓
𝑑𝑥 =
−𝜓
𝑑𝑥 =
−𝜓
2𝑚 −∞
2𝑚 −∞ 𝜕𝑥
𝜕𝑥
𝜕𝑥
𝜕𝑥 2
𝜕𝑥 2
𝑖ℏ
𝜕𝜓
𝜕𝜓 ∗ 𝑥=+∞
=
= 0,
(2.72)
𝜓∗
−𝜓
2𝑚
𝜕𝑥
𝜕𝑥 𝑥=−∞
𝑑
𝑑𝑡
∫
2
∫
+∞
where the last equality comes from the fact, that 𝜓| 𝑥−→±∞ = 0.
Note, that differently from the case of an electromagnetic wave, the Schrodinger
equation does not allow a solution in form of the non-dispersive wavepacket, 𝜓(𝑥, 𝑡) =
𝜓(𝑥 − 𝑉𝑡), where 𝑉 is a velocity of the particle. This is consequence of quadratic
dependence of energy on momentum, and related difference between phase and
group velocities.
Although, knowing the -K.R. wavefunction we can’t calculate the exact position
of the particle 𝑥, we can calculate its mean value. Let us understand, how this can be
done.
We start from a discrete random process, at which we measure some value 𝑥,
𝑀 . Suppose, that we made measurement
which can take discrete set of the values {𝑥𝑖 }𝑖=1
𝑁 −→ ∞ times, and values 𝑥𝑖 were realized 𝑁𝑖 times each. Then, mean value < 𝑥 >
is:
𝑀
𝑀
𝑀
Õ
1 Õ
𝑁𝑖 Õ
< 𝑥 >=
𝑥 𝑖 𝑁𝑖 =
𝑥𝑖
=
𝑥 𝑖 𝑃𝑖 ,
(2.73)
𝑁 𝑖=1
𝑁
𝑖=1
𝑖=1
if 𝑁 −→ ∞, where 𝑃𝑖 = lim
𝑁𝑖
𝑁 →∞ 𝑁
are probabilities of the outcomes.
Now, if instead of the discrete value 𝑥𝑖 we have a -K.R. continuous variable
𝑥 ∈ [−∞; +∞], we should replace summation by integration, and discrete probability
𝑃𝑖 by probability density 𝜌 = |𝜓| 2 . Thus, the -K.R. mean value of the coordinate 𝑥
is:
∫ +∞
∫ +∞
< 𝑥 >=
𝑥 𝜌(𝑥, 𝑡)𝑑𝑥 =
𝑥𝜓 ∗ (𝑥, 𝑡)𝜓(𝑥, 𝑡)𝑑𝑥.
(2.74)
−∞
−∞
Note, that as 𝜓 depends on time, mean value < 𝑥 >=< 𝑥(𝑡) > changes with time
as well. This change naturally corresponds to the motion of a quantum particle.
For 3D particles, we can write a -K.R. similar formula for the -K.R. mean value
of the radius-vector:
∫
< r(𝑡) >=
r𝜓 ∗ (𝑥, 𝑡)𝜓(𝑥, 𝑡)𝑑 3 r,
(2.75)
2.10 Tasks
41
where we have 3D integration over all space, instead of 1D integration.
2.10 Tasks
Find the energy levels for the following 1D potentials, using the Bohr-Sommerfeld
quantization condition
• 𝑉 (𝑥) =
𝑘 𝑥2
(2
+∞
• 𝑉 (𝑥) =
𝑘𝑥
𝑥<0
𝑥≥0
The Bohr-Sommerfeld quantization condition is
∮
𝐻 ( 𝑥, 𝑝)=𝐸
2
𝑝𝑑𝑥 = 2
∫𝑏
𝑝(𝑥)𝑑𝑥 = 𝑛ℎ,
(2.76)
𝑎
𝑝
where 𝑛 ∈ N, 𝐻 (𝑥, 𝑝) = 2𝑚
+ 𝑉 (𝑥) is the classical Hamiltonian (should not be
b and 𝑎, 𝑏 are the turning points
confused with a quantum Hamiltonian operator 𝐻),
of the potential 𝑉 (𝑥), i.e. 𝑉 (𝑎) = 𝑉 (𝑏) = 𝐸.
Chapter 3
Physical observables in quantum mechanics
3.1 Operators of physical observables
calculating the mean value of the position of the particle < 𝑥(𝑡) >=
∫After
+∞
∗ (𝑥, 𝑡)𝜓(𝑥, 𝑡)𝑑𝑥, let us construct the mean value of the momentum, defined
𝑥𝜓
−∞
as the time derivative of the mean value of the coordinate (velocity) and mass of the
particle 𝑚.
∫ +∞
𝑑
𝑥𝜓 ∗ (𝑥, 𝑡)𝜓(𝑥, 𝑡)𝑑𝑡 =
< 𝑝(𝑡) >= 𝑚 < 𝑉𝑥 (𝑡) >= 𝑚
𝑑𝑡 −∞
∫ +∞ ∫
𝜕2𝜓
𝜕𝜓 ∗
𝜕2𝜓∗
𝑖ℏ +∞ 𝜕
∗ 𝜕𝜓
𝑥 𝜓∗ 2 − 𝜓
𝑥
=𝑚
𝜓
−
𝜓
𝑑𝑥 =
𝑑𝑥
=
2 −∞ 𝜕𝑥
𝜕𝑥
𝜕𝑥
𝜕𝑥
𝜕𝑥 2
−∞
∫ +∞ 𝑖ℏ
𝜕𝜓 ∗ 𝑥=+∞
𝜕𝜓 ∗
∗ 𝜕𝜓
∗ 𝜕𝜓
−
𝑥 𝜓
−𝜓
𝜓
−𝜓
𝑑𝑥 =
2
𝜕𝑥
𝜕𝑥 𝑥=−∞
𝜕𝑥
𝜕𝑥
−∞
∫ +∞
∫ +∞
𝜕𝜓
𝑖ℏ
𝜕𝜓 ∗
𝜓∗
−
𝑑𝑥 −
𝜓
𝑑𝑥 =
2 −∞
𝜕𝑥
𝜕𝑥
−∞
∫ +∞
∫ +∞
∫
𝑥=+∞
+∞
𝜕
𝑖ℏ
∗ 𝜕𝜓
∗
∗ 𝜕𝜓
∗
𝜓
𝑑𝑥 − 𝜓𝜓
+
𝜓
𝑑𝑥 =
𝜓 −𝑖ℏ
𝜓𝑑𝑥,
−
2 −∞
𝜕𝑥
𝜕𝑥
𝜕𝑥
−∞
−∞
𝑥=−∞
(3.1)
∫
∫
where we used twice the rule of the integration by parts 𝑈𝑑𝑉 = 𝑈𝑉 − 𝑉 𝑑𝑈, and
accounted for fast decay of
the wavefunction at 𝑥 −→ ±∞ : 𝜓| 𝑥−→±∞ = 0 =⇒ <
∫ +∞
𝜕
𝑝 𝑥 (𝑡) >= −∞ 𝜓 ∗ −𝑖ℏ 𝜕𝑥
𝜓𝑑𝑥. Compare this expression with the one for the mean
value of 𝑥:
∫ +∞
∫ +∞
𝑥|𝜓| 2 𝑑𝑥 =
< 𝑥(𝑡) >=
−∞
𝜓 ∗ 𝑥𝜓𝑑𝑥.
(3.2)
−∞
They look similar, but instead of multiplication by 𝑥 in the second, one has −𝑖ℏ times
derivative on 𝑥. From this, one concludes, that in quantum mechanics momentum
an operator.
43
44
3 Physical observables in quantum mechanics
𝜕
, 𝑝 𝑥 −→ 𝑝ˆ 𝑥 .
(3.3)
𝜕𝑥
After the introduction of 𝑝ˆ 𝑥 , we can easily understand, what is the physical meaning
of the complexity of the wavefunction. Let us use the so called Madelung representation:
p
(3.4)
𝜓(𝑥, 𝑡) = 𝜌(𝑥, 𝑡)𝑒 𝑖 𝜃 ( 𝑥,𝑡) ,
𝑝ˆ 𝑥 = −𝑖ℏ
where 𝜌(𝑥, 𝑡) = |𝜓(𝑥, 𝑡)| 2 , 𝜌(𝑥, 𝑡) = |𝜓(𝑥, 𝑡)| is an amplitude, and 𝜃 (𝑥, 𝑡) is a phase
of the wavefunction. This is nothing more, than Euler representation of any arbitrary
quantum number 𝑧 = |𝑧|𝑒 𝑖 𝜃 .
Now, let us calculate the mean value of the momentum:
∫ +∞
∫ +∞
𝜕
√ −𝑖 𝜃 𝜕 √ 𝑖 𝜃
𝜓𝑑𝑥 = −𝑖ℏ
𝜌𝑒
𝜌𝑒 𝑑𝑥 =
< 𝑝 𝑥 >=
𝜓 ∗ −𝑖ℏ
𝜕𝑥
𝜕𝑥
−∞
−∞
∫ +∞
1
𝜕 𝜌 √ 𝑖 𝜃 𝜕𝜃
√ −𝑖 𝜃
𝑑𝑥 =
= −𝑖ℏ
𝜌𝑒
+ 𝑖 𝜌𝑒
√ 𝑒𝑖 𝜃
2 𝜌
𝜕𝑥
𝜕𝑥
−∞
∫
∫ +∞
∫ +∞
𝑖ℏ +∞ 𝜕 𝜌
𝜕𝜃
𝑖ℏ 𝑥=+∞
𝜕𝜃
=−
𝑑𝑥 + ℏ
𝜌 𝑑𝑥 = − 𝜌
+ℏ
𝜌 𝑑𝑥 =
2 −∞ 𝜕𝑥
𝜕𝑥
2
𝜕𝑥
−∞
−∞
𝑥=−∞
∫ +∞
𝜕𝜃
ℏ
𝜌 𝑑𝑥.
(3.5)
𝜕𝑥
−∞
From here, one sees, that < 𝑝 𝑥 >≠ 0, which corresponds to the moving particle,
only if 𝜕𝜃
=⇒ 𝜃 ≠ 0. The presence of non-zero phase 𝜃 means that
𝜕𝑥 ≠ 0
wavefunction
p
√
𝜓(𝑥, 𝑡) = 𝜌(𝑥, 𝑡)𝑒 𝑖 𝜃 ( 𝑥,𝑡) = 𝜌(cos 𝜃 + 𝑖 sin 𝜃)
(3.6)
should necessary contain an imaginary part.
Now, after having constructed operators of the coordinate 𝑥 (in 3D case r =
𝜕
e 𝑥 𝑥 + e 𝑦 𝑦 + e𝑧 𝑧) and 𝑝ˆ 𝑥 = −𝑖ℏ 𝜕𝑥
, we can write a 3D momentum operator:
p̂ = e 𝑥 𝑝ˆ 𝑥 + e 𝑦 𝑝ˆ 𝑦 + e𝑧 𝑝ˆ 𝑧 = −𝑖ℏ∇.
(3.7)
We can construct an operator of any physically observable quantity, which in
classical mechanics expresses as some function of coordinates and momenta 𝐴(r, p)
by substitution:
𝐴 = 𝐴(r, p) =⇒
ˆ p̂) = 𝐴(r,
ˆ −𝑖ℏ∇).
𝐴ˆ = 𝐴(r,
Let us consider some particular examples.
1) Angular momentum, defined in classical physics as
(3.8)
(3.9)
3.1 Operators of physical observables
45
e 𝑥 e 𝑦 e𝑧
L = r × p = 𝑥 𝑦 𝑧 = e 𝑥 𝐿 𝑥 + e 𝑦 𝐿 𝑦 + e𝑧 𝐿 𝑧 ,
𝑝𝑥 𝑝𝑦 𝑝𝑧
(3.10)
𝐿 𝑥 = 𝑦 𝑝𝑧 − 𝑧 𝑝𝑦, 𝐿𝑦 = 𝑧 𝑝𝑥 − 𝑥 𝑝𝑧 , 𝐿𝑧 = 𝑦 𝑝𝑥 − 𝑥 𝑝𝑦 .
(3.11)
In quantum mechanics, substituting 𝑝 𝑗 −→ 𝑝ˆ 𝑗 = −𝑖ℏ 𝜕𝑥𝜕 𝑗 , one gets:
𝜕
𝜕
,
𝐿ˆ 𝑥 = 𝑖ℏ 𝑧
−𝑦
𝜕𝑦
𝜕𝑧
𝜕
𝜕
𝐿ˆ 𝑦 = 𝑖ℏ 𝑥
,
−𝑧
𝜕𝑧
𝜕𝑥
𝜕
𝜕
𝐿ˆ 𝑧 = 𝑖ℏ 𝑥
−𝑦
.
𝜕𝑦
𝜕𝑥
(3.12)
(3.13)
(3.14)
2) Energy, or Hamiltonian
𝐸 = 𝐸 (r, p) = 𝐻 (r, p) = 𝑇 (p) + 𝑈 (r) =
p2
+ 𝑈 (r),
2𝑚
p̂2
ℏ2 2
𝐻ˆ = 𝐻ˆ (r, p̂) =
+ 𝑈 (r) = −
∇ + 𝑈 (r),
2𝑚
2𝑚
2
2
(3.15)
(3.16)
2
𝜕
𝜕
𝜕
where ∇2 = ∇ · ∇ = 𝜕𝑥
2 + 𝜕𝑦 2 + 𝜕𝑧 2 .
Note, that the Schrodinger equation can be written with use of the operator of the
Hamiltonian:
2
𝜕𝜓
ℏ2 2
ℏ 2
ˆ
𝑖ℏ
=−
∇ 𝜓 + 𝑈𝜓 = −
∇ + 𝑈 𝜓 = 𝐻𝜓.
(3.17)
𝜕𝑡
2𝑚
2𝑚
Knowing the wavefunction, one can compute the mean value of any physical observable characterized by the operator 𝐴ˆ as:
∫ +∞
ˆ >=
ˆ
< 𝐴(𝑡)
𝜓 ∗ (𝑥, 𝑡) 𝐴𝜓(𝑥,
𝑡)𝑑𝑥,
(3.18)
−∞
or in a 3D case:
ˆ >=
< 𝐴(𝑡)
∫
ˆ
𝜓 ∗ (r, 𝑡) 𝐴𝜓(r,
𝑡)𝑑 3 r.
(3.19)
𝐴𝑙𝑙 𝑠 𝑝𝑎𝑐𝑒
Besides the mean value of physical observables, let us introduce the important
quantity of the mean square deviation, defined as:
𝜎𝐴2 =< 𝐴ˆ 2 > − < 𝐴ˆ >2 .
(3.20)
What is the physical meaning of this quantity? Consider a discrete random process,
𝑀 and 𝐴2 , thus, { 𝐴2 } 𝑀 .
when 𝐴 can take set of values { 𝐴𝑖 }𝑖=1
𝑖 𝑖=1
46
3 Physical observables in quantum mechanics
The mean square deviation will be:
𝜎𝐴2
=
𝑀
Õ
𝑃𝑖 𝐴𝑖2
−
𝑀
Õ
!2
𝑃 𝑖 𝐴𝑖
.
(3.21)
𝑖=1
𝑖=1
Now, suggest that the value 𝑖 = 𝑖0 comes always, i. e. 𝑃𝑖0 = 1, 𝑃𝑖 = 0 if 𝑖 ≠ 𝑖0 we are then not dealing with random, but with deterministic process, and know the
value of 𝐴 for sure. In this case 𝜎𝐴2 = 0.
We can thus conclude, that 𝜎𝐴 gives us a quantitative measure of an uncertainty
of a variable 𝐴.
Let us go back to the operators of 𝑥 and 𝑝 𝑥 , suppose, that we have some wavefunction 𝜓(𝑥), and act on it by product of these two operators, in two different orders
𝑥 𝑝ˆ 𝑥 and 𝑝ˆ 𝑥 𝑥. One has:
𝜕𝜓
,
𝜕𝑥
𝜕𝜓
𝜕
𝑝ˆ 𝑥 𝑥𝜓(𝑥) = −𝑖ℏ𝑥 (𝑥𝜓) = −𝑖ℏ𝜓 − 𝑖ℏ𝑥
𝜕𝑥
𝜕𝑥
𝑥 𝑝ˆ 𝑥 𝜓(𝑥) = −𝑖ℏ𝑥
(3.22)
(3.23)
- the result is different, and the order of 𝑥 and 𝑝 𝑥 in their products really matters in
quantum mechanics. In striking contrast to classical mechanics, where 𝑥 and 𝑝 𝑥 are
just numbers, and 𝑥 𝑝ˆ 𝑥 = 𝑝ˆ 𝑥 𝑥. It is said, that in quantum mechanics, observables 𝑥
and 𝑝 𝑥 do not commute. This noncommutativity can be mathematically expressed
in the following form:
𝜕𝜓
𝜕𝜓
𝑥 𝑝ˆ 𝑥 𝜓(𝑥) − 𝑝ˆ 𝑥 𝑥𝜓(𝑥) = −𝑖ℏ 𝑥
−𝜓−𝑥
= −𝑖ℏ𝜓 =⇒
(3.24)
𝜕𝑥
𝜕𝑥
(𝑥 𝑝ˆ 𝑥 − 𝑝ˆ 𝑥 𝑥)𝜓 = 𝑖ℏ𝜓.
(3.25)
As 𝜓 here is arbitrary, we can write
𝑥 𝑝ˆ 𝑥 − 𝑝ˆ 𝑥 𝑥 = [𝑥; 𝑝ˆ 𝑥 ] = 𝑖ℏ ≠ 0.
(3.26)
If we now go to the 3D case introducing
r=
Õ
e𝑖 𝑥 𝑖 ,
p̂ = −𝑖ℏ
Õ
𝑖
For the different components of 𝑥 𝑗 , 𝑝ˆ𝑙 we have
(
𝑖ℏ,
[𝑥 𝑗 ; 𝑝ˆ𝑙 ] = 𝑖ℏ𝛿𝑙𝑘 =
0,
𝑖
e𝑖
Õ
𝜕
=
e𝑖 𝑝ˆ𝑖 .
𝜕𝑥 𝑖
𝑖
𝑗 = 𝑙,
𝑗 ≠ 𝑙,
(3.27)
(3.28)
- i.e. components of coordinate and momenta commute, if they correspond to orthogonal directions, and don’t commute, if they correspond to the same direction.
The obtained expression is known as the Heisenberg uncertainty relation.
3.1 Operators of physical observables
47
ˆ 𝐵]
ˆ ≠ 0, they can
If two operators of physical observables do not commute, [ 𝐴;
not be measured simultaneously (you can find this statement in almost any quantum
mechanics book or, even more frequently, in popular science books). Being more
precise, this statement can be formulated as following.
ˆ 𝐵]
ˆ ≠ 0, we can not find a quantum state described by a wavefunction 𝜓, in
If [ 𝐴;
which simultaneously 𝜎𝐴 = 0 and 𝜎𝐵 = 0 (i.e. if we construct a state with a well
ˆ 𝐵]
ˆ = 0,
defined value of 𝐴 in it 𝐵 will not be well defined) on the other hand, if [ 𝐴;
construction of such a state becomes possible.
Let us illustrate this statement using the example of 𝑥 and 𝑝 𝑥 observables.
Consider the case of a quantum particle, whose wavefunction is described by a
Gaussian wavepacket,
2
𝜓(𝑥) = 𝐴𝑒 −𝜅 𝑥 = 𝐴𝑒
2
− 𝑥2
𝑎
, 𝜅=
1
> 0.
𝑎2
(3.29)
We know already, that effective size of this packet 4𝑥 ∼ 𝑎1 , but now let’s be more
√
precise, determining 4𝑥 = 𝜎𝑥 = < 𝑥 2 > − < 𝑥 >2 . First, let us define the value of
the constant 𝐴. If 𝜓(𝑥) is a wavefunction, it should be normalized
r
∫ +∞
∫ +∞
2𝜋
−2𝜅 𝑥 2
2
2
2
𝑒
𝑑𝑥 = | 𝐴|
|𝜓(𝑥)| 𝑑𝑥 = | 𝐴|
= 1.
(3.30)
𝜅
−∞
−∞
where we used the well known value of Gaussian integral
r
∫ +∞
2
𝜋
𝐼 (𝜅) =
𝑒 −𝜅 𝑥 𝑑𝑥 =
.
𝜅
−∞
(3.31)
and thus
𝐴=
𝜋 − 14
𝜅
(3.32)
Note, that strictly speaking, we can multiply the value of the normalization constant
above by a phase factor 𝑒 𝑖 𝜃 with constant phase 𝜃. This multiplication, however,
will not change values of any physical observables, and two wavefunctions will be
equivalent to each other.
We already know, that effective size of the wavepacket in x-space is 4𝑥 ∼ 𝑎. Now,
let us be more precise, defining
p
4𝑥 = < 𝑥 2 > − < 𝑥 >2 = 𝜎𝑥 .
(3.33)
One has:
48
3 Physical observables in quantum mechanics
∫
< 𝑥 >=
< 𝑥 2 >=
+∞
𝑥|𝜓(𝑥)| 2 𝑑𝑥 =
−∞
∫ +∞
𝜅 12 ∫
𝑥 2 |𝜓(𝑥)| 2 𝑑𝑥 =
+∞
𝜋
−∞
𝜅 12 ∫
2
𝑥𝑒 −𝜅 𝑥 𝑑𝑥 = 0,
+∞
2
𝑥 2 𝑒 −𝜅 𝑥 𝑑𝑥 = −
𝜋
−∞
! −∞
1
𝜅 12
𝜅 12 𝑑 𝜋 12
𝜋2
1
𝑎2
=−
− 3 =
−
= .
𝜋 𝑑𝑥 𝜅
𝜋
2𝜅
2
2𝜅 2
(3.34)
𝜅 12 𝑑 ∫ +∞
2
𝑒 −𝜅 𝑥 𝑑𝑥 =
𝜋 𝑑𝑥 −∞
And for momentum:
∫ +∞
𝜅 12 ∫ +∞ 𝜅 𝑥2 𝑑 𝜅 𝑥2
𝑑𝜓
< 𝑝 𝑥 >= −𝑖ℏ
𝜓 ∗ (𝑥)
𝑑𝑥 = −𝑖ℏ
𝑒 − 2 𝑑𝑥 =
𝑒− 2
𝑑𝑥
𝜋
𝑑𝑥
−∞
−∞
3 ∫ +∞
2
𝑖ℏ𝜅 2
𝑥𝑒 −𝜅 𝑥 𝑑𝑥 = 0,
1
𝜋 2 −∞
∫ +∞
𝑑2𝜓
2
2
< 𝑝 𝑥 >= −ℏ
𝜓 ∗ (𝑥) 2 𝑑𝑥 =
𝑑𝑥
−∞
"
2 #
∫ +∞
∫ +∞ 𝑑2𝜓
𝑑𝜓
𝑑𝜓 +∞
2
2
𝜓(𝑥) 2 𝑑𝑥 = −ℏ 𝜓(𝑥)
= −ℏ
−
𝑑𝑥 =
𝑑𝑥 −∞
𝑑𝑥
𝑑𝑥
−∞
−∞
∫ +∞ 1 ∫ +∞ 𝑑 𝜅 𝑥2 2
𝑑𝜓
2
2 𝜅 2
=ℏ
𝑑𝑥 = ℏ
𝑒− 2
𝑑𝑥 =
𝑑𝑥
𝜋
𝑑𝑥
−∞
−∞
∫
𝜅 12
+∞
2
1
ℏ2 𝜅
ℏ2
𝑥 2 𝑒 −𝜅 𝑥 𝑑𝑥 = ℏ2 𝜅 2 < 𝑥 2 >= ℏ2 𝜅 2
= ℏ𝜅 2
=
= 2 =⇒
𝜋
2𝜅
2
2𝑎
−∞
p
𝑎
2
2
𝜎𝑥 = < 𝑥 > − < 𝑥 > = √ ,
2
q
ℏ
𝜎𝑝 = < 𝑝 2𝑥 > − < 𝑝 𝑥 >2 = √ ,
𝑎 2
ℏ
𝜎𝑥 𝜎𝑝 = 4𝑥4𝑝 = .
2
(3.35)
(3.36)
(3.37)
(3.38)
(3.39)
(3.40)
We got the famous Heisenberg uncertainty relation for the Gaussian wavepacket
4𝑥4𝑝 =
ℏ
,
2
(3.41)
which resembles very much (up to the factor 21 ) to what we know already, from
simple Fourier analysis.
If turns out, that for Gaussian wavepackets the product of the uncertainties is the
minimal possible, and in general
4𝑥4𝑝 ≥
ℏ
.
2
(3.42)
3.2 Stationary Schrodinger equation, eigenvalues and eigenfunctions of the operators of physical observables
49
3.2 Stationary Schrodinger equation, eigenvalues and
eigenfunctions of the operators of physical observables
The Schrodinger equation
𝑖ℏ
𝜕𝜓(r, 𝑡)
ˆ
= 𝐻𝜓(r,
𝑡)
𝜕𝑡
(3.43)
with the Hamiltonian
ℏ2 2
𝑝ˆ2
+ 𝑈 (r, 𝑡) = −
∇ + 𝑈 (r, 𝑡)
𝐻ˆ =
2𝑚
2𝑚
and added initial condition
𝜓(r, 𝑡)|𝑡=0 = 𝜓0 (r)
(3.44)
(3.45)
allows, in principle, to solve the dynamic problem for a quantum particle, i.e. determine the wavefunction 𝜓(r, 𝑡) at any arbitrary time 𝑡:
From all types of the Hamiltonians, there is an important class, corresponding to
stationary Hamiltonians, from which
p̂2
p̂2
𝐻ˆ (r, 𝑡) =
+ 𝑈 (r, 𝑡) =
+ 𝑈 (r),
2𝑚
2𝑚
(3.46)
i.e. the Hamiltonians, which do not depend on time. Examples of stationary Hamiltonians are ubiquitous - almost always the interaction between elementary particles
depend on their relative position only, not on time.
In the stationary case, one can write a solution of the time dependent Schrodinger
equation using the method of the separation of variables.
Let us represent a wavefunction, depending on spatial coordinates and time as a
product of spatial dependent, and time dependent parts:
𝜓(r, 𝑡) = 𝜒(𝑡)𝜙(r).
And put it into Schrodinger equation
𝜕
𝜕𝜒
ˆ
𝑖ℏ [ 𝜒(𝑡)𝜙(r)] = 𝜙(r) 𝑖ℏ
= 𝐻ˆ (r) [ 𝜒(𝑡)𝜙(r)] = 𝜒(𝑡) 𝐻𝜙(r).
𝜕𝑡
𝜕𝑡
(3.47)
(3.48)
Now, divide this equation by 𝜓(r, 𝑡) = 𝜒(𝑡)𝜙(r). One gets:
1 ˆ
𝑖ℏ 𝑑𝜒(𝑡)
=
𝐻𝜙(r).
𝜒(𝑡) 𝑑𝑡
𝜙(r)
(3.49)
We have in the left hand side only the functions of time, and in the right hand
side - only functions of spatial coordinates. For them to be equal to each other, the
only possible choice is to make these functions constant:
50
3 Physical observables in quantum mechanics
𝑖ℏ 𝑑𝜒(𝑡)
1 ˆ
=
𝐻𝜙(r) = 𝐸,
𝜒(𝑡) 𝑑𝑡
𝜙(r)
(3.50)
So that
𝑑𝜒(𝑡)
= 𝐸 𝜒,
𝑑𝑡
ˆ
𝐻𝜙(r)
= 𝐸 𝜙(r).
(3.51)
𝑖ℏ
(3.52)
We got two independent equations for 𝜒(𝑡) and 𝜙(r), which we should solve
separately.
Let us clarify the physical meaning of the function 𝜙(r) and constant 𝐸.
First, note that 𝐸 has the dimensionality of energy. Second, imagine that equation
ˆ
𝐻𝜙(r)
= 𝐸 𝜙(r) is satisfied. Calculate the mean square deviation for energy in this
state. One has:
∫
2
∫
3
ˆ
𝜎𝐸2 =< 𝐻ˆ 2 > − < 𝐻 >2 =
𝜙∗ (r) 𝐻ˆ 2 𝜙(r)𝑑 3 r −
𝜙∗ (r) 𝐻𝜙(r)𝑑
r =
∫
=
3
ˆ
𝜙 (r) 𝐻ˆ · 𝐻𝜙(r)𝑑
r−
∗
∫
=𝐸
∫
3
ˆ
𝜙 (r) 𝐻𝜙(r)𝑑
r
∗
2
=
∫
2
∫
3
ˆ
𝜙∗ (r) 𝐻𝜙(r)𝑑
r− 𝐸
𝜙∗ (r)𝜙(r)𝑑 3 r = 𝐸 2
𝜙∗ (r)𝜙(r)𝑑 3 r − 𝐸 2 = 0
(3.53)
This means that in the considered state we know the value of the energy exˆ = 𝐸 𝜙 represents an elliptic
actly. From a mathematical point of view, equation 𝐻𝜙
differential equation. In order to be able to solve it, one needs to add some additional constraints to the possible form of the function 𝜙(r), i.e. formulate so called
boundary conditions. As total wavefunction is normalized, we have
∫
∫
∗
3
𝜓 (r, 𝑡)𝜓(r, 𝑡)𝑑 r =
𝜒∗ (𝑡) 𝜒(𝑡)𝜙∗ (r)𝜙(r)𝑑 3 r =
∫
∫
= | 𝜒(𝑡)| 2
𝜙∗ (r)𝜙(r)𝑑 3 r =
|𝜙(r)| 2 𝑑 3 r = 1,
(3.54)
Therefore, the wavefunction 𝜙(r) should decay to zero when |r| −→ ∞, in order to
guarantee the convergence of the normalization integral.
ˆ = 𝐸 𝜙 does
It turns out, that if we demand 𝜙(r)|𝑟 −→∞ = 0, the equation 𝐻𝜙
not always have a solution. Allowed values of energy correspond to discrete energy
states of a quantum system. They are known as its eigenenergies. The corresponding
function 𝜙(r) are known as eigenfunctions of the Hamiltonian.
In 1D case:
ℏ2 𝜕 2 𝜙
+ 𝑈 (𝑥)𝜙 = 𝐸 𝜙
(3.55)
−
2𝑚 𝜕𝑥 2
and
3.2 Stationary Schrodinger equation, eigenvalues and eigenfunctions of the operators of physical observables
51
𝜙(𝑥)| 𝑥−→±∞ = 0.
(3.56)
Solution of this equation allows to determine possible values of the energy exactly,
and not approximately, as in the method using Bohr-Sommerfeld quantization rule
∫
𝑥2 (𝐸)
∫
𝑥2 (𝐸)
𝑝(𝑥)𝑑𝑥 =
𝑥1 (𝐸)
p
2𝑚 [𝐸 − 𝑈 (𝑥)]𝑑𝑥 = 𝜋ℏ𝑛,
(3.57)
𝑥1 (𝐸)
where 𝑛 = 0, 1, 2 . . .
Eigenfunctions of the Hamiltonian 𝐻ˆ satisfy the following orthogonality relation:
(
∫
1, 𝑛 = 𝑚,
∗
3
𝜙 𝑛 (r)𝜙 𝑚 (r)𝑑 (r) = 𝛿 𝑛𝑚 =
(3.58)
0, 𝑛 ≠ 𝑚,
ˆ 𝑛 = 𝐸 𝑛 𝜙 𝑛 , 𝐻𝜙
ˆ 𝑚 = 𝐸 𝑚 𝜙 𝑚 , 𝐸 𝑛 ≠ 𝐸 𝑚 . The set {𝐸 𝑛 } forms a spectrum of
where 𝐻𝜙
the system.
If energy 𝐸 𝑛 and 𝜙 𝑛 are known, we can solve the time-dependent equation for
𝜒(𝑡),
𝜓 𝑛 (r, 𝑡) = 𝜒𝑛 (𝑡)𝜙 𝑛 (r),
𝑑𝜒𝑛
𝑖ℏ
= 𝐸 𝑛 𝜒𝑛 ,
𝑑𝑡
∫
∫
𝑑𝜒𝑛
𝑖𝐸 𝑛
𝑖𝐸 𝑛
𝑑𝜒𝑛
=−
𝑑𝑡 =⇒
=−
𝑑𝑡 =⇒
𝜒𝑛
ℏ
𝜒𝑛
ℏ
𝑖𝐸 𝑛
ln 𝜒𝑛 = −
𝑡 + 𝐴,
ℏ
(3.59)
(3.60)
(3.61)
(3.62)
where 𝐴 is an arbitrary constant. Therefore
𝑖𝐸𝑛 𝑡
𝑖𝐸𝑛 𝑡
𝜒𝑛 = 𝑒 𝐴 𝑒 − ℏ = 𝐶𝑒 − ℏ ,
(3.63)
∫
∫
𝜓 𝑛∗ (r, 𝑡)𝜓 𝑛 (r)𝑑 3 r = |𝐶 | 2
𝜓 𝑛∗ (r)𝜓 𝑛 (r)𝑑 3 r = |𝐶 | 2 = 1 =⇒ 𝐶 = 1,
and
𝑖𝐸𝑛 𝑡
𝜓 𝑛 = 𝑒 − ℏ 𝜙 𝑛 (r),
ˆ 𝑛 (r) = 𝐸 𝑛 𝜙 𝑛 (r),
𝐻𝜙
(3.64)
(3.65)
where 𝐸 𝑛 can be any value from the spectrum.
Note, that the solution in the form above is not the most general one. Indeed,
Schrodinger equation is linear, and any linear combination of its two solutions is a
solution itself. Therefore, the most general solution can be represented as a sum of
solutions, corresponding to all possible values of 𝐸 𝑛 :
52
3 Physical observables in quantum mechanics
𝜓(r, 𝑡) =
Õ
𝐶𝑛 𝑒 −
𝑖𝐸𝑛 𝑡
ℏ
𝜙 𝑛 (r), 𝜓 ∗ (r, 𝑡) =
Õ
𝑛
∫
=
𝐶𝑛∗ 𝑒
𝑖𝐸𝑛 𝑡
ℏ
𝑛
𝜓 ∗ (r, 𝑡)𝜓(r)𝑑 3 r =
!
∫
Õ
𝐶𝑛∗ 𝑒
𝑖𝐸𝑛 𝑡
ℏ
𝜙 𝑛 (r) ×
∫ Õ
𝐶𝑛∗ 𝐶𝑚 𝑒
𝑖 (𝐸𝑛 −𝐸𝑚 ) 𝑡
ℏ
𝜙 𝑛 (r) × 𝜙 𝑚 (r)𝑑 3 r =
𝑛,𝑚
=
Õ
!
Õ
𝐶𝑚 𝑒
− 𝑖𝐸ℏ𝑚 𝑡
𝜙 𝑚 (r) 𝑑 3 r =
𝑚
𝑛
=
(3.66)
𝜙∗𝑛 (r) =⇒
Õ
𝐶𝑛∗ 𝐶𝑚 𝑒
𝑖 (𝐸𝑛 −𝐸𝑚 ) 𝑡
ℏ
∫
𝜙∗𝑛 (r)𝜙 𝑚 (r)𝑑 3 r =
𝑛,𝑚
𝑖 (𝐸𝑛 −𝐸𝑚 ) 𝑡
ℏ
𝛿 𝑛𝑚
𝐶𝑛∗ 𝐶𝑚 𝑒
=
Õ
𝐶𝑛∗ 𝐶𝑛
=
Õ
𝑛
𝑛,𝑚
|𝐶𝑛 | 2 = 1
(3.67)
𝑛
- this is the condition for coefficients {𝐶𝑛 }, necessary for the proper normalization
of the wavefunction.
Where from we get coefficients 𝐶𝑛 ? We should use here an initial condition:
!
Õ
Õ
𝑖𝐸𝑛 𝑡
=
𝐶𝑛 𝜙 𝑛 (r)
(3.68)
𝜓0 (r) = 𝜓(r, 𝑡)|𝑡=0 =
𝐶𝑛 𝑒 − ℏ 𝜙 𝑛 (r)
𝑡=0
𝑛
𝑛
Now, take some eigenfunction 𝜙 𝑚 , multiply the equation above 𝜙∗𝑚 (r) and integrate over all space. One then gets:
∫
Õ ∫
Õ
𝜓0 (r)𝜙∗𝑚 (r)𝑑 3 r =
𝐶𝑛
𝜙∗𝑚 (r)𝜙 𝑛 (r)𝑑 3 r =
𝐶𝑛 𝛿 𝑛𝑚 = 𝐶𝑚
(3.69)
𝑛
𝑛
- which gives us a corresponding coefficient 𝐶𝑚 . Let us now clarify the physical
meaning of these coefficients. Let us calculate the mean value of the energy:
∫
ˆ
< 𝐸 >=
𝜓 ∗ (r, 𝑡) 𝐻𝜓(r,
𝑡)𝑑 3 r =
!
!
∫ Õ
Õ
𝑖𝐸𝑛 𝑡
𝑖𝐸𝑚 𝑡
ˆ 𝑚 (r) 𝑑 3 r =
=
𝐶𝑛∗ 𝑒 ℏ 𝜙∗𝑛 (r)
𝐶𝑚 𝑒 − ℏ 𝐻𝜙
𝑛
∫
=
𝑚
!
Õ
𝑖𝐸𝑛 𝑡
𝐶𝑛∗ 𝑒 ℏ
𝜙∗𝑛 (r)
!
Õ
𝑛
=
Õ
𝐶𝑛∗ 𝐶𝑚 𝑒
𝐶𝑚 𝑒
− 𝑖𝐸ℏ𝑚 𝑡
𝐸 𝑚 𝜙 𝑚 (r) 𝑑 3 r =
𝑚
𝑖 (𝐸𝑛 −𝐸𝑚 ) ℏ𝑡
∫
𝐸𝑛
𝜙∗𝑛 (r)𝜙 𝑚 (r)𝑑 3 r =
𝑛,𝑚
Õ
𝑛,𝑚
𝑡
𝐶𝑛∗ 𝐶𝑚 𝑒 𝑖 (𝐸𝑛 −𝐸𝑚 ) ℏ 𝐸 𝑛 𝛿 𝑛𝑚 =
Õ
|𝐶𝑛 | 2 𝐸 𝑛 .
(3.70)
𝑛
Compare it Í
with the standard expression for the mean value of the random variable < 𝐴 >= 𝑛 𝑃𝑛 𝐴𝑛 , where 𝐴𝑛 are possible values of the variable, and 𝑃𝑛 are
corresponding probabilities of their realization. We immediately conclude that
|𝐶𝑛 | 2 = 𝑃𝑛
(3.71)
3.3 Supplementary material and tasks
53
are the probabilities to get a value 𝐸 = 𝐸 𝑛 in an experiment, when energy of the
particle is measured in a quantum state, characterized by the wavefunction
Õ
𝑖𝐸𝑛 𝑡
(3.72)
𝜓(r, 𝑡) =
𝐶𝑛 𝑒 − ℏ 𝜙 𝑛 (r).
𝑛
ˆ correIn analogy with energy, we can determine the eigenstates of any operator 𝐴,
sponding to some physical observable as:
ˆ 𝑛 = 𝑎 𝑛 𝜙𝑛 ,
𝐴𝜙
(3.73)
which should be again supplemented by appropriate boundary conditions. {𝑎 𝑛 } will
form a spectrum of an observable, and {𝜙 𝑛 } correspond to its eigenstates, for which
the value of 𝐴 is well defined:
2
∫
∫
3
∗ ˆ
3
∗ ˆ2
2
2
2
ˆ
(3.74)
𝜙 𝑛 𝐴𝜙 𝑛 𝑑 r = 0.
𝜙𝑛 𝐴 𝜙𝑛 𝑑 r −
𝜎𝐴 =< 𝐴 > − < 𝐴 > =
As physical observables are functions of r and p̂ = −𝑖ℏ∇, from mathematical point
of viewthe equation for corresponding eigenstates is a stationary partial differential
equation.
3.3 Supplementary material and tasks
3.3.1 Differential operators in different coordinate systems
Before formulating the tasks, we would like to give a reminder on transformation the
differential operators between different coordinate systems. The angular momentum
operator is defined as
L̂ = [r̂ × p̂] = −𝑖ℏ[r̂ × ∇]
(3.75)
For example, in Cartesian coordinates
𝜕
𝜕
𝐿ˆ 𝑧 = −𝑖ℏ(𝑥
− 𝑦 ).
𝜕𝑦
𝜕𝑥
(3.76)
Our goal is to compute 𝐿ˆ 𝑥,𝑦,𝑧 in spherical coordinates. The transformation from
Cartesian to spherical coordinates is given by
𝑥 = 𝑟 sin 𝜃 cos 𝜙
𝑦 = 𝑟 sin 𝜃 sin 𝜙
𝑧 = 𝑟 cos 𝜃,
(3.77)
54
3 Physical observables in quantum mechanics
thus we know how to transfrom the operators 𝑥,
ˆ 𝑦ˆ and 𝑧ˆ, it is trivial. Let us discuss
the transformation of differential operators, entering 𝐿ˆ 𝑥,𝑦,𝑧 . From (3.77) it follows
that
𝜕𝑥
𝜕𝑥
𝜕𝑥
𝑑𝑟 +
𝑑𝜃 +
𝑑𝜙
𝜕𝑟
𝜕𝜃
𝜕𝜙
𝜕𝑦
𝜕𝑦
𝜕𝑦
𝑑𝑟 +
𝑑𝜃 +
𝑑𝜙
𝑑𝑦 =
𝜕𝑟
𝜕𝜃
𝜕𝜙
𝜕𝑧
𝜕𝑧
𝜕𝑥
𝑑𝑧 =
𝑑𝑟 +
𝑑𝜃 +
𝑑𝜙,
𝜕𝑟
𝜕𝜃
𝜕𝜙
𝑑𝑥 =
(3.78)
which can be rewritten in a matrix form
 𝑑𝑟 
 𝑑𝑥 
 
