THE TIME- INDEPENDENT SHCRODINGER EQUATION (TISE)
ћ
,
,
ћ
=
=
̅ + ( )
ћ
, ………………………………………1
…………………………………………………………2
Potential does not depend explicitly on the time.
Equation (2) is a solution to (1) so substitute (2) into equation (1) in other to verify.
ћ
=
ћ
̅ +
ћ
Eq. (2) is a solution to (1) if
ћ
̅ +
is a solution to the
=
.
………………………………………….3
Eqn. (3) is the time-independent Schrödinger equation and
Schrödinger equation. The function
called the Eigen of Ĥ =
ћ
̅ +
is called the study-state solution to the
is will normally satisfy (3) for particular values of E. these values of E are
OR the energy eigenvalues.
To each eigenvalue will correspond one or more functions satisfying equation (3) such functions are called Eigen
functions of Hamiltonian Ĥ corresponding to the particular energy eigenvalues. The solution of equation (3)
therefore consists of finding the eigenfunctions of Ĥ. Such a problem is called eigenvalue problem. The operator Ĥ
is Hermitian and must therefore have real equation that is what we want; for the energy of the system can not be
complex or imaginary. If Eand E1 are two different eigenvalues Ĥ and
eigenfunctions corresponding to E and E1 respectively.
Then
The functions
=
and
are the numerical
= constant value
are said to describe the steady-states of the quantum mechanical system.
Infinite wire
Capacitor without edges
The Step Potential (Energy Less Than step Height)
These potential change in value in going from one range. Of course potentials which change abruptly (ie is
continuous functions of ) do not really exist in nature. Nevertheless, these idealised potentials are used frequently
1
in quantum mechanics to approximate real situations because, being constant in each range, they are easy to treat
mathematically. The results we obtain for these potentials will allow us to illustrate a number of characteristic
quantum mechanical phenomena. An analogy, that is surely familiar to students is found in the procedure used in
studying electromagnetism. This involves treating many idealised systems like the infinite wire, the capacitor
without edges etc. these systems studied because they are relatively easy to handle because they are excellent
approximations to ones and because real systems are usually complicated to treat mathematically since they have
complicated geometries.
The simplest care is the step potential:
=
> 0
0
where V0= constant
< 0
Example:
Illustrating a physical system with a potential energy function that can be approximated to a step potential. A
charge particle moves along the axis of two cylindrical electrodes held at different voltages. Its potential energy is
constant when it is inside either electrodes but it changes very rapidly when passing from one to the other.
To determine the motion of the particle according to quantum mechanics, we must find a wave function which is a
solution for total Energy E<
, The Schrödinger equation for a step potential. Since this potential is independent
of time, the actual problem is to solve the time-independent Schrodinger equation.
For < 0 (left of the step)
= 0,
So the eigenfunction that will tell us about the behaviour of the particle is a solution to simple TISE.
ћ
=
………………………………………………..(1)
> 0 (right of the step)
We have
=
2
ћ
+
=
…………………………………….(2)
The two equations are solved separately and then an eigenfunction valid for the entire range of x is constructed by
joining the two solutions together at = 0 in such a way to satisfy the requirements.
We assume the solution (general solution traveling wave eigenfunction).
=
=
Where
=
For
=
Let
√
ћ
+
………………………………………(3)
ћ
and
x< 0
A and B are constants.
<
we assume a decrease in exponential function
=> 0
−ћ
2
=
=
+ (
2
−
−
ћ
)
…………………………………………….(4)
ћ
We also assume a solution
C=0
=
+
………………………………….(5) where
generally increases without limit that
∴ equation (5) becomes
At point
= 0 the two forms of
For a continuous derivative,
and
∴
→ +∞
=
must be continuous
=
∴ =
+
3
+
= 0
(
=
)
ћ
=
√
ћ
= −
And
= 0
At
∴
=
−
2 1+
=
−
−
=
−
…………………………………….(7)
Adding equation. (6) and (7)
=
=
…………………………………(8)
2 1+
We have now determined A,B and C in terms of D. the eigenfunction for the step potential and energy ECV0 is
given by;
=
≤ 0,
2 1+
+
≥ 0
2 1−
…………………. (9)
The remaining arbitrary constant, D, determines the amplitude of the equation.
ћ
=
ћ
ћ
,
=
+
ћ
≤ 0
≥ 0
ћ
ћ
for
< 0
First term is the wave travelling in the increasing x direction and the second term is the wave traveling in
decreasing x direction (ie ≤ 0) . We associate this first term with incidence of the particle and the second term
reflection of the particle.
We now use this association to calculate the probability that the incidence of the particle is reflected which we call
the reflection coefficient R. R depends on the ratio
which specifies the amplitude of the reflected part of the
wave functionrelative to the amplitude of the incident part. But in Quantum mechanics probabilities depend on
intensities such as B*B and A*A not an amplitude.
Thus
=
∗
∗
4
=
B∗B
1−
=
A∗A
1+
=
1−
1+
⁄
⁄
∗
∗
⁄
⁄
1−
1+
∗
∗
1−
1+
⁄
⁄
⁄
⁄
= 1
The fact that the ratio equals one ( 1) means that any particle which incidents on the potential step with total energy
less than the height of the step has probability one of being reflected. It is always reflected, this is in agreement
with the predictions of classical mechanics.
SIMPLE HARMONIC OSCILATION
The simple harmonic oscillation is of much importance in physics and all fields based on physics. It is a protype of
any system involving oscillation. It is used to study the vibration of atoms in diatomic molecules the acoustic and
thermal properties of solids which arise from atomic vibrations, magnetic properties of solids that involve
vibrations in the orientation of the nuclei.
SHM is used to describe almost any system in which an entity is executing small vibrations about a point of stable
equilibrium. Application in molecular spectroscopy, solid state physics, nuclear structure, quantum field theory,
quantum optics, quantum statistical mechanics and so forth.
A particle of mass in one-dimensional harmonic oscillation is subjected to a potential
=
But
…………………………..(1)
=
in classical mechanics
resistivity force
But force
=
= −
=
=
=
…………………(2)
5
the
= −
∴
=
= −
( )= −
A particle of mass m subject to a linear restoring force
k is a force constant resulting.
