A brief introduction to quantum mechanics
M. Saffman
Department of Physics, University of Wisconsin-Madison,
1150 University Avenue, Madison, Wisconsin 53706
(Dated: May 6, 2015)
Abstract
Some notes on the basic formalism of quantum mechanics.
1
CONTENTS
I. Wave function and Schrödinger equation
4
A. Position measurement
4
B. deBroglie waves
5
C. More on free particle deBroglie waves
7
D. Schrödinger equation
8
1. Conservation of probability
9
II. States and operators
10
A. Observables
12
B. Degeneracy
14
III. Continuous Basis
15
A. Representation of Derivatives
16
B. Position and momentum representations
17
IV. Commuting observables
18
V. Evolution in time
19
A. Schrödinger and Heisenberg pictures
21
B. Quantum - Classical correspondence and Ehrenfest’s theorem
22
VI. Measurements
23
VII. Principles of quantum mechanics
25
VIII. Density matrix theory
26
A. Composite systems
29
IX. Time evolution of open quantum systems
X. One-dimensional Harmonic Oscillator
31
37
A. One-dimensional Quartic Oscillator
XI. Two dimensional harmonic oscillator
37
39
2
A. Two dimensional quartic oscillator
41
References
44
3
I.
WAVE FUNCTION AND SCHRÖDINGER EQUATION
Associating a definite trajectory (x(t), p(t)) with a particle is not consistent with experimental data from, e.g. the two-slit experiment. In quantum mechanics we instead describe
the state of a particle in terms of a wavefunction ψ(r, t). The interpretation is then that
P (r) = |ψ(r, t)|2
is the probability to find the particle at position r at time t and
d3 P (r) = |ψ(r, t)|2d3 r
is the probability to find the particle in an infinitesimal volume d3~r centered at ~r at time t.
Note the probability amplitude is complex: ψ(r, t) = ψr (r, t) + iψi (r, t) and
|ψ(r, t)|2 = ψ(r, t)ψ ∗(r, t) = [ψr (r, t) + iψi(r, t)][ψr(r, t) − iψi (r, t)] = ψr (r, t)2 + ψi (r, t)2.
A quantum state is completely determined1 by the wavefunction ψ(r, t). The wavefunction
is arbitrary up to a global phase. Making the change ψ → eıαψ does not change the
probability distribution:
P 0 (r) = |eıαψ|2 = |ψ|2 = P (r),
since |eıα|2 = 1 for any real α.
To calculate probabilities of events when there are multiple pathways we add the amplitudes and then square. So if ψ = ψ1 + ψ2 + ... the probability of an event is
P = |ψ|2 = |ψ1 + ψ2 + ..|2
and interference occurs due to the cross terms ψ1 ψ2∗ + ψ1∗ψ2 , etc. This is the wave-like
interference of particles travelling through slit one or two in the two-slit experiment.
A.
Position measurement
Let’s assume some classical apparatus measures a particle at time t to be in some region
δ 3r around r. The precision of the apparatus is δ 3 r. If we prepare N particles all in the same
state ψ the distribution of ri , i = 1...N will be given by
d3 P (r) = |ψ(r, t)|2d3 r
1
When there are additional degrees of freedom such as spin we need additional quantum numbers to fully
describe the state.
4
and the expected value of the measurement of r will be
Z
Z
Z
3
3
hri = d P (r) r = d r P (r)r = d3 r |ψ(r)|2r.
The variance or mean square deviation of the measurement is
h(r − hri)2 i = hr · r − 2r · hri + hri · hrii
= hr2 i − 2hri · hri + hri · hri
= hr2 i − hri2
≡ (∆r)2.
We will consistently use the notation (∆f)2 to denote the variance of a quantity f. The
standard deviation is then given by
∆f =
p
(∆f)2 .
Thus if we are interested in the variance of a measurement of say the x coordinate of a
particle, it is given by
2
2
2
(∆x) = hx i − hxi =
Z
3
2 2
d r |ψ(r)| x −
Z
3
2
d r |ψ(r)| x
2
.
Note that there are two different integrals here: one has x2 as part of the integrand and the
other one has only x.
B.
deBroglie waves
Photons have energy (Einstein, 1905) E = hν = h̄ω, with h̄ = h/2π,
ω = 2πν and
momentum (Einstein, 1917) p = h/λ = h̄k, with k = 2π/λ. The phase velocity is given by
vφ = λν = ω/k.
Photons have vφ = c the speed of light.
deBroglie suggested that p = h/λ = h̄k, or in vectorial form p = h̄k, is also true for
massive particles. Consider a particle of mass m and velocity v, so classically p = mv.