 
𝑑𝑦  = 𝐴  𝑑𝜃 
 
 
𝑑𝜙
 𝑑𝑧 
 
 
(3.79)
where the matrix 𝐴 is defined as
 𝜕𝑥
 𝜕𝑟
 𝜕𝑦
𝐴 =  𝜕𝑟
 𝜕𝑧
 𝜕𝑟

𝜕𝑥
𝜕𝜃
𝜕𝑦
𝜕𝜃
𝜕𝑧
𝜕𝜃
𝜕𝑥 
𝜕𝜙 
𝜕𝑦 
𝜕𝜙 
𝜕𝑧 
𝜕𝜙 .
(3.80)
Since we assume that the inverse transformation exists, namely, there are 𝑟 =
𝑟 (𝑥, 𝑦, 𝑧), 𝜃 = 𝜃 (𝑥, 𝑦, 𝑧) and 𝜙 = 𝜙(𝑥, 𝑦, 𝑧) we can write
𝜕𝑟
𝜕𝑟
𝜕𝑟
𝑑𝑥 +
𝑑𝑦 +
𝑑𝑧
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝜃
𝜕𝜃
𝜕𝜃
𝑑𝜃 =
𝑑𝑥 +
𝑑𝑦 +
𝑑𝑧
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝜙
𝜕𝜙
𝜕𝜙
𝑑𝜙 =
𝑑𝑥 +
𝑑𝑦 +
𝑑𝑧
𝜕𝑥
𝜕𝑦
𝜕𝑧
,
𝑑𝑟 =
(3.81)
which can be rewritten in a matrix form
 𝑑𝑟 
 𝑑𝑥 
 
 
 𝑑𝜃  = 𝐵  𝑑𝑦 
 
 
𝑑𝜙
𝑑𝜃 
 
 
where 𝐵 is defined as
 𝜕𝑟
 𝜕𝑥
 𝜕𝜃
𝐵 =  𝜕𝑥
 𝜕𝜙
 𝜕𝑥

𝜕𝑟
𝜕𝑦
𝜕𝜃
𝜕𝑦
𝜕𝜙
𝜕𝑦
𝜕𝑟 
𝜕𝑧 
𝜕𝜃 
𝜕𝑧 
𝜕𝜙 
𝜕𝑧 
(3.82)
(3.83)
3.3 Supplementary material and tasks
55
from comparison (3.82) with (3.79) we clearly see that 𝐵 = 𝐴−1 . The differential
operators 𝜕/𝜕𝑥, 𝜕/𝜕𝑦 and 𝜕/𝜕𝑧 can be expressed in 𝜕/𝜕𝑟, 𝜕/𝜕𝜃 and 𝜕/𝜕𝜙 using the
chain rule for a probe function 𝑓 (𝑥, 𝑦, 𝑧) ≡ 𝑓˜(𝑟 (𝑥, 𝑦, 𝑧), 𝜃 (𝑥, 𝑦, 𝑧), 𝜙(𝑥, 𝑦, 𝑧)) as
𝜕𝑟 𝜕
𝜕𝜃 𝜕
𝜕𝜙 𝜕
𝜕
=
+
+
𝜕𝑥 𝜕𝑥 𝜕𝑟 𝜕𝑥 𝜕𝜃 𝜕𝑥 𝜕𝜙
𝜕
𝜕𝑟 𝜕
𝜕𝜃 𝜕
𝜕𝜙 𝜕
=
+
+
𝜕𝑦 𝜕𝑦 𝜕𝑟 𝜕𝑦 𝜕𝜃 𝜕𝑦 𝜕𝜙
𝜕𝑟 𝜕
𝜕𝜃 𝜕
𝜕𝜙 𝜕
𝜕
=
+
+
𝜕𝑧 𝜕𝑧 𝜕𝑟 𝜕𝑧 𝜕𝜃 𝜕𝑥 𝜕𝜙
(3.84)
from here we clearly see that the corresponding matrix of this matrix equation is
nothing but 𝐵𝑇 = ( 𝐴−1 )𝑇 in other words
𝜕 
 𝜕𝑥
𝜕
  = ( 𝐴−1 )𝑇
 𝜕𝑦 
𝜕
 𝜕𝑧 
𝜕 
 𝜕𝑟
 𝜕
 
 𝜕𝜃

 𝜕
 𝜕𝜙 
(3.85)
thus we need to calculate ( 𝐴−1 )𝑇 and the matrix 𝐴 is found easily since we know the
explicit expressions (3.77). The inverse matrix 𝐴−1 can be found using the following
formula
1
adj𝐴,
(3.86)
𝐴−1 =
det 𝐴
where we introduced the adjugate matrix adj𝐴 as
adj𝐴 = 𝐶 𝑇 ,
(3.87)
where 𝐶 is the matrix which matrix elements 𝐶𝑖 𝑗 are defined as
𝐶𝑖 𝑗 = (−1) 𝑖+ 𝑗 𝑀𝑖 𝑗 ,
(3.88)
where 𝑀𝑖 𝑗 is the determinant of the matrix 𝐴 with the 𝑖-th row and 𝑗-th column
erased. For example
𝑀11 =
𝜕𝑦
𝜕𝜃
𝜕𝑧
𝜕𝜃
𝜕𝑦
𝜕𝜙
𝜕𝑧
𝜕𝜙 .
(3.89)
Thus, we now able to calculate 𝐴−1 and the transpose ( 𝐴−1 )𝑇 , which, in its turn,
allows one to express 𝜕/𝜕𝑥, 𝜕/𝜕𝑦, 𝜕/𝜕𝑧 in terms of 𝜕/𝜕𝑟, 𝜕/𝜕𝜃, 𝜕/𝜕𝜙 using (3.85).
Having transformed all the derivatives from Cartesian to spherical coordinates,
it is now straightforward to calculate the angular momentum operators in spherical
coordinates
56
3 Physical observables in quantum mechanics
𝜕
𝜕
b
+ cos 𝜙 cot 𝜃 𝜕𝜙
𝐿 𝑥 = 𝑖ℏ sin 𝜙 𝜕𝜃
𝜕
𝜕
b
− sin 𝜙 cot 𝜃 𝜕𝜙
𝐿 𝑦 = −𝑖ℏ cos 𝜙 𝜕𝜃
b
𝐿 𝑧 = −𝑖ℏ 𝜕
(3.90)
𝜕𝜙
3.3.1.1 Tasks
• Calculate the commutator [b
𝑥 2 , 𝑝b2𝑥 ]
• Knowing the angular momentum operators in spherical coordinates, calculate the
following commutators in spherical coordinates (not in Cartesian): [ b
𝐿𝑖 , b
𝐿 𝑗 ], for
𝑖, 𝑗 = 1, 2, 3.
Chapter 4
Free particles and quantum tunneling
4.1 States of a propagating quantum particle
Let us consider in more detail the states of a free quantum particle. We will start
with 1D case, and then generalize the obtained result for 3D case.
1D time dependent Schrodinger equation for a free particle reads:
𝑖ℏ
2 2
𝜕𝜓
ˆ =−ℏ 𝜕 𝜓
= 𝐻𝜓
𝜕𝑡
2𝑚 𝜕𝑥 2
(4.1)
The Hamiltonian is stationary, so we can apply the procedure of the separation of
variables:
(4.2)
𝜓(𝑥, 𝑡) = 𝜒(𝑡)𝜙(𝑥),
𝑑𝜒
𝑖ℏ
= 𝐸 𝜒,
𝑑𝑡
ℏ2 𝜕 2 𝜙
−
= 𝐸 𝜙.
2𝑚 𝜕𝑥 2
(4.3)
(4.4)
Let us consider the equation for 𝜙(𝑥) Eq. (4.4).
This is a second order differential equation, whose general solution is a linear
combination of two independent solutions, whose form depends on the sign of the
energy 𝐸. Let us consider the cases of negative and positive energies separately.
a) 𝐸 < 0
ℏ2 𝜕 2 𝜙
𝑑 2 𝜙 2𝑚|𝐸 |
−
= −|𝐸 |𝜙 =⇒
=
𝜙 = 𝜅 2 𝜙, 𝜅 =
2
2𝑚 𝜕𝑥
𝑑𝑥 2
ℏ2
𝜙(𝑥) = 𝐶1 𝑒 −𝜅 𝑥 + 𝐶2 𝑒 𝜅 𝑥 .
r
2𝑚|𝐸 |
=⇒
ℏ2
(4.5)
(4.6)
Is this a good solution? Remember, that |𝜓(𝑥, 𝑡)| 2 = |𝜙(𝑥)| 2 - is the probability
density, which should always remain finite in a physically real situation. This is not
57
58
4 Free particles and quantum tunneling
the case with wavefunction Eq. (4.6), which grows to infinity at 𝑥 −→ −∞ (term
𝑒 −𝜅 𝑥 ) or 𝑥 −→ +∞ (term 𝑒 𝜅 𝑥 ). We can conclude from this, that negative values of 𝐸
are not possible for a free quantum particle. Same result, of course, holds in classical
mechanics, where for a free particle
𝐸 =𝑇 +𝑈 =𝑇 =
𝑚𝑣 2
≥ 0.
2
(4.7)
b) 𝐸 > 0
𝑑2 𝜙
2𝑚𝐸
ℏ2 𝜕 2 𝜙
= 𝐸 𝜙 =⇒
= − 2 𝜙 = −𝑘 2 𝜙, 𝑘 =
−
2
2
2𝑚 𝜕𝑥
𝑑𝑥
ℏ
r
2𝑚𝐸
.
ℏ2
(4.8)
So that general solution can be written in the following form:
𝜙 = 𝐶1 𝑒 𝑖𝑘 𝑥 + 𝐶2 𝑒 −𝑖𝑘 𝑥
(4.9)
or, using 𝑒 ±𝑖𝑘 𝑥 = cos 𝑘𝑥 ± 𝑖 sin 𝑘𝑥:
𝜙 = 𝐶1 (cos 𝑘𝑥 + 𝑖 sin 𝑘𝑥) + 𝐶2 (cos 𝑘𝑥 − 𝑖 sin 𝑘𝑥) =
f1 cos 𝑘𝑥 + 𝐶
f2 sin 𝑘𝑥,
= (𝐶1 + 𝐶2 ) cos 𝑘𝑥 + 𝑖(𝐶1 − 𝐶2 ) sin 𝑘𝑥 = 𝐶
(4.10)
f1 = 𝐶1 + 𝐶2 , 𝐶
f2 = 𝑖(𝐶1 − 𝐶2 ).
𝐶
(4.11)
Linear independent solutions can be thus represented either in terms of complex
exponents or in terms of trigonometric functions. Note, that for any value of 𝐸 > 0
2 2
there exist two linearly independent wavefunctions (𝐸 = ℏ2𝑚𝑘 )
𝜙+𝑘 = 𝑒 𝑖𝑘 𝑥 , 𝜙−𝑘 = 𝑒 −𝑖𝑘 𝑥 ,
(4.12)
𝜙 𝑠 = cos 𝑘𝑥, 𝜙 𝑎 = sin 𝑘𝑥.
(4.13)
or
This means, that energy 𝐸 is twice degenerate. This is not surprising, and the
analogical effect holds in classical mechanics - the value of kinetic energy does not
depend on the direction of the propagation of a particle.
Note, that besides the condition 𝐸 > 0, there are no other limitations for the
energy 𝐸, i.e. the spectrum is continuous. This is always the case for the particles
which are not confined in the real space: infinite motion in classical physics will
correspond to a continuous energy spectrum in a quantum problem.
The states 𝜙±𝑘 = 𝐴𝑒 ±𝑖𝑘 𝑥 always remain finite. However, there is still a problem
with their proper normalization. Indeed, one needs that
4.1 States of a propagating quantum particle
1=
∫
+∞
−∞
𝜙∗±𝑘 (𝑥)𝜙±𝑘 (𝑥)𝑑𝑥 = | 𝐴| 2
59
∫
+∞
𝑒 ∓𝑖𝑘 𝑥 𝑒 ±𝑖𝑘 𝑥 𝑑𝑥 = | 𝐴| 2
∫
−∞
+∞
𝑑𝑥 = ∞
−∞
(4.14)
- the integral with infinite limit always diverges.
To get rid of this unwelcomed divergency, one can introduce a fictitious finite
interval [− 𝐿2 ; 𝐿2 ], replacing
∫
+∞
∫
𝑑𝑥 −→
𝑑𝑥
(4.15)
− 𝐿2
−∞
.
𝐿
2
In this case one gets:
∫
𝐿
2
− 𝐿2
𝜙∗±𝑘 (𝑥)𝜙±𝑘 𝑑𝑥 = | 𝐴| 2
1
𝜙±𝑘 (𝑥) = √ 𝑒 ±𝑖𝑘 𝑥 .
𝐿
∫
𝐿
2
− 𝐿2
1
𝑑𝑥 = | 𝐴| 2 𝐿 = 1 =⇒ 𝐴 = √ ,
𝐿
(4.16)
(4.17)
The mentioned formal divergency of normalization integral and the necessity to
introduce the fictitious finite length scale 𝐿 is the general property of the states,
corresponding to a continuous energy spectrum. Of course, the final result of the
calculation of any physical observable should not contain 𝐿, thus limit 𝐿 −→ ∞
should be taken, when calculating values of physical observables.
The obtained expressions 𝜙±𝑘 (𝑥) = √1 𝑒 𝑖𝑘 𝑥 can be combined with the solution
𝐿
of the equation
𝑖ℏ
𝑖𝐸𝑘 𝑡
𝑖ℏ𝑘 2 𝑡
ℏ2 𝑘 2
𝑑𝜒
= 𝐸𝜒 =
𝜒 =⇒ 𝜒(𝑡) = 𝑒 − ℏ = 𝑒 − 2𝑚
𝑑𝑡
2𝑚
(4.18)
to give:
1
ℏ𝑘 2
𝜓±𝑘 (𝑡) = √ 𝑒 𝑖 (±𝑘 𝑥−𝜔𝑘 𝑡) , 𝜔 𝑘 =
.
(4.19)
2𝑚
𝐿
These solutions correspond to the plane waves, propagating in positive (sign "+")
and negative (sign "-") directions of the axis X.
Let us calculate the mean value of the momentum in these states:
∫ 𝐿
∫ 𝐿
2
2
𝜕
𝜕
∗
< 𝑝ˆ 𝑥 >= −𝑖ℏ
𝜓±𝑘
(𝑥, 𝑡) 𝜓±𝑘 (𝑥, 𝑡)𝑑𝑥 = −𝑖ℏ
𝜙∗±𝑘 (𝑥) 𝜙±𝑘 (𝑥)𝑑𝑥 =
𝐿
𝐿
𝜕𝑥
𝜕𝑥
−2
−2
∫ 𝐿
∫ 𝐿
√
2
2
𝜕
− 𝑖ℏ
𝑒 ∓𝑖𝑘 𝑥 𝑒 ±𝑖𝑘 𝑥 (𝑥)𝑑𝑥 = ±ℏ𝑘
𝑑𝑥 = ±ℏ𝑘 = ± 2𝑚𝐸
(4.20)
𝜕𝑥
− 𝐿2
− 𝐿2
- we got an expression for momentum of a classical particle with kinetic energy 𝐸,
propagating in positive or negative direction.
Let us now generalize the obtained result for the case of 3D propagation, when
60
4 Free particles and quantum tunneling
𝜕𝜓(r, 𝑡)
ℏ2 2
=−
∇ 𝜓(r, 𝑡),
𝜕𝑡
2𝑚
𝜓(r, 𝑡) = 𝜒(𝑡)𝜙(r),
𝑑𝜒
𝑖ℏ
= 𝐸 𝜒,
𝑑𝑡
ℏ2 2
−
∇ 𝜙 = 𝐸 𝜙.
2𝑚
𝑖ℏ
(4.21)
(4.22)
(4.23)
(4.24)
The solution of the equation for 𝜙 can be also searched by applying the method
of the separation of the variables x, y, z. Indeed:
ℏ2 𝜕 2
ℏ2 𝜕 2
ℏ2 𝜕 2
ℏ2 2
∇ =−
−
−
= 𝐻ˆ 𝑥 + 𝐻ˆ 𝑦 + 𝐻ˆ 𝑧
𝐻ˆ = −
2
2
2𝑚
2𝑚 𝜕𝑥
2𝑚 𝜕𝑦
2𝑚 𝜕𝑧 2
(4.25)
and therefore we can write:
𝜙(r) = 𝜙 𝑥 (𝑥)𝜙 𝑦 (𝑦)𝜙 𝑧 (𝑧),
(4.26)
ˆ = ( 𝐻ˆ 𝑥 + 𝐻ˆ 𝑦 + 𝐻ˆ 𝑧 )𝜙 𝑥 (𝑥)𝜙 𝑦 (𝑦)𝜙 𝑧 (𝑧) =
𝐻𝜙
𝜙 𝑦 (𝑦)𝜙 𝑧 (𝑧) 𝐻ˆ 𝑥 𝜙 𝑥 (𝑥) + 𝜙 𝑥 (𝑥)𝜙 𝑧 (𝑧) 𝐻ˆ 𝑦 𝜙 𝑦 (𝑦) + 𝜙 𝑥 (𝑥)𝜙 𝑦 (𝑦) 𝐻ˆ 𝑧 𝜙 𝑧 (𝑧) =
𝐸 𝜙 = 𝐸 𝜙 𝑥 (𝑥)𝜙 𝑦 (𝑦)𝜙 𝑧 (𝑧).
(4.27)
Dividing this by 𝜙 = 𝜙 𝑥 𝜙 𝑦 𝜙 𝑧 , one gets:
1 ˆ
1 ˆ
1 ˆ
𝐻 𝑥 𝜙 𝑥 (𝑥) +
𝐻 𝑦 𝜙 𝑦 (𝑦) +
𝐻 𝑧 𝜙 𝑧 (𝑧) = 𝐸 =⇒
𝜙 𝑥 (𝑥)
𝜙 𝑦 (𝑦)
𝜙 𝑧 (𝑧)
𝐻ˆ 𝑥 𝜙 𝑥 (𝑥) = 𝐸 𝑥 𝜙 𝑥 ,
(4.28)
(4.29)
𝐻ˆ 𝑦 𝜙 𝑦 (𝑦) = 𝐸 𝑦 𝜙 𝑦 ,
𝐻ˆ 𝑧 𝜙 𝑧 (𝑧) = 𝐸 𝑧 𝜙 𝑧 ,
(4.30)
𝐸𝑥 + 𝐸𝑦 + 𝐸𝑧 = 𝐸
(4.32)
(4.31)
- we managed to reduce the problem of a solution of 3D partial differential equation
to a solution of three independent ordinary differential equations for 𝜙 𝑥 , 𝜙 𝑦 , 𝜙 𝑧 .
We get:
ℏ2 𝑑 2 𝜙𝑖
= 𝐸 𝑖 𝜙𝑖 (𝑖 = 1, 2, 3; 𝑥1 = 𝑥, 𝑥2 = 𝑦, 𝑥 3 = 𝑧) =⇒
2𝑚 𝑑𝑥 𝑖2
1
𝜙𝑖 = √ 𝑒 ±𝑖𝑘𝑖 𝑥𝑖 ,
𝐿𝑖
−
where
ℏ2 𝑘𝑖2
2𝑚
(4.33)
(4.34)
= 𝐸𝑖
𝜙k (r) = p
1
1
𝑒 ±𝑖𝑘 𝑥 𝑥 𝑒 ±𝑖𝑘𝑦 𝑦 𝑒 ±𝑖𝑘𝑧 𝑧 = √ 𝑒 𝑖kr ,
𝐿𝑥 𝐿𝑦 𝐿𝑧
𝑉
(4.35)
4.1 States of a propagating quantum particle
61
where k = e 𝑥 𝑘 𝑥 + e 𝑦 𝑘 𝑦 + e𝑧 𝑘 𝑧 , 𝑉 = 𝐿 𝑥 𝐿 𝑦 𝐿 𝑧 - fictitious volume introduced to
normalize the wavefunction,
𝐸=
ℏ2 k2
p2
ℏ2 2
(𝑘 𝑥 + 𝑘 2𝑦 + 𝑘 2𝑧 ) =
=
2𝑚
2𝑚
2𝑚
(4.36)
- kinetic energy is the same as for classical particles and represent the sum of the
terms, related to the motion along x, y and z axes.
2 2
Note, that for 1D motion, the states of a given energy 𝐸 = ℏ2𝑚𝑘 were twice
degenerate with respect to the direction of the motion. In the 3D case, they will
be infinitely degenerate, as there are infinite number of non-equivalent possible
directions of the propagation. Combining:
1
𝜓k (r, 𝑡) = 𝜒(𝑡)𝜙k (r) = √ 𝑒 𝑖 (kr−𝜔k 𝑡) ,
𝑉
2
ℏk
ℏ 2
𝜔𝑘 =
=
(𝑘 + 𝑘 2𝑦 + 𝑘 2𝑧 )
2𝑚
2𝑚 𝑥
(4.37)
(4.38)
- we got an expression for a 3D plane wave.
Let us calculate the mean value of the vector of the momentum p̂ = −𝑖ℏ∇ in the
constructed quantum state:
∫
∫
𝑖ℏ
𝑒 −𝑖 (kr−𝜔𝑘 𝑡) ∇𝑒 𝑖 (kr−𝜔𝑡) 𝑑 3 r =
< p >= −𝑖ℏ
𝜓k∗ (r, 𝑡)∇𝜓k (r, 𝑡) = − √
𝑉 𝑉
𝑉
∫
∫
√
𝑖ℏ
ℏk
−√
𝑒 −𝑖kr ∇𝑒 𝑖kr 𝑑 3 r = √
𝑑 3 r = ℏk = n 2𝑚𝐸
(4.39)
𝑉 𝑉
𝑉 𝑉
- we got an expression for the momentum of a free particle propagating in the
direction given by vector n, (k = n𝑘), and having the energy 𝐸.
Let us now introduce a very important concept of probability density current.
Suppose, that you have some conserving quantity, such as mass, charge, etc. By
conserving, we mean, that corresponding density 𝜌(r, 𝑡) can change with time, but
the total value, calculated as thr volume integral over all space remains constant:
∫
𝜌(r, 𝑡)𝑑 3 r = 𝑐𝑜𝑛𝑠𝑡.
(4.40)
𝐴𝑙𝑙 𝑠 𝑝𝑎𝑐𝑒
If we speak about mass density, this will correspond to the conservation of the
total mass, if about charge density - to the conservation of the total charge, and in
quantum mechanics we can put 𝜌(r, 𝑡) = |𝜓(r, 𝑡)| 2 - probability density, for which
normalization does not change with time,
∫
∫
𝜌(r, 𝑡)𝑑 3 r =
|𝜓(r, 𝑡)| 2 𝑑 3 r = 1.
(4.41)
𝐴𝑙𝑙 𝑠 𝑝𝑎𝑐𝑒
For conserving quantity, the continuity equation holds:
62
4 Free particles and quantum tunneling
𝜕𝜌
+ div j = 0,
𝜕𝑡
(4.42)
where j = 𝜌V is the density current (V is the velocity).
Indeed, consider the finite volume Ω bounded by the surface Σ. For conserving
Fig. 4.1
∫
quantity, the change of the 𝑄(𝑡) = Ω 𝜌(r, 𝑡)𝑑 3 r can only occur due to the flux
through the boundary:
∫
∫
∮
𝑑𝑄(𝑡)
𝑑
𝜕𝜌 3
𝜌(r, 𝑡)𝑑 3 r =
=
𝑑 r = − j · 𝑑s.
(4.43)
𝑑𝑡
𝑑𝑡 Ω
Ω 𝜕𝑡
Σ
The sign "-" appears, because if j · 𝑑s > 0, the flux goes outside of Ω, and 𝑄
decreases.
Using Gauss theorem we have:
∮
∫
j · 𝑑s =
div j𝑑 3 r,
(4.44)
Σ
and thus:
∫ Ω
Ω
𝜕𝜌
+ div j 𝑑 3 r = 0
𝜕𝑡
given that Ω is arbitrary, we get the aforementioned continuity equation.
(4.45)
4.2 Scattering of 1D quantum particles, tunneling effect
63
(4.46)
𝜌(r, 𝑡) = 𝜓 ∗ (r, 𝑡)𝜓(r, 𝑡),
ℏ2
𝜕𝜓
=−
∇2 𝜓 + 𝑈𝜓 =⇒
(4.47)
𝜕𝑡
2𝑚
∗
𝑖ℏ 2
𝑖
𝜕𝜓
𝑖ℏ 2 ∗ 𝑖
𝜕𝜓
=
∇ 𝜓 − 𝑈𝜓,
=−
∇ 𝜓 + 𝑈𝜓 ∗
(4.48)
𝜕𝑡
2𝑚
ℏ
𝜕𝑡
2𝑚
ℏ
∗
𝜕𝜌
𝜕
𝜕𝜓
𝜕𝜓
𝑖ℏ
𝑖
𝑖ℏ ∗ 2
𝑖
= 𝜓∗𝜓 = 𝜓
+ 𝜓∗
=−
𝜓∇2 𝜓 ∗ + 𝑈𝜓𝜓 ∗ +
𝜓 ∇ 𝜓 − 𝑈𝜓 ∗ 𝜓 =
𝜕𝑡
𝜕𝑡
𝜕𝑡
𝜕𝑡
2𝑚
ℏ
2𝑚
ℏ
𝑖ℏ 2 ∗
𝑖ℏ
∗
∗
∗ 2
−
div (𝜓∇𝜓 − 𝜓 ∇𝜓) = − div j,
(4.49)
𝜓∇ 𝜓 − 𝜓 ∇ 𝜓 = −
2𝑚
2𝑚
𝑖ℏ
where j =
𝑖ℏ
2𝑚
(𝜓∇𝜓 ∗ − 𝜓 ∗ ∇𝜓)
𝜕𝜌
+ div j = 0.
𝜕𝑡
The expression for j above describes the probability current.
Let us calculate it for the plane wave with 𝜓 = 𝜓k =
√1
𝑉
(4.50)
√1
𝑉
𝑒 𝑖 (kr−𝜔𝑘 𝑡) , 𝜓k∗ =
𝑒 −𝑖 (kr−𝜔𝑘 𝑡) , (𝑉 - volume, not velocity). We get:
𝑖k𝑒 −𝑖 (kr−𝜔𝑘 𝑡)
𝑖k𝑒 𝑖 (kr−𝜔𝑘 𝑡)
∇𝜓k∗ = −
, ∇𝜓k =
=⇒
√
√
𝑉
𝑉
𝑖ℏ
1
1
ℏk
j=
−𝑖k + 𝑖k
=
2𝑚
𝑉
𝑉
𝑚𝑉
(4.51)
(4.52)
- the current is directed parallel to k = n𝑘, i.e. parallel to the direction of the
propagation.
We can relate this to the classical expression for current: j = 𝜌V (V- velocity, 𝑉 p
ℏk
volume). Indeed: 𝜌 = 𝜓 ∗ 𝜓 = 𝑉1 , ℏk
𝑚 = 𝑚 = V, j = 𝑚𝑉 = 𝜌V.
If one uses Madelung representation
p
𝜓(r, 𝑡) = 𝜌(r, 𝑡)𝑒 𝑖 𝜃 (r,𝑡) ,
(4.53)
one gets:
j=
ℏ
𝜌(r, 𝑡)∇𝜃 (r, 𝑡)
𝑚
(4.54)
- check this yourselves.
4.2 Scattering of 1D quantum particles, tunneling effect
After having considered the states of a quantum particle in one dimension, let
us consider, how the presence of external potentials will affect its motion. We start
from simple qualitative discussion, which will allow us to relate the types of classical
motion with the types of quantum energy spectrum.
64
4 Free particles and quantum tunneling
Consider a 1D quantum particle of a mass 𝑚, moving in the external potential
𝑈 (𝑥). Suppose, that 𝑈 (𝑥) = 0 at 𝑥 −→ ±∞. There are two possible situations.
1) 𝑈 (𝑥) < 0 - potential well. Suppose, that the potential reaches its minimum 𝑉0 at 𝑥 = 0:
Fig. 4.2
The total energy of the particle: 𝐸 = 𝑇 + 𝑈, 𝑇 > 0, 𝑈 > −𝑈0 =⇒ 𝐸 > 𝑈0 .
Depending on the sign of the energy 𝐸 in classical physics the cases of finite and
infinite motion can be realized.
a) 𝐸 < 0. There are points, at which total energy is equal to potential energy:
𝑈 (𝑥1,2 ) = 𝐸. The particle can not penetrate beyond these points, as for 𝑥 < 𝑥 1 , 𝑥 >
𝑥2 , 𝐸 < 𝑈 (𝑥), which is not possible, as kinetic energy should be positive.
Points 𝑥 1,2 correspond thus to the turning points. When a classical particle reaches
them, it will be reflected. The motion of the classical particle thus occurs in the
interval 𝑥 ∈ [𝑥1 ; 𝑥2 ], and particle is bouncing between two turning points. The
bigger the energy is, the wider is the region of the classically allowed motion.
For a quantum particle, the presence of confinement will mean the appearance
of the discrete energy states at 𝐸 < 0. This can be seen from the Bohr-Somerfeld
quantization rule:
∮
𝑝(𝑥)𝑑𝑥 = 2
∫
𝑥2 (𝐸)
p
2𝑚 [𝐸 − 𝑈 (𝑥)]𝑑𝑥 = 2𝜋ℏ𝑛.
(4.55)
𝑥1 (𝐸)
In a more precise way, we will need to solve the 1D Schrodinger equation
+ 𝑈 (𝑥)𝜙 = 𝐸 𝜙, with boundary condition 𝜙(𝑥)| 𝑥−→±∞ = 0. The corresponding solution exists only for discrete set of energies (we will consider several
examples later).
2
ℏ2 𝑑 𝜙
− 2𝑚
𝑑 𝑥2
4.2 Scattering of 1D quantum particles, tunneling effect
65
b) E>0. Always E>U(x), and there are thus no turning points. The motion of a
particle is infinite - it comes from 𝑥 = −∞ at 𝑡 = −∞, and goes to 𝑥 = +∞ at 𝑡 = +∞
(or vise versa if velocity is negative).
In quantum physics, infinite motion will correspond to acontinuous energy spectrum, as for the case of ree particle with 𝑈 (𝑥) = 0, considered in detail in the previous
lecture. Indeed, if one would apply the Bohr-Sommerfeld quatization rule,
∫
𝑥2 (𝐸)
𝑝(𝑥, 𝐸)𝑑𝑥 = 𝜋ℏ𝑛,
(4.56)
𝑥1 (𝐸)
one would face an evident problem of the absence of the turning points. Solving
Schrodinger equation with boundary condition 𝜙(𝑥)| 𝑥−→±∞ = 0 will neither give
any solutions for 𝐸 > 0.
2) Now, consider the case of a potential barrier
Fig. 4.3
Again, suppose that maximal value of potential energy 𝑈0 is achieved at 𝑥 = 0.
And again, consider two intervals of the energy.
a) 𝐸 < 𝑈0 . For classical particle, there will be again two turning points, defined
by the equation 𝑈 (𝑥1,2 ) = 𝐸. But now, a classically forbidden region lies in the
interval 𝑥 ∈ [𝑥1 ; 𝑥 2 ]. The particle can thus be located either at 𝑥 < 𝑥 1 , or at 𝑥 > 𝑥 2 .
If a particle propagates from 𝑥 = −∞ at 𝑡 = −∞ in positive direction of x-axis, it
will be reflected back at 𝑥 = 𝑥1 , and go to 𝑥 = −∞ at 𝑡 = +∞ as well. In a similar
way, if a particle propagates from 𝑥 = +∞ at 𝑡 = −∞ in the negative x direction, it
will be reflected at 𝑥 = 𝑥2 and go back to 𝑥 = +∞ at 𝑡 = +∞. Reflection is ideal: for
66
4 Free particles and quantum tunneling
𝐸 < 𝑈0 , the particles can not penetrate from two classically allowed regions, as on
the way they should pass through a region 𝑥 ∈ [𝑥1 ; 𝑥2 ], where their energy should
formally become negative, which is not possible.
b) 𝐸 > 𝑈0 . The equation 𝑈 (𝑥) = 𝐸 has no solution, thus there are no turning
points. The particle propagates from 𝑥 = −∞ at 𝑡 = −∞, to 𝑥 = +∞ at 𝑡 = +∞ without
being stopped by potential barrier (or vice versa). Transmission of a classical particle
for 𝐸 > 𝑈0 will be ideal.
Now, let us show how this situation would change in quantum physics.
Consider the case of a rectangular potential barrier, defined as:


0,



𝑈 (𝑥) = 𝑈0 > 0,


 0,

𝑥 < 0,
𝑥 ∈ [0, 𝐿],
𝑥 > 𝐿.
(4.57)
Fig. 4.4
We need to solve the Schrodinger equation
−
ℏ2 𝑑 2 𝜙
+ 𝑈 (𝑥)𝜙 = 𝐸 𝜙, 𝑥 ∈ [−∞; +∞].
2𝑚 𝑑𝑥 2
(4.58)
Note, that 𝑈 (𝑥) is not a continuous function over the whole interval 𝑥 ∈
[−∞; +∞]. We should thus divide it into 3 parts:
a) Interval 1, 𝑥 < 0.
b) Interval 2, 𝑥 ∈ [0, 𝐿].
c) Interval 3, 𝑥 > 𝐿.
We can solve the Schrodinger equation at all three intervals, and then combine
together the solutions using appropriate boundary conditions, defined as conditions
of the continuity of the wavefunction and its derivative at the points 𝑥 = 0 and 𝑥 = 𝐿.
Continuity of 𝜙(𝑥) and 𝑑𝑑 𝜙𝑥 everywhere is necessary, because Schrodinger equation
contains
𝑑2 𝜙
,
𝑑 𝑥2
which will not be defined if the function 𝜙(𝑥) is not smooth.
4.2 Scattering of 1D quantum particles, tunneling effect
67
Now, let us write solutions of Schrodinger equation at the regions 1,2 and 3.
Let’s start with the case 𝐸 < 𝑈0 . This corresponds to the situation, when a
classical particle can not penetrate through a potential barrier.
Region 1:
ℏ2 𝑑 2 𝜙
= 𝐸 𝜙, 𝐸 > 0,
2𝑚 𝑑𝑥 2
𝑑2 𝜙
= −𝑘 2 𝜙,
𝑑𝑥 2
−
(4.59)
(4.60)
q
where 𝑘 = 2𝑚𝐸
. The general solution of this equation is linear combination of two
ℏ2
complex exponents:
𝜙1 = 𝐶1 𝑒 𝑖𝑘 𝑥 + 𝐵𝑒 −𝑖𝑘 𝑥 ,
(4.61)
where 𝐶1 and 𝐵 are arbitrary (for now) coefficients (complex in general).
Region 2:
ℏ2 𝑑 2 𝜙
+ 𝑈0 𝜙 = 𝐸 𝜙, 𝐸 ∈ [0; 𝑈0 ] =⇒
2𝑚 𝑑𝑥 2
𝑑2 𝜙
= 𝜅 2 𝜙,
𝑑𝑥 2
−
(4.62)
(4.63)
q
where 𝜅 = 2𝑚(𝑈ℏ20 −𝐸) .
The general solution of this equation is a linear combination of real growing and
decaying exponents:
𝜙1 = 𝐶𝑒 𝜅 𝑥 + 𝐷𝑒 −𝜅 𝑥 .
(4.64)
Region 3:
ℏ2 𝑑 2 𝜙
= 𝐸 𝜙, 𝐸 > 0,
2𝑚 𝑑𝑥 2
𝑑2 𝜙
= −𝑘 2 𝜙,
𝑑𝑥 2
(4.65)
−
(4.66)
r
𝜙 = 𝐴𝑒 𝑖𝑘 𝑥 + 𝐶2 𝑒 −𝑖𝑘 𝑥 , 𝑘 =
2𝑚𝐸
.
ℏ2
(4.67)
Before we move further, lets discuss the physical meaning of the solutions in the
regions 1 and 3.
In region 1, 𝜙1 = 𝐶1 𝑒 𝑖𝑘 𝑥 + 𝐵𝑒 −𝑖𝑘 𝑥 corresponds to the linear combination of a wave
propagating from left to right, 𝐶1 𝑒 𝑖𝑘 𝑥 , i.e. the wave hitting the potential barrier, and
wave propagating from right to left, i.e. wave reflected from the barrier. Coefficients
𝐶1 and 𝐵 give the amplitude of these waves and are related to the fluxes = probability
currents falling to the potential barrier:
68
4 Free particles and quantum tunneling
ℏ𝑘
𝑑 ∗ −𝑖𝑘 𝑥 𝑖ℏ
𝑑 𝐶1 𝑒 𝑖𝑘 𝑥
𝐶1 𝑒
− 𝐶1∗ 𝑒 −𝑖𝑘 𝑥
𝐶1 𝑒 𝑖𝑘 𝑥 =
|𝐶1 | 2 ,
2𝑚
𝑑𝑥
𝑑𝑥
𝑚
𝑑 ∗ 𝑖𝑘 𝑥 ℏ𝑘
𝑖ℏ
𝑑 −𝑖𝑘 𝑥 𝐵𝑒 −𝑖𝑘 𝑥
= − |𝐵| 2 .
𝑗𝑟 =
𝐵 𝑒
− 𝐵∗ 𝑒 𝑖𝑘 𝑥
𝐵𝑒
2𝑚
𝑑𝑥
𝑑𝑥
𝑚
𝑗 𝑖𝑛 =
(4.68)
(4.69)
In the same way in region 3 there is 𝜙3 = 𝐴𝑒 𝑖𝑘 𝑥 + 𝐶2 𝑒 −𝑖𝑘 𝑥 , i.e. wave propagating
from left to right, 𝐴𝑒 𝑖𝑘 𝑥 , which should be the transmitted wave, and wave propagating
from right to left, 𝐶2 𝑒 −𝑖𝑘 𝑥 . But if particles are hitting the barrier from the left, there
should be no particles in region 3 propagating from right to left, and thus we should
put 𝐶2 = 0. The transmitted flux will be:
𝑑 ∗ −𝑖𝑘 𝑥 ℏ𝑘 2
𝑑 𝑖𝑘 𝑥 𝑖ℏ
𝐴𝑒 𝑖𝑘 𝑥
=
𝐴 𝑒
− 𝐴∗ 𝑒 −𝑖𝑘 𝑥
𝐴𝑒
| 𝐴| .
(4.70)
𝑗𝑡 =
2𝑚
𝑑𝑥
𝑑𝑥
𝑚
One can introduce the qualitative measure of the transparency of the barrier,
writing the corresponding transmission and reflection coefficients:
𝑇=
𝐴
𝑗𝑡
=
𝑗𝑖𝑛
𝐶1
𝑅=
𝑗𝑟
𝐵
=
𝑗 𝑖𝑛
𝐶1
2
2
= | 𝐴| 2 if 𝐶1 = 1,
(4.71)
= |𝐵| 2 if 𝐶1 = 1.
(4.72)
Transmission and reflection coefficients should not depend on the amplitude of the
incoming wave (as Schrodinger equation is linear) and in all further discussion we
can put 𝐶1 = 1.
As current should be conserved, | 𝑗𝑖𝑛 | = | 𝑗 𝑡 | + | 𝑗𝑟 |, and 𝑇 + 𝑅 = 1. For classical
particles 𝑇 = 0, 𝑅 = 1 if 𝐸 < 𝑈0 .
Now, let us do the math and calculate 𝑇 (𝐸), 𝑅(𝐸) for a quantum particle. We
put 𝐶1 = 1 and 𝐶2 = 0, and write:


𝜙 (𝑥) = 𝑒 𝑖𝑘 𝑥 + 𝐵𝑒 −𝑖𝑘 𝑥 ,

 1

𝜙(𝑥) = 𝜙2 (𝑥) = 𝐶𝑒 𝜅 𝑥 + 𝐷𝑒 −𝜅 𝑥 ,


 𝜙3 (𝑥) = 𝐴𝑒 𝑖𝑘 𝑥 ,

𝑥 < 0,
𝑥 ∈ [0, 𝐿],
𝑥 > 𝐿.
The boundary conditions at 𝑥 = 0 are:
1)
𝜙1 (0) = 𝜙2 (0) =⇒ 1 + 𝐵 = 𝐶 + 𝐷.
2)
𝜙10 (0) = 𝜙20 (0) =⇒ 𝑖𝑘 (1 − 𝐵) = 𝜅(𝐶 − 𝐷).
At 𝑥 = 𝐿 we also have similar boundary conditions:
3)
𝜙2 (𝐿) = 𝜙3 (𝐿) =⇒ 𝐶𝑒 𝜅 𝐿 + 𝐷𝑒 −𝜅 𝐿 = 𝐴𝑒 𝑖𝑘 𝐿 .
4)
(4.73)
(4.74)
(4.75)
(4.76)
4.2 Scattering of 1D quantum particles, tunneling effect
𝜙20 (𝐿) = 𝜙30 (𝐿) =⇒ 𝜅(𝐶𝑒 𝜅 𝐿 − 𝐷𝑒 −𝜅 𝐿 ) = 𝑖𝑘 𝐴𝑒 𝑖𝑘 𝐿 .
69
(4.77)
Combining them together, and replacing 𝐴𝑒 𝑖𝑘 𝐿 −→ 𝐴 (𝑇 = | 𝐴| 2 , and thus phase
factor will not affect the transmission coefficient), one gets the following system of
4 linear algebraic equations for 4 unknown parameters, which can be written in the
following form:

𝐶 + 𝐷 = 1 + 𝐵,




𝐶 − 𝐷 = 𝑖𝑘 (1 − 𝐵),

𝜅
(4.78)
𝐶𝑒 𝜅 𝐿 + 𝐷𝑒 −𝜅 𝐿 = 𝐴,



𝐶𝑒 𝜅 𝐿 − 𝐷𝑒 −𝜅 𝐿 = 𝑖𝑘 𝐴.