The total energy of the Harmonic oscillator is given by the Hamiltonian
=
X and P are operators and are Hermitian
=
2
=
2
+
+
+
1
2
1
2
We now define two non-Hermitian operators
=
ћ
=
( +
)
( −
)
2ћ
Which are known as the annihilation (lowering) and creation (raising) operators respectively
The commutation relation between them is given by
We also define the number operator
[ ,
`∗
]=
,
ћ
2ћ
,
= 1
=
Which is also Hermitian. It can be shown that
=
+
+
+
6
2ћ
[ , ]
And
=
ћ
= ћ
+
ћ
=
1
2
−
+
1
2
1
2
4
Because H is just a linear function of N, N can be diagonalized simultaneously with H. we denote an energy
eigenket of N by its eigenvalue n, so
N|n > =
H|n
>=
|n >
=
but n is nonnegative integer and from (4)
+ 1 2 ћ |n >
+ 12 ћ
Let us show the physical significant of
,
And
=
,
,
=
=
,
and the energy eigenvalues are given by
,
+
,
and N.
,
= 1
= −
As a result we have
=
N
=
ln > =
+ 1
ln > =
− 1
,
ln >
,
+
+
ln >
ln >
ln > …………………..(5)
Thus the relation implies that
ln >
ln > is also an eigenket of N with eigenvalue increased (decreased)
by one. Because the increase (decrease) of n by one amount to the creation (annihilation) of one quantum unit of
7
energy ћw, the term creation operation (annihilation) operator for
ln > =
ln − 1 > where c is a numerical constant to be determined from the fact that both ln > and ln − 1 >
equation (5) it implies that
normalised.
ln − 1 >
is decreased appropriately. From
are the same up to a multiplicative constant.we write
First we note that
<
|
∗
| >= | |
6
We can evaluate the left hand side of equation (6) by noting that
If c is real and positive by convention then we obtain
Similarly;
ln > = √
is just the number operator so
= | |
− 1>
ln > = √ + 1
Suppose we keep on apply the annihilation operator
7
− 1>
to both sides of (7) we have
| > = √ |n-1>
| >=
| >=
− 1 |n-2>
( − 1)( − 2) |n-3>
We have
=<
| | >= <
8
|
∗
.( | >) ≥ 0
Which implies that n can never be negative therefore we conclude that the sequence must terminate with n=0 and
allowed values of n are non-negative. The smallest possible value of n=0 the ground state of the harmonic oscillator
has
= 1 2ћ
We thus now apply the creation
to the ground state |0 > using
|2 > =
3 >=
|1 > =
√2
√3
|0 >
|1 > = [
2 >=
| >=
√ !
(
√2
)
√3!
| > = √ + 1| + 1 > we obtain
]|0 >
|0 >
|0>
We succeeded in constructing simultaneous eigenkets of N and H with eigenvalue
n=0,1,2,3…
9
=
+ 12 ћ ,
Step I potential (Energy Greater than Height)
Example:
An electron in the cathode of photoelectric cell, which has received energy as a result of absorption of photon and
it is trying to escape from the surface of the metallic cathode. If the energy of the electron is not much higher than
the height of the potential step in the potential that it feels at the surface of the metal it may reflect back and would
not be able to escape.
A particle motion under the influence of a step potential is given by
,
satisfies TISE
=
The potential is different in the regions and left and right of the step potential.
ћ
=
ћ
< 0
………………………… (1)
= ( −
)
…………………………(2)
> 0
For free particle of momentum ρ1, the solution of equation (1) is given by
……………………(3)
< 0
=
√2
ћ
=
Eqn.(2) also describes the motion of a free particle of momentum
=
=
(
ћ
+
)
=
ρ
ћ
at
= 0,
+
. the general solution is given by
……………….. (4)
ћ
…………………….(5)
>
We know that there is only transmitted wave that travels in the region
reflection so
=
> 0 and there is nothing to cause
= 0. A, B and C which are constants must be chosen such that
10
and
are continuous
+
Therefore
∴
+
=
…………………………….(6)
The derivatives should also be satisfied for
−
=
at the same
−
=
= 0
=
……………………..(7)
From (6) and (7)
=
and
=
It is not necessary to evaluate A that determines the amplitude of the eigen function. The reflection coefficient is
defined as :
=
∗
=
∗
=
………………………….. (8)
>
< 1 when >
ie whether the total energy is greater than the height of the
potential step and this in contrast to
< 1 when >
From the above equation
We now want to define the transmission coefficient T which specifies the probability that a particle will be
transmitted past the potential step from the region
> 0
The expression for transmission coefficient T is slightly different from R because the velocity in the two regions is
different.
Transmission reflection coefficients are actually defined in terms of the ratio of the probability fluxes. A probability
flux is defined as the probability per second that a particle will be found crossing some reference point traveling in
a particular direction.
Thus
=
=
∗
∗
∗
∗
=
=
∗
∗
=VELOCITY OF PARTICLE IN REGION
V2=VELOCITY OF PARTICLE IN REGION
11
< 0
> 0
=
=
ћ
∴ T is given by
=
and
=
=
=
ћ
>
……………………. (9)
It can be shown from (8) and (9) that:
+
= 1
This is the reason why the reflection and transmission coefficients were defined in terms of the probability fluxes.
Thus when the probability flux incident upon the potential step is split into a transmitted flux and reflected flux, and
0.
their sum equals incident. From equation (10) We can say that when the particles’ flux incidents on the potential
step, it does not vanish or split at A reflection and transmission coefficient for particle incident upon a potential
step.
= 1−
=
1−
1+
12
1−
1−
EXAMPLE:
When a neutron enters a nucleus, it experiences a potential energy which drops at the nuclear surface very rapidly
from a constant internal value of about V=-50MeV.The decrease in potential is what makes it possible for a neutron
to be bound in a nucleus. Consider a neutron incident with incident with external kinetic energy of V=5MeV, which
is typical for a neutron that has just been emitted form a nuclear fusion. Estimate the probability that the neutron
will be reflected at the nuclear surface, thereby failing to enter and have its chance at inducing another fusion.
Soln.
Neutron –nuclear potential to be 1-D step potential.
V0=50MeV
E=55MeV
= 0.29
R=
Particle of energy E,1-D. dimensional rect po base mass m, incident wave of amplitude A=1 and transmitted wave
of amplitude F. for a particle of energy
propagation.
in region 1.
<
,
Region (2).
, the solution function in region (1) and (2) contains right and left
=
=
2
13
+
ћ
=
+
where
=
2
−
ћ
Note that the particle wave function has left & right propagation exponentially decaying solution. This is required if
a tunneling will flow through the barrier.
Because we will be considering a right particle incident from the left in region 3, we needed only to consider a right
propagating wave. Boundary conditions (wave function and its derivation)
A+B= C+D……………….(1)
−
+
=
−
−
=
=
………………………… (2)
………………………. (3)
……………………… (4)
Since tunneling transmission probability is
, we eliminate other coefficients.
Adding (1) and (2) gives
2A=
1−
Adding (3) and (4)
2
(3)- (4)
= 1+
+ 1+
…………………………..(5)
…………………………………… (6)
14
= 1+
2D
………………………………….( 7)
Using eqn. (6) and (7) to eliminate C and D from equation (5) gives;
2 = 1−
1+
=
+ 1+
2
1−
2
……………………. (8)
=
………………………………………. (9)
Dividing through (9) by
=
gives;
|
/
*
/
……………………………. (10)
Magnitude of eqution (10) square
∗
∗
|
=
=
=
⁄
/
……………(11)
⁄
⁄
………….. (12)
`
/
……………………..(13)
Dealing with the last two terms in parenthesis in the denominator, it is easy to check that
15
1−
+ 1+
= − 16
Substituting (14) into (13)
=
=
⁄
⁄
/
⁄
⁄
/
+ 2 1+
……………………….. (15)
⁄
…………………………………… (16)
So the tunneling transmission probability is
=
… …. …….. (14)
…………………………………….(17)
16
Particle in an Infinite Deep Square Well Potential
∞
∞
X=0
For the region 0 <
The solution is
( )=
< , ( )= 0
ѱ
Outside the well
= ∞
ѱ
=
=
= 0
2
ћ
sin
Wavefunction must be continuous, hence if ѱ
ѱ
= 0
ѱ = 0
The condition at
= 0, since
X
X=
= 0 but
∴
00< <
∞ < 0, >
+
1
2
= 0 outside the well, it must be zero at
0 +
= 1
0 = 0
17
= 0
= 0 and
=
and
Equation 1 now becomes
ѱ
Considering the boundary conditions
=
ѱ= 0
0= ѱ
=
= 0
The sine is zero for angles0, , 2 , 3 , … and thus
=
But
=
ћ
=
ћ
=
=
= 1, 2, 3, …
squaring both sides gives
=
8
=
ℎ
ℎ
2
ћ
= 1, 2, 3, …
A particle trapped in a rigid box thus can only have certain quantized energies.
For
= 1
=
=
=
8
ℎ
4ℎ
= 4
8
9ℎ
= 9
8
=
18
∞
∞
= 9
= 4
is the zero-point energy. At absolute zero (0K), quantum mechanics predicts that the particle in a box would not
= ћ =
rest but would have a zero-point energy. Also
energy).
The wave function ѱ =
ћ
, the smaller the with the larger the momentum (and
for each of the quantum states is (since
ѱ =
)
can be determined using normalization condition
1=
ѱ
=
The integral is equal to
19
=
)
1
ѱ
2
2
2
sin
The Figures above show the wave function and probability function for n=1, 2, 3, …,10 for a particle in a box
(Courtesy Giancoli).
Question
An electron is trapped in an infinitely deep square well potential of length
i)
Calculate the three lowest energy levels.
20
1.0
10
.
ii)
If a photon were emitted when an electron jumps from the
wavelength?
2 state to
Solution
i)
1
In the ground state
8
6.63
8 9.11 10
6.03
1.60
2
ii)
150.8
37.7
113.1
3
10
1.0
10
10
10
6.03
37.7
150.8
339.3
1240
Tunneling through a Barrier
21
1240
113.1
9.31
10
1 state, what would be its
In quantum mechanics, a particle such as an electron can penetrate a barrier into region forbidden by classical
mechanics. There are a number of applications that would be discussed later.
Consider a particle of mass and energy traveling to right of the axis in a free space where the potential is
= 0 so the energy is all kinetic ( = ). The particle encounters a narrow potential barrier of height
(in
energy units) is greater than and thickness (distance units).
Since <
, we expect from classical mechanics that the particle could not penetrate the barrier, but would
simply be reflected and would return in the opposite direction. This happens to macroscopic objects. Quantum
mechanics predicts n nonzero probability for finding the particle on the other side of the barrier.
The particle approaches with a sinusoidal wave but within the potential it is an exponential decay. However, the
exponential decay dies away to zero, at the end of the barrier ( = ) and > there is again a sinusoidal wave
function.
But it is a sine wave of greatly reduced amplitude. Since |ѱ| is nonzero beyond the barrier that is a nonzero
probability that the particle penetrates the barrier. The process is called tunneling through a barrier or barrier
penetration.
We describe the tunneling probability with transmission coefficient and reflection coefficient . If = 0.04 then
= 0.96. Note that + = 1 since an incident particle must either reflect or tunnel through. The transmission
coefficient can be defined as
=
≪ 1 and
For
=
Question
A 50
i)
ii)
electron approaches a square barrier 70
2 (
ћ
−
)
high and
1.0
thick
0.10
thick
What is the probability that the electron will tunnel through?
Solution
i)
22
−
iii)
For
= 0.10
= (70 − 50) × (1.6 × 10
= 3.2 × 10
2(9.1 × 10
2
= 2
,2
= 4.6
=
=
) × (3.2 × 10
1.055 × 10
=
) × (1.0 × 10 )
= 1 × 10
.
)
= 46
0.010 %
= 1.0 % chance of penetration
Examples of application of tunneling
i)
Tunnel diode
ii)
Scanning tunneling microscope
iii)
Radioactive decay
iv)
Atomic force microscope
Angular momentum
An electron of mass m in a circular Bohr orbit of radius r with a velocity v. The orbital angular momentum
eelectron of linear velocity v.
of an
See diagram on the board (Already drawn)
m= mass of electron
e= charge on electron
Orbital magnetic momemt
Dipole moment of electron of mass m, and charge (-e) in a circular pole of radius r, orbiting with a velocity v.