𝜅
One can now express 𝐶, 𝐷 from the first two equations (taking their sum and
difference):
1
𝑖𝑘
𝑖𝑘
𝐶=
1+
+ 1−
𝐵 ,
(4.79)
2
𝜅
𝜅
1
𝑖𝑘
𝑖𝑘
𝐷=
1−
+ 1+
𝐵 .
(4.80)
2
𝜅
𝜅
And then put these expressions in the third and fourth equations, which will give:
𝐶𝑒 𝜅 𝐿 + 𝐷𝑒 −𝜅 𝐿 =
1
𝑖𝑘 𝜅 𝐿 1
𝑖𝑘 𝜅 𝐿
1
𝑖𝑘 −𝜅 𝐿 1
𝑖𝑘 −𝜅 𝐿
=
1+
𝑒 +
1−
𝑒 𝐵+
1−
𝑒
+
1+
𝑒
𝐵=
2
𝜅
2
𝜅
2
𝜅
2
𝜅
𝑖𝑘 1 1
𝑖𝑘 1 𝜅 𝐿
1 𝜅𝐿
𝑒 + 𝑒 −𝜅 𝐿 +
𝑒 𝜅 𝐿 − 𝑒 −𝜅 𝐿 +
𝑒 𝜅 𝐿 + 𝑒 −𝜅 𝐿 𝐵 −
𝑒 − 𝑒 −𝜅 𝐿 𝐵 =
=
2
𝜅 2
2
𝜅 2
𝑖𝑘
𝑖𝑘
= ch 𝜅𝐿 + sh 𝜅𝐿 + ch 𝜅𝐿 − sh 𝜅𝐿 𝐵,
(4.81)
𝜅
𝜅
where we introduced hyperbolic cosines and sines as:
1 𝜅𝐿
𝑒 + 𝑒 −𝜅 𝐿 = cos(𝑖𝜅𝐿),
2
1 𝜅𝐿
sh 𝜅𝐿 =
𝑒 − 𝑒 −𝜅 𝐿 = −𝑖 sin(𝑖𝜅𝐿)
2
=⇒ ch2 𝜅𝐿 − sh2 𝜅𝐿 = 1
𝑖𝑘
𝑖𝑘
ch 𝜅𝐿 + sh 𝜅𝐿 + ch 𝜅𝐿 − sh 𝜅𝐿 𝐵 = 𝐴.
𝜅
𝜅
ch 𝜅𝐿 =
In a similar way, Eq. 4 in the system (4.78) gives:
𝑖𝑘
𝑖𝑘
𝑖𝑘
sh 𝜅𝐿 + ch 𝜅𝐿 + sh 𝜅𝐿 − ch 𝜅𝐿 𝐵 = 𝐴.
𝜅
𝜅
𝜅
Dividing Eq. (4.86) by Eq. (4.85) we get:
(4.82)
(4.83)
(4.84)
(4.85)
(4.86)
70
4 Free particles and quantum tunneling
ch 𝜅𝐿 +
ch 𝜅𝐿 + 𝑖𝑘𝜅 sh 𝜅𝐿 +
sh 𝜅𝐿 +
𝑖𝑘
𝜅
ch 𝜅𝐿 𝐵 𝑖𝑘
= .
𝜅
ch 𝜅𝐿 − 𝑖𝑘𝜅 sh 𝜅𝐿 𝐵
sh 𝜅𝐿 −
𝑖𝑘
𝜅
(4.87)
From where we can express 𝐵 as:
2
1 + 𝑘𝜅 2 sh 𝜅𝐿
.
− 1 sh 𝜅𝐿 + 2𝑖𝑘
𝜅 ch 𝜅𝐿
𝐵= 𝑘2
𝜅2
(4.88)
From where we can express the reflection coefficient:
2 2
1 + 𝑘𝜅 2 sh2 𝜅𝐿
,
2
2
2
− 1 sh2 𝜅𝐿 + 4𝑘
ch
𝜅𝐿
𝜅2
𝑅= 𝑘2
𝜅2
(4.89)
which gives for the transmission probability:
2 2
1 + 𝑘𝜅 2 sh2 𝜅𝐿
𝑇 =1−𝑅 =1− =
2 2
2
2
ch
𝜅𝐿
1 − 𝑘𝜅 2 sh2 𝜅𝐿 + 4𝑘
2
𝜅
2
2
2
2
2
ch2 𝜅𝐿
sh2 𝜅𝐿 + 4𝑘
1 − 𝑘𝜅 2 − 1 + 𝑘𝜅 2
𝜅2
=
2 2
2
2
1 − 𝑘𝜅 2 sh2 𝜅𝐿 + 4𝑘
ch
𝜅𝐿
𝜅2
2
4 𝑘𝜅 2 ch2 𝜅𝐿 − sh2 𝜅𝐿
1
=
,
2
2
2
2
2
2
2
2
𝑘
1 𝑘
𝜅
4𝑘
1 − 𝜅 2 sh 𝜅𝐿 + 𝜅 2 ch 𝜅𝐿 ch 𝜅𝐿 + 4 𝜅 − 𝑘 sh 𝜅𝐿
(4.90)
q
q
2𝑚(𝑈0 −𝐸)
,
𝜅
=
.
as 𝑘 = 2𝑚𝐸
ℏ2
ℏ2
Note, that:
1) 𝑇 (𝐸) ≠ 0 - which means, that quantum particle can penetrate through a
potential barrier even if its energy is below the height of the barrier, which does not
happen for classical particles. This remarkable property is known as tunneling effect
and is directly related to the wave nature of quantum particles.
Note, that there is an analog of tunneling effect in classical optics: light can not
propagate inside metals, but can nevertheless, penetrate through thin metallic films.
2) 𝑇 decreases with increase of the thickness of the barrier 𝐿. Indeed, in the
expression for 𝑇, one has ch2 (𝜅𝐿), sh2 (𝜅𝐿) in the denominator, which are increasing
functions of their arguments.
4.2 Scattering of 1D quantum particles, tunneling effect
71
At the same
as 𝜅 decays with growth
q time, 𝑇 is an increasing function of energy,
2
2𝑚(𝑈0 −𝐸)
1 𝑘
𝜅
.
At
𝐸
=
0,
𝑇
=
0
(as
the
term
of 𝐸, 𝜅 =
−
sh2 𝜅𝐿 −→ ∞ at
4 𝜅
𝑘
ℏ2
q
𝐸 −→ 0, as 𝑘 = 2𝑚𝐸
, in the denominator).
ℏ2
2
At 𝑈0 = 𝐸 we get 𝑥 −→ 0, and thus cos(𝜅𝐿) −→ 0, and 41 𝑘𝜅 − 𝑘𝜅 sh 𝜅𝐿 −→
1 𝑘2
4 𝜅2
1
1+
1 2 2
4 𝑘 𝐿 as sh 𝜅𝐿
2
0
= 2𝑚𝑈
.
𝑘 = 2𝑚𝐸
ℏ2
ℏ2
sh 𝜅𝐿 −→
𝑚𝑈0 𝐿 2
2ℏ2
as
' 𝜅𝐿 when 𝜅 −→ 0. Thus 𝑇 (𝑈0 ) =
1
1+ 41 𝑘 2 𝐿 2
=
For large values of 𝐿, when 𝜅𝐿 0 one can obtain approximate expres
sion for transmission coefficient, making replacement ch 𝜅𝐿 = 12 𝑒 𝜅 𝐿 + 𝑒 −𝜅 𝐿 −→
1
1 𝜅𝐿
𝜅 𝐿 − 𝑒 −𝜅 𝐿 −→ 1 𝑒 𝜅 𝐿 =⇒
2 𝑒 , sh 𝜅𝐿 = 2 𝑒
2
𝑇=
4𝑒 −2𝜅 𝐿
2 .
1 + 14 𝑘𝜅 − 𝑘𝜅
(4.91)
Now, let us consider the situation when 𝐸 > 𝑈0 . The only difference with the
case with 𝐸 < 𝑈0 will be the equation in the region 2, which will read:
−
ℏ2 𝑑 2 𝜙
+ 𝑈0 𝜙 = 𝐸 𝜙,
2𝑚 𝑑𝑥 2
(4.92)
which now can be rewritten as:
𝑑 2 𝜙 2𝑚
= 2 (𝑈0 − 𝐸) 𝜙 = −𝜅 2 𝜙,
𝑑𝑥 2
ℏ
where 𝜅 =
q
2𝑚
ℏ2
(4.93)
(𝑈0 − 𝐸), and thus


𝜙 (𝑥) = 𝑒 𝑖𝑘 𝑥 + 𝐵𝑒 −𝑖𝑘 𝑥 ,

 1

𝜙(𝑥) = 𝜙2 (𝑥) = 𝐶𝑒 𝑖𝜅 𝑥 + 𝐷𝑒 −𝑖𝜅 𝑥 ,


 𝜙3 (𝑥) = 𝐴𝑒 𝑖𝑘 𝑥 ,

𝑥 < 0,
𝑥 ∈ [0, 𝐿],
𝑥 > 0.
(4.94)
q
q
2𝑚
(𝑈0 − 𝐸).
𝑘 = 2𝑚𝐸
,
𝜅
=
2
ℏ
ℏ2
One can then do all the same algebra as for 𝐸 < 𝑈0 case, using the same boundary
conditions 𝜙1 (0) = 𝜙2 (0), 𝜙10 (0) = 𝜙20 (0), 𝜙2 (𝐿) = 𝜙3 (𝐿), 𝜙20 (𝐿) = 𝜙30 (𝐿). But
we can also note, comparing the Eq. (4.94) for the wavefunction for 𝐸 > 𝑈0 with
those Eq. (4.78) for 𝐸 < 𝑈0 , that one can get the former making the replacement
𝜅 −→ 𝑖𝜅 in the latter. One can thus directly write the reflection amplitude for 𝐸 > 𝑈0
case, replacing 𝜅 −→ 𝑖𝜅 in the Eq. (4.90) for 𝑇 (𝐸):
72
4 Free particles and quantum tunneling
1
𝑇 (𝐸) =
(4.95)
,
2
ch2 (𝑖𝜅𝐿) + 41 𝑖𝜅𝑘 − 𝑖𝜅𝑘 sh2 (𝑖𝜅𝐿)
1 𝑖𝜅 𝐿
ch(𝑖𝜅𝐿) =
𝑒
+ 𝑒 −𝑖𝜅 𝐿 = cos 𝜅𝐿,
2
1 𝑖𝜅 𝐿
sh(𝑖𝜅𝐿) =
𝑒
− 𝑒 −𝑖𝜅 𝐿 = 𝑖 sin 𝜅𝐿
2
1
=⇒ 𝑇 (𝐸) =
2
cos2 (𝜅𝐿) + 14 𝑘𝜅 + 𝑘𝜅 sin2 (𝑖𝜅𝐿)
𝑘=
1
4
q
2𝑚𝐸
,
ℏ2
𝜅=
q
2𝑚
ℏ2
(4.96)
(4.97)
(4.98)
(𝑈0 − 𝐸). One can note the following:
1) when 𝜅 −→ 0, which means that 𝐸 −→ 𝑈0 , one gets: cos2 (𝜅𝐿) −→ 1, and
2
2
𝑘
𝜅
+
sin2 𝜅𝐿 −→ 14 𝑘𝜅 2 sin2 𝜅𝐿 −→ 14 𝑘 2 𝐿 2 , as sin 𝜅𝐿 ' 𝜅𝐿 when 𝑥 −→ 0.
𝜅
𝑘
Thus 𝑇 (𝑈0 ) =
1
1+ 14 𝑘 2 𝐿 2
=
1
1+
𝑚𝑈0 𝐿 2
2ℏ2
which coincides with the expression obtained at
page 71 for 𝑇 (𝑈0 ).
2) 𝑇 is periodically turning to one, if sin 𝜅𝐿 = 0, 𝜅𝐿 = 𝜋𝑛, 𝑛 = 1, 2, 3 . . .
I.e. transmission of a quantum particle through a rectangular potential barrier for
the case, when 𝐸 > 𝑈q
0 has resonant character, with perfect transmission occurring
when 𝜅𝐿 = 𝜋𝑛, 𝜅 =
2𝑚
ℏ2
(𝑈0 − 𝐸). Using de Broglie relation 𝑝 = ℏ𝜅 =
2𝐿
𝜆
2 𝜋ℏ
𝜆 ,
we
𝜆
2 𝑛, i.e. when one
conclude, that perfect transmission occurs, when
= 𝑛 =⇒ 𝐿 =
can put an integer number of the de Broglie half waves at the length of a barrier. The
situation is equivalent to those observed when light passes through thin film, where
one has the alternating minima and maxima in transmission due to the interference
effect.
As for the classical particles, the situation will be very different, as 𝑇 (𝐸) = 1 for
𝐸 > 𝑈0 for them. Dependence 𝑇 (𝐸) is given by the following plot:
4.3 Supplementary material and tasks
73
Fig. 4.5
4.3 Supplementary material and tasks
4.3.1 Evolution of a wave function from the initial condition
We know that in 1D the evolution of a wave function in the absence of any forces
(i.e. a free particle) obeys the free Schrodinger equation
𝑖ℏ
𝜕𝜓(𝑥, 𝑡)
ℏ2 𝜕 2 𝜓(𝑥, 𝑡)
.
=−
𝜕𝑡
2𝑚 𝜕𝑥 2
(4.99)
It is not difficult to demonstrate, that the general solution of this equation is given by
∫+∞
𝜓(𝑥, 𝑡) =
−∞
here 𝐸 (𝑘) =
(ℏ𝑘) 2
2𝑚
𝑖𝐸 (𝑘) 𝑡
𝑑𝑘
𝑐(𝑘)𝑒 𝑖𝑘 𝑥 𝑒 − ℏ ,
2𝜋
(4.100)
is the energy dispersion of a free particle. Indeed, it is clearly
𝑖𝐸 (𝑘) 𝑡
seen that 𝜙 𝑘 (𝑥, 𝑡) = 𝑐(𝑘)𝑒 𝑖𝑘 𝑥 𝑒 − ℏ does satisfy the Schrodinger equation above
(you can check it by direct substitution) if you consider 𝑘 as just a parameter (and
thus 𝑐(𝑘) is just a constant). If you convinced yourselves that 𝜙 𝑘 (𝑥, 𝑡) is a solution,
∫+∞ 𝑑𝑘
then it is evident that 𝜓(𝑥, 𝑡) =
2 𝜋 𝜙 𝑘 (𝑥, 𝑡) is a solution as well, since one can
−∞
interchange
∫+∞ 𝑑𝑘
−∞
2 𝜋 ...
with the differential operators
𝜕
𝜕𝑡
and
𝜕2
𝜕𝑥 2
in LHS and RHS
of the Schrodinger equation (4.99). In other words, if you plug 𝜙 𝑘 (𝑥, 𝑡) in the
74
4 Free particles and quantum tunneling
Schrodinger equation above and apply
∫+∞ 𝑑𝑘
−∞
2𝜋
to the both parts, the equality will still
hold. We haven’t shown that 𝜓(𝑥, 𝑡), given by Eq. (4.100), is the general solution of
the Schrodinger equation Eq. (4.99), but we have shown that it is a solution. A proof
of the fact that it is indeed general (i.e. any solution has this form) is out of the scope
of the present course.
How to find the coefficients 𝑐(𝑘)? They can be found from the initial condition,
i.e. there is a one-to-one map between 𝑐(𝑘) and 𝜓(𝑥, 𝑡 = 0):
∫+∞
𝑐(𝑘) =
𝑑𝑥𝑒 −𝑖𝑘 𝑥 𝜓(𝑥, 𝑡 = 0).
(4.101)
−∞
The proof of this formula was discussed previously for a slightly different setting.
4.3.1.1 Tasks
Let’s choose as an 2initial condition the well studied wave function of the form
− 𝑥
𝜓0 (𝑥) = 𝜋 1/41√ 𝜎 𝑒 2𝜎2 , i.e. 𝜓(𝑥, 𝑡 = 0) = 𝜓0 (𝑥).
• Find 𝑐(𝑘) for this 𝜓(𝑥, 𝑡 = 0) using Eq. (4.101)
• Knowing 𝑐(𝑘), find 𝜓(𝑥, 𝑡) by plugging 𝑐(𝑘) in Eq. (4.100)
p
• Having found 𝜓(𝑥, 𝑡), calculate for this state 𝜎𝑥 (𝑡) = hb
𝑥 2 i − hb
𝑥 i 2 and 𝜎𝑝𝑥 (𝑡) =
p
∫+∞
b =
b
b is a physical
hb
𝑝 2𝑥 i − hb
𝑝 𝑥 i 2 . A reminder: h 𝐴i
𝑑𝑥𝜓 ∗ (𝑥, 𝑡) 𝐴𝜓(𝑥,
𝑡), where 𝐴
−∞
𝜕
operator of interest, for example b
𝑥 = 𝑥, 𝑝b𝑥 = −𝑖ℏ 𝜕𝑥
.
• Prove that 𝜎𝑥 (𝑡)𝜎𝑝𝑥 (𝑡) ≥ ℏ/2 for any 𝑡 ≥ 0.
4.3.2 Quantum tunneling through a rectangular potential
Consider the following potential


0,



𝑉 (𝑥) = 𝑉1 ,


𝑉2 ,

𝑥<0
0<𝑥<𝐿
𝐿<𝑥
(4.102)
The transmission coefficient is defined as 𝑇 = | 𝑘𝑘31 ||𝑡| 2 and the reflection coefficient
is 𝑅 = |𝑟 | 2 and of course 𝑅 + 𝑇 = 1. As we saw in the chapter above the transmission
coefficient 𝑇 exhibits the Fabry-Perot oscillations (the peaks corresponding to a
resonance tunneling) for 𝐸 > 𝑉1 .
4.3 Supplementary material and tasks
75
Derive the transmission coefficient for the cases 𝑉2 < 𝐸 < 𝑉1 and 0 < 𝐸 < 𝑉2
and collect the three cases all together on a single plot. For this task you may use
Mathematica or any other tools to do the algebra. Assume that 𝑉2 < 𝑉1 .
4.3.3 Quantum tunneling through a 𝜹-barrier
We remind the readers, that the Dirac 𝛿-function satisfying the following properties
∫ +∞
𝛿(𝑥 − 𝑥0 )𝑑𝑥 = 1
(4.103)
−∞
and
∫
+∞
−∞
.
𝛿(𝑥 − 𝑥0 )𝑔(𝑥)𝑑𝑥 = 𝑓 (𝑥0 )
(4.104)
If we consider a potential barrier of the form 𝑉 (𝑥) = 𝛼𝛿(𝑥) it can be thought of as
an infinitelly thin and infinitely high version of the ordinary potential barrier, given
by
(
𝑉0 , −𝐿/2 < 𝑥 < 𝐿/2
𝑉 (𝑥) =
(4.105)
0,
|𝑥| > 𝐿/2
For this potential barrier we have found the transmission coefficient in lectures, for
𝐸 > 𝑉0 it reads
𝑇=
8𝐸 (𝐸 − 𝑉0 )
8𝐸 2 − 8𝐸𝑉0 + 𝑉02 − 𝑉02 cos
√ √
2 2𝐿 𝑚(𝐸−𝑉0 )
ℏ
(4.106)
Task: Consider the delta-potential 𝑉 (𝑥) = 𝛼𝛿(𝑥 − 𝑥0 ), where 𝛼 > 0 and find the
transmission coefficient 𝑇 for positive energies. Show that it yields the same result
for 𝑇 as in Eq. (4.106) in the limit when the barrier (4.105) is infinitelly thin and
infinitelly tall, but the the area beneath the barrier is preserved when taking the limit.
Hint: when solving the scattering problem, you will need to derive a system of
two linear equations. One of them can be obtained from the continuity condition, the
other one – by integrating the stationary Schrodinger equation from 𝑥0 − 𝜖 to 𝑥0 + 𝜖
and then sending 𝜖 to zero.
4.3.4 Task: reflection from a two dimensional step
Derive the analogue of Snell’s law for a plane matter wave hitting a 2D step-potential
𝑉 (𝑥, 𝑦) = 0 for 𝑥 < 0 and 𝑉 (𝑥, 𝑦) = −𝑉0 for 𝑥 > 0.
Chapter 5
Bound states of quantum 1D and 2D particles
5.1 Bound states in 1D square potential well
After having considered the case of a rectangular potential barrier, let us turn to the
situation of a potential well, where there is in classics an interval of the energies, for
which the motion is finite, and thus one can expect discretization of the energy in
the quantum case.
We start from infinitely deep potential well, for which


∞,



𝑈 (𝑥) = 0,


 ∞,

𝑥 < −𝑎,
𝑥 ∈ [−𝑎, 𝑎],
𝑥 > +𝑎.
(5.1)
Fig. 5.1
The motion of a particle is thus restricted in the region 𝑥 ∈ [−𝑎; 𝑎] of the length
𝐿 = 2𝑎, by the presence of two hard walls at 𝑥 = −𝑎 and 𝑥 = +𝑎, beyond which the
77
78
5 Bound states of quantum 1D and 2D particles
particle can not penetrate, as its energy 𝑈 becomes infinite. Classical particle of any
energy will bounce between the walls, so we expect that all energy spectrum of a
quantum system will be discrete.
To find the spectrum, we need to solve the Schrodinger equation
−
ℏ2 𝑑 2 𝜙
= 𝐸 𝜙 for 𝑥 ∈ [−𝑎; 𝑎]
2𝑚 𝑑𝑥 2
(5.2)
(note that 𝑈 = 0 in this region), imposing the following boundary conditions:
𝜙(𝑥)| 𝑥=±𝑎 = 0. Rewriting Schrodinger equation in the form:
r
𝑑2 𝜙
2𝑚𝐸
2
= −𝑘 𝜙, 𝑘 =
,
(5.3)
2
𝑑𝑥
ℏ2
one gets:
(5.4)
𝜙 = 𝐴𝑒 𝑖𝑘 𝑥 + 𝐵𝑒 −𝑖𝑘 𝑥 ,
where constants 𝐴, 𝐵, together with the possible values of 𝑘 (and energy, as 𝐸 =
should be defined from the boundary conditions. One gets:
𝜙(𝑥)| 𝑥=−𝑎 = 𝐴𝑒 −𝑖𝑘 𝑎 + 𝐵𝑒 𝑖𝑘 𝑎 = 0,
𝜙(𝑥)| 𝑥=𝑎 = 𝐴𝑒
𝑖𝑘 𝑎
+ 𝐵𝑒
−𝑖𝑘 𝑎
= 0.
ℏ2 𝑘 2
2𝑚 ,
(5.5)
(5.6)
(5.7)
For parameters 𝐴, 𝐵 one thus has system of two linear homogenous equations
−𝑖𝑘 𝑎 𝑖𝑘 𝑎 𝑒
𝑒
𝐴
= 0,
(5.8)
𝑒 𝑖𝑘 𝑎 𝑒 −𝑖𝑘 𝑎 𝐵
which has nontrivial solution (𝐴 ≠ 0, 𝐵 ≠ 0) only if its determinant is zero. One
thus has:
𝑒 −𝑖𝑘 𝑎 𝑒 𝑖𝑘 𝑎
= 𝑒 −2𝑖𝑘 𝑎 − 𝑒 2𝑖𝑘 𝑎 = −2𝑖 sin(2𝑘𝑎) = 0 =⇒
𝑒 𝑖𝑘 𝑎 𝑒 −𝑖𝑘 𝑎
sin(2𝑘𝑎) = sin(𝑘 𝐿) = 0 =⇒ 𝑘 𝐿 = 𝜋𝑛,
(5.9)
(5.10)
where 𝑛 = 1, 2, . . . (note, that 𝑛 = 0 is not possible, as in this case 𝑒 𝑖𝑘 𝑎 = 𝑒 −𝑖𝑘 𝑎 =
1, 𝜙 = 𝐴 + 𝐵 = 𝑐𝑜𝑛𝑠𝑡, and boundary conditions demand that 𝐴 = −𝐵 so that 𝜙 = 0).
The energy spectrum is thus given by:
𝐸𝑛 =
ℏ2 𝑘 𝑛2 ℏ2 𝜋 2 𝑛2
=
.
2𝑚
2𝑚𝐿 2
(5.11)
Note, that this result coincides with those obtained earlier by use of Bohr-Sommerfeld
quantization rule:
5.1 Bound states in 1D square potential well
∫
𝑥2 (𝐸)
∫
𝑎
𝑝(𝑥, 𝐸)𝑑𝑥 =
𝑥1 (𝐸)
𝐸𝑛 =
ℏ2 𝜋 2 𝑛2
.
2𝑚𝐿 2
√
√
√
2𝑚𝐸 𝑑𝑥 = 2𝑎 2𝑚𝐸 = 𝐿 2𝑚𝐸 = 𝜋ℏ𝑛 =⇒
79
(5.12)
−𝑎
(5.13)
Let us now define the wavefunctions. Consider separately the cases of odd and
even 𝑛.
1) Odd 𝑛 = 1, 3, 5, . . . 2𝑘𝑎 = 𝜋𝑛, 𝑘𝑎 = 𝜋𝑛
2 - half integer number of 𝜋. In this
case: −𝑒 𝑖𝑘 𝑎 = 𝑒 −𝑖𝑘 𝑎 , and Eq. (5.8) reduced to: 𝐴 − 𝐵 = 0 =⇒ 𝐴 = 𝐵 =⇒ 𝜙(𝑥) =
𝜋𝑛
𝐴(𝑒 −𝑖𝑘 𝑥 + 𝑒 𝑖𝑘 𝑥 ) = 2𝐴 cos(𝑘𝑥) = 𝐴ˆ cos(𝑘𝑥) with 𝑘 = 𝑘 𝑛 = 2𝑎
= 𝜋𝑛
𝐿 . Normalization
ˆ
constant 𝐴 = 2𝐴 can be determined from the condition:
∫ +𝑎
∫ +𝑎
∫ +𝑎
𝜋𝑛𝑥 cos2
𝜙∗ (𝑥)𝜙(𝑥)𝑑𝑥 =
𝜙2 (𝑥)𝑑𝑥 = 𝐴ˆ 2
𝑑𝑥 =
2𝑎
−𝑎
−𝑎
−𝑎
∫ +𝑎
𝜋𝑛𝑥 𝑥=𝑎 1 + cos 𝜋𝑛𝑥
𝑎
𝑎
=
𝐴ˆ 2
𝑑𝑥 = 𝐴ˆ 2 𝑎 +
sin
2
𝜋𝑛
𝑎
−𝑎
𝑥=−𝑎
1
𝐴ˆ 2 𝑎 = 1 =⇒ 𝐴ˆ = √ =⇒
(5.14)
𝑎
𝜋𝑛𝑥 r 2
𝜋𝑛𝑥 1
𝜙 𝑛 (𝑥) = √ cos
=
cos
,
(5.15)
2𝑎
𝐿
𝐿
𝑎
with 𝑛 = 1, 3, 5, . . .
Note, that these functions correspond to the standing waves formed between
the walls, and are symmetric (even) with respect to the inversion of the sign of x:
𝜙 𝑛 (𝑥) = 𝜙 𝑛 (−𝑥).
Now, consider the case when 𝑛 = 2, 4, 6, . . . - even number. One gets: 2𝑘 𝑛 𝑎 =
𝜋𝑛, 𝑘 𝑛 𝑎 = 𝜋𝑛
2 - integer number of 𝜋. In this case, 𝐴 + 𝐵 = 0 =⇒ 𝐴 = −𝐵, and
𝜋𝑛
ˆ
𝜙(𝑥) = 𝐴(𝑒 𝑖𝑘 𝑥 − 𝑒 −𝑖𝑘 𝑥 ) = 2𝑖 𝐴 sin(𝑘𝑥) = 𝐴ˆ sin(𝑘𝑥), 𝑘 = 𝑘 𝑛 = 2𝑎
=∫ 𝜋𝑛
𝐿 . Again, 𝐴 =
𝑎 2
1
√ , from the condition of the normalization of the wavefunction,
𝜙 (𝑥)𝑑𝑥 = 1.
−𝑎
𝑎
Thus:
𝜋𝑛𝑥 𝜋𝑛𝑥 1
1
= √ sin
, 𝑛 = 2, 4, 6, . . .
(5.16)
𝜙 𝑛 (𝑥) = √ sin
2𝑎
𝐿
𝐿
2𝑎
We got again expressions, describing the standing waves between the walls, but
now inversion of the sign of x leads to inversion of the sign of 𝜙(𝑥) as well, i.e.
functions 𝜙 𝑛 (𝑥) for even 𝑛 are antisymmetric (odd): 𝜙 𝑛 (𝑥) = −𝜙 𝑛 (−𝑥).
Note, that potential 𝑈 (𝑥) itself is even, 𝑈 (𝑥) = 𝑈 (−𝑥). Let us summarize the
properties of 1D wavefunction for this particular case of symmetric (even) potential
𝑈 (𝑥):
1) Wavefunctions 𝜙 𝑛 (𝑥) can be chosen to be real, i.e. they correspond to nonpropagating states (standing waves).
2) Wavefunctions 𝜙 𝑛 (𝑥) are either symmetric (even), 𝜙 𝑛 (𝑥) = 𝜙 𝑛 (−𝑥) or antisymmetric (odd), 𝜙 𝑛 (𝑥) = −𝜙 𝑛 (−𝑥).
80
5 Bound states of quantum 1D and 2D particles
3) The wavefunction corresponding to the ground state (state with lowest energy),
𝑛 = 1, is symmentric, 𝜙1 (𝑥) = 𝜙1 (−𝑥). These three statements hold for any 1D
symmetric potential 𝑈 (𝑥) = 𝑈 (−𝑥).
Now, let us consider the more realistic case of a rectangular potential well of the
finite depth, i.e. the case


0,



𝑈 (𝑥) = −𝑈0 ,


 0,

𝑥 < −𝑎,
𝑥 ∈ [−𝑎, 𝑎],
𝑥 > 𝑎.
(5.17)
Fig. 5.2
This kind of a model potential will describe, e.g. the situation when a layer of
semiconductor material of one sort is placed between materials of another sort, i.e.
semiconductor quantum well. We now need to solve a Schrodinger equation
−
ℏ2 𝑑 2 𝜙
+ 𝑈 (𝑥)𝜙 = 𝐸 𝜙
2𝑚 𝑑𝑥 2
(5.18)
with boundary conditions 𝜙(𝑥)| 𝑥−→±∞ = 0.
Note, that as a classical particle reveals finite motion only for −𝑈0 < 𝐸 < 0, we
expect that solutions with this boundary condition exist only for a discrete set of
energies, lying in the same interval −𝑈0 < 𝐸 < 0.
Also, note, that in the regions 𝑥 < −𝑎, 𝑥 > 𝑎, which are classically forbidden,
we can not demand that 𝜙(𝑥) = 0, we can only say that 𝜙(𝑥) should decay in them
at infinities. This makes the problem different from the case of hard wall potential.
Similar to what was done for the case of a rectangular potential barrier, let us
divide the space into three regions (region 1: 𝑥 < −𝑎; region 2: 𝑥 ∈ [−𝑎; 𝑎];
region 3: 𝑥 > 𝑎), solve Schrodinger equation in all of them, and then combine the
5.1 Bound states in 1D square potential well
81
Fig. 5.3
three solutions using the boundary conditions at 𝑥 = ±𝑎 points, which demand that
𝜙(𝑥) and its first derivative 𝜙 0 (𝑥) are continuous: 𝜙1 (−𝑎) = 𝜙2 (−𝑎), 𝜙10 (−𝑎) =
𝜙20 (−𝑎), 𝜙2 (𝑎) = 𝜙3 (𝑎), 𝜙20 (𝑎) = 𝜙30 (𝑎).
Region 1: 𝑈 (𝑥) = 0, (𝐸 ∈ [−𝑈0 ; 0])
ℏ2 𝑑 2 𝜙1
= 𝐸 𝜙1 ,
2𝑚 𝑑𝑥 2
𝑑 2 𝜙1
= 𝜅 2 𝜙1 , 𝑥 < −𝑎,
𝑑𝑥 2
−
(5.19)
(5.20)
q
q
2𝑚|𝐸 |
where 𝜅 = − 2𝑚𝐸
=
=⇒ 𝜙1 = 𝐶𝑒 𝜅 𝑥 + 𝐶1 𝑒 −𝜅 𝑥 - general solution. Now let
ℏ2
ℏ2
us impose boundary condition 𝜙1 (𝑥)| 𝑥−→−∞ = 0. The term 𝐶1 𝑒 −𝜅 𝑥 grows to infinity
when 𝑥 −→ −∞, so, to satisfy the boundary condition we need to put 𝐶1 = 0, and
write: 𝜙1 = 𝐶𝑒 𝜅 𝑥 .
Region 2: 𝑈 (𝑥) = −𝑈0 (𝐸 ∈ [−𝑈0 ; 0])
ℏ2 𝑑 2 𝜙2
− 𝑈0 𝜙 2 = 𝐸 𝜙 2 ,
2𝑚 𝑑𝑥 2
𝑑 2 𝜙2
= −𝑘 2 𝜙2 , 𝑥 ∈ [−𝑎; 𝑎],
𝑑𝑥 2
−
(5.21)
(5.22)
q
q
2𝑚(𝑈0 −|𝐸 |)
0)
where 𝑘 = 2𝑚(𝐸+𝑈
=
. The general solution is: 𝜙2 = 𝐴𝑒 𝑖𝑘 𝑥 + 𝐵𝑒 −𝑖𝑘 𝑥 ,
ℏ2
ℏ2
and we keep both terms, as non of them violates the boundary conditions.
Region 3: 𝑈 (𝑥) = 0 (𝐸 ∈ [−𝑈0 ; 0])
82
5 Bound states of quantum 1D and 2D particles
ℏ2 𝑑 2 𝜙3
= 𝐸 𝜙3 ,
2𝑚 𝑑𝑥 2
𝑑 2 𝜙3
= 𝜅 2 𝜙3 , 𝑥 > 𝑎,
𝑑𝑥 2
(5.23)
−
(5.24)
q
q
2𝑚 |𝐸 |
where 𝜅 = − 2𝑚𝐸
=
.
2
ℏ
ℏ2
The general solution is: 𝜙3 (𝑥) = 𝐷𝑒 −𝜅 𝑥 + 𝐷 1 𝑒 𝜅 𝑥 , but the term 𝐷 1 𝑒 𝜅 𝑥 grows to
infinity at 𝑥 −→ +∞, so to satisfy the boundary condition 𝜙(𝑥)| 𝑥−→+∞ = 0, we need
to put 𝐷 1 = 0 and write: 𝜙3 = 𝐷𝑒 −𝜅 𝑥 . Thus we get:


𝜙 = 𝐶𝑒 𝜅 𝑥 ,

 1

𝜙 = 𝜙2 = 𝐴𝑒 𝑖𝑘 𝑥 + 𝐵𝑒 −𝑖𝑘 𝑥 ,


 𝜙3 = 𝐷𝑒 −𝜅 𝑥 ,

𝑥 < −𝑎,
𝑥 ∈ [−𝑎; 𝑎],
𝑥 > 𝑎.
(5.25)
To
coefficients 𝐴, 𝐵, 𝐶, 𝐵 together with energy 𝐸 (it enters into 𝜅 =
q define the q
2𝑚|𝐸 |
, 𝑘 = 2𝑚(𝑈ℏ02− |𝐸 |) ) we use boundary conditions at 𝑥 = ±𝑎:
ℏ2
(5.26)
𝜙1 (−𝑎) = 𝜙2 (−𝑎) =⇒ 𝐶𝑒 −𝜅 𝑎 = 𝐴𝑒 −𝑖𝑘 𝑎 + 𝐵𝑒 𝑖𝑘 𝑎 ,
𝜙10 (−𝑎)
=
𝜙20 (−𝑎)
=⇒ 𝜅𝐶𝑒
𝜙2 (𝑎) = 𝜙3 (𝑎) =⇒ 𝐷𝑒
−𝜅 𝑎
−𝜅 𝑎
= 𝑖𝑘 ( 𝐴𝑒
= 𝐴𝑒
𝑖𝑘 𝑎
−𝑖𝑘 𝑎
+ 𝐵𝑒
− 𝐵𝑒
−𝑖𝑘 𝑎
𝑖𝑘 𝑎
), ,
,
𝜙20 (𝑎) = 𝜙30 (𝑎) =⇒ −𝜅𝐷𝑒 −𝜅 𝑎 = 𝑖𝑘 ( 𝐴𝑒 𝑖𝑘 𝑎 − 𝐵𝑒 −𝑖𝑘 𝑎 ), .
(5.27)
(5.28)
(5.29)
Dividing Eq. (5.27) by Eq. (5.26) in this system, one gets:
𝑖𝑘 ( 𝐴𝑒 −𝑖𝑘 𝑎 − 𝐵𝑒 𝑖𝑘 𝑎 )
= 𝜅 =⇒
𝐴𝑒 −𝑖𝑘 𝑎 + 𝐵𝑒 𝑖𝑘 𝑎
(𝜅 − 𝑖𝑘)𝑒 −𝑖𝑘 𝑎 𝐴 + (𝜅 + 𝑖𝑘)𝑒 𝑖𝑘 𝑎 𝐵 = 0.
(5.30)
(5.31)
And dividing Eq. (5.29) by Eq. (5.28):
𝑖𝑘 ( 𝐴𝑒 𝑖𝑘 𝑎 − 𝐵𝑒 −𝑖𝑘 𝑎 )
= −𝜅 =⇒
𝐴𝑒 𝑖𝑘 𝑎 + 𝐵𝑒 −𝑖𝑘 𝑎
(𝜅 + 𝑖𝑘)𝑒 𝑖𝑘 𝑎 𝐴 + (𝜅 − 𝑖𝑘)𝑒 −𝑖𝑘 𝑎 𝐵 = 0.
(5.32)
(5.33)
Thus, for coefficients 𝐴, 𝐵, one has the following system of linear homogeneous
algebraic equations:
(𝜅 − 𝑖𝑘)𝑒 −𝑖𝑘 𝑎 (𝜅 + 𝑖𝑘)𝑒 𝑖𝑘 𝑎
𝐴
= 0,
(5.34)
(𝜅 + 𝑖𝑘)𝑒 𝑖𝑘 𝑎 (𝜅 − 𝑖𝑘)𝑒 −𝑖𝑘 𝑎 𝐵
This system has non-trivial solution (𝐴 ≠ 0, 𝐵 ≠ 0) only when its determinant is
zero, which gives:
5.1 Bound states in 1D square potential well
83
(𝜅 − 𝑖𝑘)𝑒 −𝑖𝑘 𝑎 (𝜅 + 𝑖𝑘)𝑒 𝑖𝑘 𝑎
= (𝜅 − 𝑖𝑘) 2 𝑒 −2𝑖𝑘 𝑎 − (𝜅 + 𝑖𝑘) 2 𝑒 2𝑖𝑘 𝑎 = 0
(5.35)
(𝜅 + 𝑖𝑘)𝑒 𝑖𝑘 𝑎 (𝜅 − 𝑖𝑘)𝑒 −𝑖𝑘 𝑎
2
𝜅 − 𝑖𝑘
=⇒
= 𝑒 4𝑖𝑘 𝑎 .
(5.36)
𝜅 + 𝑖𝑘
q
q
|
2𝑚(𝑈0 −|𝐸 |)
,
𝑘
=
, so it is an equation for the single
In this equation 𝜅 = 2𝑚ℏ|𝐸
2
ℏ2
unknown variable 𝐸.
Now, use the Euler representation of complex numbers:
p
𝑘 𝜅 − 𝑖𝑘
𝑘
𝜅 ± 𝑖𝑘 = 𝜅 2 + 𝑘 2 𝑒 ±𝑖 arctan 𝜅 ,
= 𝑒 −2𝑖 arctan 𝜅 =⇒
𝜅 + 𝑖𝑘
2
−2𝑖 arctan 𝑘𝜅
4𝑖𝑘 𝑎
𝑒
=𝑒
.
(5.37)
(5.38)
Taking the square root one gets:
𝑒 −2𝑖 arctan 𝜅 = ±𝑒 2𝑖𝑘 𝑎 .
𝑘
(5.39)
The sign ± here defines two sets of the solutions, corresponding to symmetric and
antisymmetric states, as we will see.
1) Class 1, take minus sign
𝑒 −2𝑖 arctan 𝜅 = −𝑒 2𝑖𝑘 𝑎 = 𝑒 𝑖 (2𝑘 𝑎+ 𝜋) =⇒
𝑘
𝑘
𝜋
− 2 arctan = 2𝑘𝑎 + 𝜋 =⇒ − arctan = 𝑘𝑎 +
𝜅
𝜅
2
1
𝑘
𝜋
= cotan(𝑘𝑎) =
= − tan 𝑘𝑎 +
=⇒
𝜅
2
tan(𝑘𝑎)
𝑘 tan(𝑘𝑎) = 𝜅,
𝑘
(5.40)
(5.41)
(5.42)
(5.43)
q
q
|
2𝑚(𝑈0 −|𝐸 |)
,
𝑘
=
. Let us analyze the solutions of this equation
where 𝜅 = 2𝑚ℏ|𝐸
2
ℏ2
graphically. Express first 𝜅 via 𝑘. One has:
2𝑚𝑈0 2𝑚|𝐸 |
−
= 𝛾2 − 𝜅2,
(5.44)
ℏ2
ℏ2
p
p
0
𝛾 2 − 𝑘 2 , and we get 𝑘 tan(𝑘𝑎) = 𝛾 2 − 𝑘 2 =⇒
where 𝛾 2 = 2𝑚𝑈
2 . Then 𝜅 =
ℏ
q
2
tan(𝑘𝑎) = 𝛾𝑘 2 − 1. This is a transcendental equation, which can not be solved
analytically. But its graphical analysis will allow us to understand how many discrete
states exists depending on the depth of the well 𝑈0 and its width 𝐿 = 2𝑎.
Graphically, solutions
are defined by the intersections of the graphs for 𝑓 =
q
𝑘2 =
tan(𝑘𝑎) and 𝑓 =
𝛾2
𝑘2
− 1. The latter function is real only for 𝑘 ≤ 𝛾, which means
84
5 Bound states of quantum 1D and 2D particles
q
q
2𝑚𝑈0
(𝑈
that 2𝑚
+
𝐸)
≤
=⇒ 𝐸 < 0 - this is how it should be, there are no
0
2
ℏ
ℏ2
bound states for 𝐸 > 0 as classical motion for this case is infinite.
q
2
Fig. 5.4 Curve 1 for 𝑓 = 𝛾𝑘 2 − 1 : 𝛾 <
q
2
for 𝑓 = 𝛾𝑘 2 − 1 : 𝛾 ∈ 2𝑎𝜋 ; 3𝑎𝜋 .
𝜋
𝑎.
Curve 2 for 𝑓 =
q
𝛾2
𝑘2
−1 : 𝛾 ∈
2𝜋
𝑎; 𝑎
𝜋
. Curve 3
q Let us see, what happens with the solutions if one changes the parameter 𝛾 =
2𝑚𝑈0
. For a situation of a small 𝛾, when 𝛾 < 𝑎𝜋 , so that 𝛾𝑎
𝜋 < 1 there is only one
ℏ2
q
2
intersection between the curve 𝑓 = 𝛾𝑘 2 − 1 and one branch of the curve 𝑓 = tan(𝑘𝑎)
q
2𝑚𝑈0
𝑎
situation of small 𝛾𝑎
correspond to shallow (small 𝑈0 ) and narrow (small
𝜋 = 𝜋
ℏ2
𝑎) wells. Note, that it does not matter how small 𝛾𝑎 is, one intersection always exist,
which means that a rectangular quantum well always has at least one bound state.
This statement can be generalized for any symmetric potential
well with arbitrary
q
2
𝛾
𝑈 (𝑥) = 𝑈 (−𝑥) if 𝑎𝜋 < 𝛾 < 2𝑎𝜋 , 1 ≤ 𝛾𝑎
− 1 intersects with two
𝜋 ≤ 2 the curve
𝑘2
branches of tan(𝑘𝑎), so that there are two bound states
of
the
considered class 1.
q
2𝑚𝑈0
𝑎
one will get more and
Increasing the dimensionless parameter 𝛾𝑎
𝜋 = 𝜋
ℏ2
more and more states.
It can be easily demonstrated, that solutions belonging to class 1, when tan(𝑘𝑎) =
𝜅
correspond
to symmetric (even) wavefunctions with 𝜙(𝑥) = 𝜙(−𝑥). Indeed, in
𝑘
the region 2 we have: 𝜙2 (𝑥) = 𝐴𝑒 𝑖𝑘 𝑥 + 𝐵𝑒 −𝑖𝑘 𝑥 , 𝜙2 (−𝑥) = 𝐴𝑒 −𝑖𝑘 𝑥 + 𝐵𝑒 𝑖𝑘 𝑥 , so that
symmetric solutions correspond to 𝜙2 (𝑥) = 𝜙2 (−𝑥) =⇒ 𝐴 = 𝐵. According to
Eq. (5.34) on page 82, this will mean that
5.1 Bound states in 1D square potential well
85
(𝜅 − 𝑖𝑘)𝑒 −𝑖𝑘 𝑎 + (𝜅 + 𝑖𝑘)𝑒 𝑖𝑘 𝑎 = 0 =⇒
𝜅 − 𝑖𝑘
𝜅
= −𝑒 2𝑖𝑘 𝑎 tan(𝑘𝑎) = .
𝜅 + 𝑖𝑘
𝑘
(5.45)
(5.46)
Class 2 solutions.
Now, let us take sign "+" in the equation on the page 83:
𝑒 −2𝑖 arctan 𝜅 = 𝑒 2𝑖𝑘 𝑎 − arctan
𝑘
𝑘
= 𝑘𝑎 =⇒
𝜅
𝑘
tan(𝑘𝑎) = − .
𝜅
(5.47)
(5.48)
This can be rewritten as:
r
cotan(𝑘𝑎) = −
𝛾2
− 1,
𝑘2
(5.49)
which can be again analyzed graphically. Note, that differently from the solutions of
q
2
Fig. 5.5 Curves 1, 2, 3 in this plot for the function 𝑓 = − 𝛾𝑘 2 − 1 correspond to the increasing
values of the parameter 𝛾.
q
𝜋
2𝑎 2𝑚𝑈0
class 1, solutions of class 2 do not always exist, if 𝛾 < 2𝑎
=⇒ 2𝛾𝑎
=
≤
𝜋
𝜋
ℏ2
q
2
1. The curves 𝑓 = cotan(𝑘𝑎) and 𝑓 = − 𝛾𝑘 2 − 1 do not intersect. Increasing
of the parameter 𝛾𝑎 will lead to the appearance of more and more intersections,
corresponding to the bound states of class 2. One can show, that solutions of class
2 correspond to antisymmetric wavefunctions. Indeed, in the region 2, 𝜙2 (𝑥) =
86
5 Bound states of quantum 1D and 2D particles
𝐴𝑒 𝑖𝑘 𝑥 + 𝐵𝑒 −𝑖𝑘 𝑥 , 𝜙2 (−𝑥) = 𝐴𝑒 −𝑖𝑘 𝑥 + 𝐵𝑒 𝑖𝑘 𝑥 , and condition 𝜙2 (𝑥) = −𝜙2 (−𝑥) means
that 𝐴 = −𝐵. According to the equation at the page 82, this means that
(𝜅 − 𝑖𝑘)𝑒 −𝑖𝑘 𝑎 − (𝜅 + 𝑖𝑘)𝑒 𝑖𝑘 𝑎 = 0 =⇒
𝜅 − 𝑖𝑘
= 𝑒 2𝑖𝑘 𝑎 =⇒
𝜅 + 𝑖𝑘
𝜅
cotan(𝑘𝑎) = − .
𝑘
(5.50)
(5.51)
(5.52)
Let us summarize the properties of the solutions:
1) They are either symmetric or antisymmetric.
2) The ground state is symmetric.
3) Symmetric and antisymmetric solutions alternate.
5.2 Quantum harmonic oscillator
In classical physics, one dimensional harmonic oscillator is realized, when the returning force acting on the mass 𝑚 is proportional to the displacement of a particle
from its equilibrium position and is -A.O.E. directed opposite to it. The high school
example is a mass 𝑚 on a spring.
Fig. 5.6
In this case, 𝐹 = −𝑘𝑥, where 𝑘 is a constant characterizing the spring, and
equation of motion (classical) reads:
𝑚𝑤 = 𝑚
𝑑2𝑥
= 𝐹 = −𝑘𝑥 =⇒
𝑑𝑡 2
𝑑2𝑥
= −𝜔2 𝑥,
𝑑𝑡 2
q
where 𝜔 = 𝑚𝑘 .
Potential energy corresponding to 𝐹 = −𝑘𝑥 is:
(5.53)
(5.54)
5.2 Quantum harmonic oscillator
87
𝑑𝑈
= −𝑘𝑥 =⇒
𝑑𝑥
𝑘𝑥 2 𝑚𝜔2 𝑥 2
𝑈=
=
,
2
2
𝑈 (𝑥)| 𝑥→±∞ −→ 0.
(5.55)
−
(5.56)
(5.57)
It does not matter what the energy of the particle is, the classical motion is always
Fig. 5.7
2 2
finite, as the equation for the turning points 𝑈 (𝑥) = 𝐸 = 𝑚𝜔2 𝑥 has always solutions,
q
2𝐸
𝑥1,2 = ± 𝑚𝜔
2 . Classical equation of motion can be easily solved:
𝑑2𝑥
= −𝜔2 𝑥 =⇒
𝑑𝑡 2
𝑥(𝑡) = 𝐴 cos(𝜔𝑡) + 𝐵 sin(𝜔𝑡),
(5.58)
(5.59)
where coefficients 𝐴, 𝐵 are defined by initial displacement of the particle at 𝑡 =
0, 𝑥(0), and its initial velocity, 𝑉 (0) =
= 𝜔(−𝐴 sin(𝜔𝑡) + 𝐵 cos(𝜔𝑡))|𝑡=0 =
𝑑𝑥
𝑑𝑡
𝑡=0
𝜔𝐵 =⇒ 𝐵 = 𝑉 𝜔(0) , and 𝑥(𝑡) = 𝑥(0) cos(𝜔𝑡) + 𝑉 𝜔(0) sin(𝜔𝑡).
The particle thus experiences harmonic oscillations with a period 𝑇 = 2𝜔𝜋 which
is not depending on the energy of the particle. The latter property is unique to a
harmonic oscillator, for any other confining potential the period will depend on
energy.
In a quantum world, of course, it is hard to imagine masses on springs. The
problem of a harmonic oscillator, however, is ubiquitous in quantum mechanics.
2
2
One of the reasons is, that the harmonic potential 𝑈 (𝑥) = 𝑚𝜔
2 𝑥 gives good energy
approximation for almost any potential 𝑈 (𝑥) which has a minimum. Indeed, suppose
that potential 𝑈 (𝑥) has minimum at 𝑥 = 0. If 𝐸 is small, then 4𝑥 is small as well
88
5 Bound states of quantum 1D and 2D particles
Fig. 5.8
(the region of classically allowed motion), and for 𝑈 (𝑥) one can use the Taylor
decomposition
𝑈 (𝑥) ≈ 𝑈 (0) +
where
𝑑𝑈
𝑑𝑥
𝑥+
𝑥=0
1 𝑑 2𝑈
2 𝑑𝑥 2
𝑥 2 = 𝑈 (0) +
𝑥=0
1 𝑑 2𝑈
2 𝑑𝑥 2
𝑥2 =
𝑥=0
1 𝑑 2𝑈
2 𝑑𝑥 2
𝑥2,
𝑥=0
(5.60)
= 0, as 𝑈 (𝑥) has minimum at that point, and we dropped 𝑈 (0) as it
𝑑𝑈
𝑑𝑥
𝑥=0
is just a constant addition to potential energy which does not affect any physically
observable quantity.
We said that energy 𝐸 should be "small", but energy is a dimensional quantity,
so we need to specify, with respect to what other characteristic parameter it is small,
i.e. introduce a dimensionless small parameter which controls the goodness of our
harmonic approximation. It can be introduced in the following way.
In harmonic potential approximation, turning points are:
1 𝑑 2𝑈
𝑥 2 = 𝐸,
(5.61)
2 𝑑𝑥 2 𝑥=0
v
u
u
u
2𝐸
t
𝑥=± .
(5.62)
2
𝑑 𝑈
𝑑 𝑥2
𝑥=0
In turning points all the energy is potential. On the other hand, next term in the
Taylor expansion of 𝑈 (𝑥) will be:
1 𝑑 2𝑈
1 𝑑 3𝑈
2
𝑈 (𝑥) ≈
𝑥 +
𝑥3 .
(5.63)
2 𝑑𝑥 2 𝑥=0
3! 𝑑𝑥 3 𝑥=0
5.2 Quantum harmonic oscillator
89
The ratio of the third term to the second term should be small, which gives
𝑑 3𝑈
𝑑 𝑥3
𝑥
𝑥=0
3
𝑑 2𝑈
𝑑 𝑥2
𝑑 3𝑈
𝑑 𝑥3
√
2𝐸
1,
𝑥=0
≈
3
𝑑 2𝑈
𝑑 𝑥2
23
𝑥=0
(5.64)
𝑥=0
which means that approximation works, if
𝑑 2𝑈
𝐸
𝑑𝑥 2
If potential 𝑈 (𝑥) is symmetric and
3
𝑥=0
𝑑 3𝑈
𝑑 𝑥3
𝑑 3𝑈
𝑑𝑥 3
−2
(5.65)
𝑥=0
= 0, then one needs to retain the 𝑥 4
𝑥=0
term,
1 𝑑 2𝑈
2 𝑑𝑥 2
𝑈 (𝑥) ≈
𝑥2 +
𝑥=0
1 𝑑 4𝑈
4! 𝑑𝑥 4
𝑥4,
(5.66)
𝑥=0
and similar analysis will give the condition for energy
𝑑 4𝑈
𝑑 𝑥4
𝑥=0
12
𝑑 2𝑈
𝑑 𝑥2
𝐸
𝑥2
=
6
𝑥=0
2 2
𝑑 𝑈
𝑑𝑥 2
𝑑 4𝑈
𝑑 𝑥4
𝑥=0
𝑑 2𝑈
𝑑 𝑥2
𝐸
𝑥=0
2
1 =⇒
(5.67)
.
(5.68)
𝑥=0
−1
4
𝑑 𝑈
𝑑𝑥 4
𝑥=0
Now, having discussed how harmonic potentials appear in the quantum case, we can
pass to the solution of the Schrodinger equation and determination of the energy
spectrum. As in classics we always have finite motion, we expect that the spectrum
will be fully discrete.
The Schrodinger equation reads:
−
ℏ2 𝑑 2 𝜙 𝑚𝜔2 𝑥 2
+
𝜙 = 𝐸 𝜙.
2𝑚 𝑑𝑥 2
2
(5.69)
It should be completed by the following boundary conditions: 𝜙(𝑥)| 𝑥→±∞ = 0.
Let us rewrite the Schrodinger equation in the dimensionless
form. For that,
q
introduce first characteristic length scale of the problem, 𝑥0 =
ℏ
𝑚𝜔 ,
𝑥 02 =
can convince oneself that 𝑥0 is length just checking its dimensionality: [𝑥0
[ℏ]
[𝐸] [𝑡]
=
=
[𝑚] [𝜔]
[𝑚] [𝑡 −1 ]
[𝐸] [𝑡] 2 [𝑚] [𝑉] 2 [𝑡] 2
=
= [𝐿] 2 − 0𝑘.
[𝑚]
[𝑚]
ℏ
𝑚𝜔 . One
]2
[𝑥0 ] 2 =
(5.70)
90
5 Bound states of quantum 1D and 2D particles
And then do with Schrodinger equation the following:
ℏ2 𝑑 2 𝜙 𝑚𝜔2 𝑥 2
+
𝜙 = 𝐸 𝜙 =⇒
2𝑚 𝑑𝑥 2
2
𝑑 2 𝜙 2𝑚 𝑚𝜔2 2
− 2
𝑥 − 𝐸 𝜙 = 0.
2
𝑑𝑥 2
ℏ
(5.71)
−
Then, multiply this equation by 𝑥02 =
ℏ
𝑚𝜔
(5.72)
and get:
ℏ 2𝑚 𝑚𝜔2 2
−
𝑥 − 𝐸 𝜙 = 0 =⇒
𝑚𝜔 ℏ2
2
!
2
2
𝑥 2 2𝐸
𝑚𝜔 2 2𝐸
2𝑑 𝜙
2𝑑 𝜙
𝜙 = 𝑥0 2 − 2 −
= 0.
𝑥0 2 −
𝑥 −
ℏ
ℏ𝜔
𝑑𝑥
𝑑𝑥
𝑥0 ℏ𝜔
𝑑2 𝜙
𝑥02 2
𝑑𝑥
And introducing dimensionless parameter 𝜉 =
𝑥
𝑥0
=
p 𝑚𝜔
ℏ
(5.73)
(5.74)
𝑥, one can rewrite this as:
𝑑2 𝜙
− (𝜉 2 − 𝜆)𝜙 = 0,
𝑑𝜉 2
where 𝜆 =
2𝐸
ℏ𝜔 .
𝜙 = 𝑒−
(5.75)
Let us now make the following substitution:
𝜉2
2
𝑈,
2
2
2
𝑑𝜙
𝑑𝑈
𝑑
− 𝜉2
− 𝜉 2 𝑑𝑈
− 𝜉2
− 𝜉2
= −𝜉𝑒
𝑈+𝑒
=
𝑈 (𝜉)𝑒
=𝑒
− 𝑈𝜉
𝑑𝜉
𝑑𝜉
𝑑𝜉
𝑑𝜉
2
2
2
𝑑𝑈
𝑑2𝜉
𝑑𝑈
𝑑 𝑈
− 𝜉2
− 𝜉2
=
−𝜉𝑒
−
𝑈
−
𝜉
−
𝑈𝜉
+
𝑒
=
𝑑𝜉
𝑑𝜉
𝑑𝜉 2
𝑑𝜉 2
2
𝜉2
𝑑𝑈
𝑑 𝑈
2
−
2𝜉
=𝑒 − 2
−
(𝜉
−
1)𝑈
.
𝑑𝜉
𝑑𝜉 2
(5.76)
(5.77)
(5.78)
Placing this to the Schrodinger equation, one gets:
2
2
𝑑 𝑈
𝑑𝑈
𝑑2𝜉
2
2
2
− 𝜉2
−
(𝜉
−
1)𝑈
−
(𝜉
−
𝜆)𝑈
=
−
(𝜉
−
𝜆)𝜙
=
𝑒
−
2𝜉
𝑑𝜉
𝑑𝜉 2
𝑑𝜉 2
2
𝜉2
𝑑 𝑈
𝑑𝑈
𝑒− 2
−
2𝜉
−
(𝜆
−
1)𝑈
= 0 =⇒
(5.79)
𝑑𝜉
𝑑𝜉 2
𝑑 2𝑈
𝑑𝑈
− 2𝜉
− (𝜆 − 1)𝑈 = 0.
𝑑𝜉
𝑑𝜉 2
(5.80)
𝜉2
This should be completed with the boundary condition 𝜙(𝜉)| 𝜉 →±∞ = 𝑒 − 2 𝑈 (𝜉)| 𝜉 →±∞ =
0 i.e. the function 𝑈 (𝜉) can itself grow to infinity at 𝑥 → ±∞, but they can not grow
𝜉2
𝜉2
faster than 𝑒 2 , so that 𝜙(𝜉) = 𝑈 (𝜉)𝑒 − 2 decay to zero at 𝜉 → ±∞.
Let us search for the solution of the Eq. (5.80) in the form of an infinite series:
5.2 Quantum harmonic oscillator
91
∞
Õ
𝑈 (𝜉) =
𝑑𝑈
=
𝑑𝜉
𝑏𝑘 𝜉 𝑘 ,
(5.81)
𝑘 𝑏 𝑘 𝜉 𝑘−1 ,
(5.82)
𝑘=0
∞
Õ
𝑘=0
∞
Õ
𝑑 2𝑈
=
𝑑𝜉 2
𝑘 (𝑘 − 1)𝑏 𝑘 𝜉 𝑘−2 .
(5.83)
𝑘=0
Putting this into Eq. (5.80) one gets:
∞
Õ
𝑘 (𝑘 − 1)𝑏 𝑘 𝜉 𝑘−2 − 2
∞
Õ
𝑘=0
=
∞
Õ
𝑘 (𝑘 − 1)𝑏 𝑘 𝜉 𝑘−2 +
𝑘=0
𝑘 𝑏 𝑘 𝜉 𝑘 + (𝜆 − 1)
𝑘=0
∞
Õ
∞
Õ
𝑏𝑘 𝜉 𝑘 =
𝑘=0
(𝜆 − 2𝑘 − 1)𝑏 𝑘 𝜉 𝑘 = 0.
(5.84)
𝑘=0
Consider the first sum in this expression. Renaming the index 𝑘 0 = 𝑘 − 2, 𝑘 =
𝑘 0 + 2, 𝑘 − 1 = 𝑘 0 + 1, we can rewrite it as:
∞
Õ
𝑘 (𝑘 − 1)𝑏 𝑘 𝜉 𝑘−2 =
∞
Õ
0
(𝑘 0 + 2) (𝑘 0 + 1)𝑏 𝑘 0 +2 𝜉 𝑘 =
𝑘 0 =0
𝑘=0
∞
Õ
(𝑘 + 2) (𝑘 + 1)𝑏 𝑘+2 𝜉 𝑘
𝑘=0
(5.85)
(we replaced 𝑘 0 → 𝑘, as it is just a summation index which we can call as we like.
We then get:
∞
Õ
[(𝑘 + 2) (𝑘 + 1)𝑏 𝑘+2 − (1 + 2𝑘 − 𝜆)𝑏 𝑘 ] 𝜉 𝑘 = 0.
(5.86)
𝑘=0
As this sum should be zero at any 𝜉, each of its terms should be zero, which allows
to get the following recurrent relation between the coefficients 𝑏 𝑘+2 and 𝑏 𝑘 :
𝑏 𝑘+2 =
1 + 2𝑘 − 𝜆
𝑏𝑘 .
(𝑘 + 2) (𝑘 + 1)
(5.87)
Let us analyze the asymptotic behavior when 𝑘 → ∞. In this case, one should keep
only the term 2𝑘 in the numerator, and 𝑘 2 in denominator, which gives:
𝑏 𝑘+2 ≈
2
𝑏𝑘 .
𝑘
Compare the established asymptotic behavior with those for the function 𝑒
know, that 𝑈 (𝜉) should not grow faster the 𝑒
𝜉2
2
(5.88)
𝜉2
2
. We
, thus in order to satisfy the requested
92
5 Bound states of quantum 1D and 2D particles
boundary condition, the decay of 𝑏 𝑘 with 𝑘 given by Eq. (5.88) should be faster,
than decay of coefficients of decomposition of 𝑒
For this latter we get:
𝑒
𝜉2
2
=
𝜉2
2
into Taylor series.
𝑛 Õ
∞
∞
Õ
1 𝜉2
1 2𝑛
=
𝜉 .
𝑛
𝑛!
2
𝑛!2
𝑛=1
𝑛=1
(5.89)
Therefore: (2𝑛 = 𝑘)
𝑏 2𝑛 =
1
1
= 𝑏𝑘 = 𝑘 ,
𝑛!2𝑛
𝑘
!2 2
(5.90)
2
1
𝑘 ,
+ 1 !2 2 +1
𝑘
2 !
1
1
1
1
= =
= ∼
2 𝑘 +1 ! 2 𝑘 +1
𝑘 +2 𝑘
2
2
𝑏 𝑘+2 = 𝑏 𝑘+2
𝑏𝑘
(5.91)
𝑘
2
(5.92)
for 𝑘 → ∞. This decay is faster then those given
therefore the function
Í by Eq. (5.88),
1+2𝑘−𝜆
𝑘 with 𝑏
𝑈 (𝜉) given by the Taylor expansion 𝑈 (𝜉) = ∞
𝑏
𝜉
𝑘
𝑘+2 = (𝑘+2) (𝑘+1) 𝑏 𝑘
𝑘=0
grows faster then function 𝑒
𝜉2
2
, and thus 𝑈 (𝜉)𝑒
−𝜉2
2
−→ ∞, which contradicts
𝑥→±∞
the requested boundary condition.
The only way to avoid this divergence and put 𝜙(𝜉) to zero at 𝜉 → ±∞ is to
demand, that Taylor expansion for 𝑈 (𝜉) is reduced to finite polynomial in stead of an
𝜉2
𝜉2
infinite series, as 𝑒 − 2 decays faster then any polynomial, 𝜙(𝜉) = 𝑈 (𝜉)𝑒 − 2 −→ 0
in this case.
We need thus to put 𝑏 𝑘+2 = 0 at some value of 𝑘, which gives the condition for
2𝐸
𝜆 = ℏ𝜔
:
2𝐸
𝜆=
= 2𝑛 + 1
(5.93)
ℏ𝜔
where 𝑛 = 0, 1, 2, 3.... is an integer number, which means that the spectrum is given
by the following expression:
1
(5.94)
𝐸 𝑛 = ℏ𝜔 𝑛 +
2
Note, that the spectrum of quantum harmonic oscillator is equidistant, i.e. the
distance between two neighbouring levels corresponding to 𝑛, 𝑛 + 1, 𝐸 𝑛+1 − 𝐸 𝑛 =
4𝐸 𝑛 = ℏ𝜔 - does not depend on 𝑛.
The obtained result can be compared with those following from Bohr-Sommerfeld
quantization
5.2 Quantum harmonic oscillator
𝑈 (𝑥) =
𝑚𝜔2 𝑥 2
,
2
2
𝑚𝜔2 𝑥1,2
2
93
(5.95)
r
= 𝐸 =⇒ 𝑥 1,2 (𝐸) = ±
2𝐸
= ±𝑎,
𝑚𝜔2
(5.96)
𝑝 2 𝑚𝜔2 𝑥 2
+
= 𝐸 =⇒
(5.97)
2𝑚
2
s
r
𝑥 2
√
𝑚𝜔2 2
𝑝(𝑥) = 2𝑚 𝐸 −
𝑥 = 2𝑚𝐸 1 −
,
(5.98)
2
𝑎
∫ 𝑎r
∫ 1q
∮
𝑥 2
√
√
1−
1 − 𝜉 2 𝑑𝑥 = (5.99)
𝑝(𝑥)𝑑𝑥 = 2 2𝑚𝐸
𝑑𝑥 = 2𝑎 2𝑚𝐸
𝑎
−𝑎
−1
q
4𝐸 𝜉
2𝜋𝐸
1
+1
2
=
=
= 2𝜋ℏ𝑛
(5.100)
1 − 𝜉 + arcsin 𝜉 k−1
𝜔 2
2
𝜔
From which it follows that
Fig. 5.9
(5.101)
𝐸 𝑛 = ℏ𝜔𝑛.
This differs from the exact equation 𝐸 𝑛 = ℏ𝜔 𝑛 + 12 by the term ℏ𝜔
2 , which corresponds to so called zero oscillation energy. The reason of the difference between
semiclassical and quantum result is the possibility of the penetration of the wavefunction into classically forbidden regions, which is not accounted for in the approach
based on Bohr-Sommerfeld quantization.
Now, let us establish the wavefunctions corresponding to different 𝑛:
94
5 Bound states of quantum 1D and 2D particles
𝜙(𝜉) = 𝑒 −
∞
Õ
𝜉2
2
(5.102)
𝑏𝑘 𝜉 𝑘 ,
𝑛=0
1 + 2𝑘 − 𝜆
𝑘 −𝑛
𝑏𝑘 = 2
𝑏𝑘 ,
(𝑘 + 2) (𝑘 + 1)
(𝑘 + 2) (𝑘 + 1)
𝑏 𝑘+2 =
(5.103)
where 𝑛 = 0, 1, 2, 3 . . . depending on the number of state.
a) 𝑛 = 0 one gets 𝑏 2 = 0, and one need to put 𝑏 1 = 𝑏 3 = . . . = 0 as well (otherwise
we will have infinite number of odd in 𝑘 terms). Then,
2
𝜙0 (𝜉) = 𝑏 0 𝑒
− 𝜉2
=⇒ 𝜙(𝑥) = 𝑏 0 𝑒
− 12
𝑥
𝑥0
2
(5.104)
The value of 𝑏 0 is determined from normalization condition:
∫ +∞ 𝑥 2
∫ +∞
−
𝑒 𝑥0 𝑑𝑥 =
𝜙20 (𝑥)𝑑𝑥 = 𝑏 20
−∞
−∞
∫ +∞
√
2
−𝜉2
𝑏0 𝑥0
𝑒 𝑑𝜉 = 𝑏 20 𝑥0 𝜋 = 1 =⇒
(5.105)
−∞
−1
1
𝑏0 = 𝑥0 2 𝜋− 4 =
𝜙(𝜉) =
𝑚𝜔 14
𝜋ℏ
𝑚𝜔 14
(5.106)
,
𝜋ℏ
𝑒−
𝜉2
2
, 𝜙(𝑥) =
𝑚𝜔 14
𝑒−
𝑚𝜔 𝑥 2
2ℏ
.
𝜋ℏ
(5.107)
This function corresponds to the ground state (state with minimal energy), and it is
symmetric, 𝜙(𝑥) = 𝜙(−𝑥), as expected for ground state of any arbitrary symmetric
potential 𝑈 (𝑥) = 𝑈 (−𝑥). Also, it does not have nodes, i.e. the points where 𝜙(𝑥)
q = 0,
𝜙0 (𝑥) is just a Gaussian. The particle is localized at characteristic scale 𝑥0 =
ℏ
𝑚𝜔 .
Fig. 5.10
b) 𝑛 = 1 in this case 𝑏 3 = 0, and we need to put as well 𝑏 0 = 𝑏 2 = 𝑏 4 = . . . = 0,
otherwise expression for the wavefunction will contain infinite number of terms with
5.3 Quantum particle in 1D periodic potential, Kronig-Penny model
even 𝑛. One thus has:
95
2
(5.108)
𝜙1 (𝜉) = 𝑏 1 𝜉𝑒 − 𝜉 ,
where 𝑏 1 can be again found from normalization:
∫
+∞
−∞
𝜙21 (𝑥)𝑑𝑥
=
𝑏 21
∫
+∞
−∞
𝑥 −
𝑒
𝑥0
𝑥
𝑥0
2
𝑑𝑥 =
𝑏 21 𝑥0
∫
+∞
2
𝜉 2 𝑒 − 𝜉 𝑑𝜉 =
−∞
𝑏1 𝑥0 √
𝜋,
2
(5.109)
where we used
r
∫ +∞
∫ +∞
2
2
𝜕
𝑑 𝜋 1√
𝜉 2 𝑒 − 𝜉 𝑑𝜉 = − lim
𝑒 −𝜆 𝜉 𝑑𝜉 = − lim
=
𝜋.
𝜆→0 𝜕𝜆 −∞
𝜆→1 𝑑𝜆
𝜆 2
−∞
(5.110)
In a similar way, for 𝑛 = 2 one has: 𝑏 2 = − 22 𝑏 0 = −𝑏 0 , 𝑏 4 = 0, and 𝜙2 (𝜉) =
2
𝑏 0 (𝜉 2 − 1)𝑒 − 𝜉 , etc. The general expression of properly normalized wavefunction
corresponding to an arbitrary 𝑛 is:
2
1
𝑥 − 21 𝑥𝑥
1
0
𝐻
,
(5.111)
𝑒
𝜙 𝑛 (𝑥) =
𝑛
1
1
𝑥0
(2𝑛 𝑛!) 2 (𝜋𝑥 02 ) 4
where 𝑥0 =
q
ℏ
𝑚𝜔 ,
and 𝐻𝑛 (𝜉) are Hermit polynomials, defined as:
𝐻𝑛 (𝜉) = (−1) 𝜉 𝑒 𝜉
2
𝑑𝑛 −𝜉 2
𝑒 .
𝑑𝜉 𝑛
(5.112)
For even 𝑛 the wavefunctions are symmetric, 𝜙 𝑛 (𝑥) = 𝜙 𝑛 (−𝑥), for odd n antisymmetric, 𝜙 𝑛 (𝑥) = −𝜙 𝑛 (−𝑥).The wavefunction corresponding to a given 𝑛 has 𝑛
nodes, i.e. points at which 𝜙 𝑛 (𝑥) = 0. Thus, for 𝑛 = 1 : 𝜙1 (𝜉) = 𝑏 1 𝜉𝑒 −
- it has anode at 𝜉 = 0 =⇒ √𝑥 = 0, for 𝑛 = 2 : 𝜙2 (𝜉) = 𝑏 0
nodes at 𝜉 = ±1, 𝑥 = ±𝑥0 / 2, etc.
2
𝜉
(1− 𝜉 2 )𝑒 − 2
𝜉2
2
, 𝜙1 (0) = 0
, 𝜙2 (±1) = 0,
5.3 Quantum particle in 1D periodic potential, Kronig-Penny
model
Let us consider the problem of a quantum particle, moving in 1D periodic potential,
which has the following property: 𝑈 (𝑥 + 𝑑) = 𝑈 (𝑥). This potential has sequence of
the alternating minima and maxima.
For a classical particle, the motion will be finite for 𝐸 < 𝑈0 , where 𝑈0 is the
maximal value of the potential, and infinite for 𝐸 > 𝑈0 . In the case of a finite
motion the classical particle potential will be localized in one of the potential wells.
However, in the quantum case, due to the phenomenon of the quantum tunneling,
the particle can tunnel from one well to another, and confinement will be lost.
96
5 Bound states of quantum 1D and 2D particles
Fig. 5.11 Profiles of the wavefunctions of Harmonic oscillator
Fig. 5.12 𝐸 > 𝑈0 , classical infinite motion; 𝐸 < 𝑈0 , classical finite motion is one of the well; 𝑑 period of potential
Let us now define how the energy spectrum will look like. First, let us give a
statement, which goes without proof in this course, known as Bloch theorem:
In a periodic potential 𝑈 (𝑥 +𝑑) = 𝑈 (𝑥) the solutions of the stationary Schrodinger
2