23
The orbital angular momentum
=
of the electron with linear velocity V is given by;
r= radius of orbit, p=linear momentum
=
(1)
Charge circulating current in the loop generates
=
(2)
But
∴ =
=
(3)
(4)
T= period of charge circulating a loop of magnetic field which is the same at large distance from the loop as at that
of magnetic dipole located at centre of the loop and oriented far to the plane.
The orbital magnetic dipole μ of the system is given by;
μ =
…………………………… (5)
μ =
………………………… (6)
μ =
∗
2
Dividing through equation (6) by equation (1)
=
=
This ratio is a combination of universal constants
=
ћ
Where μ =
= 0.929× 10
A = Area =
ћ
24
2
= 1
And μ
μ =
But L=
number.
and it is called orbital g-factor or gyromagnetic ratio
=
Bohr magnetic
ћ
( + 1) ћ
μ
and L are anti parallel
= the orbital angular momentum quantum number or the orbital quantum
∴μ =
=
μ =
=
μ
ћ
−
ћ
μ
=
μ
μ
ћ
=
+ 1
( + 1)ћ
μ
ћ
ћ
magnetic dipole moment is placed in an applied magnetic field B, it experiences a torque and associated
with this torque is a potential energy of orientation
= μ
∆ = −μ ∗
25
Stern-Gerlach Experiment
A beam of vaporized silver atoms is collimated and passed through an inhomogeneous magnetic field of magnetic
gradient
in the z direction. Before entering the field the beam of silver atoms were randomly oriented. After
entering the field they were split into two, the spin up were parallel to the field and the spin down anti-parallel. The
experiment was repeated for other atoms and two traces were always obtained.
S
Z
oven
N
Collimator
Magnet
Detector/Screen
The diagram showing the Stern-Gerlach Experiment
The results could only be explained by postulating that an electron has an intrinsic (built-in) magnetic dipole
moment μ and posses an intrinsic angular momentum S called spin. Spin is purely quantum mechanical concept
and has classical analog.
=
=
number
ћ
+ 1 ћ and Z- component
are not quantized s is spin quantum number and
μ = −
26
μ
ћ
is spin magnetic quantum
μ
= −
μ
=
spin g factor
= 2
is called spin or spin quantum number and has a value of
number
= ±12
=
and
is called spin magnetic quantum
= + , which is spin up parallel
= − , which spin down or anti-parallel
The transverse force FZ experience the atoms in z direction is given by;
F=μ
∴
=
…………(1) But μ =
μ
ћ
SZ=msћ and
∴
= ±1
= μ
The experiment of Stern & Gerlach which discovered the existence of spin also found space quantization i.e. the
orientation of atoms in space is quantized and only a set of allowed values are obtained.
Quantum particles may possess an orbital angular momentum and an intrinsic angular momentum called spin. The
orbital angular momentum resembles the classical particle. It is a vector with both magnitude and direction but spin
angular momentum does not have a classical manifestation.
Angular momentum in quantum in quantum physics must be specified using quantum numbers. Normally l and m l
are used to describe orbital angular momentum and s and ms are used for spin angular momentum. Quantum
numbers j and mj are used when quantum numbers arises from a combination of spin and orbital angular moment
and when general angular moment is being described. The only possible precise values for the magnitude of orbital
angular momentum are given by;
L=
( + 1)ћ
where l=0,1,2,3,4……
When we choose the orbital angular momentum in the z direction, there are 2l+1 possible outcome given by;
=
27
ћ
Total Angular Momentum
An can have both orbital and spin angular momenta. For example a 2p state of hydrogen = 1 and
state = 2 and
spin :
= . The total angular momentum is the vector sum of the orbital angular momentum
The magnitude of the total angular momentum
=
= . In the 4d
=
and the
+
is quantized
( + 1) ћ
For the single in the hydrogen atom quantum mechanics give the result that can be
Or
=
+
=
+
1
2
=
−
=
−
1
2
But can never be less than zero, just as and . For 1 state, = 0 and =
2 state, = 1
can be either
. The component for is quantized in the usual way:
= , − 1, … , −
Question
What are the possible values for in the 3 state of a hydrogen?
Solution
or
=
+
=
+
For a
=
−
=
−
state = 2
is only possibility. For
28
1
2
1
2
states, say
Therefore
=
+
=
= 2+
+
= 2−
5 3
,
2 2
=
SPECTROSCOPIC NOTATION
1 5
=
2 2
1 3
=
2 2
The specific state of an atom including the total angular momentum quantum number , using the following
spectroscopic notation
Where
is the orbital quantum number
=
Example: 2
⁄
has
= 2, = 1, =
0 1 2 3 4
=
ORBITAL INTERACTION ENERGY
When a magnetic dipole μ is placed in a magnetic field exerts a torque
energy associated with this interaction is given by
If the
μ = −
= μ.
is directed along the z axis
,
=
ћ;
= 0, ± 1, ± 2, … , ±
The interaction energy is given by
μ =
=
2
2
ћ
ћ
29
= μ×
on the dipole. The potential
=
μ
The effect of the magnetic field is to shift the energy of each orbital state by U. The interaction energy depends on
the value of
because
determines the orientation of the orbital magnetic moment relative to the magnetic field.
Without a magnetic field these states have the same energy; that is they are degenerate. The magnetic field removes
the degeneracy. In the presence of magnetic field, they are split into 2 + 1 distinct energy levels and adjacent
energy levels differ in energy by
For a hydrogen
n= 4
n= 3
n= 2
ћ
= μ
.
=3
=2
=1
=0
n= 1
No magnetic field
With magnetic field
Zeeman Effect
It is the splitting of atomic energy levels and the associated spectra when atoms are placed in a magnetic field. This
confirms space quantization
No magnetic field
With a magnetic field
30
One application of Zeeman Effect is splitting of energy levels by a magnetic field which is the basis of magnetic
resonance technique.
2
3d
0
22
Ei -Ef
2p
Selection Rules
∆
∆ = ±1
= 0, ± 1
Spin Orbit interaction
The interaction between electrons spin magnetic dipole moment and the internal magnetic field of the orbiting
electrons in the atom. The internal magnetic field is related to the electrons’ orbital angular momentum. It is a weak
interaction and it is responsible in part for the fine structure of the excited state of one-electron atom.
An analogous effect of spinning orbit interaction occurs in nuclei called nuclear spin-orbit interaction but in such
atoms its reasonable because the internal magnetic fields are strong nucleus of the atom has a magnetic dipole
31
moment that interacts with the orbital and /or spin magnetic dipole moment of electrons. These effects are called
hyperfine structure radio astronomers use this wavelength to map clouds in interstellar hydrogen gas that are too
cold to emit visible.
Larmor Precession
Magnetic moments are not statically aligned exactly parallel or antiparallels to an external magnetic field. They are
forced to align at angle to the magnetic field B and this causes them to wobble around the axis of the field at a fixed
freq.
Example of this precession “wobble” is a spinning top. It is the spin angular momentum of the top that prevents it
from falling over and causing it wobble in addition to spinning.
[adakato]
The periodic wobbling mention that the top assumes in a gravitational field is called precession. The Earth also
precesses on its axis but much slower.
In an exactly analogous way, magnetic moment vector of atoms in a magnetic field precesses with a characteristic
frequency called the Larmor frequency
.
ћ
=
ћ
=
μ
ћ
Questions
1. Consider an electron in hydrogen having total energy -0.5440 eV.
A) What are the possible values of its orbital angular momentum (in terms of ћ)?
B) What wavelength of light would it take to excite this electron to the next higher
shell? Is this photon visible to humans?
2. A quantum state is four fold degenerate. If = 2 and|
> is used. If is the
principal quantum number, orbital quantum number and magnetic
quantum number. Write down the degenerate states.
32
3. How many different states are possible for an electron whose principal
quantum number is = 3.
Solution
n
l
ml
ms
N
l
ml
ms
3
2
2
1/2
3
1
1
1/2
3
2
2
-1/2
3
1
1
-1/2
3
2
1
1/2
3
1
0
1/2
3
2
1
-1/2
3
1
0
-1/2
3
2
0
1/2
3
1
-1
1/2
3
2
0
-1/2
3
1
-1
-1/2
3
2
-1
1/2
3
0
0
1/2
3
2
-1
-1/2
3
0
0
-1/2
3
2
-2
1/2
3
2
2
1/2
Today, we consider the picture of a spinning electron as not legitimate. We cannot view an electron as localized
object, much less a spinning one. What is important is that the electron can have two different states due to some
33
intrinsic property that behaves like angular momentum called spin. Electron has spin quantum number
produces angular momentum
=
√
( + 1)ћ=
The z component
Example;
=
ћ
ћ and
=
,−
=
which
i.e. spin up and spin down respectively.
A boson W is spin-one particle with s=1 and ms=+1,0,-1
Electron is spin-half (fermion) with s=1/2 and ms=± 1/2
Orbital Interaction Energy
When a magnetic dipole μ is placed in a magnetic field
= μ
u= μ.
If the
on the dipole. The potential Energy associated with this interaction is given
is directed along the z-axis
μ =
=
Interacting Energy
μ = −μ
, the field exerts a torque
= ћ2
=
ћ
ћ
μ =
−
2
= −
2
ћ
= 0, ± 1, + 2
(orbital magnetic interacting energy)
(Bohr magneton)
34
±
μ=
μ
(orbital magnetic interaction)
the effect of the magnetic field is to shift the energy of each orbital state by an
amount U. The interacting energy u depends on the value of the orbital magnetic
moment relative to the magnetic field
is the magnetic quantum number.
= 2 + 1
=
Without a magnetic field these states have the same energy; that is they are
degenerated. The magnetic field removes the degeneracy. In the presence of
magnetic field they are split into 2 + 1 distinct energy levels; adjacent energy levels
differ in energy by
Show that
,
ћ
2
= 0 light.
= μ
.
A particle has orbital angular momentum given by the quantum number = 3 and
spin angular momentum given by the quantum number = 1.
a) How many distinct states are there with different values for z components of
the orbital and spin angular moments?
b) What are the possible values for the quantum number j that describes the total
angular momentum of the particle?
c) How many distinct states are there with different values for the magnitude and
z component of total angular momentum?
A chemical electron moves in a circle of radius 1mm with velocity 1ms-1.
What is the value of the quantum number which gives a quantized angular
moment close to the angular momentum of the classical electron?
35
Angular Momentum and Commutations and Eigenvalues
As has already been mentioned, in classical mechanics angular momentum is given
by = × . But in quantum mechanics, L, r and p are operators which have the
Cartesian coordinates as show below
= (
,
= (
)
,
= ( , , )
Thus
=
=
Also
,
,
=
)
−
= − ћ(
−
)
−
= − ћ(
−
)
=
+
−
= − ћ(
+
−
)
In the Cartesian coordinates, the commutation relations between
[
[
,
,
]= ћ
]= ћ
36
( =
, , ) are
[ ,
It can be shown that
]= ћ
[ ,
Which implies that
[ ,
]= 0
]= [ ,
]= [ ,
]= 0
Lowering and Raising Operators
The raising operator is defined as
=
+
=
−
The lowering operator
and
∴
are not Hermitian.