ℏ2 𝑑 𝜙
+ 𝑈 (𝑥)𝜙 = 𝐸 𝜙 can be written in the following form: 𝜙(𝑥) =
equation − 2𝑚
𝑑 𝑥2
𝑒 𝑖𝑘 𝑥 𝑉 (𝑥), where the function 𝑉 (𝑥) is periodic, with the same period of the -A.O.E.
external potential, 𝑉 (𝑥 + 𝑑) = 𝑉 (𝑥), i.e. the wavefunction is a product of the term,
corresponding to a plane wave, 𝑒 𝑖𝑘 𝑥 , and so called Bloch function 𝑉 (𝑥). 𝑘 plays role
of momentum, and is known as quasimomentum. The energy 𝐸 is a function of it,
2 2
but differently from the case of a free particle, 𝐸 ≠ ℏ2𝑚𝑘 .
5.3 Quantum particle in 1D periodic potential, Kronig-Penny model
97
For detailed analysis, let us consider the example of Kronig-Penny model, where
periodic potential 𝑈 (𝑥) is formed by a sequence of rectangular barriers of the height
𝑈0 and width 𝑏, and take 𝑎 to be the distance between the barriers: The period of
Fig. 5.13
a structure is 𝑑 = 𝑎 + 𝑏. As 𝑈 (𝑥) is not a continuous function, we need to solve
separately the Schrodinger equation in the region of the barrier (region 1 in the
figure, 𝑥 ∈ [−𝑏; 0]) and the region of the well (region 2 in the figure, 𝑥 ∈ [0; 𝑎])
and then connect the two solutions together, using the condition of the continuity of
the wavefucntion 𝜙(𝑥) and its first derivative 𝜙 0 (𝑥) at the boundary between the two
regions, at 𝑥 = 0.
We have:
Region 1:
ℏ2 𝑑 2 𝜙1
−
+ 𝑈0 𝜙 1 = 𝐸 𝜙 1 .
(5.113)
2𝑚 𝑑𝑥 2
Suppose, that 𝐸 < 𝑈0 . Then we have:
𝑑 2 𝜙1 2𝑚
= 2 (𝑈0 − 𝐸)𝜙1 = 𝛽2 𝜙1 ,
𝑑𝑥 2
ℏ
where 𝛽 =
q
2𝑚(𝑈0 −𝐸)
,
ℏ2
(5.114)
so that
h
i
𝜙1 (𝑥) = 𝐴𝑒 −𝛽 𝑥 + 𝐵𝑒 𝛽 𝑥 = 𝑒 𝑖𝑘 𝑥 𝑒 −𝑖𝑘 𝑥 𝐴𝑒 −𝛽𝑥 + 𝐵𝑒 𝛽 𝑥 = 𝑒 𝑖𝑘 𝑥 𝑉1 (𝑥),
(5.115)
where 𝑉1 (𝑥) is a Bloch part of the wavefunction, 𝑉1 (𝑥) = 𝑒 −𝑖𝑘 𝑥 𝐴𝑒 −𝛽 𝑥 + 𝐵𝑒 𝛽𝑥 .
For the case 𝐸 > 0, we would have
𝑑 2 𝜙1
= −𝛽2 𝜙1 ,
𝑑𝑥 2
(5.116)
98
5 Bound states of quantum 1D and 2D particles
q
𝛽 = 2𝑚(𝑈ℏ20 −𝐸) , and solution for this case is obtained by simple substitution 𝛽 → 𝑖𝛽.
Region 2:
ℏ2 𝑑 2 𝜙 1
= 𝐸 𝜙1 ,
(5.117)
−
2𝑚 𝑑𝑥 2
where 𝛼 =
q
2𝑚𝐸
ℏ2
𝑑 2 𝜙2
= −𝛼2 𝜙2 ,
𝑑𝑥 2
(5.118)
𝐶𝑒 −𝑖 𝛼𝑥 + 𝐷𝑒 𝑖 𝛼𝑥 𝑒 −𝑖𝑘 𝑥 = 𝑒 𝑖𝑘 𝑥 𝑉2 (𝑥),
(5.119)
and
𝜙2 = 𝐶𝑒 −𝑖 𝛼𝑥 + 𝐷𝑒 𝑖 𝛼𝑥 = 𝑒 𝑖𝑘 𝑥
where 𝑉2 = 𝑒 −𝑖𝑘 𝑥 𝐶𝑒 −𝑖 𝛼𝑥 + 𝐷𝑒 𝑖 𝛼𝑥 .
Now, we can write the conditions at 𝑥 = 0:
𝜙1 (0) = 𝜙2 (0) =⇒ 𝐴 + 𝐵 = 𝐶 + 𝐷,
(5.120)
𝜙10 (0)
(5.121)
𝜙20 (0)
1
2
1
𝐵=
2
𝐴=
=⇒ 𝛽(−𝐴 + 𝐵) = 𝑖𝛼(−𝐶 + 𝐷) =⇒
𝑖𝛼
𝑖𝛼
1+
𝐶+ 1−
𝐷 ,
𝛽
𝛽
𝑖𝛼
𝑖𝛼
1−
𝐶+ 1+
𝐷 .
𝛽
𝛽
=
(5.122)
(5.123)
Additional pair of the equations follows from the condition of the periodicity of the
Bloch part of the wavefunction 𝑉 (𝑥) = 𝑉 (𝑥 + 𝑑), which gives:
(5.124)
𝑉1 (−𝑏) = 𝑉2 (𝑎) =⇒
𝐴𝑒
(𝛽+𝑖𝑘)𝑏
𝑉10 (−𝑏)
=
+ 𝐵𝑒
(−𝛽+𝑖𝑘)𝑏
𝑉20 (𝑎) =⇒
(𝛽+𝑖𝑘)𝑏
− (𝛽 + 𝑖𝑘) 𝐴𝑒
= 𝐶𝑒
−𝑖 ( 𝛼+𝑘) 𝑎
+ 𝐷𝑒
𝑖 ( 𝛼−𝑘) 𝑎
(5.125)
,
(5.126)
+ (𝛽 − 𝑖𝑘)𝐵𝑒
(−𝛽+𝑖𝑘)𝑏
= −𝑖(𝛼 + 𝑖𝑘)𝐶𝑒
−𝑖 ( 𝛼+𝑘) 𝑎
+ 𝑖(𝛼 − 𝑘)𝐷𝑒 𝑖 ( 𝛼−𝑘) 𝑎 .
(5.127)
Placing in these two equations expressions for 𝐴, 𝐵 from Eq. (5.122), (5.123) at page
98, and writing the condition of the existence of nontrivial solution demanding that
the system is zero, one gets the following equation which gives a relation between
the energy 𝐸 and quasimomentum 𝑘:
𝛽2 − 𝛼2
sinh(𝛽𝑏) sin(𝛼𝑎) + cosh(𝛽𝑏) cos(𝛼𝑎) = cos(𝑑𝑘),
2𝛼𝛽
(5.128)
q
𝑧
−𝑧
𝑒 𝑧 +𝑒−𝑧
2𝑚𝐸
where sinh 𝑧 = sh 𝑧 = 𝑒 −𝑒
,
𝑑
=
𝑎
+
𝑏,
cosh
𝑧
=
ch
𝑧
=
,
𝛼
=
, 𝛽=
2
ℏ2
q
q 2
2𝑚(𝑈0 −𝐸)
2𝑚(𝑈0 −𝐸)
, if 𝐸 > 0, making a replacement 𝛽 −→ 𝑖𝛽 = 𝑖
one gets:
ℏ2
ℏ2
5.3 Quantum particle in 1D periodic potential, Kronig-Penny model
99
𝛽2 + 𝛼2
sin(𝛽𝑏) sin(𝛼𝑎) + cos(𝛽𝑏) cos(𝛼𝑎) = cos(𝑑𝑘),
2𝛼𝛽
(5.129)
where we used: sh(𝑖𝑧) = 𝑖 sin 𝑧, ch(𝑖𝑧) = cos 𝑧.
Let us analyze the solution of this equation graphically writing
r
r
q
2𝑚(𝑈0 − 𝐸)
2𝑚𝑈0
2
=
− 𝛼 = 𝛾 2 − 𝛼2 ,
𝛽=
ℏ2
ℏ2
q
0
, we can write (for 𝐸 < 𝑈0 ):
where 𝛾 = 2𝑚𝑈
ℏ2
(5.130)
−
q
q
𝛾 2 − 2𝛼2
2
2
sh(𝑏
𝛾
−
𝛼
)
sin(𝛼𝑎)
+
ch(𝑏
𝛾 2 − 𝛼2 ) cos(𝛼𝑎) =
p
2𝛼 𝛾 2 − 𝛼2
p
p
1 − 2𝑧 2
sh(𝛾𝑏 1 − 𝑧2 ) sin(𝛾𝛼𝑧) + ch(𝛾𝑏 1 − 𝑧 2 ) cos(𝛾𝑎𝑧) =
√
2𝑧 1 − 𝑧2
𝛼
𝐹 (𝑧) = cos(𝑘 𝑑), 𝑧 = .
(5.131)
𝛾
Graphically, the energy will be determined by an intersection of the plot of 𝐹 (𝑧), with
a horizontal line corresponding to cos(𝑘 𝑑). This horizontal line is lying between the
points 𝐹 = −1 and 𝐹 = +1. Looking at the plot of the function 𝐹 (𝑧), one sees, that
Fig. 5.14
100
5 Bound states of quantum 1D and 2D particles
it has the regions for which |𝐹 (𝑧)| > 1. In the corresponding intervals of energies
there are no solutions of the equation 𝐹 (𝑧) = cos(𝑘 𝑑) for any 𝑘. This means, that
the energy of these states is not possible for a quantum particle, they correspond
to so called energy stop bands, or gaps. On the other hand, if |𝐹 (𝑧)| < 1 , there
are intersection between the lines 𝐹 (𝑧), cos(𝑘 𝑑) and energies in these intervals are
possible. It is easy to see that
𝐹 (0) =
𝛾𝑎
sh(𝛾𝑏) + ch(𝛾𝑏) > 1,
2
(5.132)
and thus first stop band starts from 𝐸 = 0.
For each value of 𝑘 there is an infinite number of the intersections, corresponding
to different allowed energy bands. Note, that the shift 𝑘 −→ 𝑘 + 2𝑑𝜋 does not change
the value of cos(𝑘 𝑑), so quasimomentum 𝑘 is defined up to the factor 2𝑑𝜋 . Usually,
the region of meaningful 𝑘 is chosen as 𝑘 ∈ − 𝑑𝜋 ; 𝑑𝜋 , which correspond to so called
first Brillovin zone.
Note, that in the first allowed zone, if we increase 𝑘, the position of the intersection
point shifts from left to right, i.e. energy is increasing function of quasimomentum.
However, in the second allowed band, intersection point moves from right to left with
increase of 𝑘, which means that energy is decreasing function of 𝑘. Qualitatively the
Fig. 5.15 Structure of energy bands
same structure of energy bands will occur in any periodic potential 𝑈 (𝑥 + 𝑑) = 𝑈 (𝑥).
The main features of the energy spectrum of quantum particle in periodic potentials
are thus:
5.4 Quantum particle in a magnetic field and Landau quantization
101
1) Spectrum reveal band structure, where gaps=stop bands=regions with prohibited energies alternate with allowed bands.
2) In the allowed bands the energy is a function of quasimomentum 𝑘. It can be
an increasing or decreasing function of 𝑘.
3) By analogy with mass of the free particle,
𝐸=
1
1 𝑑2 𝐸
ℏ2 𝑘 2
=⇒
= 2 2,
2𝑚
𝑚 ℏ 𝑑𝑘
(5.133)
one can introduce the concept of an effective mass of a particle in a periodic potential
by same expression:
1
1 𝑑2 𝐸
=
(5.134)
𝑚 ∗ ℏ2 𝑑𝑘 2
Note, that as dispersion is not any more parabolic, effective mass depends on 𝑘:
𝑚 ∗ = 𝑚(𝑘). Moreover, it can even become negative, as it happens near 𝑘 = ± 𝑎𝜋 in
the first allowed band and near 𝑘 = 0 in the second allowed band.
In physics, the most important example of the formation of band structure in
periodic potentials is the formation of bands in solids, which result from the periodic
potential of a crystalline lattice.
5.4 Quantum particle in a magnetic field and Landau
quantization
In classical physics charged particle, which is place into external magnetic field B
is subjected to the force F = 𝑞 [V × B], which is proportional to the cross product
of its velocity and magnetic field itself. As force is velocity dependent, we can not
represent it in the form F = −∇𝑈 (r), and use the expression for Hamiltonian in the
𝑝2
form we are used to, 𝐻 = 𝑇 + 𝑈 = 2𝑚
+ 𝑈 (r), for which transition to quantum case
is done using substitution p −→ p̂ = −𝑖ℏ∇.
We need, thus, to construct classical Hamiltonian for a particle in magnetic field,
which will give us correct value of the Lorenz force and results in the following
equation of motion
𝑑V
𝑚
= F = 𝑞[V × B].
(5.135)
𝑑𝑡
Let us start from presenting the Lagrangian of a classical particle with charge 𝑞,
moving in electromagnetic field, characterized by electric field, E = −∇𝜙 − 𝜕A
𝜕𝑡 , and
magnetic field, B = ∇ × A, where 𝜙(r, 𝑡) and A(r, 𝑡) are scalar and vector potentials.
We know from experiment, that total force acting on the particle will be:
F = 𝑞𝐸 + 𝑞[V × B]
and thus 3𝑟 𝑑 law of Newton will give:
(5.136)
102
5 Bound states of quantum 1D and 2D particles
𝑚
𝑑V
𝑑2r
=𝑚
= 𝑞(E + [V × B]).
𝑑𝑡
𝑑𝑡 2
(5.137)
Projections of this equation on axis 𝑥 = 𝑥1 , 𝑦 = 𝑥2 and 𝑧 = 𝑥3 should be equivalent
to Lagrange equations:
𝜕𝐿
𝑑 𝜕𝐿
−
= 0,
(5.138)
𝑑𝑡 𝜕 𝑥¤𝑖
𝜕𝑥𝑖
where 𝑥¤𝑖 = 𝑉𝑖 . Let us show that this is the case for the following Lagrangian:
𝐿=
𝑚V2
𝑚 2
+ 𝑞VA − 𝑞𝜙 =
𝑥¤ + 𝑦¤ 2 + 𝑧¤2 + 𝑞( 𝑥¤ 𝐴 𝑥 + 𝑦¤ 𝐴 𝑦 + 𝑧¤ 𝐴 𝑧 ) − 𝑞𝜙. (5.139)
2
2
Take, for example 𝑥𝑖 = 𝑥1 = 𝑥. We have:
𝜕𝐿
𝜕𝐿
=
= 𝑚 𝑥¤ + 𝑞 𝐴 𝑥 .
𝜕 𝑥¤1
𝜕 𝑥¤
(5.140)
We thus get:
𝑑
𝑑𝐴 𝑥
𝑑 𝜕𝐿
(𝑚 𝑥¤ + 𝑞 𝐴 𝑥 ) = 𝑚 𝑥¥ + 𝑞
=
=
𝑑𝑡 𝜕 𝑥¤
𝑑𝑡
𝑑𝑡
𝜕 𝐴𝑥 𝜕 𝐴𝑥
𝜕 𝐴𝑥
𝜕 𝐴𝑥
= 𝑚 𝑥¥ + 𝑞
+
𝑥¤ +
𝑦¤ +
𝑧¤ ,
𝑑𝑡
𝑑𝑥
𝑑𝑦
𝑑𝑧
(5.141)
where we accounted, that 𝐴 𝑥 is function of all coordinates and time, 𝐴 𝑥 =
𝐴𝑥
. (Note
𝐴 𝑥 (𝑥(𝑡), 𝑦(𝑡), 𝑧(𝑡), 𝑡) and used the chain rule to calculate the derivative 𝑑𝑑𝑡
𝑑 𝐴𝑥
𝑑A
the difference between full derivative 𝑑𝑡 and partial derivative 𝑑𝑡 . The former is
not zero even if A is stationary, i.e. does not depend on time explicitly.)
𝜕𝐿
𝜕
=𝑞
𝑥¤ 𝐴 𝑥 + 𝑦¤ 𝐴 𝑦 + 𝑧¤ 𝐴 𝑧 − 𝜙(𝑥, 𝑦, 𝑧, 𝑡) =
𝜕𝑥
𝜕𝑥
𝜕 𝐴𝑦
𝜕 𝐴𝑥
𝜕 𝐴 𝑧 𝜕𝜙
𝑞 𝑥¤
+ 𝑦¤
+ 𝑧¤
−
.
𝜕𝑥
𝜕𝑥
𝜕𝑥
𝜕𝑥
(5.142)
From these we get the following equation of motion for 𝑥 coordinate:
𝜕 𝐴𝑦
𝜕 𝐴𝑥 𝜕 𝐴𝑥
𝜕 𝐴𝑥
𝜕 𝐴𝑥
𝜕 𝐴𝑥
𝜕 𝐴 𝑧 𝜕𝜙
𝑚 𝑥¥ + 𝑞
+
𝑥¤ +
𝑦¤ +
𝑧¤ − 𝑞 𝑥¤
+ 𝑦¤
+ 𝑧¤
−
= 0,
𝜕𝑡
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑥
𝜕𝑥
𝜕𝑥
𝜕𝑥
(5.143)
𝜕 𝐴𝑦 𝜕 𝐴𝑥
𝜕𝜙 𝜕 𝐴 𝑥
𝜕 𝐴𝑧 𝜕 𝐴 𝑥
𝑚 𝑥¥ = 𝑞 −
−
+ 𝑞 𝑦¤
−
+ 𝑧¤
−
.
(5.144)
𝜕𝑥
𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜕𝑥
𝜕𝑧
And now use relations, connecting potentials 𝜙, A with fields E, B:
5.4 Quantum particle in a magnetic field and Landau quantization
103
𝜕A
−𝜕𝜙 𝜕 𝐴 𝑥
=⇒ 𝐸 𝑥 = −
−
,
(5.145)
𝜕𝑡
𝜕𝑥
𝜕𝑡
e 𝑥 e 𝑦 e𝑧
𝜕 𝐴𝑦 𝜕 𝐴𝑥
𝜕 𝐴𝑧 𝜕 𝐴 𝑦
𝜕 𝐴 𝑥 𝜕 𝐴𝑧
𝜕 𝜕 𝜕
+ e𝑦
+ e𝑧
=⇒
B = 𝜕𝑥 𝜕𝑦 𝜕𝑧 = e 𝑥
−
−
−
𝜕𝑦
𝜕𝑧
𝜕𝑧
𝜕𝑥
𝜕𝑥
𝜕𝑦
𝐴 𝑥 𝐴 𝑦 𝐴𝑧
(5.146)
𝜕 𝐴𝑦 𝜕 𝐴𝑥
−
= 𝐵𝑧 ,
(5.147)
𝜕𝑥
𝜕𝑦
𝜕 𝐴 𝑥 𝜕 𝐴𝑧
−
= 𝐵𝑦 .
(5.148)
𝜕𝑧
𝜕𝑥
E = −∇𝜙 −
And we get:
𝑚 𝑥¥ = 𝑞𝐸 𝑥 + 𝑞(𝑉𝑦 𝐵 𝑧 − 𝑉𝑧 𝐵 𝑦 ),
(5.149)
2𝑛𝑑
which coincides with x - projection of the equation of motion written as
law of
Newton:
𝑑2r
(5.150)
𝑚 2 = F = 𝑞(E + [V × B]).
𝑑𝑡
Now, capitalising on the Lagrangian, let us introduce classical canonical momentum
and Hamiltonian. We define them as:
𝑝𝑖 =
𝜕𝐿
𝜕𝐿
=
, 𝑖 = 1, 2, 3
𝜕 𝑥¤𝑖 𝜕𝑉𝑖
(5.151)
- components of the canonical momentum, and
𝐻 (p, r) = p · V − 𝐿.
(5.152)
Let us check, that for the situation of a particle in the external potential, 𝐿 = 𝑇 − 𝑈 =
𝑚V2
2 − 𝑈 (r), we get well known expression p = 𝑚V, 𝐻 = 𝑇 + 𝑈. One has in this
case:
𝑚 2
𝑥¤1 + 𝑥¤22 + 𝑥¤32 − 𝑈 (𝑥 1 , 𝑥2 , 𝑥3 ),
𝐿=
(5.153)
2
𝜕𝐿
𝑝𝑖 =
= 𝑚𝑥 𝑖 =⇒
(5.154)
𝜕 𝑥¤𝑖
p = 𝑚V,
(5.155)
𝑚V2
−𝑈 =
𝐻 = p · V − 𝐿 = 𝑚V2 −
2
𝑚V2
p2
+ 𝑈 (r) =
+ 𝑈 (r),
2
2𝑚
(5.156)
and transition to quantum mechanics, as we know, is performed via substitution
p −→ p̂ = −𝑖ℏ∇.
Now, for Lagrangian Eq. (5.139) at the page 102 we get:
104
5 Bound states of quantum 1D and 2D particles
𝜕𝐿
= 𝑚 𝑥¤𝑖 + 𝑞 𝐴𝑖 =⇒
𝜕 𝑥¤𝑖
p = 𝑚V + 𝑞A.
𝑝𝑖 =
(5.157)
(5.158)
Note, that in the presence of the external vector potential canonical momentum contains addition 𝑞A to the usual term 𝑚V, known as kinetic momentum. Importantly,
when one goes from classical to quantum mechanics using p −→ p̂ = −𝑖ℏ∇, one
should use canonical, and not kinetic momentum.
Now, for the Hamiltonian we get:
𝑚V2
+ 𝑞A − 𝑞𝜙 =
𝐻 = pV − 𝐿 = V(𝑚V + 𝑞A) −
2
𝑚V2
1
(p − 𝑞A) 2 + 𝑞𝜙.
+ 𝑞𝜙 =
2
2𝑚
(5.159)
Note, that written as function of velocity, kinetic energy in magnetic field is the
2
p2
same, as for the case of free particle, 𝑇 = 𝑚V
2 , but this can not be rewritten as 2𝑚 ,
because p ≠ 𝑚V any more.
Now, transition to quantum Hamiltonian from classical one is obvious, and we
get
1
1
( p̂ − 𝑞A) 2 + 𝑞𝜙 =
(𝑖ℏ∇ + 𝑞A) 2 + 𝑞𝜙.
𝐻ˆ =
2𝑚
2𝑚
(5.160)
Let us consider the states of a quantum particle, placed in the magnetic field 𝐵
directed along z-axis (choice of axis is of course arbitrary), B = e𝑧 𝐵. As usual,
even before solving the Schrodinger equation, we can try to predict the type of the
spectrum, analyzing classical motion. In classical physics, we get for F = 𝑞 [V × B]:
𝑑𝑉𝑥
= 𝑞𝑉𝑦 𝐵,
𝑑𝑡
𝑑𝑉𝑦
= −𝑞𝑉𝑥 𝐵,
𝑚
𝑑𝑡
𝑑𝑉𝑧
𝑚
= 0.
𝑑𝑡
𝑚
(5.161)
(5.162)
(5.163)
As 𝐹𝑧 = 0, the motion along z axis is free (infinite motion), so that 𝑉𝑧 = 𝑉0𝑧 , 𝑧 =
𝑧0 + 𝑉0𝑧 𝑡, where 𝑉0𝑧 , 𝑧0 are initial values of velocity along z axis (which does not
change), and coordinate 𝑧.
As for the motion in x-y plane perpendicular to the magnetic field, multiplying
equation for 𝑉𝑦 by 𝑖 and adding to equation for 𝑉𝑥 , one gets:
𝑑𝑉𝑦
𝑑𝑉𝑥
𝑚
+𝑖
= 𝑞𝐵(𝑉𝑦 − 𝑖𝑉𝑥 ),
(5.164)
𝑑𝑡
𝑑𝑡
which can be rewritten as:
5.4 Quantum particle in a magnetic field and Landau quantization
105
𝑑𝜉
= −𝑖𝑞𝐵𝜉,
𝑑𝑡
where 𝜉 = 𝑉𝑥 + 𝑖𝑉𝑦 its solution reads:
(5.165)
𝑚
𝜉 = 𝜉 (0)𝑒 −𝑖
where 𝜔 𝐵 =
𝑞𝐵
𝑚
𝑞𝐵
𝑚 𝑡
(5.166)
= 𝜉0 𝑒 −𝑖 𝜔𝐵 𝑡 ,
is known as cyclotron frequency. Thus:
𝜉 = 𝑉𝑥 + 𝑖𝑉𝑦 = (𝑉𝑥0 + 𝑖𝑉𝑦0 ) (cos(𝜔 𝐵 𝑡) − 𝑖 sin(𝜔 𝐵 𝑡)) =
𝑉𝑥0 cos(𝜔 𝐵 𝑡) + 𝑉𝑦0 sin(𝜔 𝐵 𝑡) + 𝑖(𝑉𝑦0 cos(𝜔 𝐵 𝑡) + 𝑉𝑦0 − 𝑉𝑥0 sin(𝜔 𝐵 𝑡)) =⇒
(5.167)
𝑑𝑥
,
(5.168)
𝑉𝑥 = 𝑉𝑥0 cos(𝜔 𝐵 𝑡) + 𝑉𝑦0 sin(𝜔 𝐵 𝑡) =
𝑑𝑡
𝑑𝑦
𝑉𝑦 = 𝑉𝑦0 cos(𝜔 𝐵 𝑡) + 𝑉𝑦0 − 𝑉𝑥0 sin(𝜔 𝐵 𝑡) =
.
(5.169)
𝑑𝑡
And integrating those two equations we get:
𝑉𝑦0
𝑉𝑥0
(1 − cos(𝜔 𝐵 𝑡)),
sin(𝜔 𝐵 𝑡) +
𝜔𝐵
𝜔
𝑉𝑦0
𝑉𝑥0
sin(𝜔 𝐵 𝑡) +
(cos(𝜔 𝐵 𝑡) − 1).
𝑦(𝑡) = 𝑦 0 +
𝜔𝐵
𝜔
(5.170)
𝑥(𝑡) = 𝑥0 +
(5.171)
These two equations can be combined together to give:
𝑥(𝑡) − 𝑥0 −
𝑉𝑦0
𝜔
2
2 𝑉2 + 𝑉2
𝑉𝑥0
𝑥0
𝑦0
+ 𝑦(𝑡) − 𝑦 0 +
=
,
𝜔
𝜔2𝐵
𝑉 2 +𝑉 2
(5.172)
which corresponds to a circle of a radius 𝑅 2 = 𝑥0𝜔𝐵 𝑦0 , centered at 𝑋0 = 𝑥0 + 𝜔𝑦0 , 𝑌0 =
𝑦 0 − 𝑉𝜔𝑥0 . The particle in xy plane rotates around this circle with frequency 𝜔 𝐵 = 𝑒𝐵
𝑚 .
The motion an x-y plane is thus finite, and in quantum case one should expect the
appearance of discretization of energies corresponding to it.
The trajectory of a classical particle in magnetic field is thus a spiral:
Let us now consider the solution of a quantum problem. We have
1
(𝑖ℏ∇ − 𝑞A) 2 𝜙(r) = 𝐸 𝜙(r),
2𝑚
𝑉
(5.173)
where ∇ × A = B = e𝑧 𝐵.
First thing we need to do, is to define vector potential A corresponding to B = e𝑧 𝐵.
One has:
106
5 Bound states of quantum 1D and 2D particles
Fig. 5.16
∇ × A = e𝑥
𝜕 𝐴𝑦 𝜕 𝐴𝑥
𝜕 𝐴 𝑥 𝜕 𝐴𝑧
𝜕 𝐴𝑧 𝜕 𝐴 𝑦
−
+ e𝑦
−
+ e𝑧
−
= e𝑧 𝐵 =⇒
𝜕𝑦
𝜕𝑧
𝜕𝑧
𝜕𝑥
𝜕𝑥
𝜕𝑦
(5.174)
𝜕 𝐴𝑧 𝜕 𝐴 𝑦
−
= 0,
𝜕𝑦
𝜕𝑧
𝜕 𝐴 𝑥 𝜕 𝐴𝑧
−
= 0,
𝜕𝑧
𝜕𝑥
𝜕 𝐴𝑦 𝜕 𝐴𝑥
−
= 𝐵.
𝜕𝑥
𝜕𝑦
(5.175)
(5.176)
(5.177)
These equations can be satisfied if one chooses for example: 𝐴 𝑦 = 𝐵 𝑥 , 𝐴 𝑥 = 𝐴 𝑧 = 0.
Note, that this choice is not unique, and correct vector potential can be chosen in
infinite number of possible ways. This is consequence of the fact, that so called
gradient transformation: A0 = 𝐴 + ∇ 𝑓 , where 𝑓 is an arbitrary scalar function, does
not change the value of the magnetic field.
Indeed:
B0 = ∇ × A0 = ∇ × (A + ∇ 𝑓 ) = ∇ × A0 + ∇ × ∇ 𝑓 = ∇ × A = B,
(5.178)
as ∇ × ∇ 𝑓 ≡ 0 (curl of any gradient is always zero).
Let us show that in quantum mechanics, although it is vector potential itself, and
not magnetic field, enters into Schrodinger equation, gradient transformation (also
known as gauge transformation) does not change the energy spectrum. We have:
5.4 Quantum particle in a magnetic field and Landau quantization
107
1
(𝑖ℏ∇ + 𝑞A) 2 𝜙 = 𝐸 𝜙,
2𝑚
1
(𝑖ℏ∇ + 𝑞A + 𝑞∇ 𝑓 ) 2 𝜙 0 = 𝐸 0 𝜙 0 .
2𝑚
Let us make the following substitution: 𝜙 0 = 𝑒
(𝑖ℏ∇ + 𝑞A + 𝑞∇ 𝑓 )𝜙 0 = (𝑖ℏ∇ + 𝑞A + 𝑞∇ 𝑓 )𝑒
= (𝑞A + 𝑞∇ 𝑓 )𝑒
𝑖𝑞 𝑓
ℏ
𝜙 − 𝑞∇ 𝑓 𝑒
𝑖𝑞 𝑓
ℏ
+ 𝑖ℏ𝑒
𝑖𝑞 𝑓
ℏ
𝑖𝑞 𝑓
ℏ
𝑖𝑞 𝑓
ℏ
(5.179)
(5.180)
𝜙. We have:
𝜙 = (𝑞A + 𝑞∇ 𝑓 )𝑒
∇𝜙 = 𝑒
𝑖𝑞 𝑓
ℏ
𝑖𝑞 𝑓
ℏ
h 𝑖𝑞 𝑓 i
𝜙 + 𝑖ℏ∇ 𝑒 ℏ 𝜙 =
(𝑖ℏ∇ + 𝑞A) 𝜙,
(5.181)
and
(𝑖ℏ∇ + 𝑞A + 𝑞∇ 𝑓 ) 2 𝜙 0 = (𝑖ℏ∇ + 𝑞A + 𝑞∇ 𝑓 ) 𝑒
=𝑒
𝑖𝑞 𝑓
ℏ
𝑖𝑞 𝑓
ℏ
(𝑖ℏ∇ + 𝑞A) 𝜙 =
2
(5.182)
(𝑖ℏ∇ + 𝑞A) 𝜙.
1
(𝑖ℏ∇ + 𝑞A) 2 𝜙 =
Placing it into Eq. (5.180) above, and canceling the term 𝑒 ℏ we get 2𝑚
0
0
𝐸 𝜙 =⇒ 𝐸 = 𝐸 .
For the case of a constant magnetic field with B = e𝑧 𝐵, the introduced at the page
106 vector potential with 𝐴 𝑦 = 𝐵 𝑥 𝐴 𝑥 = 𝐴 𝑧 = 0 corresponds to the so called Landau
𝐵
gauge. Another widely used choice 𝐴 𝑥 = − 2𝑦 , 𝐴 𝑦 = 𝐵2𝑥 , 𝐴 𝑧 = 0 corresponds to
the symmetric gauge.
The Hamiltonian in the Landau gauge reads:
𝑖𝑞 𝑓
2
𝑝ˆ2𝑧
𝑝ˆ2
1
1
(p − 𝑞A) 2 = 𝑥 +
𝐻ˆ =
+
𝑝ˆ 𝑦 − 𝑞𝐵 𝑥 ,
2𝑚
2𝑚 2𝑚 2𝑚
which can be rewritten as:
"
#
2
ℏ2 𝜕 2
𝜕
𝑞𝐵 𝑥
𝜕2
− 𝑖
−
+
+ 2 𝜙 = 𝐸 𝜙.
2𝑚 𝜕𝑥 2
𝜕𝑦
ℏ
𝜕𝑧
(5.183)
(5.184)
ˆ 𝑝ˆ 𝑧 ] = [ 𝐻;
ˆ 𝑝ˆ 𝑦 ] = 0 As only x coordinate enters to the Hamiltonian, we have [ 𝐻;
Hamiltonian commutes with 𝑝ˆ 𝑧 , 𝑝ˆ 𝑦 , which means that its eigenfunction 𝜙(𝑥, 𝑦, 𝑧)
are in the same time eigenfunctions of 𝑝ˆ 𝑧 , 𝑝ˆ 𝑦 . But the latter are plane waves:
𝑝ˆ 𝑧 𝑒 𝑖𝑘𝑧 𝑧 = −𝑖ℏ
𝜕 𝑖𝑘𝑧 𝑧
𝑒
= ℏ𝑘 𝑧 𝑒 𝑖𝑘𝑧 𝑧 = 𝑝 𝑧 𝑒 𝑖𝑘𝑧 𝑧
𝜕𝑧
(5.185)
This means that we can search the solution in the following form:
𝜙(𝑥, 𝑦, 𝑧) = 𝑒 𝑖 (𝑘 𝑦 𝑦+𝑘𝑧 𝑧) 𝜙(𝑥).
Placing this expression into Schrodinger equation, one gets:
(5.186)
108
5 Bound states of quantum 1D and 2D particles
2
ℏ2 𝑘 2𝑧
ℏ2 𝑑 2 𝜙 ℏ2
𝑞𝐵 𝑥
+
−
𝑘
−
𝜙
+
𝜙 = 𝐸 𝜙.
𝑦
2𝑚 𝑑𝑥 2 2𝑚
ℏ
2𝑚
ℏ2 𝑘 2
(5.187)
ℏ𝑘
Introducing the parameters 𝜀 = 𝐸 − 2𝑚𝑧 , 𝑥0 = 𝑒𝐵𝑦 , this can be rewritten in the
following form:
ℏ2 𝑑 2 𝜙 𝑚𝜔2𝐵
−
+
(𝑥 − 𝑥0 ) 2 𝜙 = 𝜀𝜙.
(5.188)
2𝑚 𝑑𝑥 2
2
2
2
ℏ 𝑑 𝜙
+
This coincides with an equation for 1D quantum Harmonic oscillator, − 2𝑚
𝑑 𝑥2
𝑚𝜔 2 2
2 𝑥 𝜙
= 𝜀𝜙 with frequency 𝜔 = 𝜔 𝐵 =
from zero to 𝑥0 =
ℏ𝑘
− 𝑒𝐵𝑦 .
𝑒𝐵
𝑚 ,
and with equilibrium position shifted
But this shift of course does not change energy: and thus
Fig. 5.17
for 𝜀 we can immediately write:
1
𝜀 = ℏ𝜔 𝐵 𝑛 +
,
2
(5.189)
which gives for the energy
𝐸 = ℏ𝜔 𝐵
ℏ2 𝑘 2𝑧
1
𝑛+
+
.
2
2𝑚
(5.190)
This expression contains two terms. The second one corresponds to the free motion
of the particle along z axis and just gives the corresponding kinetic energy. Naturally,
as motion along z is infinite,
this term is continuous.
1
The first term ℏ𝜔 𝐵 𝑛 + 2 corresponds to the quantization of the circular motion
in xy plane. It describes the formation of a discrete set of so called Landau levels.
Note, that Landau levels are infinitely degenerate. Indeed, if we change 𝑘 𝑦 in
𝜙(𝑥, 𝑦, 𝑧) = 𝑒 𝑖 (𝑘 𝑦 𝑦+𝑘𝑧 𝑧) 𝜙(𝑥), the wavefunction is definitely changing in non-trivial
5.5 Supplementary material and tasks
109
way (remember, that phase matters). However, then 𝑘 𝑦 is only entering into parameter
𝑥0 , which does not affect energy.
The fact that quantum problem of a charged particle in magnetic field reduce
to the problem of Harmonic oscillator is remarkable. This analogy holds in the
classical case as well. Indeed, solution of classical equation of motion is very similar
in both cases - we have harmonic dependence of coordinates of time, with period
independent on energy.
5.5 Supplementary material and tasks
5.5.1 Infinite Square Well
We considered an infinite square well potential
(
+∞ |𝑥| > 𝐿/2
𝑉 (𝑥) =
0
|𝑥| ≤ 𝐿/2
(5.191)
and solved the stationary Schrodinger equation
ˆ
𝐻𝜓(𝑥)
= 𝐸𝜓(𝑥).
where
2
(5.192)
2
ℏ 𝜕
+ 𝑉 (𝑥).
(5.193)
𝐻ˆ = −
2𝑚 𝜕𝑥 2
We reduced this problem to the free stationary Schrodinger equation with the boundary conditions
( 2 2
ℏ 𝜕
− 2𝑚
𝜓(𝑥) = 𝐸𝜓(𝑥)
𝜕𝑥 2
(5.194)
𝜓(𝑥 = 𝐿/2) = 𝜓(𝑥 = −𝐿/2) = 0
The corresponding eigenenergies are given by
𝐸𝑛 =
ℏ2 𝜋 2 𝑛2
,
2𝑚𝐿 2
(5.195)
where 𝑛 = 1, 2, 3, ... and they coincide with the ones we obtained using the BohrSommerfeld quantization! The case 𝑛 = 0 is trivial since for 𝑛 = 0 the corresponding
eigenfunction 𝜓 𝑛=0 (𝑥) = 0 for all 𝑥 which implies the absence of a particle so we
ℏ2 𝜕2
𝜓 (𝑥) = 𝐸 𝑛 𝜓 𝑛 (𝑥)) are given
exclude it. The corresponding eigenfunctions (− 2𝑚
𝜕𝑥 2 𝑛
by
q