+
2
=
=
Show that
[ ,
[
,
,
±
±]
= 0
= ±
] = 2ћ
ALGEBRA OF ANGULAR MOMENTUM
37
±
−
2
The operators
operators, that is
describe physical quantities hence they must be Hermitian
(
) =
⟹( ) =
We now find the simultaneous eigen functions of both
comprise a complete orthogonal basis:
|
>=
|
>=
+ 1 ћ |
ћ|
When the lowering and raising operators act on |
=
|
=
>=
Differential Representations
> ) which
> , they give the following
+ 1 ћ ,
+ 1>
− 1 ћ ,
− 1>
)( +
+ 1)ћ| ,
+ 1>
( +
)( −
+ 1)ћ| ,
− 1>
+ 1 −
(|
>
( −
>=
> is an eigenvector of
+ 1 −
>
and
with eigenvalue ( + 1)
The representation of eigenvectors and eigenvalues is often more convenient using
spherical coordinates
=
=
∅
∅
=
The operators representing angular momentum in spherical coordinates are
38
= ћ(
∅
= ћ(−
= −ћ (
= ћ
= ћ
∅
+
∅
∅
1
(
+
(−
+
∅
∅
+
= − ћ
∅
∅
+
+
)
∅
)
1
∅
∅
∅
)
The eigenvectors of and are functions that depend on angles
hence we can represent the wave function as
ѱ , ,∅ =
For a central potential ( ) =
where | > =
( , ∅)
( )
( ). We find hat
Commutation Relation (Continued)
,
,
= iћJ
= ћ
J ,J
,
= 0
= ћ
39
( , ∅)
∅ only
( , ∅) are spherical harmonics
,
= iћJ
,
= 2ћ
+
,
−
| , ⟩=
| , ⟩=
|,
−
⟩=
Show that
,
+
,
− ћ
,
ћ
,
,
+
+
±
+ 1⟩
− 1⟩
+ 1 ћ |, ⟩
Example
+
+ 1 ћ| ,
−
ћ| , ⟩
|, ⟩=
= ±ћ
+ 1 ћ| ,
+
+
|, ⟩=
±
= − ћJ
+ 1 ћ |, ⟩
ћ| , ⟩
⟩=
|,
,
J ,J
,
+
= 0
,
,
,
+
,
+ − ћ
+ ћ
,
= 0
40
,
= yp − Zp
L ,L
= iћL
= xp − yp
L ,L
= iћL
S ,S
= iћS
= Zp − Xp
L ,L
= − iћ y
−
= − iћ x
− y
= − iћ Z
,
− x
= 2ћJ
J = J − J + ћJ
J = J − J
− ћJ
= iћL
S ,S
= iћS
S ,S
= iћS
L = L
J = J
S = S
+ L
+ J
+ S
= J + iJ
=
−
=
−
=
+
=
=
+
41
−
+ L
+ J
+ S
,
=
,
+
,
−
,
=
=
=
Example;
−
,
,
,
+
+
−
,
,
,
Show that:
,
,
= ћ
= +ћ
,
= 0
Spherical Harmonics
The spherical polar coordinates (r, , ∅) of point P.
42
=
sin cos ∅
=
cos
=
sin sin ∅
We have considered wave functions in Cartesian coordinates x,y and z. it is more
useful to consider wave function in spherical polar coordinates r, and ∅.
A quantum state represented by a wave function ψ(r, , ∅), the dependence on
∅
Specifies an angular shape that determines the orbital angular momentum properties
of the state. All possible angular momentum properties can be described using the
simultaneous eigen functions are called spherical harmonics and denoted by
,
.
=
, ∅ and they satisfy eigen value equations.
+ 1 ћ
.
and
.
=
Where the quantum numbers and
= − …,+
ћ
.
can take values of = 0,1,2, … and
These eigen functions are orthogonal because they satisfy
If
≠
,
= 0
= and they are usually normalized so that
is the solid angle
,
|
= sin
43
= 1
∅
The orbital angular momentum L and the orbital magnetic dipole moment μ of an
electron –e moving in a Bohr orbit. The magnetic field β produced by the circulating
charge is indicated by the curved lines. The fictitious magnetic dipole that would
produce an identical field from the loop is indicated by its poles N, S.
, ,
=
If we express them in spherical coordinate the difficulty is removed.
=
…………………… (1)
=
We express the TISE
ћ
μ , ,
+
, ,
=
And the application operator is given by;
=
+
, ,
………………………… (2)
……………………….(3)
+
We assume a solution to the TISE
, ,
=
-
)∅(
There are three solutions
functions
…………………… (4)
, ,
to the equation (2) that split into parts of the
44
R(r), ( ) and ∅
We substitute equation (4) and (3) into (2)
ћ
∅
∅
+
∅
+
When we carry out particle partial differentiation
ћ
∅
∅
∅
+
Multiplying through by 1
∅
∅
∅
= −
=
∅ћ
−
+
∅
+
− ∅=
+
−
−
( )
− ∅
∅=
2μ
ћ
∅
−
( )
The left side of the equation does not depend on the on R and and the right hand
side does not also depend on and their common value does not depend on any
variable and therefore must be a constant.
∴
And
−
1
1
∅ ……………………….. (5)
−
+
2μ
ћ
1
−
−
= −
2μ
ћ
−
−
1
= −
Left does not depend on the right side variables therefore both must be equal to
( + 1)
45
−
+
ћ
+
−
=
+ 1
………………………….. (6)
……………………… (7)
= ( + 1)
Solution of equation (5) is given by;
∅
1=Cos
|
and ψ=0 and ψ=2
=
∴∅ 0 = ∅ 2
2 +
| = 0,1,2,3 …
=
2
+ve or –ve integer
∅
=
In solving for the function H( ) we obtain acceptable solutions only if is equal to one
of the integers
= |
|, |
| + 1, |
| + 2, |
The acceptable solutions can be written as
And
and |
|
|
|
cos
= sin|
−
|
|
|
| + 3…
cos
are polynomials in cos that depends on the quantum number
cos
= ℎ
R(r) has also accepted solution only if the total energy E has one of the values of
=
4
−μ
2ћ
46
=
− 13.6
N=quantum number
n= + 1, + 2, + 3
The acceptable solution must be conveniently be written as
( )
And
ћ
=
EXAMPLES OF Rne(r)
ℎ
1
0
2
2
2
=
0
0
1
1
=
≅ 0.51
0
=
0
±1
=
±
=
1
=
4√2
1
1
√
2−
4√2
1
8√
±
n specifies the total energy of the atom hence the name principal quantum number.
Azimuthal Quantum number or orbital angular momentum depends
is also called azimuthal quantum number.
Because of the split in energy level it is called magnetic qu..
There are also situations with completely different eigen functions corresponding to
exactly the same total eigen value En. As the eigen function describes the behviour of
47
the atom which have states with completely different behavior but however, the
same total energy. This phenomenon is called degeneracy.
IDENTICAL PARTICLES
In classical mechanics, we can keep track of two particles 1 and 2 anywhere these
particles go. We can even colour the as blue and red. We can examine how the red
particle moves as time passes.
In QM however, identical particles are truly indistinguishable. Because we can not
specify them by colour ie red and blue. We cannot also trace their trajectory because
that requires the position measurement at an instant of
time
. The time independent schrodinger equation of our
system of two nonintrating particles in three dimensions can be written as;
ћ
=
+
= ℎ
+
ℎ
= ℎ
…
= ℎ
−
ћ
+
+
1
2
ℎ
48
+
=
……….. (1)
…
= ℎ
…
=
ℎ
ET= the total energy for the total system since there is no interaction between the
two particles, they are more independent.
VT
,
,
+
,
,
The total eigen function can be written as;
…
=
,
Where
, ,
and
schrodinger equations.
,
,
,
,
,
satisfy identical one-particle time independent
We have the quantum numbers specifying the three space coordinates. There is also
one or more quantum numbers to specify the orientation of the spin of the particle.
We use α, β and to specify quantum numbers ie space and spin of a particle.
,
,
,
=
,
=
(1)
(2)
Eigen function particle 1 is state α and 2 in state β
Total parts
…
=
1
(2) …………….. (2)
…
=
1
(2) ………………… (3)
Total equation eigen function to specify particle 1 in state β and 2 in state α.
We consider measurable quantity, the probability density
For equation (2)
∗
=
∗
1
∗
2
(1)
(2) ……………………… (4)
49
And for equation (3) we have
∗
∗
=
∗
1
2
(1)
(2) ……………………….. (5)
The particles are identical and indistinguishable so we should be able to exchange
their labeling without changing a measurable quantity such as probability density.
1→ 2
∗
=
2→ 1
∗
1
∗
2
(1)
(2) →→
∗
=
∗
1
∗
2
(1)
(2)
Left changes to right, 1 changes to 2, 2 changes to1.
The relabelled probability density function is not equal to the original one. It must be
emphasized that these are not acceptable eigenfunctions for the accurate description
of a system of identical particles.
Eigenfunction which satisfy the TISE
=
=
√
√
(1)
1
(2) +
2 −
(1)
(1)
(2) …………………….. (5)
(2) ……………………… (6)
EQN. (5) Symmetric total eigenfunction
(6) antisymmetric total eigenfunction
Tutorials
a) What is the lowest possible energy ( in eV) of an electron in hydrogen if its
orbital angular momentum is √12ћ
b) What are the largest and smallest values of the Z-component of the orbital
angular momentum (in terms of ћ) for the electron in part (a)?
50
c) What are the smallest largest values of the spin angular momentum (in terms of
ћ) for the electron in part (a)?
d) What are the smallest largest values of the orbital angular momentum (in terms
of ћ) for an electron in the M shell of hydrogen?
Consider an electron in hydrogen having total energy of 0.5440 eV.
a) what are the possible values of its orbital angular momentum (in terms of
ћ)?
b) What wavelength of light would it take to excite this electron to the next
higher shell? Is this photon visible to humans?
Thus the total energy of the system containing a particle in quantum state
α and another in quantum state β will not depend on which particle is in
which state if the particles are identical.
Therefore both
=
(1) (2) and
=
(1) (2) are solutions to
TISE equation (1). Corresponding to the same value of the total energy E I.
Because the quantum is linear in
, it follows immediately that the
linear combinations
of the two forms of
are also solutions.
They correspond to the same value of EP, therefore they are degenerate
solutions that is
are different eigenfunctions corresponding to
precisely the same eigen value. The phenomenon is called exchange
degeneracy since the difference between the degenerate eigen functions
has to do with exchange of the particle labels.
We now set out to show that the probability density function is
unchanged. By changing the particles label we have;
√
=
√
(1)
(2) +
(1)
(2) →→→
=
√
1
2 −
(1)
(2) →→→
(2)
(1) +
(2)
(1) =
51
……………………… (7)
√
1
2 −
(1)
(2) = −
………………… (8)
From the above equation (7) and (8) we see that the symmetric total
equation function
is unchanged by the exchange of particles labels and
that the antisymmetric total eigen function
is a multiplied by minus (-)
by an exchange the particle labels. The properties give rise to their names.
We now show the probability densities;
∗
→→ ∗
∗
∗
∗
→→ (− 1)
−
Electrion, protons, neutrons have spin 521 2, the deuterium nucleus 2H
has spin=1 and Helium nucleus 4He has spin s=0 he nce for both
symmetric and antisymmetric total eigenfunctions, the probability density
functions are not changed by the exchange of the particle labels.
We can also use derterminants to find the antisymmetric eigen functions.
Eg.
1
=
1
2 −
1
2
√2
So the later determinants can be used as follows;
1
2
=
√
1
2
Where 2!= 2x1=2
For three particles we have
1
2
3
1
1
2
1
=
√3!
1
2
3
Where 3!= 3x2x1 = 6
When we expand the determinant is given by;
52
=
1
√3!
−
1
1
2
2
3 +
3 −
1
1
2
2
3 +
3 −
1
1
2
2
3
3
It can be written for symmetric wave function. We use the linear combination.
THE EXCLUSIVE PRINCIPLE.
In a multielectron atom there can never be more than one electron in the same
quantum state. The antisymmetric total eigenfunction which is given by;
1
=
1
2 −
(1) (2) ≡ 0
√2
When both particles are in the same space and spin quantum state.
Hence if two particles are described by the antisymmetric total eigenfunction, they
cannot both be a state with the same space and spin quantum numbers.
Bosons & Fermions
Bosons are particles with inter spin and system of identical bosons must have
quantum states that are symmetric when the two particles are exchanged ie must
have total symmetric eigenfunction.
Eg α-particle, 4He, photon, deuterium, phonons(spin as)
Fermions on the other hand are particles with half-integer spin and systems of
identical fermions must have quantum states which are antisymmetric (ie total
antisymmetric eigenfunctions) when two particles are exchanged.
Because electrons are spin half particles, they are indistinguishable in the fermion
way. They never have symmetric states but only antisymmetric states eg. Photons,
neutrons, electrons, 3H.