 𝐿2 cos(𝑘 𝑛 𝑥) 𝑛 = 1, 3, 5...

𝜓 𝑛 (𝑥) = q
(5.196)

 𝐿2 sin(𝑘 𝑛 𝑥) 𝑛 = 2, 4, 6...

110
5 Bound states of quantum 1D and 2D particles
√
where 𝑘 𝑛 = 2𝑚𝐸 𝑛 /ℏ = 𝜋𝑛/𝐿.
The stationary solutions are useful but nothing interesting happens when the
system is in one of the eigenstates 𝜓 𝑛 (𝑥) since for them any observables (such as h𝑥i,
ˆ
h 𝑝i
ˆ etc) are time-independent. We know from lectures that the general solution of the
𝜕
ℏ2 𝜕2
𝜓(𝑥, 𝑡) = 𝑖ℏ 𝜕𝑡
𝜓(𝑥, 𝑡)
corresponding time-dependent Schrodinger equation − 2𝑚
𝜕𝑥 2
is given by
Õ
𝜓(𝑥, 𝑡) =
𝑐 𝑛 𝜓 𝑛 (𝑥)𝑒 −𝑖𝐸𝑛 𝑡/ℏ
(5.197)
𝑛
where it is assumed that we got rid of the potential by "hiding" it in the boundary
conditions (𝜓(𝑥 = 𝐿/2, 𝑡) = 𝜓(𝑥 = −𝐿/2, 𝑡) = 0 for all 𝑡) exactly as we did for the
stationary equation. The time-independent (!) coefficients 𝑐 𝑛 are arbitrary, with the
only restriction that the wave function 𝜓(𝑥, 𝑡) must be normalized.
p
• Show that the normalization coefficient 2/𝐿 is the same for 𝜓 𝑛 (𝑥) ∼ cos(𝑘 𝑛 𝑥)
(corresponding to 𝑛 = 1, 3, 5...) and for 𝜓 𝑛 (𝑥) ∼ sin(𝑘 𝑛 𝑥) (corresponding to
𝑛 = 2, 4, 6...).
• Consider the following time-dependent solution
𝜓(𝑥, 𝑡) = 𝑐 1 𝜓1 (𝑥)𝑒 −𝑖𝐸1 𝑡/ℏ + 𝑐 2 𝜓2 (𝑥)𝑒 −𝑖𝐸2 𝑡/ℏ
2
(5.198)
2
𝜕
ℏ 𝜕
𝜓(𝑥, 𝑡) = 𝑖ℏ 𝜕𝑡
𝜓(𝑥, 𝑡).
of the corresponding full Schrodinger equation − 2𝑚
𝜕𝑥 2
Here 𝜓 𝑛 (𝑥) are given by Eq.(5.196).
ℏ2 𝜕 2
𝜕
(a)Make sure that it does satisfy − 2𝑚
𝜓(𝑥, 𝑡).
𝜓(𝑥, 𝑡) = 𝑖ℏ 𝜕𝑡
𝜕𝑥 2
(b)Find the
∫ ∞relation between the coefficients 𝑐 1 , 𝑐 2 , using the normalization
condition −∞ 𝜓 ∗ (𝑥, 𝑡)𝜓(𝑥, 𝑡)𝑑𝑥 = 1 for all 𝑡.
(c)Find how the average position h𝑥i(𝑡)
ˆ
depends on time for this state.
(d)(additional questions) What is the probability that the particle has energy 𝐸 1 ?
What is the probability that the particle has energy 𝐸 2 ? What is the probability
that the particle has energy, that is neither 𝐸 1 nor 𝐸 2 ?
5.5.2 Dirac-Kronig-Penney model. Bound states in a 𝜹-potential
If the width of each barrier tend to zero while its heights tends to infinity, than the
Kronig-Penney model turns into the Dirac-Kronig-Penney model with the periodic
𝑚=+∞
Í
potential 𝑉 (𝑥) = 𝑉0
𝛿(𝑥 − 𝑚𝑎) with the following dispersion equation for
𝑚=−∞
dimensionless 𝑞˜ = 𝑞𝑎 and 𝑘˜ = 𝑘𝑎:
cos 𝑞˜ + 𝑢 0
sin 𝑞˜
= cos 𝑘˜
𝑞˜
(5.199)
here √we introduced the dimensionless quantity 𝑢 0 = 𝑚𝑉0 𝑎/ℏ2 , 𝑞 is defined as
𝑞 = 2𝑚𝐸/ℏ and the quasi-momentum 𝑘 comes from the Bloch theorem stating
5.5 Supplementary material and tasks
111
that eigenstates have the form 𝜓(𝑥) = 𝑒 𝑖𝑘 𝑥 𝑈 (𝑥), where 𝑈 (𝑥 + 𝑛𝑎) = 𝑈 (𝑥) for any
𝑛 ∈ Z. The graphical solution to Eq. (5.199) is shown in Fig. 5.5.2 (see below).
Fig. 5.18 Graphical representation Eq. (5.199). The red curve represents the LHS cos 𝑞˜ + 𝑢0 sin𝑞˜ 𝑞˜
at 𝑢0 = 10. The green and yellow lines depict the upper (+1) and lower (-1) border that the RHS
cos 𝑘˜ can take.
• Task 1. Solve Eq. (5.199) in the vicinity of zeros of the LHS and find 𝐸 = 𝐸 (𝑘).
Finding the zeros of the LHS can not be done analytically since cos 𝑞˜ + 𝑢 0 sin𝑞˜ 𝑞˜ = 0
is a transcendental equation. But you have to assume that the solutions are known,
the roots are given 𝑞˜ 1 , 𝑞˜ 2 , 𝑞˜ 3 , .... For example, I found the first root 𝑞˜ 1 ≈ 2.865
numerically for 𝑢 0 = 10. Plot the first band 𝐸 1 (𝑘) for this values (you won’t need
a PC for that, it has a very simple form).
• Task 2. (a) Find the bound states for the following potential
𝑉 (𝑥) = −𝛼𝛿(𝑥 − 𝑥0 ),
(5.200)
∫ +∞
here 𝛼 > 0. A bound state is a state that is normalizable, i.e. −∞ |𝜓(𝑥)| 2 𝑑𝑥 is
finite (in contrast to 𝜓(𝑥) ∼ 𝑒 ±𝑖𝑘 𝑥 for which the integral diverges). Physically,
that means that the wave functions are localized ("bound" to the potential). It is
clear that in our case (i.e. 𝑉 (𝑥) = −𝛼𝛿(𝑥 − 𝑥 0 ) the bound states can only exists for
𝐸 < 0 since for 𝐸 > 0 the general solution is represented by linear combinations
of 𝑒 ±𝑖𝑘 𝑥 ).
• (b) Plot all the eigenstates (you won’t need a PC, they have a simple form).
• (c) Show that for 𝛼 < 0, i.e. when 𝑉 (𝑥) = |𝛼|𝛿(𝑥 − 𝑥0 ) there are no bound state
solutions, which is quite intuitive since in that case we deal with a potential barrier
112
5 Bound states of quantum 1D and 2D particles
rather than with a potential well and we know from classical mechanics that finite
trajectories in 1D are only possible in well-like potentials (bound states in QM
are direct analogs of finite trajectories in classical mechanics).
Hint: when solving the bound states problem, you will need to derive a system
of two linear equations and find the condition for solvability. One of them can
be obtained from the continuity condition, the other one – by integrating the
stationary Schrodinger equation from 𝑥0 − 𝜖 to 𝑥0 + 𝜖 and then sending 𝜖 to zero.
• (d)Find the bound states of the following potential 𝑉 (𝑥) = −𝛼(𝛿(𝑥 −𝑎) +𝛿(𝑥 +𝑎)).
Is the ground state even or odd? Plot the wave functions of the ground state and
the first excited state.
5.5.3 Classical and Semi-classical Electrodynamics
We introduced a vector potential A(r, 𝑡) and a scalar potential 𝜙(r, 𝑡) such that an
electric and a magneic fields are expressed as follows
E(r, 𝑡) = −grad𝜙(r, 𝑡) −
𝜕A(r, 𝑡)
𝜕𝑡
B(r, 𝑡) = curlA(r, 𝑡)
(5.201)
In classical mechanics one may introduce the (classical!) Hamiltonian for a particle
of mass 𝑚 and charge 𝑞 in the presence of an electromagnetic field as
H (r, p) =
(p − 𝑞A(r, 𝑡)) 2
+ 𝑞𝜙(r, 𝑡)
2𝑚
(5.202)
The time evolution of the system is uniquely defined by Hamilton’s equations
𝑑p
𝜕H (r, p)
=−
𝑑𝑡
𝜕r
𝑑r 𝜕H (r, p)
=
𝑑𝑡
𝜕p
(5.203)
Now let’s consider the semi-classical case, when the particle is treated using
Quantum Mechanics but the electromagnetic field is treated classically. Then the
(quantum!) Hamiltonian is given by
( p̂ − 𝑞A(r, 𝑡)) 2
𝐻ˆ =
+ 𝑞𝜙(r, 𝑡)
2𝑚
(5.204)
• For our classical Hamiltonian (5.202) derive the equations of motion (which are
given by Eq. (5.203)) and show that they can be reduced to a single second-order
differential equation which is nothing but Newton’s second law for a particle in
the presence of an electromagnetic field.
5.5 Supplementary material and tasks
113
• In lectures we derived the probability current from the Schrodinger equation for
a free particle in 1D. Derive (you can do the same steps as we did in lectures) the
probability current from the Schrodinger with the Hamiltonian (5.204) (which
describes a charged particle in 3D space in the presence of an electromagnetic
field).
5.5.4 Quantum harmonic oscillator in an electric field
Let’s consider the quantum harmonic oscillator in an uniform electric field. The
Hamiltonian of the system is
𝑝ˆ2 𝑚𝜔2 𝑥 2
𝐻ˆ =
+
− 𝑞𝐸𝑥,
2𝑚
2
(5.205)
where 𝑞 is the charge of the oscillating particle and 𝐸 is the strength of the electric
field.
• Find the eigenstates and eigenenergies of the system. Hint: you can reduce this
problem to a simple quantum harmonic oscillator that you studied in lectures by
making an appropriate change of variables.
• For the ground state of the system calculate the mean value of the position operator
h𝑥i and the mean value of the momentum operator h 𝑝i.
ˆ
5.5.5 2D quantum harmonic oscillator in a magnetic field
Let’s suppose that motion of the charged particle is restricted to 2D plane which can
be realized in real life as a two-dimensional electron gas (2DEG) at the interface of
semiconductors, in other words let’s assume that the kinetic energy operator is merely
𝑇ˆ = 𝑝ˆ2𝑥 /(2𝑚)+ 𝑝ˆ2𝑦 /(2𝑚). Now let’s confine the particle in the x-direction by introducing the confinement potential 𝑉 (𝑥) = 𝑚𝜔20 𝑥 2 /2. Why does it "confine" the particle in
the x-direction? Because in the Hamiltonian 𝐻ˆ = 𝑇ˆ + 𝑉ˆ (𝑥) = 𝑝ˆ2𝑥 /(2𝑚) + 𝑝ˆ2𝑦 /(2𝑚) +
𝑚𝜔20 𝑥 2 /2 the x-component and the y-component do not mix and the variables can be
separated 𝜓(𝑥, 𝑦) = 𝜓 𝑥 (𝑥)𝜓 𝑦 (𝑥). Next, as you remember the eigenstates of 1D quantum harmonic oscillator 𝐻ˆ 𝑄𝐻 𝑂 𝜓 𝑥 (𝑥) = ( 𝑝ˆ2𝑥 /(2𝑚) + 𝑚𝜔20 𝑥 2 /2)𝜓 𝑥 (𝑥) = 𝐸𝜓 𝑥 (𝑥) are
localized in the vicinity of 𝑥 = 0. Thus now the motion is restricted to a thin (but not
infinitely thin as it is in the z-direction!) line in the x-direction, so we have a so-called
quantum wire. Now let’s add a magnetic field pointing out in the z-direction. Such a
field can be described by A = (0, 𝑥𝐵, 0)𝑇 (this is called the Landau gauge) since the
corresponding magnetic field is B = curlA = (0, 0, 𝐵)𝑇 . We assume that there is no
electric field, so 𝜙 = 0. Thus, the Hamiltonian of the system is (here p̂ = ( 𝑝ˆ 𝑥 , 𝑝ˆ 𝑦 )𝑇 )
114
5 Bound states of quantum 1D and 2D particles
2 2
( 𝑝ˆ 𝑦 − 𝑞𝑥𝐵) 2 𝑚𝜔20 𝑥 2
𝑝ˆ2
( p̂ − 𝑞A) 2 𝑚𝜔0 𝑥
𝐻ˆ =
+
+ 𝑞𝜙 = 𝑥 +
+
2𝑚
2
2𝑚
2𝑚
2
(5.206)
which is derived from the Hamiltonian of an ordinary quantum harmonic oscillator
𝐻ˆ = p̂2 /2𝑚 + 𝑚𝜔20 𝑥 2 /2 using the so-called minimal coupling scheme, namely
p̂ → p̂ − 𝑞A and 𝐻ˆ → 𝐻ˆ + 𝑞𝜙.
The eigenstates of the quantum wire described by the Hamiltonian (5.206) can be
ˆ = 0 and hence the eigenstates of 𝐻ˆ are simultaneously
found by noting that [ 𝑝ˆ 𝑦 , 𝐻]
eigenstates of 𝑝ˆ 𝑦 and can be written as 𝜓(𝑥, 𝑦) = 𝜓 𝑥 (𝑥) exp (−𝑖𝑘 𝑦 𝑦) since
𝑝ˆ 𝑦 𝜓(𝑥, 𝑦) = 𝑝ˆ 𝑦 𝜓 𝑥 (𝑥) exp (−𝑖𝑘 𝑦 𝑦) = 𝑝 𝑦 𝜓(𝑥, 𝑦)
(5.207)
where 𝑝 𝑦 = ℏ𝑘 𝑦 .
ˆ = 0 and find [ 𝑝ˆ 𝑥 , 𝐻].
ˆ
• Show that [ 𝑝ˆ 𝑦 , 𝐻]
• Solve the stationary equation with the Hamiltonian (5.206), and find the average
position h𝑥i
ˆ in the ground state. Does it depend on the sign of 𝐵?
Chapter 6
Quantum particle in a central potential
6.1 Stationary Schrodinger equation in a central potential,
separation of variables
Let us consider a problem of a quantum particle, moving in three dimensions in
the central potential, when potential energy depends only on the distance to a given
attraction center, which can always be placed into the origin of the coordinate system.
The Hamiltonian of the system is:
2
ℏ 2
𝐻ˆ = 𝑇ˆ + 𝑈ˆ = −
∇ + 𝑈 (𝑟), 𝑟 =
2𝑚
q
𝑥 2 + 𝑦2 + 𝑧2 .
(6.1)
And the energy spectrum shouldpbe found from the stationary Schrodinger equation
ˆ
𝐻𝜓(r)
= 𝐸𝜓(r). As 𝑈 (𝑟) = 𝑈 ( 𝑥 2 + 𝑦 2 + 𝑧 2 ) can not be represented as a sum of the
terms dependent on x, y, z (the latter is possible only for the case of a 3D Harmonic
2
𝑚𝜔 2
2
oscillator with 𝑈 (𝑟) = 𝑚𝜔
𝑥 2 + 𝑦 2 + 𝑧2 ), the variables in Cartesian
2 𝑟 =
2
coordinates can not be separated, and one needs to use spherical coordinates, which
property account for the spherical symmetry of the problem, i.e. the invariance of
the Hamiltonian under rotations around any axis passing through the origin of the
coordinate system
𝑧 = 𝑟 cos 𝜃,
𝑦 = 𝑟 sin 𝜃 sin 𝜙,
𝑥 = 𝑟 sin 𝜃 cos 𝜙,
q
𝑟 = 𝑥 2 + 𝑦2 + 𝑧2 ,
p
𝑥2 + 𝑦2
𝜃 = arctan
,
𝑧
𝑦
𝜙 = arctan .
𝑥
(6.2)
115
116
6 Quantum particle in a central potential
Fig. 6.1
The expression for a Hamiltonian in spherical coordinates can be obtained via its
expression in Cartesian coordinates
∇2 =
𝜕2
𝜕2
𝜕2
+
+
𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2
(6.3)
by using expression of 𝑟, 𝜃, 𝜙 as functions of 𝑥, 𝑦, 𝑧, and applying the Chain rule,
i.e.
𝜕
𝜕𝑟 𝜕
𝜕𝜃 𝜕
𝜕𝜙 𝜕
=
+
+
(6.4)
𝜕𝑥 𝜕𝑥 𝜕𝑟 𝜕𝑥 𝜕𝜃 𝜕𝑥 𝜕𝜙
and similar expressions for
∇2 =
𝜕
𝜕
𝜕𝑦 , 𝜕𝑧 .
After performing this procedure, we get:
1 𝜕 2 𝜕
1
𝜕
𝜕
1
𝜕2
,
𝑟
+
sin
𝜃
+
𝜕𝑟
𝜕𝜃
𝑟 2 𝜕𝑟
𝑟 2 sin2 𝜃 𝜕𝜃
𝑟 2 sin2 𝜃 𝜕𝜙2
(6.5)
which allows one to rewrite the Schrodinger equation in the following form:
ℏ2 1 𝜕 2 𝜕𝜓
1
𝜕𝜓
1
𝜕𝜓
𝜕2𝜓
−
𝑟
+
sin 𝜃
+
+ 𝑈 (𝑟)𝜓 = 𝐸𝜓.
2𝑚 𝑟 2 𝜕𝑟
𝜕𝑟
𝜕𝜃
𝑟 2 sin2 𝜃 𝜕𝜃
𝑟 2 sin2 𝜃 𝜕𝜙2
(6.6)
2
Multiplying by 2𝑚𝑟
we
get:
ℏ2
−
𝜕 2 𝜕𝜓
1 𝜕
𝜕𝜓
1 𝜕 2 𝜓 2𝑚𝑟 2
𝑟
+
sin 𝜃
+
[𝑈 (𝑟) − 𝐸]𝜓 = 0, (6.7)
𝜕𝑟
𝜕𝑟
sin 𝜃 𝜕𝜃
𝜕𝜃
sin2 𝜃 𝜕𝜙2 ℏ2
and thus, collecting terms with 𝑟 on the left, and terms with angles 𝜃, 𝜙 on the right,
we get:
6.1 Stationary Schrodinger equation in a central potential, separation of variables
−
2𝑚𝑟 2
1 𝜕
𝜕𝜓
1 𝜕2𝜓
𝜕 2 𝜕𝜓
𝑟
+ 2 [𝑈 (𝑟) − 𝐸]𝜓 =
sin 𝜃
+
.
𝜕𝑟
𝜕𝑟
sin 𝜃 𝜕𝜃
𝜕𝜃
ℏ
sin2 𝜃 𝜕𝜙2
117
(6.8)
Now, we can separate the variable 𝑟 from the variables 𝜃 and 𝜙, representing the
wavefunction 𝜓(r) as:
𝜓(r) = 𝑅(𝑟)𝑌 (𝜃, 𝜓).
(6.9)
Placing this into the Schrodinger equation and dividing it by 𝜓 = 𝑅𝑌 , one gets:
1
𝑑 2 𝑑𝑅
2𝑚𝑟 2
−
𝑟
+ 2 [𝑈 (𝑟) − 𝐸] 𝑅 =
𝑅
𝑑𝑟
𝑑𝑟
ℏ
1
𝜕𝑌
1 𝜕 2𝑌
1 𝜕
=
sin 𝜃
+
= −𝜆,
(6.10)
𝑌 sin 𝜃 𝜕𝜃
𝜕𝜃
sin2 𝜃 𝜕𝜙2
where constant of separation 𝜆 is just a number not depending on any coordinates,
as we have function of only 𝑟 in the left, and function of only 𝜃, 𝜙 in the right. We
therefore get two independent equations for 𝑅 = 𝑅(𝑟) and 𝑌 = 𝑌 (𝜃, 𝜙):
𝜕𝑌
1 𝜕 2𝑌
1 𝜕
= −𝜆𝑌
(6.11)
sin 𝜃
+
sin 𝜃 𝜕𝜃
𝜕𝜃
sin2 𝜃 𝜕𝜙2
- angular part of Schrodinger equation which is universal, i.e. does not contain the
potential 𝑈 (𝑟) (the only important thing is that it is central), and
ℏ2 1 𝑑 2 𝑑𝑅
ℏ2 𝜆
−
𝑅 = 𝐸 𝑅.
(6.12)
𝑟
+ 𝑈 (𝑟) +
2𝑚 𝑟 2 𝑑𝑟
𝑑𝑟
2𝑚𝑟 2
The term in the square brackets corresponds to some effective potential, corresponding to the radial motion in the central field. Let us clarify the physical meaning
ℏ2 𝜆
of the term 2𝑚𝑟
2 in it. For that, let us consider the corresponding classical problem.
It is well known, that the trajectory of a classical particle moving in any central
potential 𝑈 (𝑟) lies in plane, which can be chosen to coincide with x-y plane. This is
consequence of the conservation of angular momentum L = [r × p] = 𝑚 [r × V] in
central potentials. For the motion in x-y plane, L = e𝑧 𝐿, with 𝐿 = 𝐿 𝑧 = 𝑥 𝑝 𝑦 − 𝑦 𝑝 𝑥 =
𝑚(𝑥 𝑦¤ − 𝑦 𝑥)
¤ = 𝑐𝑜𝑛𝑠𝑡. Introducing polar coordinates 𝑟, 𝜃, 𝑥 = 𝑟 cos 𝜃, 𝑦 = 𝑟 sin 𝜃,
we get:
𝑑
𝑑
𝐿 = 𝑚 𝑟 cos 𝜃 (𝑟 sin 𝜃) − 𝑟 sin 𝜃 (𝑟 cos 𝜃) =
𝑑𝑡
𝑑𝑡
2¤
2
¤
𝑚 𝑟 𝑟¤ cos 𝜃 sin 𝜃 + 𝑟 𝜃 cos 𝜃 − 𝑟 𝑟¤ sin 𝜃 cos 𝜃 + 𝑟 2 𝜃¤ sin2 𝜃 = 𝑚𝑟 2 𝜃,
(6.13)
from where we get 𝜃¤ = 𝑚𝑟𝐿 2 .
Now, let us write the classical Hamiltonian (=total energy):
118
6 Quantum particle in a central potential
Fig. 6.2
𝑚
𝐻 = 𝐸 = 𝑇 + 𝑈 = ( 𝑥¤ 2 + 𝑦¤ 2 ) + 𝑈 (𝑟) =
2
𝑚 ¤
(𝑟¤ cos 𝜃 − 𝑟 𝜃 sin 𝜃) 2 + (𝑟¤ sin 𝜃 + 𝑟 𝜃¤ cos 𝜃) 2 + 𝑈 (𝑟) =
=
2
𝑚 2
=
𝑟¤ cos2 𝜃 − 2𝑟 𝑟¤ cos 𝜃 sin 𝜃 + 𝑟 2 𝜃¤2 sin2 𝜃 + 𝑟¤2 sin2 𝜃 + 2𝑟 𝑟¤ cos 𝜃 sin 𝜃 + 𝑟 2 𝜃¤2 cos2 𝜃 + 𝑈 (𝑟) =
2
𝑚𝑟¤2
𝐿2
𝑚 2
+
+ 𝑈 (𝑟) =
= (𝑟¤ + 𝑟 2 𝜃¤2 ) + 𝑈 (𝑟) =
2
2
2𝑚𝑟 2
𝑚𝑟¤2
𝐿2
+ 𝑈𝑒 𝑓 𝑓 (𝑟), 𝑈𝑒 𝑓 𝑓 = 𝑈 (𝑟) +
.
(6.14)
=
2
2𝑚𝑟 2
Comparing this classical expression with what we got in the radial part of Schrodinger
𝜆ℏ2
equation, 𝑈𝑒 𝑓 𝑓 = 𝑈 (𝑟) + 2𝑚𝑟
2 , one concludes that the last term there is simply
centrifugal energy (same as in classics), and 𝜆ℏ2 is related to the square of the
angular momentum of a quantum particle. Looking at the equation for 𝑌 at page 117,
and multiplying it by ℏ2 :
1 𝜕
𝜕𝑌
1 𝜕 2𝑌
= 𝜆ℏ2𝑌 .
(6.15)
−ℏ2
sin 𝜃
+
sin 𝜃 𝜕𝜃
𝜕𝜃
sin2 𝜃 𝜕𝜙2
One can guess that the operator in the square brackets in the left hand side here
should of the angular momentum,
1 𝜕
𝜕
1 𝜕2
𝐿ˆ 2 = −ℏ2
sin 𝜃
+
.
(6.16)
sin 𝜃 𝜕𝜃
𝜕𝜃
sin2 𝜃 𝜕𝜙2
Then, this equation can be rewritten in the form 𝐿ˆ 2𝑌 = ℏ2 𝜆𝑌 , and thus ℏ2 𝜆 will
correspond to the eigenvalues of the square of angular momentum, corresponding
to its values that can be detected in the experiment.
6.2 Operators of angular momentum
119
6.2 Operators of angular momentum
Let us show, that our guess is correct. For that, construct an operator of the quantum
angular momentum, using classical relation for it L = [r × p], and applying the
substitution p → p̂ = −𝑖ℏ∇. One gets:
e 𝑥 e 𝑦 e𝑧
L̂ = [r × p̂] = 𝑥 𝑦 𝑧 =
𝑝ˆ 𝑥 𝑝ˆ 𝑦 𝑝ˆ 𝑧
e 𝑥 (𝑦 𝑝ˆ 𝑧 − 𝑧 𝑝ˆ 𝑦 ) + e 𝑦 (𝑧 𝑝ˆ 𝑥 − 𝑥 𝑝ˆ 𝑧 ) + e𝑧 (𝑥 𝑝ˆ 𝑦 − 𝑦 𝑝ˆ 𝑥 ) =
e 𝑥 𝐿ˆ 𝑥 + e 𝑦 𝐿ˆ 𝑦 + e𝑧 𝐿ˆ 𝑧 =⇒
ˆ𝐿 𝑥 = 𝑦 𝑝ˆ 𝑧 − 𝑧 𝑝ˆ 𝑦 = 𝑖ℏ 𝑧 𝜕 − 𝑦 𝜕 ,
𝜕𝑦
𝜕𝑧
𝜕
𝜕
ˆ𝐿 𝑦 = 𝑧 𝑝ˆ 𝑥 − 𝑥 𝑝ˆ 𝑧 = 𝑖ℏ 𝑥
−𝑧
,
𝜕𝑧
𝜕𝑥
𝜕
𝜕
𝐿ˆ 𝑧 = 𝑥 𝑝ˆ 𝑦 − 𝑦 𝑝ˆ 𝑥 = 𝑖ℏ 𝑦
−𝑥
.
𝜕𝑥
𝜕𝑦
(6.17)
(6.18)
(6.19)
(6.20)
From these expression, one can easily see, that components of the operator of angular
momentum corresponding to the orthogonal axes are not commuting. For example:
[ 𝐿ˆ 𝑥 ; 𝐿ˆ 𝑦 ] = [𝑦 𝑝ˆ 𝑧 − 𝑧 𝑝ˆ 𝑦 ; 𝑧 𝑝ˆ 𝑥 − 𝑥 𝑝ˆ 𝑧 ] =
[𝑦 𝑝ˆ 𝑧 ; 𝑧 𝑝ˆ 𝑥 ] − [𝑦 𝑝ˆ 𝑧 ; 𝑥 𝑝ˆ 𝑧 ] − [𝑧 𝑝ˆ 𝑦 ; 𝑧 𝑝ˆ 𝑥 ] + [𝑧 𝑝ˆ 𝑦 ; 𝑥 𝑝ˆ 𝑧 ] =
𝑦 𝑝ˆ 𝑥 [ 𝑝ˆ 𝑧 ; 𝑧] + 𝑥 𝑝ˆ 𝑦 [𝑧; 𝑝ˆ 𝑧 ] = 𝑖ℏ(𝑥 𝑝 𝑦 − 𝑦 𝑝 𝑥 ) = 𝑖ℏ𝐿 𝑧 .
(6.21)
Commutators for other components can be obtained in the similar way, which will
give us: [ 𝐿ˆ 𝑥 ; 𝐿ˆ 𝑦 ] = 𝑖ℏ 𝐿ˆ𝑧 ; [ 𝐿ˆ 𝑧 ; 𝐿ˆ 𝑥 ] = 𝑖ℏ 𝐿ˆ𝑦 ; [ 𝐿ˆ 𝑦 ; 𝐿ˆ 𝑧 ] = 𝑖ℏ 𝐿ˆ𝑥 . We can also construct
the operator of the square of the angular momentum as: 𝐿ˆ 2 = 𝐿ˆ 2𝑥 + 𝐿ˆ 2𝑦 + 𝐿ˆ 2𝑧 . This
operator commutes with any of the projections 𝐿ˆ 𝑥 , 𝐿ˆ 𝑦 , 𝐿ˆ 𝑧 . For example:
[ 𝐿ˆ 𝑥 ; 𝐿ˆ 2 ] = [ 𝐿ˆ 𝑥 ; 𝐿ˆ 2𝑥 ] + [ 𝐿ˆ 𝑥 ; 𝐿ˆ 2𝑦 ] + [ 𝐿ˆ 𝑥 ; 𝐿ˆ 2𝑧 ] =
𝐿ˆ 𝑦 [ 𝐿ˆ 𝑥 ; 𝐿ˆ 𝑦 ] + [ 𝐿ˆ 𝑥 ; 𝐿ˆ 𝑦 ] 𝐿ˆ 𝑦 + 𝐿ˆ 𝑧 [ 𝐿ˆ 𝑥 ; 𝐿ˆ 𝑧 ] + [ 𝐿ˆ 𝑥 ; 𝐿ˆ 𝑧 ] 𝐿ˆ 𝑥 =
𝑖ℏ( 𝐿ˆ 𝑦 𝐿ˆ 𝑧 + 𝐿ˆ 𝑧 𝐿ˆ 𝑦 − 𝐿ˆ 𝑧 𝐿ˆ 𝑦 − 𝐿ˆ 𝑦 𝐿ˆ 𝑧 ) = 0.
(6.22)
And similarly [ 𝐿ˆ 𝑦 ; 𝐿ˆ 2 ] = [ 𝐿ˆ 𝑧 ; 𝐿ˆ 2 ] = 0. The fact, that different projections of quantum angular momentum do not commute with each other, but all of them commute
with square angular momentum means that different components of angular momentum can not be measured simultaneously, i.e. do not have common eigenfunctions,
while any components of angular momentum can be measured simultaneously with
its operators 𝐿ˆ 𝑥 , 𝐿ˆ 𝑦 , 𝐿ˆ 𝑧 has the same eigenfunctions as the operator 𝐿ˆ 2 . Using
Eq. (6.18,6.19,6.20) at page 119, relations between 𝑥, 𝑦, 𝑧 and 𝑟, 𝜃, 𝜙 given at
page 115 and chain rule, one can express the operators of the components of the
120
6 Quantum particle in a central potential
angular momentum 𝐿ˆ 𝑗 and operator 𝐿ˆ 2 in spherical coordinates
𝜕
𝜕
+ cos 𝜙 cotan 𝜃
,
𝐿ˆ 𝑥 = 𝑖ℏ sin 𝜙
𝜕𝜃
𝜕𝜙
𝜕
𝜕
𝐿ˆ 𝑦 = 𝑖ℏ − cos 𝜙
,
+ sin 𝜙 cotan 𝜃
𝜕𝜃
𝜕𝜙
𝜕
𝐿ˆ 𝑧 = −𝑖ℏ ,
𝜕𝜙
2
ˆ𝐿 2 = −ℏ2 1 𝜕 sin 𝜃 𝜕 + 1 𝜕
.
sin 𝜃 𝜕𝜃
𝜕𝜃
sin2 𝜃 𝜕𝜙2
(6.23)
(6.24)
(6.25)
(6.26)
The last expression indeed coincides with one given of page 118. The Hamiltonian
of the system (see page 116) can be rewritten as:
ℏ2 1 𝜕 2 𝜕
ℏ2
1 𝜕
𝜕
1 𝜕2
𝐻ˆ = −
𝑟
−
sin
𝜃
+
+ 𝑈 (𝑟) =
2𝑚 𝑟 2 𝜕𝑟
𝜕𝑟
2𝑚 sin 𝜃 𝜕𝜃
𝜕𝜃
sin2 𝜃 𝜕𝜙2
ℏ2 𝜕 2 𝜕
𝐿ˆ 2
+ 𝑈 (𝑟).
(6.27)
−
𝑟
+
𝜕𝑟
2𝑚𝑟 2 𝜕𝑟
2𝑚𝑟 2
This Hamiltonian commutes with both 𝐿ˆ 2 , 𝐿ˆ 𝑧 (or any other projection 𝐿ˆ 𝑥 . 𝐿ˆ 𝑦 ):
ˆ 𝐿ˆ 2 ] = 0, [ 𝐻;
ˆ 𝐿ˆ 𝑧 ] = 0.
[ 𝐻;
this means, that the energy of a quantum particle can be measured simultaneously
with absolute value of its angular momentum and its 𝑧 projection.
ˆ 𝐿ˆ 2 , 𝐿ˆ 𝑧 has common sot set of the eigenIn the other words, the operators 𝐻,
functions.
6.3 Spherical harmonics
Let us analyze the solutions of the angular part of the Schrodinger equation. We
have:
1 𝜕
𝜕𝑌
1 𝜕 2𝑌
sin 𝜃
+
= −𝜆𝑌 ,
(6.28)
sin 𝜃 𝜕𝜃
𝜕𝜃
sin2 𝜃 𝜕𝜙2
which coincides with eigenvalue equation for 𝐿ˆ 2 , 𝐿ˆ 2𝑌 = ℏ2 𝜆𝑌 . Multiplying this
equation by sin2 𝜃, we get:
𝜕
𝜕𝑌
𝜕 2𝑌
sin 𝜃
sin 𝜃
+
= −𝜆 sin2 (𝜃)𝑌 =⇒
(6.29)
𝜕𝜃
𝜕𝜃
𝜕𝜙2
𝜕
𝜕𝑌
𝜕 2𝑌
sin 𝜃
sin 𝜃
+ 𝜆 sin2 (𝜃)𝑌 = − 2 .
(6.30)
𝜕𝜃
𝜕𝜃
𝜕𝜙
6.3 Spherical harmonics
121
Here, the left hand side contains only 𝜃, and right hand side contains only 𝜙,
which allows us to separate the variables and write 𝑌 (𝜃, 𝜙) = 𝑃(𝜃)Φ(𝜙), which
gives after division by 𝑌 :
𝑑
𝑑𝑃
1 𝑑2Φ
1
sin 𝜃
sin 𝜃
+ 𝜆 sin2 (𝜃)𝑃 = − 2
=𝜉
(6.31)
𝑃
𝑑𝜃
𝑑𝜃
Φ 𝑑𝜙2
- constant of the separation. Thus:
𝑑
𝑑𝑃
sin 𝜃
sin 𝜃
+ (𝜆 sin2 𝜃 − 𝜉)𝑃 = 0,
𝑑𝜃
𝑑𝜃
(6.32)
𝑑2Φ
= −𝜉Φ.
𝑑𝜙2
(6.33)
Let us analyze the second equation. In order to determine the possible values of
the constant of separation 𝜉, it should be completed by the boundary condition:
Φ(𝜙 + 2𝜋) = Φ(𝜙).
Indeed, changing of the angular variable 𝜙 would mean that we made a round trip
in 𝑥 − 𝑦 plane, returning to the same point (see plot in the page 118, and, obviously,
the wave function should remain the same.
Let us now analyze solutions of the equations for Φ with 2𝜋 periodic boundary
condition. Consider the cases of positive and negative 𝜉 separately.
2
a) 𝜉 < 0, 𝑥𝑖 = −𝑚 2 =⇒ 𝑑𝑑 𝜙Φ2 = 𝑚 2 Φ, Φ(𝜙) = 𝐴𝑒 𝑚𝜙 + 𝐵𝑒 −𝑚𝜙 , Φ(𝜙 + 2𝜋) =
𝐴𝑒 2𝑚 𝜋 𝑒 𝑚𝜙 + 𝐵𝑒 −2𝑚 𝜋 𝑒 −𝑚𝜙 ≠ Φ(𝜙) except for the case when 𝐴 = 𝐵 = 0.
2
b) Therefore 𝜉 should be chosen positive, 𝜉 = 𝑚 2 , 𝑑𝑑 𝜙Φ2 = −𝑚 2 Φ which will give
us:
(6.34)
Φ(𝜙) = 𝐴𝑒 𝑖𝑚𝜙 + 𝐵𝑒 −𝑖𝑚𝜙 ,
Φ(𝜙 + 2𝜋) = 𝐴𝑒
2 𝜋𝑖𝑚 𝑖𝑚𝜙
𝑒
+ 𝐵𝑒
−2 𝜋𝑖𝑚 −𝑖𝑚𝜙
𝑒
.
(6.35)
Using Φ(𝜙) = Φ(𝜙 + 2𝜋) we get:
𝑒 ±2 𝜋𝑖𝑚 = 1,
(6.36)
which holds for integer 𝑚 = 0, ±1, ±2, etc.
Solutions can be thus written in the following form:
1
Φ𝑚 (𝜙) = √ 𝑒 𝑖𝑚𝜙 , 𝑚 = 0, ±1, ±2 . . .
(6.37)
2𝜋
∫ 2𝜋
where √1 is a normalization constant, chosen so, that 0 Φ∗𝑚 (𝜙)Φ𝑚 (𝜙)𝑑𝜙 = 1. Of
2𝜋
course, linear combinations of solutions Eq. (6.37) corresponding to Φ𝑚 , Φ−𝑚 will
be also good solutions, and one can choose, e.g., √1 (Φ𝑚 (𝜙) +Φ−𝑚 (𝜙)) = 𝜋1 cos(𝑚𝜙)
and
1
√
(Φ𝑚 (𝜙)
𝑖 2
− Φ−𝑚 (𝜙)) =
1
𝜋
2
sin(𝑚𝜙) instead.
122
6 Quantum particle in a central potential
However, solutions written in the form Eq. (6.37) has those advantage, that they
are corresponding to the eigenfunctions of 𝐿ˆ 𝑧 . Indeed, the eigenvalue problem for
𝐿ˆ 𝑧 reads:
𝐿ˆ 𝑧 Φ = 𝑙 𝑧 Φ.
(6.38)
Putting here the expression for the operator 𝐿ˆ 𝑧 given at page 120, one gets:
−𝑖ℏ
𝑑Φ
= 𝑙 𝑧 Φ,
𝑑𝜙
(6.39)
𝑖𝑙𝑧
which gives Φ(𝜙) = 𝐴𝑒 ℏ𝜙 . Application of periodic boundary condition Φ(𝜙 + 2𝜋) =
Φ(𝜙) will give 𝑙 𝑧 = 𝑚ℏ, and Φ(𝜙) = √1 𝑒 𝑖𝑚𝜙 , which coincides with Eq. (6.37).
2𝜋
Now, let us turn to the equation for 𝑃(𝜃) (see page 121):
𝑑
𝑑𝑃
sin 𝜃
sin 𝜃
+ (𝜆 sin2 𝜃 − 𝑚 2 )𝑃 = 0
(6.40)
𝑑𝜃
𝑑𝜃
(we put 𝜉 = 𝑚 2 ).
Let us introduce there the variable 𝑥 = cos 𝜃 (this is not x coordinate). We have:
𝑑
𝑑𝑥 𝑑
𝑑
=
= − sin 𝜃 ,
𝑑𝜃 𝑑𝜃 𝑑𝑥
𝑑𝑥
𝑑
𝑑
𝑑
= − sin2 𝜃
= (𝑥 2 − 1) .
sin 𝜃
𝑑𝜃
𝑑𝑥
𝑑𝑥
(6.41)
(6.42)
Thus, equation for 𝑃 can be rewritten as:
𝑑𝑃
𝑑
(1 − 𝑥 2 )
+ [𝜆(1 − 𝑥 2 ) − 𝑚 2 ]𝑃 = 0 =⇒
𝑑𝑥
𝑑𝑥
𝑑
𝑚2
2 𝑑𝑃
𝑃 = 0.
(1 − 𝑥 )
+ 𝜆−
𝑑𝑥
𝑑𝑥
1 − 𝑥2
(1 − 𝑥 2 )
(6.43)
(6.44)
This equation should be completed by boundary conditions. Points 𝑥 ± 1 (𝜃 = 0, 𝜋)
𝑚2
2
are special the term 1−𝑥
2 in them diverges, and the term (1 − 𝑥 ) before the derivative
goes to zero. We thus need to demand, that 𝑃(𝑥) remains finite at 𝑥 → ±1.
Solutions, satisfying this boundary condition exist only for 𝜆 = 𝑙 (𝑙 + 1) for
𝑙 = 0, 1, 2, . . . and |𝑚| ≤ 𝑙. The latter condition means, that for 𝑙 = 0 𝑚 = 0, for
𝑙 = 1 𝑚 = 0, ±1, for 𝑙 = 2 𝑚 = 0, ±1, ±2 etc. The functions 𝑃(𝑥) = 𝑃𝑙𝑚 (𝑥) are
given by so called associated Legendre polynomials, which are defined as:
For 𝑙 = 0:
1 𝑑𝑙 2
𝑃𝑙0 (𝑥) = 𝑃𝑙 (𝑥) 𝑙
(𝑥 − 1) 𝑙 .
(6.45)
2 𝑙! 𝑑𝑥 𝑙
For 𝑙 ≠ 0:
6.3 Spherical harmonics
123
𝑃𝑙𝑚 (𝑥) = (−1) 𝑚 (1 − 𝑥 2 )
|𝑚|
2
|𝑚| 𝑑 𝑙+|𝑚|
𝑑 |𝑚|
(−1) 𝑚
𝑃𝑙 (𝑥) = 𝑙 (1 − 𝑥 2 ) 2
(𝑥 2 − 1) 𝑙 .
|𝑚|
2 𝑙!
𝑑𝑥
𝑑𝑥 𝑙+|𝑚 |
(6.46)
Combining 𝑌 = 𝑃(𝜃)Φ(𝜙) we get for the angular part of the wavefunction the
following expression:
s
(2𝑙 + 1) (𝑙 − 𝑚)!
𝑌𝑙𝑚 (𝜃, 𝜙) =
𝑃𝑙𝑚 (cos 𝜃)𝑒 𝑖𝑚𝜙 .
(6.47)
4𝜋 (𝑙 + 𝑚)!
These functions are known as spherical harmonics. Spherical harmonics form an
orthonormal basis for functions of angular variables, i.e. any function 𝑓 (𝜃, 𝜙) can
be represented as linear combination of spherical harmonics,
Õ
𝑓 (𝜃, 𝜙) =
𝑓𝑙𝑚𝑌𝑙𝑚 (𝜃, 𝜙),
(6.48)
𝑙, |𝑚| ≤𝑙
∫
∗
𝑑Ω𝑌𝑙𝑚
(𝜃, 𝜙)𝑌𝑙0 𝑚0 (𝜃, 𝜙) = 𝛿𝑙𝑙0 𝛿 𝑚𝑚0 ,
∫
∗
𝑓𝑚𝑙 =
𝑑Ω𝑌𝑙𝑚
(𝜃, 𝜙) 𝑓 (𝜃, 𝜙).
(6.49)
(6.50)
Examples of spherical harmonics are:
1
𝑌00 = √
2 𝜋
(6.51)
- just a constant.
1
𝑌10 (𝜃, 𝜙) =
2
r
3
cos 𝜃,
𝜋
r
1 3
sin 𝜃𝑒 ±𝑖 𝜙 ,
𝑌1,±1 (𝜃, 𝜙) = ∓
2 2𝜋
√
5
𝑌20 (𝜃, 𝜙) = √ (3 cos2 𝜃 − 1),
4 𝜋
√
15
𝑌2,±1 (𝜃, 𝜙) = ∓ √ sin 2𝜃𝑒 ±𝑖 𝜙 ,
2 2𝜋
√
15
𝑌2,±2 (𝜃, 𝜙) = √ sin 2𝜃𝑒 2±𝑖 𝜙 .
2 2𝜋
(6.52)
(6.53)
(6.54)
(6.55)
(6.56)
Now, let us turn to the radial part of Schrodinger equation, which can be written
as (see page 117, where we put 𝜆 = 𝑙 (𝑙 + 1))
ℏ2 1 𝑑 2 𝑑𝑅
ℏ2 𝑙 (𝑙 + 1)
𝑟
+
𝑈
(𝑟)
+
𝑅 = 𝐸 𝑅.
(6.57)
−
2𝑚 𝑟 2 𝑑𝑟
𝑑𝑟
2𝑚𝑟 2
124
6 Quantum particle in a central potential
First thing, that we note, that while the quantum number 𝑙, characterizing the value
of the square of the angular momentum ℏ2 𝑙 (𝑙 + 1), 𝑙 = 0, 1, 2 . . . is entering into
2 𝑙 (𝑙+1)
effective potential 𝑈𝑒 𝑓 𝑓 = 𝑈 (𝑟)+ ℏ 2𝑚𝑟
2 , and thus, generally, the energy will depend
on it, the quantum number 𝑚, characterizing projection of angular momentum on z
axis, ℏ𝑚 is not entering into this equation, and thus does not influence the energy. As
for 𝑙 ≥ 1 several values of 𝑚 are possible, this means that energy states in the central
potential with 𝑙 ≥ 1 are 2𝑙 + 1 times degenerate. This is the consequence of the
spherical symmetry of the problem: while absolute value of the angular momentum
changes centrifugal potential, its orientation can not - as all directions are equivalent.
This is one of the illustration of a general rule, stating that presence of the symmetries
leads to degeneracies of energy levels in quantum mechanics (in classical mechanics
it leads to the appearance of the conservation laws). Eq. (6.57) should be completed
by boundary conditions which for the bound states in an attractive potential read:
a) 𝑅(𝑟)|𝑟 →∞ = 0
b) 𝑅(𝑟)|𝑟 →0 remains finite.
The latter condition is necessary, because Eq. (6.57) contains terms with 𝑟12 in
the denominator.
Making the substitution 𝑅 = 𝜒𝑟 , we get:
𝑑𝑅
𝑑 𝜒 1 𝑑𝜒
𝜒
=
=
− ,
𝑑𝑟
𝑑𝑟 𝑟
𝑟 𝑑𝑟 𝑟 2
𝑑 2 𝑑𝑅
𝑑
𝑑𝜒
𝑑 2 𝜒 𝑑𝜒 𝑑𝜒
𝑑2 𝜒
𝑟
=
𝑟
−𝜒 =𝑟 2 +
−
=𝑟 2,
𝑑𝑟
𝑑𝑟
𝑑𝑟 𝑑𝑟
𝑑𝑟
𝑑𝑟
𝑑𝑟
𝑑𝑟
(6.58)
(6.59)
which being placed in Eq. (6.57) will give:
𝜒
ℏ2 1 𝑑 2 𝜒
𝜒
=⇒
+ 𝑈𝑒 𝑓 𝑓 (𝑟) = 𝐸
2
2𝑚 𝑟 𝑑𝑟
𝑟
𝑟
ℏ2 𝑑 2 𝜒
−
+ 𝑈𝑒 𝑓 𝑓 (𝑟) 𝜒 = 𝐸 𝜒,
2𝑚 𝑑𝑟 2
−
(6.60)
(6.61)
and we need to add boundary conditions: 𝜒(𝑟)|𝑟 →∞ = 0, and 𝜒(𝑟)|𝑟 →0 = 0 (so that
𝑅 = 𝜒𝑟 remain finite).
The equation for 𝜒 completely coincides with equation of a one dimensional
2 𝑙 (𝑙+1)
quantum particle in the effective potential 𝑈𝑒 𝑓 𝑓 (𝑟) = 𝑈 (𝑟) + ℏ 2𝑚𝑟
2 . Condition
𝜒(𝑟)|
=
0
means,
that
the
particle
can
not
go
to
the
region
of
negative
𝑟 (indeed,
p𝑟 →0
2
2
2
𝑟 = 𝑥 + 𝑦 + 𝑧 is always positive), and is equivalent to the placing of a hard wall
at 𝑟 = 0.
Effective potential depends on 𝑙, and have set of energy levels characterized by the
integer number 𝑛. The quantum states of a particle in a central potential are thus characterized by set of 3 quantum numbers, 𝜓 𝑛𝑙𝑚 (r) = 𝑅𝑛𝑙 (𝑟)𝑌𝑙𝑚 (𝜃, 𝜙) where 𝑛, known
as a principal quantum number, corresponds to the number of a state in the effective
potential 𝑈𝑒 𝑓 𝑓 , 𝑙 known as orbital quantum number, characterizing the square of
angular momentum, 𝑙 = 0, 1, 2 . . . and 𝑚, known as magnetic quantum number,
characterizing z - projection of an angular momentum, 𝑚 = 0, ±1, ±2, |𝑚| ≤ 𝑙.
6.3 Spherical harmonics
125
States with different 𝑙 corresponds to orbitals of different type.
𝑙 = 0 corresponds to spherically symmetric orbitals, for which the wavefunction
does not depend on the angles 𝜃, 𝜙.These are so called s-states. S states are not
degenerate as for them 𝑚 = 0.
l=1 corresponds to p orbitals. They are triply degenerate, as now 𝑚 = 0, ±1, and
profiles of the probability density 𝜌 = 𝜓 ∗ 𝜓 become angular dependent, according to
expressions for spherical harmonics 𝑌10 , 𝑌1,±1 given at page 123.
l=2 corresponds to d - orbitals. They are 5 times degenerate, 𝑚 = 0, ±1, ±2 . . ..
Angular dependence is given by 𝑌20 , 𝑌2,±1 , 𝑌2,±2 .
l=3 corresponds to f - orbitals, they are 7 times degenerate, etc.
Profiles of different orbitals are schematically shown below:
Fig. 6.3
126
6 Quantum particle in a central potential
6.4 Hydrogen atom
Hydrogen atom represents one of the most important problems in quantum mechanics. One of the motivations for the creation of quantum theory was discovery of the
discrete nature of the emission spectrum of atomic hydrogen. It was discovered, that
emission frequencies are well approximated by the following formula:
#
"
1
1
(6.62)
ℏ𝜔 = 𝑅𝑦 2 − 2 ,
𝑛1 𝑛2
where 𝑅𝑦 - is a Rydberg constant, 𝑅𝑦 ≈ 13, 6 eV and 𝑛1 , 𝑛2 are positive integer
numbers, 𝑛1,2 = 1, 2, 3 . . .. This experimental discovery can be explained, if one
suggests, that the energies of electron in an atom are quantized, and the energy of
these quantized levels are given by:
𝐸𝑛 = −
𝑅𝑦
,
𝑛2
(6.63)
where sign "-" accounts that hydrogen atom is a bound state in attractive. Coulomb
2
potential, 𝑈 (𝑟) = − 4 𝜋𝑒𝜖0 𝑟 with 𝑟 being distance between the proton and electron.
Formula Eq. (6.62) is then easily understandable: emission of photons of energy
𝐸 = ℏ𝜔 goes when electron goes down from the state with 𝐸 𝑖 = − 𝑅𝑦
𝑛2 (initial energy)
to the state with 𝐸 𝑓 = − 𝑅𝑦
𝑛1 (final energy): The first simplest theory, explaining the
Fig. 6.4
Eq. (6.62) and giving theoretical estimate of the Rygberg constant was proposed by
Niels Bohr, and is based on Bohr-Sommerfeld quantization rule. The logic was the
2
following. We know that in attractive potential 𝑈 (𝑟) = − 4 𝜋𝑒𝜖0 𝑟 = − 𝛼𝑟 the classical
finite motion occurs for 𝐸 < 0, and corresponding trajectories are elliptic. Keeping
in mind that with any quantum particle one can associate a De Broglie wave, and that
in a stationary state one can only put integer number of wavelengths the trajectory,
6.4 Hydrogen atom
with 𝜆 =
2 𝜋ℏ
𝑝 ,
127
one can write:
∮
p(l)𝑑l = 2𝜋ℏ𝑛.
(6.64)
Let us consider the simplest case of circular trajectory:
Fig. 6.5
Fig. 6.6
2
Second law of Newton with 𝑤 = 𝑉𝑟 , with 𝑉 being velocity of electron, 𝑟 - radius
of the trajectory, gives: (coulomb Force)
𝑒2
𝛼
= 2 =⇒
2
4𝜋𝜖0 𝑟
𝑟
𝑉2
𝛼
𝑚𝑒
= 2 =⇒ 𝑚 𝑒𝑉 2 𝑟 = 𝛼.
𝑟
𝑟
𝐹=
(6.65)
(6.66)
128
6 Quantum particle in a central potential
And, as for circular trajectory the absolute value of momentum 𝑝 remains constant
and it is always parallel to elementary displacement along the trajectory, one gets for
Bohr-Sommerfeld quantization rule:
∮
p𝑑l = 𝑝 · 2𝜋𝑟 = 2𝜋𝑚 𝑒𝑉𝑟 = 2𝜋𝑛ℏ =⇒
(6.67)
(6.68)
𝑚 𝑒𝑉𝑟 = 𝑛ℏ,
where 𝑛 = 1, 2, 3 . . .. Note, that Bohr-Sommerfeld quantization rule in the considered case is nothing but condition of the quantization of z projection of the angular
momentum,
𝑚 𝑒𝑉𝑟 = 𝐿 𝑧 = 𝑛ℏ.
(6.69)
Placing this into Eq. (6.62) one gets:
𝑚 𝑒𝑉 2 𝑟 = 𝑚 𝑒𝑉𝑟 · 𝑉 = 𝑛ℏ𝑉 = 𝛼 =⇒
𝛼
𝑉 = 𝑉𝑛 =
𝑛ℏ
(6.70)
(6.71)
2 2
𝑛 ℏ
- velocity can get only quantized values, and 𝑟 𝑛 = 𝑚𝑛ℏ
= 𝑚
- the radii of the
𝑒 𝑉𝑛
𝑒𝛼
permitted orbitals are also quantized. The energy corresponding to such an orbital
can be then expressed as:
𝑚 𝑒𝑉𝑛2
𝛼
𝑚 𝑒 𝛼2 𝑚 𝑒 𝛼2
−
= 2 2− 2 2 =
2
𝑟 𝑛 2𝑛 ℏ
𝑛 ℏ
𝑚 𝑒 𝑒4 1
𝑅𝑦
𝑚 𝑒 𝛼2
=− 2,
− 2 2 =−
2𝑛 ℏ
𝑛
32𝜋 2 ℏ2 𝜖 02 𝑛2
𝐸 𝑛 = 𝑇𝑛 + 𝑈𝑛 =
(6.72)
where we got expression of the Rydberg constant,
𝑅𝑦 =
𝑚 𝑒 𝑒4
,
32𝜋 2 ℏ2 𝜖02
(6.73)
which gives almost exactly the experimental value of ≈ 13, 6 eV. Correction can be
done in order to account for the finite mass of a proton by substituting electron mass
𝑚 𝑒 by reduced mass
!
𝑚𝑒 𝑚 𝑝
1
𝜇=
= 𝑚𝑒
≈ 𝑚𝑒 ,
(6.74)
𝑚𝑒
𝑚𝑒 + 𝑚 𝑝
1+ 𝑚
𝑝
as 𝑚 𝑒 ≈ 1800𝑚 𝑝 .
Now, let us consider what will happen if we solve the problem exactly, i.e. find
the eigenvalues of stationary Schrodinger equation
ℏ2 2 𝛼
∇ −
𝜓(r) = 𝐸𝜓(r).
(6.75)
−
2𝑚 𝑒
𝑟
6.4 Hydrogen atom
129
As coulomb potential is central, we can write 𝜓(r) = 𝑅(𝑟)𝑌 (𝜃, 𝜙), and for radial
part use the following equation (see page 117)
ℏ2 1 𝑑 2 𝑑𝑅
𝛼 ℏ2 𝑙 (𝑙 + 1)
−
𝐸
𝑅 = 0,
(6.76)
−
𝑟
+
−
+
2𝑚 𝑒 𝑟 2 𝑑𝑟
𝑑𝑟
𝑟
2𝑚𝑟 2
where 𝛼 =
−
𝑒2
4 𝜋 𝜖0 .Dividing
this equation by 4|𝐸 |, one gets:
ℏ2
𝛼 1 ℏ2 𝑙 (𝑙 + 1) 1
1 𝑑 2 𝑑𝑅
−
𝑟
+
−
𝑅 = 0,
−
8𝑚 𝑒 |𝐸 | 𝑟 2 𝑑𝑟
𝑑𝑟
4|𝐸 | 𝑟 8𝑚 𝑒 |𝐸 |𝑟 2 4
(6.77)
𝐸
where we accounted for that 𝐸 < 0, and thus |𝐸
| = −1.
Let us rewrite this equation
in
the
dimensionless
form, introducing the following
q
characteristic length: 𝑟 0 =
ℏ2
8𝑚𝑒 |𝐸 | .
Let us check the dimensionality:
ℏ2
[ℏ] 2
[𝐸] 2 [𝑡] 2 [𝐸] [𝑡] 2 [𝑚] [𝑉] 2 [𝑡] 2
=
=
=
=
= [𝐿] 2 .
8𝑚 𝑒 |𝐸 |
[𝑚] [𝐸]
[𝑚] [𝐸]
[𝑚]
[𝑚]
Introducing dimensionless radius 𝜌 =
part of wavefunction in the form:
𝑟
𝑟0 ,
(6.78)
one can rewrite the equation for the radial



2
1 𝑑
𝑙
(𝑙
+
1)
𝛼
ℏ
𝑑𝑅
1
1


2
−
−
𝜌
+

𝑅=
𝑟
2
2
2


𝑑𝜌
𝑑𝜌
4|𝐸
|𝑟
4
𝜌
8𝑚|𝐸 |𝑟 0 𝑟
0 𝑟0


𝑟0


𝛽 1 𝑙 (𝑙 + 1)
1 𝑑
𝑑𝑅
𝜌2
+
− −
𝑅 = 0,
𝑑𝜌
𝜌 4
𝜌 2 𝑑𝜌
𝜌2
q
where 𝛽 = 4 |𝐸𝛼|𝑟0 = 𝛼ℏ 2𝑚|𝐸𝑒 | .
(6.79)
𝜌
Now, we can make a substitution: 𝑅(𝜌) = 𝑒 − 2 𝐹 (𝜌), then:
𝑑𝑅
𝑑 −𝜌
1
− 𝜌2 𝑑𝐹
2
=
𝑒 𝐹 (𝜌) = 𝑒
− 𝐹 ,
𝑑𝜌 𝑑𝜌
𝑑𝜌 2
2
𝑑 − 𝜌 𝑑𝐹 1
𝑑𝐹 1
𝑑2 𝑅
− 𝜌2 𝑑 𝐹
2
=
−
𝑒
− 𝐹 =𝑒
+ 𝐹 ,
𝑑𝜌
𝑑𝜌 2
𝑑𝜌 4
𝑑𝜌 2
𝑑𝜌 2
(6.80)
(6.81)
and then:
2
1 𝑑
1
𝑑𝑅
𝑑 2 𝑅 2 𝑑𝑅
2 𝑑𝑅
2𝑑 𝑅
𝜌
=
2𝜌
+
𝜌
=
+
=
2
2
2
𝑑𝜌
𝑑𝜌
𝜌 𝑑𝜌
𝜌
𝑑𝜌
𝑑𝜌 2 𝜌 𝑑𝜌
2
2
𝜌
𝑑 𝐹 𝑑𝐹 1
2 𝑑𝐹 1
2
𝑑𝐹
1 1
− 𝜌2 𝑑 𝐹
+
𝐹
+
−
𝐹
=
𝑒
−
1
+
−
𝐹 .
= 𝑒− 2
−
+
𝑑𝜌 4
𝜌 𝑑𝜌 𝜌
𝜌
𝑑𝜌
4 𝜌
𝑑𝜌 2
𝑑𝜌 2
(6.82)
130
6 Quantum particle in a central potential
Placing this into Schrodinger equation, one gets:
𝑑2 𝐹
2
𝛽 1 𝑙 (𝑙 + 1)
𝑑𝐹
1 1
+
𝐹
+
𝐹 = 0,
−
1
+
−
−
−
𝜌
𝑑𝜌
4 𝜌
𝜌 4
𝑑𝜌 2
𝜌2
𝑑2 𝐹
2
𝑑𝐹
𝛽 − 1 𝑙 (𝑙 + 1)
+
−
1
+
−
𝐹 = 0.
𝜌
𝑑𝜌
𝜌
𝑑𝜌 2
𝜌2
(6.83)
(6.84)
In this equation let us perform one more substitution:
𝐹 (𝜌) = 𝜌 𝑠 𝐿 (𝜌),
𝑑𝐹
𝑑𝐿
= 𝑠𝜌 𝑠−1 𝐿 + 𝜌 𝑠 ,
𝑑𝜌
𝑑𝜌
2
2
𝑑 𝐹
𝑠−2
𝑠−1 𝑑𝐿
𝑠−1 𝑑𝐿
𝑠𝑑 𝐿
=
𝑠(𝑠
−
1)
𝜌
𝐿
+
𝑠𝜌
+
𝑠𝜌
+
𝜌
=
𝑑𝜌
𝑑𝜌
𝑑𝜌 2
𝑑𝜌 2
𝑑𝐿
𝑑2 𝐿
= 𝑠(𝑠 − 1) 𝜌 𝑠−2 𝐿 + 2𝑠𝜌 𝑠−1
+ 𝜌𝑠 2 .
𝑑𝜌
𝑑𝜌
(6.85)
(6.86)
(6.87)
Putting this to Schrodinger equation, we get:
𝑑𝐿
𝑑2 𝐿
𝑠(𝑠 − 1) 𝜌 𝑠−2 𝐿 + 2𝑠𝜌 𝑠−1
+ 𝜌𝑠 2 +
𝑑𝜌
𝑑𝜌
2
2
𝛽 − 1 𝑙 (𝑙 + 1) 𝑠
𝑑𝐿
𝑠−1
𝑠
+
− 1 𝑠𝜌
+
−1 𝜌
+
−
𝜌 𝐿 = 0.
𝜌
𝜌
𝑑𝜌
𝜌
𝜌2
(6.88)
Multiplying this equation by 𝜌 2 , we get:
𝑑2 𝐿
𝑑𝐿
+ 𝜌 𝑠+2 2 +
𝑠(𝑠 − 1) 𝜌 𝑠 𝐿 + 2𝑠𝜌 𝑠+1
𝑑𝜌
𝑑𝜌
2
𝑑𝐿
+
− 1 𝑠𝜌 𝑠 𝐿 + (2 − 𝜌) 𝜌 𝑠+1
+ [(𝛽 − 1) 𝜌 − 𝑙 (𝑙 + 1)] 𝜌 𝑠 𝐿 =
𝜌
𝑑𝜌
𝑑𝐿
𝑑2 𝐿
= 𝜌 𝑠 𝜌 2 2 + [2𝑠𝜌 + (2 − 𝜌) 𝜌]
+ [𝑠(𝑠 − 1) + 𝑠(2 − 𝜌) + (𝛽 − 1) 𝜌 − 𝑙 (𝑙 + 1)] 𝐿 = 0.
𝑑𝜌
𝑑𝜌
(6.89)
From where we get:
𝜌2
𝑑2 𝐿
𝑑𝐿
+ 𝜌[2(𝑠 + 1) − 𝜌]
+ [(𝛽 − 𝑠 − 1) 𝜌 + 𝑠(𝑠 + 1) − 𝑙 (𝑙 + 1)] 𝐿 = 0. (6.90)
𝑑𝜌
𝑑𝜌 2
𝑑𝐿
Putting here 𝜌 → 0, and demanding that 𝑑𝐿
𝑑𝜌 and 𝑑𝜌 remain finite we get: 𝑠(𝑠 + 1) −
𝑙 (𝑙 + 1) = 0, which gives us 𝑠 = 𝑙 (another possible solution is 𝑠 = −(𝑙 + 1), but it is
not good, since 𝐹 (𝜌) = 𝜌 𝑠 𝐿(𝜌) will have divergency at 𝜌 → 0 for 𝑠 = −(𝑙 + 1) < 0).
Taking 𝑠 = 𝑙, we get
6.4 Hydrogen atom
131
𝜌
𝑑𝐿
𝑑2 𝐿
+ [2(𝑙 + 1) − 𝜌]
+ (𝛽 − 𝑙 − 1)𝐿 = 0.
𝑑𝜌
𝑑𝜌 2
(6.91)
Now, as in the case of the problem of quantum harmonic oscillator, we can search
the solution in the form a Taylor series:
𝐿=
∞
Õ
(6.92)
𝑎𝑛 𝜌𝑛,
𝑛=0
∞
Õ
𝑑𝐿
=
𝑑𝜌
(6.93)
𝑎 𝑛 𝑛𝜌 𝑛−1 ,
𝑛=0
∞
Õ
𝑑2 𝐿
=
𝑑𝜌 2
𝑎 𝑛 𝑛(𝑛 − 1) 𝜌 𝑛−2 .
(6.94)
𝑛=0
Putting this into Eq. (6.91), we get:
∞
Õ
𝑎 𝑛 𝑛(𝑛 − 1) 𝜌 𝑛−1 + 2(𝑙 + 1)
=
𝑎 𝑛 𝑛𝜌 𝑛−1 −
𝑎 𝑛 𝑛[𝑛 − 1 + 2(𝑙 + 1)] 𝜌 𝑛−1 −
𝑛=0
∞
Õ
𝑎 𝑛 𝑛𝜌 𝑛 + (𝛽 − 𝑙 − 1)
𝑛=0
𝑛=0
𝑛=0
∞
Õ
∞
Õ
∞
Õ
∞
Õ
𝑎𝑛 𝜌𝑛 =
𝑛=0
(𝑛 + 𝑙 + 1 − 𝛽)𝑎 𝑛 𝜌 𝑛 = 0.
(6.95)
𝑛=0
In the first sum let us make the change 𝑛 0 = 𝑛 − 1, 𝑛 = 𝑛 0 + 1, which will give:
∞
Õ
𝑎 𝑛 𝑛[𝑛 − 1 + 2(𝑙 + 1)] 𝜌 𝑛−1 =
0
(𝑛 0 + 1) (𝑛 0 + 2𝑙 + 2)𝑎 𝑛0 +1 𝜌 𝑛 =
𝑛0 =0
𝑛=0
∞
Õ
∞
Õ
(𝑛 + 1) (𝑛 + 2𝑙 + 2)𝑎 𝑛+1 𝜌 𝑛 ,
(6.96)
𝑛=0
as 𝑛 0 is just summation index and substitution 𝑛 0 → 𝑛 can be done. We then have:
∞
Õ
{(𝑛 + 1) (𝑛 + 2𝑙 + 2)𝑎 𝑛+1 − (𝑛 + 𝑙 + 1 − 𝛽)𝑎 𝑛 } 𝜌 𝑛 = 0,
(6.97)
𝑛=0
which gives us a recurrent relation for the coefficients an:
𝑎 𝑛+1 =
𝑛+𝑙+1−𝛽
𝑎𝑛 .
(𝑛 + 1) (𝑛 + 2𝑙 + 2)
(6.98)
From this expression we get the following asymptotical behavior:
𝑎 𝑛+1 =
𝑎𝑛
𝑛
(6.99)
132
6 Quantum particle in a central potential
Note, that the same asymptotical behavior is characteristic for a growing exponent,
for which
𝑒𝜌 =
∞
Õ
1 𝑛
1 𝑎 𝑛+1
𝑛!
1
1
𝜌 , 𝑎𝑛 = ,
=
=
≈ ,
𝑛!
𝑛! 𝑎 𝑛
(𝑛 + 1)! 𝑛 + 1 𝑛
𝑛=0
(6.100)
when 𝑛 → ∞.
This means, that the function 𝐿(𝜌) at infinity is 𝐿 (𝜌) ∼ 𝑒 𝜌 , which will give for
the function
−𝜌
𝜌
−𝜌
(6.101)
𝑅(𝜌) = 𝑒 2 𝐹 (𝜌) = 𝜌 𝑙 𝑒 2 𝐿(𝜌) ∼ 𝜌 𝑙 𝑒 2 ,
so we will thus get the divergency, contrary to the required boundary condition
𝑅(𝜌)|𝜌→∞ = 0. The only way to avoid this divergency is to demand, that infinite
Taylor series becomes finite polynomial, which means, that at some point 𝑎 𝑛+1 = 0,
and thus:
𝑛 + 𝑙 + 1 − 𝛽 = 0, 𝑛 = 0, 1, 2 . . . .
(6.102)
q
𝑚𝑒
Placing here the expression for 𝛽 = 𝛼ℏ 2|𝐸
| , one gets (𝐸 = −|𝐸 |):
𝐸 =−
𝑚𝛼2
𝑅𝑦
1
𝑚 𝑒 𝑒4
=−
=−
,
2
+ 𝑙 + 1)
(𝑛 + 𝑙 + 1) 2
32𝜋 2 𝜖02 ℏ2 (𝑛 + 𝑙 + 1) 2
2ℏ2 (𝑛
(6.103)
where 𝑅𝑦 =, 𝑛 = 0, 1, 2, 3, and 𝑙 = 0, 1, 2, 3.
This formula fully coincides with the expression obtained by using of the BohrSommerfeld approach (see page 128), 𝐸 𝑛 = − 𝑅𝑦
, if one puts 𝑛 + 𝑙 + 1 → 𝑛 =
𝑛2
1, 2, 3 . . ..
However, exact consideration with use of Schrodinger equation allows to shed
more light on the nature of the quantized energy state. One can note the following:
1) States with different 𝑙 correspond to different angular dependence of the wavefunction (see pages 123, 125) the lowest energy state corresponds to
𝑛 = 0, 𝑙 = 0, 𝐸 0 = −
𝑚 𝑒 𝑒4
= −𝑅𝑦
32𝜋 2 𝜖0 ℏ2
(6.104)
which is the energy of the ionization of atomic hydrogen. Ground state corresponds
to spherically symmetric wavefunction,
1
𝜓000 (r) = 𝑅00 (𝑟)𝑌00 (𝜃, 𝜙) = √ 𝑅00 (𝑟)
2 𝜋
(6.105)
(the shape of radial part of wavefunction 𝑅00 (𝑟) will be clarified later). The ground
state is degenerate, i.e. have single corresponding wavefunction. This is 1s state.
2) The first excited state of Eq. (6.103) corresponds either to 𝑛 = 1, 𝑙 = 0 or
𝑛 = 0, 𝑙 = 1. The state with 𝑛 = 1, 𝑙 = 0 corresponds to 2s state, for which the
wavefunction is also spherically symmetric 𝜓100 (r) = 𝑅10 (𝑟)𝑌00 (𝜃, 𝜙) = 2√1 𝜋 𝑅10 (𝑟).
6.4 Hydrogen atom
133
The state with 𝑛 = 0, 𝑙 = 1, corresponding to 2p state, has, however, nontrivial
angular dependence. In fact, it is not a single state, but triplet of states, corresponding
to the three possible values of z-projection of angular momenta 𝑚, 𝑚 = 0, ±1. The
corresponding wavefunctions read: 𝜓01𝑚 (r) = 𝑅01 (𝑟)𝑌1𝑚 (𝜃, 𝜙), 𝑚 = −1, 0, +1
and expression for 𝑌1𝑚 (𝜃, 𝜙) are given at page 123.
The first excited state is thus 4 times degenerate: the combinations of quantum
numbers 𝑛, 𝑙, 𝑚 corresponding to it (see pages 125-125 about quantum numbers)
can be: (𝑛, 𝑙, 𝑚) = (1, 0, 0), (0, 1, −1), (0, 1, 0), (0, 1, +1). Note, that here there is
both fundamental degeneracy between the energies of states with different 𝑚 inside
2p triplet, characteristic to any central potential, and related to spherical symmetry
of the problem, and so called accidental degeneracy, characteristic to Coulomb
potential only (in any potential different from 𝑈 (𝑟) = − 𝛼𝑟 energies of 1s and 2s
states will be different).
3) The second excited state will correspond to the following possible combinations
of 𝑛 and 𝑙: 𝑛 = 2, 𝑙 = 0; 𝑛 = 1, 𝑙 = 1; 𝑛 = 0, 𝑙 = 2 states with 𝑛 = 2, 𝑙 = 0 are
spherically symmetric 3s states. States with 𝑛 = 1, 𝑙 = 1 corresponds again to triplet
of 3p states, the corresponding possible values of 𝑚 are 0, ±1 states with 𝑛 = 0, 𝑙 = 2
correspond to pentaplet of 3d states, with possible values of 𝑚 = 0, ±1, ±2. The
second excited state will be thus 9 times degenerate (one 3s state, three 3p states and
5 3d states). The degeneracy, again, can be fundamental (between different 𝑚 in 3p
triplet and 5d pentaplet) and accidental (between 3s, 3p and 3d states).
How we can try to understand the presence of accidental degeneracy in the case
of Coulomb potential?
We mentioned already, that in quantum physics, degeneracies are related with
symmetries. Fundamental degeneracy (equivalence of energy of states with different
𝑚 corresponding to same 𝑙) is consequence of rotational symmetry. In classical
mechanics, on the other hand, symmetries lead to the appearance of conserving
quantities. Rotational symmetry leads to the conservation of the angular momentum
L.
The presence of the accidental degeneracy can be expected to come from the
presence of some hidden symmetry for 𝑈 (𝑟) = − 𝛼𝑟 potential. If this is the case, in
classical physics we will have some additional conserving quantity. And this is indeed
like that - in 𝑟1 potentials the so called Laplace-Runge-Lenz vector is conserved:
r
A = p × L − 𝑚𝛼r0 = p × L − 𝑚𝛼 .
𝑟
Let us show, that for 𝑈 (𝑟) = − 𝛼𝑟 , so that F = − 𝑟𝛼2 r0 = − 𝛼r
,
𝑟3
(6.106)
𝑑A
𝑑𝑡
= 0. Calculate
𝑑
𝑑p
𝑑L
𝛼
𝑚𝛼
𝑑r
(p × L) =
×L+p×
= F × L = − 3r × L = − 3 r r ×
.
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑟
𝑟
(6.107)
Use here A × [B × C] = B(A · C) − C(A · B), then:
134
6 Quantum particle in a central potential
and then use
𝑑r
𝑑r
𝑑
𝑚𝛼
− 𝑟2
,
(p × L) = − 3 r r
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑟
(6.108)
𝑑
𝑑r
𝑑
𝑑𝑟
(r · r) = 2r
= 𝑟 2 = 2𝑟 ,
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑡
(6.109)
which gives:
𝑑𝑟
𝑑 r
𝑑
𝑚𝛼
1 𝑑r
r 𝑑𝑟
2 𝑑r
= 𝑚𝛼
= 𝑚𝛼
(p × L) = − 3 r · 𝑟
−𝑟
− 2
.
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑟 𝑑𝑡 𝑟 𝑑𝑡
𝑑𝑡 𝑟
𝑟
(6.110)
We then get:
𝑑A
𝑑 h
ri
=
p × L − 𝑚𝛼 =
𝑑𝑡
𝑑𝑡
𝑟
𝑑 r
𝑑 r
− 𝑚𝛼
= 0.
𝑚𝛼
𝑑𝑡 𝑟
𝑑𝑡 𝑟
(6.111)
In quantum mechanics, Runge Lenz vector should be replaced by an operator, with
p → p̂ = −𝑖ℏ∇
1
 = {[ p̂ × L̂] − [ L̂ × p̂]} − r,
(6.112)
2
where we need to write 12 {[ p̂ × L̂] − [ L̂ × p̂]} instead of classical expression [p × L]
because components 𝑝 𝑗 , 𝐿 𝑘 are not commuting, and when writing quantum analogs
of the products of the type 𝑥 𝑝 𝑥 → 12 (𝑥 𝑝ˆ 𝑥 + 𝑝ˆ 𝑥 𝑥), as in quantum physics the order of
𝑝ˆ 2 𝛼
ˆ =
ˆ 𝐴]
the non-commuting operators matters. One can show that for 𝐻ˆ = − 2𝑚
− 𝑟 , [ 𝐻;
0 - this is responsible for accidental degeneracy.
4) The shape of the radial part of an electron in the Coulomb potential is given
by the following expression:
𝑅𝑦
,
(6.113)
(𝑛 + 𝑙 + 1) 2
𝜓 = 𝜓 𝑛𝑙𝑚 (r) = 𝑅𝑛𝑙 (𝑟)𝑌𝑙𝑚 (𝜃, 𝜙),
(6.114)
1
(
)
3
2
𝜌
2
𝑛!
𝑙 − 𝑛𝑙
2 𝐿 2𝑙+1
𝑅𝑛𝑙 (𝑟) =
𝜌
𝑒
𝑛𝑙
𝑛+2𝑙+1 (𝜌 𝑛𝑙 ),
(𝑛 + 𝑙 + 1)𝑎 0 2(𝑛 + 𝑙 + 1) [(𝑛 + 2𝑙 + 1)!] 3
(6.115)
𝐸 =−
2
𝜖0 ℏ
is so called Bohr radius,𝜌 𝑛𝑙 =
where 𝑎 𝐵 = 4 𝜋𝑚𝑒
2
called Laguerre polynomials, defined as:
2𝑟
(𝑛+𝑙+1) 𝑎0 ,
and 𝐿 2𝑙+1
𝑛+2𝑙+1 stand for so
6.4 Hydrogen atom
135
𝑛
1 𝑑
𝑒 𝑥 𝑑 𝑛 −𝑥 𝑛
(𝑒
𝑥
)
=
−
1
𝑥𝑛,
𝑛! 𝑑𝑥 𝑛
𝑛! 𝑑𝑥
𝑑𝑚
𝐿 𝑛𝑚 (𝑥) = 𝑚 𝐿 𝑛𝑚 (𝑥).
𝑑𝑥
𝐿 𝑛 (𝑥) =
(6.116)
(6.117)
For 1s (𝑛 = 0, 𝑙 = 0) the radial part of the wavefunction reads:
1
𝑅00 (𝑟) =
2𝑎 0
32
𝑒
− 𝑎𝑟
0
.
(6.118)
For 2s (𝑛 = 1, 𝑙 = 0):
1
𝑅10 (𝑟) =
2𝑎 0
23 For 2p (𝑛 = 0, 𝑙 = 1):
1
𝑅01 (𝑟) =
2𝑎 0
𝑟
− 𝑟
2−
𝑒 2𝑎0 .
𝑎0
32
𝑟
− 𝑟
√ 𝑒 2𝑎0
𝑎0 3
(6.119)
(6.120)
etc. One can note the following:
a) Equation for 𝑅00 (𝑟) remains finite at 𝑟 → 0 and has no nodes, i.e. the points
where 𝑅00 (𝑟) is zero.
b) Equation for 𝑅10 (𝑟) remains finite at 𝑟 → 0, but has one node at 𝑟 = 2𝑎 0 .
c) Equation for 𝑅01 (𝑟) goes to zero when 𝑟 → 0, thus having single node at 𝑟 = 0.
In general, all states with 𝑙 = 0 (s-states) are finite at 𝑟 = 0, and the number of
their nodes is defined by principal quantum number 𝑛.
All states with 𝑙 ≠ 0 have node at 𝑟 = 0. Total number of their nodes is equal to
𝜖0 ℏ2
𝑛+𝑙. The parameter 𝑎 𝐵 = 4 𝜋𝑚𝑒
2 , introduced at page 134 (Bohr radius) coincides with
2
the radius of the smallest Bohr orbital (𝑛 = 1), obtained in the page 129 (𝛼 = 4 𝜋𝑒 𝜖0 ),
and has the physical meaning of the effective size of hydrogen atom in the ground
state (1s). It is equal to the most probable distance of an electron from a proton in the
state given by expectation value of the radius in the state given by the wavefunction
𝜓1𝑠 (r) = 𝑅00 (𝑟)𝑌00 (𝜃, 𝜙) =
1
− 𝑎𝑟
0 .
√ 𝑒
𝑎0 𝜋
3
2
(6.121)
Indeed, let us calculate the probability to find an electron at an interval of the
distances [𝑟, 𝑟 + 𝑑𝑟] from the proton:
𝑑𝑃 = |𝜓1𝑠 (r)| 2 · 4𝜋𝑟 2 𝑑𝑟 =
4 2 − 𝑎2𝑟
𝑟 𝑒 0 𝑑𝑟 = 𝜌(𝑟)𝑑𝑟,
𝑎 30
(6.122)
where the factor 4𝜋𝑟 2 𝑑𝑟 gives the volume of the spherical layer of the radius 𝑟 and
thickness 𝑑𝑟. The maximum of the radial probability density 𝜌(𝑟) is reached when
136
6 Quantum particle in a central potential
𝑑𝜌 8𝑟
𝑟
− 2𝑟
𝑒 𝑎0 = 0,
= 3 1−
𝑑𝑟
𝑎0
𝑎0
(6.123)
which gives 𝑟 = 𝑎 0 /
Note, that if we calculate the mean distance of an electron from a proton, the
result will be different:
∫
< 𝑟 >1𝑠 =
4
= 3
𝑎0
∫
∞
0
∗
𝜓1𝑠
(r)𝑟𝜓1𝑠 (𝑟)𝑑 3 r =
𝑟3𝑒
− 𝑎2𝑟
0
𝑑𝑟 =
2
𝑎0 ≠ 𝑎0
3
1
𝜋𝑎 30
∫
2𝜋
∫
𝑑𝜙
0
0
𝜋
sin 𝜃𝑑𝜃
∫
∞
𝑟3𝑒
− 𝑎2𝑟
0
𝑑𝑟 =
0
(6.124)
6.5 Hydrogen atom in the external magnetic field and Zeeman
effect
Let us consider how magnetic field will affect the states of an electron in the hydrogen
atom. The Hamiltonian of the problem can be obtained from the Hamiltonian in the
absence of magnetic field by the canonical substitution p̂ → p̂ − 𝑞A = p + 𝑒A, as for
𝑞 = −𝑒. We thus have:
1
𝛼
𝐻ˆ =
(p + 𝑒A) − ,
(6.125)
2𝑚 𝑒
𝑟
2
𝛼 = 4 𝜋𝑒 𝜖0 , ∇ × A = B. Let us direct B along z-axis, B = e𝑧 𝐵. Then, if we use
symmetric gauge (for different gauges see pages 105-??), we can write:
𝜕 𝐴𝑦 𝜕 𝐴𝑥
−
= 𝐵,
𝜕𝑥
𝜕𝑦
𝐵𝑦
𝐵𝑥
𝐴𝑥 = − , 𝐴𝑦 =
.
2
2
(6.126)
(6.127)
And then:
"
#
2 2
1
𝑒𝐵𝑥
𝛼
𝑒𝐵𝑦
2
𝑝ˆ 𝑥 −
+ 𝑝ˆ 𝑦 −
+ 𝑝𝑧 − =
2𝑚 𝑒
2
2
𝑟
2
2
2
2
𝑒 𝐵 𝑦
𝑒 𝐵2 𝑥 2
1
𝛼
2
2
𝑝ˆ − 𝑒𝐵𝑦 𝑝ˆ 𝑥 +
+ 𝑝ˆ 𝑦 + 𝑒𝐵𝑥 𝑝ˆ 𝑦 +
− =
2𝑚 𝑒 𝑥
4
4
𝑟
𝐻ˆ =
1
𝑒𝐵
𝑒 2 𝐵2 2
𝛼
( 𝑝ˆ2𝑥 + 𝑝ˆ2𝑦 + 𝑝ˆ2𝑧 ) +
(𝑥 𝑝ˆ 𝑦 − 𝑦 𝑝ˆ 𝑥 ) +
(𝑥 + 𝑦 2 ) − =
2𝑚 𝑒
2𝑚 𝑒
8𝑚 𝑒
𝑟
ℏ2 2 𝛼
𝑒𝐵 ˆ
𝑒 2 𝐵2 2 2
−
∇ − +
𝐿𝑧 +
𝑟 sin 𝜃 = 𝐻ˆ 0 + 𝐻ˆ 𝑝 + 𝐻ˆ 𝑑 ,
2𝑚 𝑒
𝑟 2𝑚 𝑒
8𝑚 𝑒
where
(6.128)
6.5 Hydrogen atom in the external magnetic field and Zeeman effect
ℏ2 2 𝛼
𝐻ˆ 0 = −
∇ −
2𝑚 𝑒
𝑟
137
(6.129)
is the Hamiltonian in the absence of the magnetic field,
𝑒𝐵 ˆ
𝐻ˆ 𝑝 =
𝐿𝑧
2𝑚 𝑒
(6.130)
is the so called paramagnetic part of the Hamiltonian, and
𝑒 2 𝐵2 2 2
𝑟 sin 𝜃
𝐻ˆ 𝑑 =
8𝑚 𝑒
(6.131)
is the diamagnetic part of the Hamiltonian.
Let us estimate the order of magnitude of the terms 𝐻ˆ 𝑝 , 𝐻ˆ 𝑑
𝑒𝐵ℏ
𝑒𝐵
< 𝐿ˆ 𝑧 >'
,
< 𝐻ˆ 𝑝 >=
2𝑚 𝑒
2𝑚 𝑒
(6.132)
𝑒 2 𝐵2 𝑎 20
𝑒 2 𝐵2
< 𝐻ˆ 𝑑 >=
< 𝑟 2 sin2 𝜃 >'
,
8𝑚 𝑒
8𝑚 𝑒
(6.133)
where 𝑎 0 - is the Bohr radius of an atom, which gives an estimate for its size.
Therefore:
2
< 𝐻 𝑑 > 𝑒𝐵𝑎 0
'
(6.134)
< 𝐻𝑝 >
4ℏ
- this value depends on magnetic field, and goes to zero when 𝐵 → 0. Let us estimate
the value of the field 𝐵 for which 𝐻 𝑝 and 𝐻 𝑑 become comparable, placing
𝑒𝐵𝑎 20
< 𝐻𝑑 >
=1=
=⇒
< 𝐻𝑝 >
4ℏ
𝐵=
4ℏ
4 · ·10−34
≈
≈ 106𝑇
2
𝑒𝑎 0 1, 6 · 10−19 · 0, 25 · 10−20
(6.135)
(6.136)
- this is extremely large value, which is not accessible in the laboratory. Therefore,
in most of the cases one can safely neglect 𝐻 𝑑 , and write:
ℏ2 2 𝛼
𝑒𝐵 ˆ
𝐻ˆ ' 𝐻ˆ 0 + 𝐻ˆ 𝑝 = −
∇ − +
𝐿𝑧 .
2𝑚 𝑒
𝑟 2𝑚 𝑒
(6.137)
Note, that the full Hamiltonian 𝐻ˆ = 𝐻ˆ 0 + 𝐻ˆ 𝑝 + 𝐻ˆ 𝑑 commutes with the operator of
z-projection of the angular momentum
𝜕
ˆ = 0,
𝐿ˆ 𝑧 = −𝑖ℏ , [ 𝐿ˆ 𝑧 ; 𝐻]
𝜕𝜙
(6.138)
as the term 𝐻ˆ 𝑑 does not contain the angle 𝜙. Therefore, eigenstates of full Hamiltonian are in the same time the eigenstates of 𝐿ˆ 𝑧 , and magnetic quantum number 𝑚
138
6 Quantum particle in a central potential
remains good quantum number. On the other hand,
ˆ = [ 𝐿ˆ 2 ; 𝐻ˆ 𝑑 ] ≠ 0,
[ 𝐿ˆ 2 ; 𝐻]
(6.139)
as the operator 𝐿ˆ 2 contains derivatives in 𝜃, and 𝐻ˆ 𝑑 is 𝜃-dependent. Therefore,
the eigenstates of the full Hamiltonian are not the eigenstates of 𝐿ˆ 2 . This is the
consequence of the breaking of the central symmetry of the Hamiltonian by an
external magnetic field.
Note, however, that simplified Hamiltonian 𝐻ˆ 0 + 𝐻ˆ 𝑝 commutes with 𝐿ˆ 2 as well:
𝑒𝐵 ˆ ˆ 2
[ 𝐿 𝑧 ; 𝐿 ].
[ 𝐻ˆ 0 + 𝐻ˆ 𝑝 ; 𝐿ˆ 2 ] = [ 𝐻ˆ 0 ; 𝐿ˆ 2 ] +
2𝑚 𝑒
(6.140)
Therefore, we can conclude that if we neglect 𝐻 𝑑 , the wavefunctions remain the
same, as for 𝐻ˆ 0 , 𝜓 𝑛𝑙𝑚 (r) = 𝑅𝑛𝑙 (𝑟)𝑌𝑙𝑚 (𝜃, 𝜙). We then get:
ˆ 𝑛𝑙𝑚 = 𝐻ˆ 0 𝜓 𝑛𝑙𝑚 + 𝐻ˆ 𝑝 𝜓 𝑛𝑙𝑚 = 𝐸 (0) 𝜓 𝑛𝑙𝑚 + 𝑒𝐵 𝐿ˆ 𝑧 𝜓 𝑛𝑙𝑚 = 𝐸 0 + 𝑒𝐵ℏ𝑚 𝜓 𝑛𝑙𝑚 ,
𝐻𝜓
𝑛𝑙
𝑛𝑙
2𝑚 𝑒
2𝑚 𝑒
(6.141)
(0)
𝑅𝑦
where 𝐸 𝑛𝑙
= − (𝑛+𝑙+1)
2 . The energy now depends not only on 𝑛, 𝑙 (in their combination 𝑛 + 𝑙 + 1), but as well on the magnetic quantum number 𝑚:
(0)
𝐸 𝑛𝑙𝑚 = 𝐸 𝑛𝑙
+
𝑅𝑦
𝑒ℏ𝐵
𝑚=−
+ 𝜇 𝐵 𝐵𝑚,
2𝑚 𝑒
(𝑛 + 𝑙 + 1) 2
(6.142)
𝑒ℏ
. Note, that the change of the sign of
where we introduced Bohr magneton 𝜇 𝐵 = 2𝑚
𝑒
𝑚, which corresponds to the change of the direction of the rotation, leads to the change
of the sign of energy correction produced by 𝐻 𝑝 . Clockwise and anticlockwise
rotations are thus become non-equivalent, which means that magnetic field breaks
time inversion symmetry, spinning the charged particle - we have this effect already
in the classical mechanics.
The result Eq. (6.142) can be interpreted as stemming from the interaction of the
magnetic moment of an atom, produced by electric current of an electron rotating
around proton, with magnetic field.
To understand this, let us consider again the Bohr model of a hydrogene atom.
around any axis passing through the origin of the coordinate system Electron rotating
around the proton produce an electric current, which can be calculated as: 𝐼 = 𝜆𝑉,
𝑒
where 𝜆 = 2 𝜋𝑟
- linear charge density (we spread the charge of an electron 𝑒 over
the circle 2𝜋𝑟). This current will produce the magnetic moment of the value
𝜇 = 𝐼𝑆 =
𝑒
𝑒
𝑒
𝑒
𝑉 𝜋𝑟 2 = 𝑉𝑟 =
(𝑚𝑉𝑟) =
𝐿,
2𝜋𝑟
2
2𝑚 𝑒
2𝑚 𝑒
(6.143)
where 𝑆 is the area of the frame with current, 𝑆 = 𝜋𝑟 2 , and 𝐿 = 𝑚𝑉𝑟 - absolute value
of the angular momentum.
In the vectorial form one can write:
6.5 Hydrogen atom in the external magnetic field and Zeeman effect
139
Fig. 6.7
𝜇 = 𝛾 𝐿 L,
(6.144)
where 𝛾 𝐿 = − 2𝑚𝑒 𝑒 - known as gyromagnetic ratio of an electron. The sign "-"
corresponds to negative electron charge.
Being placed in the magnetic field B, magnetic dipole gets an additional energy
equal to:
𝑒
𝐻 𝑑 = −𝜇 · B = −𝛾 𝐿 L · B =
L · B.
(6.145)
2𝑚 𝑒
So that the total Hamiltonian of an atom will be:
𝐻=
𝛼
p2
− + 𝐻𝑑 ,
2𝑚 𝑒 𝑟
(6.146)
which, after performing classical to quantum substitution p → p̂ = −𝑖ℏ∇, L → L̂ =
[r × p̂] and choosing the axis z to coincide with the direction of the magnetic field,
B = e𝑧 · 𝐵, gives us
ℏ2 2 𝛼
𝑒𝐵 ˆ
𝐻ˆ = −
∇ − +
𝐿𝑧,
(6.147)
2𝑚 𝑒
𝑟 2𝑚 𝑒
which corresponds to the Hamiltonian derived at the page 137.
According to the expression for the energy in magnetic field Eq. (6.142) at the
page 138,
𝑅𝑦
𝐸 𝑛𝑙𝑚 = −
+ 𝜇 𝐵 𝐵𝑚, |𝑚| ≤ 𝑙
(6.148)
(𝑛 + 𝑙 + 1) 2
the states with 𝑙 ≥ 1, which were 2𝑙 + 1 times degenerate in the absence of the
magnetic field should lose the degeneracy, and produce multiplets, corresponding
to 2𝑙 + 1 possible projections of 𝑚. This splitting is known as Zeeman splitting:
The sketch above illustrates the reshaping of the spectrum in the magnetic field,
according to the theory developed by us. Note, that as for s-states 𝑙 = 0 and 𝑚 = 0,
their energy should remain the same, and they are not splitted as 𝑝, 𝑑, 𝑓 etc. states.
However, this is not all what is seen in experiment - in reality s-states split into
energetic doublet. What is the origin of this splitting of s-states? That is due to the
presence of the spin S, internal angular momentum of an electron. As electron is a
charged particle, the presence of angular momentum S will necessarily mean, that it
140
6 Quantum particle in a central potential
Fig. 6.8
Fig. 6.9
has also magnetic moment 𝜇 = 𝛾𝑆 S, where 𝛾𝑆 is spin gyromagnetic ratio. Two facts
demonstrate, that spin is purely quantum property of elementary particles, and there
are troubles with its semiclassical representation, e.g. representing an electron as a
charged rotating sphere.
1) First, it turns out that spin is half integer in the units of ℏ, 𝑆 = 2ℏ . Therefore, it
has only two possible projections on any (say, z) axis:
ℏ
𝑆𝑧 = ± .
2
This is puzzling: indeed, for any type of orbital angular momentum we have
(6.149)
6.5 Hydrogen atom in the external magnetic field and Zeeman effect
141
Fig. 6.10
𝜕
,
𝜕𝜙
𝐿ˆ 𝑧 Φ = 𝐿 𝑧 Φ,
𝐿 𝑧 = −𝑖ℏ
Φ = 𝐴𝑒
𝑖𝐿 𝑧 𝜙
ℏ
,
(6.150)
(6.151)
(6.152)
and if we put half-integer value for 𝐿 𝑧 = ℏ2 , we get Φ(𝜙 + 2𝜋) = −Φ(𝜙), which
contradicts the requested 2𝜋 periodic boundary condition Φ(𝜙 + 2𝜋) = −Φ(𝜙) (see
page 122).
2) Second, it turns out that spin gyromagnetic ratio 𝛾𝑆 is about twice bigger then
orbital gyromagnetic ratio,
𝑒
𝛾𝑆 ' −
= 2𝛾 𝐿 .
(6.153)
𝑚𝑒
The introduction of the concept of spin will allow us to explain the reported Zeeman
splitting of s-states. Indeed, the spin of electron in these states can be oriented either
along magnetic field, for which case the energy will be
𝐸↑ = −
𝑅𝑦
ℏ
𝑅𝑦
− 𝛾𝑆 𝐵 = −
− 𝜇 𝐵 𝐵,
2
(𝑛 + 1) 2
(𝑛 + 1) 2
(6.154)
where
𝜇𝐵 =
𝑒ℏ
2𝑚 𝑒
(6.155)
or, oriented opposite to it, for which case
𝐸↓ = −
𝑅𝑦
ℏ
𝑅𝑦
+ 𝛾𝑆 𝐵 = −
− 𝜇 𝐵 𝐵.
(𝑛 + 1)
2
(𝑛 + 1) 2
(6.156)
142
6 Quantum particle in a central potential
The splitting between the components of the doublet is:
4𝐸 = 𝐸 ↓ − 𝐸 ↑ = 2𝜇 𝐵 𝐵.
(6.157)
What will happen with Zeeman splitting of the states with 𝑙 ≠ 0 if we account for
spin? They will split into states with different 𝑚, and then for each 𝑚 additionally
split by spin
Fig. 6.11
Schematically, the expected splitting of the p-states is shown at the sketch above.
Note, that in reality the structure of Zeeman splitting can be very different from
the predicted pentaplet structure due to the effects of spin-orbit interaction.
6.6 Tasks
6.6.1 Bound states in the 3D two-body problem
Consider the following interaction term between two particles of masses 𝑚 1,2
𝑉 (|r1 − r2 |) = −𝛼𝛿(|r1 − r2 | − 𝑎),
(6.158)
which is a quantum analog of the classical attractive hard sphere potential (𝛼 > 0).
Find the ground state of the system and check if it is a bound state. How to attack the
problem: (a) introduce new variables R = 𝑚 1 r1 + 𝑚 2 r2 /(𝑚 1 + 𝑚 2 ) (center of mass)
and r = r1 −r2 (relative position vector). (b) the potential is spherically symmetric so
you can pass to spherical coordinates. (c) having passed to sherical coordinates you
may separate the variables 𝑟 and 𝜃, 𝜙. For 𝜃, 𝜙 you will get the standard equations,