The pauli exclusive principle plays a governing role in determining the physical and
chemical properties of atoms and leads us to an understanding of the periodic table
53
of the elements. It has a crucial role in determining the electrical and thermal
properties of electrons in solids.
The bosm way of being indistinguishable also leads to important phenomena.
Because bosons are described by symmetric quantum states, many bosons may
occupy the same single-particle state and when this happens quantum-mechanical
behavior of macroscopic scale arises. The most important example of bosons
togetherness is the coherence arise because photons are bosons with spin one,
have probabilities to have the same energy an momentum in which the same way as
two particles with symmetric wave function have a high probability of being at the
same location. Bosom togetherness is also responsible for the superfluidity of liquid
helium consists of weakly interacting helium at temperature below 2.2K . liquid
helium consists of weakly interacting helium atoms which have like bosons because
they consists of 4He nucleus spin zero and two electrons with a combine spin of zero.
At low temperatures a considerable fraction of atoms in liquid helium ‘condense’
into the lowest energy state. They form Base-Einstein condensate in which are
coherent with each other and move collectively without friction.
Recently almost pure Bose-Einstein condensate have been produced by cooling
atoms in magnetic traps; indeed the 2001 Noble Prize in physics awarded Eric
Cornell, Wolfgang Ketterle and Carl Wiemann for their work in producing the first
pure Bose-Einstein condensates in 1995.
Surprisingly, boson-like togetherness also occurs in situations where fermion-like
behavior is expected. It occurs in the super conductivity of metals at low
temperatures because pairs of electrons act alike indistinguishable bosons. It occurs
in the superconductivity of metals at low temperatures because pairs of electrons
act like indistinguishable bosons. It also probably occurs when liquid helium-3
becomes a superfluid at very low temperatures. Helium-3 atoms become a
superfluid at very low temperatures. Helium -3 atoms, unlike the normal helium
atoms are fermions because 3He has spin half, but pairs of helium-3 atoms can act
54
like a system of indistinguishable bosons and give arise to collective motion with no
friction in liquid helium-3.
Quantum statistics
Bosons obey Bose-Einstein statistics and the probability of occupation f(E) of energy
levels (E) at absolute temperature is given by;
1
=
/
− 1
Where = ℎ
K= Boltzman constant
Fermions on the other hand obey Fermi-Dirac statistics and the probability of
occupation an energy level given by;
1
=
/
+ 1
For an electron if =
=
=
=
1
∴
=
/
+ 1
We now sketch various graphs when T increases
55
At increases, more and more of the electrons are excited to states with >
.
At absolute zero, all the states are occupied (occupation probability) at energy up to
EF0.
If an atom (not necessary in its ground state) is placed in an external field, the energy
levels shift and the wave functions are distorted. This is called the stark effect.
Power series
When C=0
[J2,Jx]=0
=
−
|,
⟩=
=
|,
=
+
−
=
+ 1 ћ |, ⟩
⟩= ћ
+
+ 1 −
56
+
−
+
+
+ 1 |,
+
. ..
+ 1⟩
−
. ..
|,
⟩= ћ
+ 1 −
+ 1 |,
− 1⟩
Starle effect is the splitting of the energy levels of the hydrogen atom (or
other/atoms) resulting from the application of an electric field. If an electric field is
applied, its effect is viewed as perturbutattion and is evaluated within the framework
of time-independent perturbutation theory for degenerate levels,
TIME INDEPENDENCE PERTURBUTATION THEORY
We assume a Hamilton H0 which is time independent and has known energy levels
arising from TISE.
|
⟩=
> ………………………………… (1)
Superscript (0) shows that the quantities are associated with the unperturbed
system. We now introduce a perturbation V which is a Hamiltonian of a weak physical
disturbance like external electric field V is Hermitian. We also introduce dimensioners
parameter ranging from 0 (no perturbation) to 1(full perturbation).
The perturbed Hamiltonian are given by
=
+
………………….. (3)
+
| ⟩=
The energy levels and eigenstates of the perturbed Hamiltonian are given by;
| ⟩ ………………… (4)
If the perturbation is sufficiently weak, we can write En and | ⟩ as a power series.
57
=
+
( )⟩
| ⟩= |
+
+ …. ……………………. (5)
+
( )⟩
⟩ + … ……………………. (6)
+
Substituting (5) and (6) into (4) we have;
|
+
( )⟩
⟩+
+ ⋯
( )
⟩+ ⋯ =
( )
+
+
+ ⋯
|
( )⟩
+
Expanding and putting λ=7
The first order equation is given by;
|
⟩+
|
⟩=
Multiplying through by ⟨
= ⟨
( )
| |
( )⟩
|
( )
⟩+
|
( )
⟨
|
|
( )⟩
⟩ …………………….. (7)
( )
= 0, ⟨
……………………………………… (8)
( )⟩
|
= 1
We use the resolution identity
= ∑
⟩= ∑
⟩⟨
|
⟩⟨
⟩+
|
⟩+ |
( )⟩
( ) ⟩⟨ ( )
|
−
|
( )⟩
( )⟩
…………………… (9)
Where | ( ) ⟩ is the othogonal complement of the |
(9) we have;
( )
|
=
( )⟩
|
⟩⟨
= ⟨
( )
and from equations (7) and
( )
|
( )⟩
Suppose the zeroth-order energy level is not degenerate and we multiply through by
⟨ ( ) | we have;
( )
−
⟨
( )
( )⟩
58
|
( )⟩
( )⟩
|
⟨
=
( )
( )
( )⟩
| |
( )
−
|
( )⟩
|n(1) is proportional to ⟨ ( ) | | ( ) ⟩ and inversely proportional to the energy
difference between eigen states k and n. therefore, perturbation deform the eigen
state to greater extent if there are more eigen states at nearby energies.
Second order
( )
=
| ⟩= |
( )⟩
+
( ) ⟩⟨ ( )
( )
−
+ ⟨
|
| |
( )⟩
( )
( )⟩
+
( )
+
( )
|
59
( )⟩
( )
|
−
⟨
( )
( )|
( )
|
−
|
( ) ⟩⟨ ( )
( )
| |
( )
−
( )
( )