whereas the main object of interest will be the equation for the radial part of the
6.6 Tasks
143
wave function 𝑅(𝑟) since it contains the eigenergies of the problem (corresponding
to internal dynamics, there is also a contribution from motion of the center of mass.).
This equation will contain 𝑙 from the angular momentum. Since we are interested in
the ground state solely, assume that 𝑙 = 0. Do all the aforementioned transformations
by hand, explicitly (passing to spherical coordinates etc).
Additional (hard) task: show that for spherically symmetric attractive potentials
the lowest bound state (if it exists) indeed corresponds to 𝑙 = 0.
6.6.2 Spherical quantum well in 3D
Find the condition of existence of bound states of a spherical quantum well of radius
𝑎 and depth 𝑉0 (i.e. 𝑉 (𝑟) = −𝑉0 for 𝑟 ≤ 𝑎 and 𝑉 (𝑟) = 0 for 𝑟 > 𝑎).
Chapter 7
Spin of electron
7.1 Mathematical description of spin of electron
We have seen, that spin of electron can have only 2 projections 𝑆 𝑧 = ± 2ℏ on any
given axis, say, on z axis. To account for both of them simultaneously, it is natural
to replace the scalar wavefunction 𝜓(r, 𝑡) by a 2-spinor
ˆ 𝑡) = 𝜓+ (r, 𝑡) .
𝜓(r, 𝑡) −→ 𝜓(r,
(7.1)
𝜓− (r, 𝑡)
The physical meaning of the components of this spinor is the following:
|𝜓+ (r, 𝑡)| 2 = 𝜓+∗ (r, 𝑡)𝜓+ (r, 𝑡) = 𝜌+ (r, 𝑡)
(7.2)
gives the probability density, corresponding to a particle with 𝑆 𝑧 = + ℏ2 projection of
spin on z axis, and
|𝜓− (r, 𝑡)| 2 = 𝜓−∗ (r, 𝑡)𝜓− (r, 𝑡) = 𝜌− (r, 𝑡)
(7.3)
gives the probability density, corresponding to a particle with 𝑆 𝑧 = − 2ℏ spin projection
on z axis.
The total probability density is 𝜌(r, 𝑡) = 𝜌+ (r, 𝑡) + 𝜌− (r, 𝑡) is normalized to unity,
and we get
∫
∫
∫
3
3
∗
∗
ˆ 𝑡),
𝜌(r, 𝑡)𝑑 r =
𝑑 r[𝜓+ 𝜓+ + 𝜓− 𝜓− ] =
𝑑 3 r𝜓ˆ + (r, 𝑡) 𝜓(r,
(7.4)
where we introduced Hermitian conjugate spinor
𝜓ˆ + (r, 𝑡) = (𝜓+∗ (r, 𝑡); 𝜓−∗ (r, 𝑡)),
(7.5)
so that
145
146
7 Spin of electron
ˆ 𝑡) = (𝜓+∗ ; 𝜓−∗ ) 𝜓+ = 𝜓+∗ 𝜓+ + 𝜓−∗ 𝜓− = 𝜌+ (r, 𝑡) + 𝜌− (r, 𝑡) = 𝜌(r, 𝑡).
𝜓ˆ + (r, 𝑡) 𝜓(r,
𝜓
(7.6)
The operators, corresponding to physical observables, can be divided into those,
𝑝ˆ 2
ℏ2 2
affecting the spatial coordinates only, such as kinetic energy 𝑇ˆ = 2𝑚
= − 2𝑚
∇ or
the part of the Hamiltonian which does not include the action of the magnetic field
1
on spin, 𝐻ˆ 0 = 2𝑚
(p− 𝑞A) 2 + 𝑞𝜙(r), and those, which affect spin degrees of freedom,
for example the operator of the energy corresponding to interaction of spin magnetic
moment with magnetic field. The corresponding classical Hamiltonian reads (page
141):
𝑒
𝐻 𝑝 = 𝛾𝑆 S · B =
S · B,
(7.7)
𝑚𝑒
and transition to quantum mechanics will need the introduction of the operators of
the components of spin:
𝑒
𝑒 ˆ
Ŝ · B =
( 𝑆 𝑥 𝐵 𝑥 + 𝑆ˆ 𝑦 𝐵 𝑦 + 𝑆ˆ 𝑧 𝐵 𝑧 ).
𝐻 𝑝 −→ 𝐻ˆ 𝑝 =
𝑚𝑒
𝑚𝑒
(7.8)
As spin is an internal angular momentum of an electron, the constructed operators should have same commutation relations, as operators of the orbital angular
momentum 𝐿ˆ 𝑥 , 𝐿ˆ 𝑦 , 𝐿ˆ 𝑧 (see page 119):
[ 𝑆ˆ 𝑥 ; 𝑆ˆ 𝑦 ] = 𝑖ℏ𝑆ˆ 𝑧 ,
[ 𝑆ˆ 𝑧 ; 𝑆ˆ 𝑥 ] = 𝑖ℏ𝑆ˆ 𝑦 ,
(7.10)
[ 𝑆ˆ 𝑦 ; 𝑆ˆ 𝑧 ] = 𝑖ℏ𝑆ˆ 𝑥 .
(7.11)
(7.9)
𝜓
ˆ
ˆ
ˆ
ˆ
On the other hand, operators 𝑆 𝑥 , 𝑆 𝑦 𝑆 𝑧 act in the space of 2-spinors 𝜓 = + , and,
𝜓−
therefore, should be represented in terms of 2 × 2 matrices, satisfying commutation
relation Eq. (7.10-7.11), where product of two operators in noting but the standard
product of two corresponding matrices.
To satisfy this demand, one can choose:
ℏ
𝑆ˆ 𝑗 = 𝜎 𝑗 , 𝑗 = 𝑥, 𝑦, 𝑧,
2
where 𝜎 𝑗 are so called Pauli matrices, given by the following expressions:
01
𝜎𝑥 =
,
10
0 −𝑖
𝜎𝑦 =
,
𝑖 0
1 0
𝜎𝑧 =
.
0 −1
(7.12)
(7.13)
(7.14)
(7.15)
7.1 Mathematical description of spin of electron
147
Let us check that such a choice given indeed correct commutation relations
Eq. (7.10-7.11). One has:
ℏ2
ℏ2 0 1 0 −𝑖
0 −𝑖 0 1
[ 𝑆ˆ 𝑥 ; 𝑆ˆ 𝑦 ] = 𝑆ˆ 𝑥 𝑆ˆ 𝑦 − 𝑆ˆ 𝑦 𝑆ˆ 𝑥 = (𝜎𝑥 𝜎𝑦 − 𝜎𝑦 𝜎𝑥 ) =
−
=
𝑖 0 10
4
4 10 𝑖 0
ℏ2 𝑖 0
ℏ 1 0
ℏ
ℏ2 𝑖 0
−𝑖 0
−
=
= 𝑖ℏ
= 𝑖ℏ 𝜎𝑧 = 𝑖ℏ𝑆ˆ 𝑧 .
(7.16)
0 𝑖
4 0 −𝑖
2 0 −𝑖
2 0 −1
2
In the similar way other two commutation relations can be checked.
Let us now construct the operator of the square of spin. We get:
ℏ2 0 1 0 1
0 −𝑖 0 −𝑖
1 0 1 0
𝑆ˆ2 = 𝑆ˆ2𝑥 + 𝑆ˆ2𝑦 + 𝑆ˆ2𝑧 =
+
+
=
𝑖 0 𝑖 0
0 −1 0 −1
4 10 10
ℏ2 1 0
3ℏ2
10
10
=
+
+
=
𝐼 = ℏ2 𝑆(𝑆 + 1)𝐼,
(7.17)
01
01
4 01
4
where 𝑆 =
1
2
- the value of the spin in unit of ℏ
10
𝐼=
01
(7.18)
is the unity matrix.
The time dependent Schrodinger equation for an electron in the electromagnetic
field reads:
𝜕 𝜓ˆ
ˆ
𝑖ℏ
= 𝐻ˆ 𝜓,
(7.19)
𝜕𝑡
where the Hamiltonian, including both orbital and spin dynamics, reads:
𝐻ˆ =
1
1
( p̂ + 𝑒A) 2 − 𝑒𝜙 − 𝛾𝑆 Ŝ · B =
( p̂ + 𝑒A) 2 − 𝑒𝜙 + 𝜇 𝐵 B · 𝜎,
2𝑚 𝑒
2𝑚 𝑒
(7.20)
where
𝜕A
, B(r, 𝑡) = ∇ × A.
(7.21)
𝜕𝑡
Consider the case, when external magnetic field is homogeneous, i.e. B = 𝑐𝑜𝑛𝑠𝑡
is not depending on the coordinates. In this case, the Hamiltonian can be represented
as sum of the terms acting on coordinates r and spin only,
E(r, 𝑡) = −∇𝜙 −
𝐻ˆ = 𝐻ˆ 𝑝 + 𝐻ˆ 0 ,
(7.22)
where
1
𝐻ˆ 0 = −
( p̂ + 𝑒A) 2 − 𝑒𝜙,
2𝑚 𝑒
𝐻ˆ 𝑝 = 𝜇 𝐵 B · 𝜎
(7.23)
(7.24)
148
7 Spin of electron
In this case the spatial and spin dynamics can be separated from each other, i.e. the
spinor wavefunction can be factorized as:
𝜓+ (r, 𝑡)
𝜒+ (𝑡)
ˆ
𝜓(r, 𝑡) =
= 𝜓(r, 𝑡)
= 𝜓(r, 𝑡) 𝜒(𝑡)
ˆ
(7.25)
𝜓− (r, 𝑡)
𝜒− (𝑡)
- product of scalar 𝜓(r, 𝑡) and spinor 𝜒(𝑡),
ˆ
which depends on time only. Normalization gives:
∫
∫
ˆ 𝑡)𝑑 3 r =
𝜓ˆ + (r, 𝑡) 𝜓(r,
𝜓 ∗ (r, 𝑡)𝜓(r, 𝑡)𝑑 3 r · 𝜒+ (𝑡) 𝜒(𝑡) = 𝜒+ (𝑡) 𝜒(𝑡) = 1
(7.26)
- the coordinate and spinor parts are independently normalized to unity, for the latter
normalization condition gives:
𝜒
𝜒+ 𝜒 = ( 𝜒+∗ ; 𝜒−∗ ) + = | 𝜒+ | 2 + | 𝜒− | 2 = 1.
(7.27)
𝜒−
Placing wavefunction Eq. (7.25) to time dependent Schrodinger equation we get:
𝜕
𝜕𝜓
𝜕𝜒
𝑖ℏ (𝜓 𝜒) = 𝑖ℏ 𝜒
+𝜓
= ( 𝐻ˆ 0 + 𝐻ˆ 𝑝 )𝜓 𝜒 = 𝜒 𝐻ˆ 0 𝜓 + 𝜓 𝐻ˆ 𝑝 𝜒.
(7.28)
𝜕𝑡
𝜕𝑡
𝜕𝑡
Multiplying this by
1
𝜓
𝜒+ we get:
1 𝜕𝜓
𝜕𝜒
1
+ 𝑖ℏ𝜒+
= 𝜒+ 𝜒 𝐻ˆ 0 𝜓 + 𝜒+ 𝐻ˆ 𝑝 𝜒 =⇒
𝜓 𝜕𝑡
𝜕𝑡
𝜓
1 𝜕𝜓 1 ˆ
𝜕𝜒
𝑖ℏ
− 𝐻0 𝜓 = −𝑖ℏ𝜒+
+ 𝜒+ 𝐻ˆ 𝑝 𝜒.
𝜓 𝜕𝑡
𝜓
𝜕𝑡
𝑖ℏ𝜒+ 𝜒
(7.29)
(7.30)
The left hand side here contains r, while the right hand side not, we can thus simply
put that
1 𝜕𝜓 1 ˆ
− 𝐻0 𝜓 = 𝜆 =⇒
𝜓 𝜕𝑡
𝜓
𝜕𝜓
𝑖ℏ
= ( 𝐻ˆ 0 + 𝜆)𝜓,
𝜕𝑡
𝜕𝜒
− 𝑖ℏ𝜒+
+ 𝜒+ 𝐻ˆ 𝑝 𝜒 = 𝜆 = 𝜆 𝜒+ 𝜒 =⇒
𝜕𝑡
𝜕𝜒
+
ˆ
𝜒 𝑖ℏ
− ( 𝐻 𝑝 − 𝜆) 𝜒 = 0 =⇒
𝜕𝑡
𝜕𝜒
𝑖ℏ
= ( 𝐻ˆ 𝑝 − 𝜆) 𝜒.
𝜕𝑡
𝑖ℏ
(7.31)
(7.32)
(7.33)
(7.34)
(7.35)
Note, that for both equations the constant of the separation 𝜆 just contributes to
constant addition to the potential energy, which can be removed by simple phase
7.1 Mathematical description of spin of electron
149
transformation:
𝜓 = 𝑒−
𝑖𝜆𝑡
ℏ
˜
𝜓,
𝑖𝜆𝑡
ℏ
𝜒,
𝜒=𝑒 e
ee
𝜓ˆ = 𝜓 𝜒 = 𝜓
𝜒,
e
𝜕𝜓
e,
𝑖ℏ
= 𝐻ˆ 0 𝜓
𝜕𝑡
𝜕e
𝜒
𝑖ℏ
= 𝐻ˆ 𝑝 e
𝜒.
𝜕𝑡
(7.36)
(7.37)
(7.38)
(7.39)
(7.40)
In the case when Hamiltonian, moreover, is not depending on time, we can reduce
both coordinate and spinor part of the Schrodinger equation to stationary Schrodinger
equation, introducing the corresponding independent energy levels:
𝑖ℏ
𝜕𝜓
= 𝐻0 𝜓,
𝜕𝑡
−𝑖𝐸0 𝑡
𝜓 = 𝑒 ℏ 𝜙(r),
𝐻ˆ 0 𝜙(r) = 𝐸 0 𝜙(r),
𝜕𝜒
𝑖ℏ
= 𝐻 𝑝 𝜒,
𝜕𝑡
−𝑖𝐸 𝑝 𝑡
𝜒 = 𝑒 ℏ 𝜒 0,
𝐻ˆ 𝑝 𝜒 0 = 𝐸 𝑝 𝜒 0,
(7.41)
(7.42)
(7.43)
(7.44)
(7.45)
(7.46)
and the total energy is:
−𝑖 (𝐸0 +𝐸 𝑝 ) 𝑡
𝜕 𝜓ˆ
𝜕
𝜕 −𝑖 (𝐸0 +𝐸 𝑝 ) 𝑡 0
ℏ
ℏ
= 𝑖ℏ 𝜓 𝜒 = 𝑖ℏ 𝑒
𝜙𝜒 = 𝑒
(𝐸 0 + 𝐸 𝑝 ) (𝜙 𝜒 0) =
𝜕𝑡
𝜕𝑡
𝜕𝑡
−𝑖 (𝐸0 +𝐸 𝑝 ) 𝑡
ℏ
( 𝐻ˆ 0 + 𝐻ˆ 𝑝 )(𝜙 𝜒 0) =⇒
(7.47)
=𝑒
0
0
0
ˆ
ˆ
ˆ
ˆ
ˆ
( 𝐻0 + 𝐻 𝑝 ) 𝜓 = (𝐸 0 + 𝐸 𝑝 ) 𝜓 = 𝐸 𝜓 ,
(7.48)
𝑖ℏ
where 𝜓ˆ 0 = 𝜙 𝜒 0 - stationary part of the wavefunction, and 𝐸 = 𝐸 0 + 𝐸 𝑝 . Consider
now the eigenstates and eigenvalues of the spinor part of the Hamiltonian, describing
the interaction between spin and magnetic field (paramagnetic Hamiltonian 𝐻ˆ 𝑝 =
𝛾𝑆 Ŝ · B = 𝜇 𝐵 B · 𝜎. One gets:
𝐵𝑧
𝐵 𝑥 − 𝑖𝐵 𝑦
ˆ
𝐻 𝑝 = 𝜇 𝐵 B · 𝜎 = 𝜇 𝐵 (𝐵 𝑥 𝜎𝑥 + 𝐵 𝑦 𝜎𝑦 + 𝐵 𝑧 𝜎𝑧 ) = 𝜇 𝐵
𝐵 𝑥 + 𝑖𝐵 𝑦 −𝐵 𝑧
𝜇 𝐵 𝐵 𝑧 − 𝐸 𝑝 𝜇 𝐵 (𝐵 𝑥 − 𝑖𝐵 𝑦 ) 𝜒+
𝐻ˆ 𝑝 𝜒 = 𝐸 𝑝 𝜒 =⇒
= 0.
(7.49)
𝜇 𝐵 (𝐵 𝑥 + 𝑖𝐵 𝑦 ) −𝜇 𝐵 𝐵 𝑧 − 𝐸 𝑝 𝜒−
This linear homogeneous equation for components 𝜒+ , 𝜒− has non-trivial solution,
only when its determinant is equal to zero:
150
7 Spin of electron
𝜇 𝐵 𝐵 𝑧 − 𝐸 𝑝 𝜇 𝐵 (𝐵 𝑥 − 𝑖𝐵 𝑦 )
= 𝐸 2𝑝 − 𝜇2𝐵 𝐵2𝑧 − 𝜇2𝐵 (𝐵 𝑥 − 𝑖𝐵 𝑦 ) (𝐵 𝑥 + 𝑖𝐵 𝑦 ) =
𝜇 𝐵 (𝐵 𝑥 + 𝑖𝐵 𝑦 ) −𝜇 𝐵 𝐵 𝑧 − 𝐸 𝑝
q
= 𝐸 2𝑝 − 𝜇2𝐵 (𝐵2𝑥 + 𝐵2𝑦 + 𝐵2𝑧 ) = 0 =⇒ 𝐸 𝑝 = ±𝜇 𝐵 𝐵 = ±𝜇 𝐵 𝐵2𝑥 + 𝐵2𝑦 + 𝐵2𝑧 (7.50)
- we get two values of the energy corresponding to two possible orientations of the
spin of electron, one along the magnetic field (sign "+"), and another opposite to it
(sign "-").
Let us now find the corresponding spinor wavefunctions. Take 𝐸 𝑝 = −𝜇 𝐵 𝐵. Then,
from the equation 𝐻ˆ 𝑝 𝜒 = 𝐸 𝑝 𝜒 = −𝜇 𝐵 𝐵 𝜒 we get:
𝐵 𝑧 + 𝐵 𝐵 𝑥 − 𝑖𝐵 𝑦 𝜒+
𝜇𝐵
= 0 =⇒
(7.51)
𝐵 𝑥 + 𝑖𝐵 𝑦 −𝐵 𝑧 + 𝐵 𝜒−
(𝐵 𝑧 + 𝐵) 𝜒+ + (𝐵 𝑥 − 𝑖𝐵 𝑦 ) 𝜒− = 0,
(7.52)
(𝐵 𝑥 + 𝑖𝐵 𝑦 ) 𝜒+ + (𝐵 − 𝐵 𝑧 ) 𝜒− = 0.
(7.53)
From second of this equation (and as 𝐸 𝑝 = −𝜇 𝐵 𝐵 is an eigenvalue, first equation is
equivalent to second, so that we can use any of them) we get:
𝜒− = −
𝐵 𝑥 + 𝑖𝐵 𝑦
,
𝐵 − 𝐵𝑧
(7.54)
which gives us:
𝜒=𝐴
e=
where 𝐴
𝐴
𝐵−𝐵 𝑧
1
−
𝐵 𝑥 +𝑖𝐵 𝑦
𝐵−𝐵 𝑧
!
e
=𝐴
𝐵 − 𝐵𝑧
,
−(𝐵 𝑥 + 𝑖𝐵 𝑦 )
(7.55)
- normalization coefficient, which can be defined as:
𝐵 − 𝐵𝑧
=
−𝐵 𝑥 − 𝑖𝐵 𝑦
e 2 [(𝐵 − 𝐵 𝑧 ) 2 + 𝐵2𝑥 + 𝐵2𝑦 ] = | 𝐴|
e 2 (2𝐵2 − 2𝐵𝐵 𝑧 ) = | 𝐴|
e 2 2𝐵(𝐵 − 𝐵 𝑧 ) = 1 =⇒
= | 𝐴|
(7.56)
1
e= p
𝐴
,
(7.57)
2𝐵(𝐵 − 𝐵 𝑧 )
e 2 (𝐵 − 𝐵 𝑧 ; −𝐵 𝑥 + 𝑖𝐵 𝑦 )
𝜒+ 𝜒 = | 𝐴|
which gives us:
q
𝐵−𝐵 𝑧
cos 𝜃2𝐵
©
ª
2𝐵
𝜒↑ = ­
=
,
®
𝐵 +𝑖𝐵
− sin 𝜃2𝐵 𝑒 𝑖 𝜙𝐵
−√ 𝑥 𝑦
« 2𝐵 (𝐵−𝐵𝑧 ) ¬
(7.58)
Let us choose the spherical system of the coordinates to characterize the orientation
of the magnetic field B in the real space,
𝐵 𝑥 = 𝐵 sin 𝜃 𝐵 cos 𝜙 𝐵 ; 𝐵 𝑦 = 𝐵 sin 𝜃 𝐵 sin 𝜙 𝐵 ; 𝐵 𝑧 = 𝐵 cos 𝜃 𝐵 .
(7.59)
7.2 Spin precession in the external magnetic field and magnetic resonance
151
Then, one has:
r
𝐵 − 𝐵𝑧
=
2𝐵
r
1 − cos 𝜃 𝐵
= sin(𝜃 𝐵 /2),
2
(7.60)
𝐵 𝑥 + 𝑖𝐵 𝑦
sin 𝜃 𝐵 𝑒 𝑖 𝜙𝐵
sin 𝜃 𝐵
−p
= −p
𝑒 𝑖 𝜙𝐵 = − cos(𝜃 𝐵 /2)𝑒 𝑖 𝜙𝐵
=
2
sin(𝜃
𝐵 /2)
2𝐵(𝐵 − 𝐵 𝑧 )
2(1 − cos 𝜃 𝐵 )
(7.61)
And thus
sin 𝜃2𝐵
,
(7.62)
𝜒↑ =
− cos 𝜃2𝐵 𝑒 𝑖 𝜙𝐵
Doing the same procedure for 𝐸 𝑝 = +𝜇 𝐵 𝐵, we will get:
𝜒↓ =
cos 𝜃2𝐵
.
sin 𝜃2𝐵 𝑒 𝑖 𝜙𝐵
(7.63)
Note, that states 𝜒↑ and 𝜒↓ are orthogonal to each other, so that their scalar product
gives:
𝜃𝐵
𝜃 𝐵 −𝑖 𝜙𝐵
cos 𝜃2𝐵
+
𝜒↑ 𝜒↓ = (sin
; − cos
𝑒
= 0.
(7.64)
)
sin 𝜃2𝐵 𝑒 𝑖 𝜙𝐵
2
2
7.2 Spin precession in the external magnetic field and magnetic
resonance
Suppose, that we have a particle with spin 21 placed in the external magnetic field.
Then, as we know (pages 148-149), the time evolution of the spinor part of its
wavefunction is given by the following equation:
𝑖ℏ
(7.65)
𝜒+ (𝑡)
, 𝐻ˆ 𝑝 = 𝜇 𝐵 B · 𝜎.
𝜒− (𝑡)
The eigenvalues and eigenvectors of corresponding stationary equation 𝐻ˆ 𝑝 𝜒 =
𝐸 𝑝 𝜒 were obtained in the previous lecture. They were corresponding to 𝐸 𝑝 = ±𝜇 𝐵 𝐵
with spin being oriented along opposite magnetic field.
Now, let us imagine that initially the spin is directed making some angle with
magnetic field. Then, we will have non-trivial evolution of the spinor wavefunction
𝜒(𝑡), which we are now going to define.
Before we analyze the quantum problem, let us consider the evolution of a classical
magnetic moment 𝜇 = 𝛾L.
Magnetic field produced a torque acting on magnetic moment: 𝜏 = 𝜇 × B. Then,
we have:
where 𝜒 =
𝑑𝜒
= 𝐻 𝑝 𝜒,
𝑑𝑡
152
7 Spin of electron
𝑑L
= 𝜏 = 𝜇 × B = 𝛾 e 𝑥 (𝐿 𝑦 𝐵 𝑧 − 𝐿 𝑧 𝐵 𝑦 ) + e 𝑦 (𝐿 𝑧 𝐵 𝑥 − 𝐿 𝑥 𝐵 𝑧 ) + e𝑧 (𝐿 𝑥 𝐵 𝑦 − 𝐿 𝑦 𝐵 𝑥 ) ,
𝑑𝑡
(7.66)
which gives:
𝑑𝐿 𝑥
= 𝛾(𝐿 𝑦 𝐵 𝑧 − 𝐿 𝑧 𝐵 𝑦 ),
𝑑𝑡
𝑑𝐿 𝑦
= 𝛾(𝐿 𝑧 𝐵 𝑥 − 𝐿 𝑥 𝐵 𝑧 ),
𝑑𝑡
𝑑𝐿 𝑧
= 𝛾(𝐿 𝑥 𝐵 𝑦 − 𝐿 𝑦 𝐵 𝑥 ).
𝑑𝑡
(7.67)
(7.68)
(7.69)
Let us orient the magnetic field along z-axis, 𝐵 𝑧 = 𝐵, 𝐵 𝑥 = 𝐵 𝑦 = 0. Then we get:
𝑑𝐿 𝑥
= 𝛾𝐵𝐿 𝑦 ,
𝑑𝑡
𝑑𝐿 𝑦
= −𝛾𝐵𝐿 𝑥 ,
𝑑𝑡
𝑑𝐿 𝑧
= 0.
𝑑𝑡
(7.70)
(7.71)
(7.72)
Introducing e
𝐿 = 𝐿 𝑥 + 𝑖𝐿 𝑦 , one gets:
𝑑
(𝐿 𝑥 + 𝑖𝐿 𝑦 ) = 𝛾𝐵(𝐿 𝑦 − 𝑖𝐿 𝑥 ) = −𝑖𝛾𝐵(𝐿 𝑥 + 𝑖𝐿 𝑦 ),
𝑑𝑡
𝑑e
𝐿
= −𝑖𝛾𝐵e
𝐿 =⇒ e
𝐿 (𝑡) = e
𝐿(0)𝑒 −𝑖𝛾𝐵𝑡 .
𝑑𝑡
(7.73)
(7.74)
Then we obtain:
𝐿 𝑥 (𝑡) + 𝑖𝐿 𝑦 (𝑡) = 𝐿 𝑥 (0) cos(𝜔 𝐿 𝑡) + 𝐿 𝑦 (0) sin(𝜔 𝐿 𝑡)+
+ 𝑖(𝐿 𝑦 (0) cos(𝜔 𝐿 𝑡) − 𝐿 𝑥 (0) sin(𝜔 𝐿 𝑡)) =⇒
(7.75)
𝐿 𝑥 (𝑡) = 𝐿 𝑥 (0) cos(𝜔 𝐿 𝑡) + 𝐿 𝑦 (0) sin(𝜔 𝐿 𝑡),
(7.76)
𝐿 𝑦 (𝑡) = 𝐿 𝑦 (0) cos(𝜔 𝐿 𝑡) − 𝐿 𝑥 (0) sin(𝜔 𝐿 𝑡),
(7.77)
𝐿 𝑧 (𝑡) = 𝐿 𝑧 (0).
(7.78)
These equation describe the precession of the magnetic moment (known as Larmor precession) around the magnetic field with the frequency
𝜔 𝐿 = 𝛾𝐵,
(7.79)
known as Larmor frequency.
Now, consider the problem using the quantum approach. Let us suggest, that
magnetic field is again directed along z-axis, and
7.2 Spin precession in the external magnetic field and magnetic resonance
153
Fig. 7.1
1 0
𝐻ˆ 𝑝 = 𝜇 𝐵 𝜎 × B = 𝜇 𝐵 𝜎𝑧 𝐵 = 𝜇 𝐵 𝐵
0 −1
(7.80)
and initially the spin is oriented, let us say, along x-axis this means, that 𝜒(0) is an
eigenvector of 𝑆 𝑥 corresponding to the eigenvalue +1/2:
𝜋
𝑆ˆ 𝑥 𝜒(0) = 𝜒(0)𝜎𝑥 𝜒(0) = 𝜒(0),
2 0 1 𝜒+ (0)
𝜒 (0)
= +
=⇒
1 0 𝜒− (0)
𝜒− (0)
1 1
𝜒− (0) = 𝜒+ (0) =⇒ 𝜒(0) = √
,
2 1
where
√1
2
(7.81)
(7.82)
(7.83)
is normalization factor, 𝜒+ (0) 𝜒(0) = 1. We then have:
𝑑𝜒
𝑑 𝜒+
1 0
𝜒+
𝑖ℏ
= 𝑖ℏ
= 𝜇𝐵 𝐵
=⇒
𝜒
0
−1
𝜒
𝑑𝑡
𝑑𝑡 −
−
𝑑𝜒+
𝑖ℏ
= 𝜇 𝐵 𝐵 𝜒+ =⇒
𝑑𝑡
𝑖 𝜇 𝐵 𝐵𝑡
𝑖 𝜇 𝐵 𝐵𝑡
1
𝜒+ (𝑡) = 𝜒+ (0)𝑒 − ℏ = √ 𝑒 − ℏ ,
2
𝑑𝜒−
𝑖ℏ
= −𝜇 𝐵 𝐵 𝜒− =⇒
𝑑𝑡
𝑖 𝜇 𝐵 𝐵𝑡
1 𝑖 𝜇𝐵 𝐵𝑡
𝜒− (𝑡) = 𝜒− (0)𝑒 + ℏ = √ 𝑒 + ℏ .
2
Noting that
𝜇𝐵 𝐵
𝑒ℏ 𝐵 1
=
= 𝛾𝑆 𝐵,
ℏ
2𝑚 𝑒 ℏ 2
(7.84)
(7.85)
(7.86)
(7.87)
(7.88)
(7.89)
154
7 Spin of electron
where 𝛾𝑆 =
𝑒
2𝑚𝑒
- spin gyromagnetic ratio for electron, we can write:
!
𝜔𝐿
1 𝑒 −𝑖 2 𝑡
𝜒+ (𝑡)
𝜔𝐿
,
𝜒(𝑡) =
=√
𝜒− (𝑡)
2 𝑒𝑖 2 𝑡
(7.90)
where 𝜔 𝐿 = 𝛾𝑆 𝐵 is the classical Larmor frequency
(see pages 152 and 153). The
4𝜋
4𝜋
wavefucntion is the periodic function of time, 𝜒 𝑡 + 𝜔
, but the period 𝑇 = 𝜔
is
𝐿
𝐿
twice the period of the classical Larmor precession.
Let us see how mean values of the spin projections change with time. We have:
!
𝜔𝐿
0 1 𝑒 −𝑖 𝜔2𝐿 𝑡
𝜔𝐿
𝜋
𝜋
𝜔
< 𝑆 𝑥 (𝑡) >= 𝜒+ (𝑡) 𝑆ˆ 𝑥 𝜒(𝑡) = 𝜒+ (𝑡)𝜎𝑥 𝜒(𝑡) =
=
𝑒 𝑖 2 𝑡 ; 𝑒 −𝑖 2 𝑡
1 0 𝑒 𝑖 2𝐿 𝑡
2
4
!
𝜔𝐿
𝜋
𝜋 𝑖 𝜔𝐿 𝑡 −𝑖 𝜔𝐿 𝑡 𝑒 𝑖 2 𝑡
𝜋 1 𝑖 𝜔𝐿 𝑡
𝜔𝐿
=
𝑒 2 ;𝑒 2
𝑒
+ 𝑒 −𝑖 𝜔𝐿 𝑡 = cos(𝜔 𝐿 𝑡) = 𝑆 𝑥 (0) cos(𝜔 𝐿 𝑡)
−𝑖
𝑡
4
22
2
𝑒 2
(7.91)
(as we directed at 𝑡 = 0 the spin along x-axis, then 𝑆 𝑥 (0) =
In the similar way, for < 𝑆 𝑦 (𝑡) > we get:
< 𝑆 𝑦 (𝑡) >= 𝜒+ (𝑡) 𝑆ˆ 𝑦 𝜒(𝑡) =
𝜋
2,
𝑆 𝑦 (0) = 𝑆 𝑧 (0) = 0).
𝜋 +
𝜒 (𝑡)𝜎𝑦 𝜒(𝑡) = 𝑆 𝑥 (0) sin(𝜔 𝐿 𝑡),
2
(7.92)
and
𝜋
< 𝑆 𝑧 (𝑡) >= 𝜒+ (𝑡) 𝑆ˆ 𝑧 𝜒(𝑡) = 𝜒+ (𝑡)𝜎𝑧 𝜒(𝑡) = 0.
2
(7.93)
Comparing the obtained equations with those, we got for the classical moment L at
page 152, we see, that they fully coincide if we put 𝐿 𝑥 (0) = 𝜋2 , 𝐿 𝑦 (0) = 𝐿 𝑧 (0) = 0.
The difference in sign in the equation for < 𝑆 𝑦 (𝑡) > comes from the fact, that for
electron the charge and thus gyromagnetic ratio are negative, and thus we should
replace 𝜔 𝐿 → −𝜔 𝐿 .
Note, that it turns out that in quantum physics the period of spin precession
coincides with classical Larmor precession, and is twice smaller then the period of
the change of wavefunction. It means, that when spin makes one round of precession
the wavefucntion itself changes its sign, and two rounds of precession are necessary
to regain its initial value. This remarkable fact is related to the half integer nature of
the spin of electron.
Can we somehow detect this change of the sign in one rotation round?
Yes, if we perform an interference experiment shown below: Suppose, that we
have electronic beamsplitter, and electrons, which have spins oriented along x-axis
are splitted into two paths. Along path 1 there is no magnetic field. Along path 2
there is a magnetic field directed along z-axis and tuned in such a way, that it turns
ones between the input and the output points. Then the fluxes which passed through
paths 1 and 2 will interfere, and as single round of spin rotation leads to the sign
7.2 Spin precession in the external magnetic field and magnetic resonance
155
Fig. 7.2
change for the wavefunction, this in the negative interference and outgoing intensity
minimum.
Now, let us consider the phenomenon of the magnetic resonance. Suppose, that
we have the following time-dependent magnetic field:
𝐵 𝑧 = 𝐵 k = 𝑐𝑜𝑛𝑠𝑡,
(7.94)
𝐵 𝑦 = 𝐵⊥ sin(𝜔𝑡),
(7.95)
𝐵 𝑥 = 𝐵⊥ cos(𝜔𝑡).
(7.96)
Such magnetic field can be created, e.g. by circular polarized electromagnetic wave
Fig. 7.3
156
7 Spin of electron
of the frequency 𝜔, propagating along the axis, along which constant magnetic field
𝐵 k is additionally applied.
Suppose, that initially the spin is aligned opposite to magnetic field 𝐵 k (i.e.
magnetic moment 𝜇 coincides with it), which corresponds to minimal energy. The
presence of the rotating field 𝐵⊥ perpendicular to it can lead to the processes of the
spin flip. Let us analyze what will be the probability to find the spin in spin up state.
For this we should define the time evolution of the spinor part of wavefunction 𝜒(𝑡),
and find
𝑃+ (𝑡) = | 𝜒+ (𝑡)| 2 = 𝜒+∗ (𝑡) 𝜒+ (𝑡).
(7.97)
Initially
𝜒(0) =
0
,
1
(7.98)
so 𝑃+ (0) = 0.
The time-dependent Hamiltonian of the system is:
𝐻ˆ 𝑝 (𝑡) = 𝜇 𝐵 B · 𝜎 = 𝜇 𝐵 (𝐵 𝑥 𝜎𝑥 + 𝐵 𝑦 𝜎𝑦 + 𝐵 𝑧 𝜎𝑧 ) = 𝜇 𝐵
𝐵 k 𝐵⊥ 𝑒 −𝑖 𝜔𝑡
, (7.99)
𝐵⊥ 𝑒 𝑖 𝜔𝑡 −𝐵 k
𝑑𝜒
= 𝐻ˆ 𝑝 (𝑡) 𝜒,
𝑑𝑡 𝑑 𝜒+
𝐵 k 𝐵⊥ 𝑒 −𝑖 𝜔𝑡 𝜒+
𝑖ℏ
= 𝜇𝐵
=⇒
𝐵⊥ 𝑒 𝑖 𝜔𝑡 −𝐵 k
𝜒−
𝑑𝑡 𝜒−
𝑑𝜒+
= 𝜇 𝐵 (𝐵 k 𝜒+ + 𝐵⊥ 𝑒 −𝑖 𝜔𝑡 𝜒− ),
𝑖ℏ
𝑑𝑡
𝑑𝜒−
𝑖ℏ
= 𝜇 𝐵 (𝐵⊥ 𝑒 𝑖 𝜔𝑡 𝜒+ − 𝐵 k 𝜒− ).
𝑑𝑡
(7.100)
𝑖ℏ
(7.101)
(7.102)
(7.103)
Let us make the following substitutions:
(7.104)
𝜔
𝜒− = 𝑒 𝑖 2 𝑡 e
𝜒− ,
𝑑𝜒−
𝑑
= 𝑒𝑖
𝑑𝑡
𝑑𝑡
𝜔
2 𝑡
𝜒− = 𝑒 𝑖
𝜔
2 𝑡
𝑑e
𝜒−
𝜔
+𝑖 e
𝜒− ,
𝑑𝑡
2
(7.105)
(7.106)
𝜔
𝜒+ ,
𝜒+ = 𝑒 −𝑖 2 𝑡 e
𝜔
𝜔
𝑑e
𝜒+
𝑑𝜒+
𝑑
𝜔
= 𝑒 −𝑖 2 𝑡 𝜒+ = 𝑒 −𝑖 2 𝑡
−𝑖 e
𝜒+ .
𝑑𝑡
𝑑𝑡
𝑑𝑡
2
Placing this into Eq. (7.102,7.103), we get:
ℏ𝜔
𝑑e
𝜒+
𝑖ℏ
= 𝜇𝐵 𝐵 k −
e
𝜒+ + 𝜇 𝐵 𝐵⊥ e
𝜒− ,
𝑑𝑡
2
𝑑e
𝜒−
ℏ𝜔
𝑖ℏ
= 𝜇 𝐵 𝐵⊥ e
𝜒+ − 𝜇 𝐵 𝐵 k −
e
𝜒− .
𝑑𝑡
2
(7.107)
(7.108)
(7.109)
7.2 Spin precession in the external magnetic field and magnetic resonance
157
From here the reason of the performing of the transformation becomes clear: we
reduced the system of linear differential equations with time-dependent coefficients
to the system with time independent coefficients, for which there is a standard method
of solution.
Introducing
ℏ𝜔
1
𝜇𝐵 𝐵𝑧 −
,
(7.110)
Ωk =
ℏ
2
𝜇 𝐵 𝐵⊥
Ω⊥ =
,
(7.111)
ℏ
we can rewrite the system of equations for e
𝜒± in the following form:
𝑑e
𝜒+
= ℏΩ k e
𝜒+ + ℏΩ⊥ e
𝜒− ,
𝑑𝑡
𝑑e
𝜒−
= ℏΩ⊥ e
𝜒+ − ℏΩ k e
𝜒− =⇒
𝑖ℏ
𝑑𝑡
𝑑e
𝜒
𝑑 e
𝜒+
ℏΩ k ℏΩ⊥
e
𝜒+
e𝑝 e
𝑖ℏ
= 𝑖ℏ
=
=𝐻
𝜒,
𝜒−
ℏΩ⊥ −ℏΩ k e
𝜒−
𝑑𝑡
𝑑𝑡 e
(7.113)
ℏΩ k ℏΩ⊥
e
𝐻𝑝 =
.
ℏΩ⊥ −ℏΩ k
(7.115)
𝑖ℏ
where
(7.112)
(7.114)
We see, that equation for e
𝜒 looks like time-dependent Schrodinger equation with
e𝑝 , and corresponding solution can be written as:
stationary Hamiltonian 𝐻
e
𝜒 (𝑡) = 𝐶1 𝜒1 𝑒
−𝑖𝐸1 𝑡
ℏ
+ 𝐶2 𝜒2 𝑒
−𝑖𝐸1 𝑡
ℏ
,
(7.116)
where 𝐸 1 , 𝐸 2 , 𝜒1 , 𝜒2 are defined from the stationary Schrodinger equation (see
pages 51, 52)
e𝑝 𝜒1,2 = 𝐸 1,2 𝜒1,2 ,
𝐻
(7.117)
and to define the coefficients 𝐶1 and 𝐶2 we use the initial condition:
e
𝜒 (0) = 𝐶1 𝜒1 + 𝐶2 𝜒2 =⇒
+
𝐶1,2 = 𝜒1,2
e
𝜒 (0),
(7.118)
(7.119)
+ 𝜒
+
+
as 𝜒1,2
1,2 = 0, 𝜒1 𝜒2 = 𝜒2 𝜒1 = 0 (eigenstates corresponding to the different
energies are orthogonal). We have from Eq. (7.117):
ℏΩ k − 𝐸
ℏΩ⊥
𝜒+
= 0.
(7.120)
ℏΩ⊥ −ℏΩ k − 𝐸 𝜒−
This equation has non-trivial solution, when
158
7 Spin of electron
ℏΩ k − 𝐸
ℏΩ⊥
= 𝐸 2 − ℏ2 (Ω2⊥ + Ω2k ) = 0 =⇒
ℏΩ⊥ −ℏΩ k − 𝐸
q
𝐸 1,2 = ±ℏ Ω2⊥ + Ω2k = ±ℏΩ.
(7.121)
(7.122)
The corresponding eigenfunctions
q read:
a) For 𝐸 = 𝐸 1 = ℏΩ, Ω = Ω2⊥ + Ω2k
Ω k − Ω Ω⊥
Ω ⊥ −Ω k − Ω
𝜒+ =
𝜒+
= 0 =⇒
𝜒−
!
Ωk + Ω
𝜒− =⇒ 𝜒1 =
Ω⊥
Ω k +Ω
Ω⊥
1
(7.123)
(7.124)
.
(𝜒1 is written up to normalization constant).
b) For 𝐸 2 = −ℏΩ we get in a similar way
𝜒2 =
Ω k −Ω
Ω⊥
1
!
(7.125)
,
Thus, the the solution for e
𝜒 (𝑡) can be written as:
!
e
𝜒 (𝑡) = 𝐶1 𝑒 −𝑖Ω𝑡
Ω k +Ω
Ω⊥
1
+ 𝐶2 𝑒 𝑖Ω𝑡
At 𝑡 = 0 we get:
e
𝜒 (𝑡) =
𝐶1
Ω k −Ω
Ω⊥
Ω k +Ω
Ω⊥
+ 𝐶2
𝐶1 + 𝐶2
Ω k −Ω
Ω⊥
1
!
.
(7.126)
!
.
But from the initial condition (page 156) we get:
0
e
𝜒 (0) = 𝜒(0) =
,
1
(7.127)
(7.128)
which gives us the following set of equations for the determination of the coefficients
𝐶1 and 𝐶2 :
( Ω +Ω
Ω k −Ω
k
Ω⊥ 𝐶1 + Ω⊥ 𝐶2 = 0,
(7.129)
𝐶1 + 𝐶2 = 1,
which gives us:
𝐶1 =
Ω − Ωk
Ω + Ωk
, 𝐶2 =
,
2Ω
2Ω
and then, equation for e
𝜒 (𝑡) can be written as:
(7.130)
7.3 Tasks
159
!
Ω + Ω k 𝑖Ω𝑡 Ωk −Ω
Ω
⊥
+
𝑒
=
2Ω
1
1
!
Ω2 −Ω2k
−𝑖Ω𝑡 − 𝑒 𝑖Ω𝑡 )
(𝑒
2ΩΩ⊥
=
=
1
−𝑖Ω𝑡 + (Ω + Ω )𝑒 𝑖Ω𝑡
(Ω
−
Ω
)𝑒
k
k
2Ω
! !
Ω2 −Ω2
−𝑖 ΩΩ⊥ sin(Ω𝑡)
e
𝜒+ (𝑡)
−𝑖 2ΩΩ⊥k sin(Ω𝑡)
=
.
=
=
Ω
Ω
e
𝜒− (𝑡)
cos(Ω𝑡) + 𝑖 Ωk sin(Ω𝑡)
cos(Ω𝑡) + 𝑖 k sin(Ω𝑡)
Ω − Ω k −𝑖Ω𝑡
e
𝜒 (𝑡) =
𝑒
2Ω
Ω k +Ω
Ω⊥
!
(7.131)
Ω
Keeping in mind that (page 156) 𝜒− = 𝑒 𝑖 2 𝑡 e
𝜒− , 𝜒+ = 𝑒 −𝑖 2 𝑡 e
𝜒+ , we can write:
𝜔
𝜔
Ω⊥ −𝑖 𝜔 𝑡
𝑒 2 sin(Ω𝑡),
Ω
Ωk
𝜔
𝜒− (𝑡) = 𝑒 𝑖 2 𝑡 (cos(Ω𝑡) + 𝑖
sin(Ω𝑡),
Ω
(7.132)
𝜒+ (𝑡) = 𝑖
(7.133)
which gives for the probability to find an electron in spin-up state, i.e. the probability
of a spin flip:
𝑃+ (𝑡) = | 𝜒+ (𝑡)| 2 =
2
𝜇2𝐵 𝐵⊥
Ω2⊥
2
sin
(Ω𝑡)
=
2 + (𝜇 𝐵 −
Ω2k + Ω2⊥
𝜇2𝐵 𝐵⊥
𝐵 k
ℏ𝜔 2
2 )
sin2 (Ω𝑡).
We see, that 𝑃+ (𝑡) is an oscillating function of time with the frequency
s
2
q
𝜇2𝐵 𝐵⊥
𝜇𝐵 𝐵 k 𝜔 2
Ω = Ω2k + Ω2⊥ =
+
−
.
ℏ
ℏ
2
(7.134)
(7.135)
The maximal value of the probability to find the electron with flipped spin (when
sin2 (Ω𝑡) = 1) is:
2
𝜇2𝐵 𝐵⊥
(7.136)
𝑃𝑚𝑎𝑥 =
2 .
ℏ𝜔
2
2
𝜇 𝐵 𝐵⊥ + 𝜇 𝐵 𝐵 k − 2
This function is 1 when ℏ𝜔 = 2𝜇 𝐵 𝐵 k , and decays when the rotation frequency 𝜔
deviates from this value: Resonant enhancement of spin flips around ℏ𝜔 = 2𝜇 𝐵 𝐵 k
is known as effect of magnetic resonance.
7.3 Tasks
Find how the average spin hŜi(𝑡) = 𝜒† (𝑡) Ŝ𝜒(𝑡) depends on time for the case of
magnetic resonance. Assume that the system is initially prepared in the state 𝜒(𝑡 =
0) = (cos 𝜉0 , 𝑒 i𝛽 sin 𝜉0 )𝑇 .
160
Fig. 7.4
7 Spin of electron