Combining these relations we see that
p = h̄k and p = mv
5
imply
k=
mv
h̄
for the effective wavevector of a particle. The deBroglie wavelength is then
λdB =
2π
2πh̄
h
=
=
.
|k|
m|v| mv
We will describe the particle by a plane wave ψ(r, t) = ψ0eı(k·r−ωt). The particle has
a definite momentum given by p = h̄k but is completely delocalized in space, since ψ is
everywhere nonzero and |ψ|2 = 1 and is constant. This is consistent with the Heisenberg
uncertainty principle (∆p = 0 so we must have ∆x = ∞).
For a free particle the energy is E = p2 /2m so ω = E/h̄ = p2 /2mh̄ and the deBroglie
wave is
ψ = ψ0 eı(k·r−ωt) = ψ0e
ı
h̄
„
2
p
p·r− 2m
t
«
.
Note that the phase velocity is
vφ =
ω
p2 /2mh̄
p
p
=
=
6= vclassical = .
k
p/h̄
2m
m
The phase velocity is not equal to the classical velocity. However the group velocity is
vg ≡
dω
d p2
p
=
=
= vclassical.
dk
dp 2m
m
As we will see later the group velocity gives the rate of transport of the probability density
which corresponds to the classical velocity.
We could use these waves to describe the results of a two-slit experiment with particles.
If ψ1 is the solution with only slit 1 open, and ψ2 is the solution with only slit 2 open then
ψc = ψ1 + ψ2 is the solution with both slits open and
|ψc(r)|2 = |ψ1 |2 + |ψ2|2 + (ψ1ψ2∗ + ψ1∗ψ2)
gives the probability distribution on the output screen . We can interpret this as the first
term being “particle like” and the second term which shows interference being “wave like”.
You might be wondering why we have to use a complex ψ instead of writing for the free
particle
ψ = A sin(kz − ωt)
6
x
V2
V1
p2
θ2
θ1
z
p1
FIG. 1. Particle with momentum p1 is incident from the left at angle θ1 . The potential step at
z = 0 results in transmission of a particle with momentum p2 at angle θ2 .
which describes a wave traveling along z. The trouble is
|ψ|2 = A2 sin2(kz − ωt)
so the probability distribution is not uniform, it is a function of z. So, although we can
prepare states like this, they do not describe a free particle. (What kind of situation does
the sin function describe? We will find out in more detail in a few lectures.)
C.
More on free particle deBroglie waves
Consider the situation shown in the figure where there is a potential step. The potential
is V1 for z < 0 and V2 for z > 0. The deBroglie wave can be written as
ψ = Aeı(p·r/h̄−Et/h̄)
p
where the energy is E = p2 /2m + V and the momentum is p = 2m(E − V ). So for z < 0,
p
p
p1 = 2m(E − V1 ) and for z > 0, p2 = 2m(E − V2 ). The relative index of refraction of
the two regions is defined as the ratio of the phase velocities or
r
vφ,2
p2
E − V2
n=
=
=
.
vφ,1
p1
E − V1
Snell’s law of refraction then says that the change in direction of the wave at the interface
is given by sin θ1 = n sin θ2 so
sin θ1
=n=
sin θ2
r
E − V2
E − V1
where n = n2/n1 is the relative index of refraction.
7
Let’s check if this is consistent with classical particle mechanics? The interface gives a
force on the particle F = −∇V and since V = V (z), F has only a z component perpendicular
to the interface. Therefore px must be constant which says p1 sin θ1 = p2 sin θ2 . Therefore
r
sin θ1
p2
E − V2
=
=
sin θ2
p1
E − V1
which is consistent with what we found using a deBroglie wave and Snell’s law. It is generally
true that for motion in a piecewise constant potential where the energy is positive the same
motion is found using Newtonian mechanics or wave mechanics. We will see however, that
there are also differences between classical and quantum mechanics. Classically the particle
is always transmitted as long as it has enough energy. Quantum mechanically there is a
finite probability of particle reflection.
D.
Schrödinger equation
If we ask the question what wave equation are the deBroglie waves the solution of we find
the Schrödinger equation (it is surprising that deBroglie did not find this equation). The
demonstration is simple. For a particle in a potential V (r), the energy is E = p2 /2m + V (r),
and the wave is ψ = ψ0 e(ı/h̄)(p·r−Et). We can check that
E
∂ψ
= −i ψ,
∂t
h̄
Thus
ih̄
so
and ∇2 ψ = −
p2
ψ.
h̄2
∂ψ
h̄2 2
p2
+
∇ ψ = Eψ −
ψ = Vψ
∂t
2m
2m
h̄2 2
∂ψ
ih̄
=−
∇ ψ+Vψ
∂t
2m
which is Schrödinger’s wave equation.
The Hamiltonian is the sum of kinetic and potential energy
H=−
h̄2 2
∇ + V (r).
2m
We can then write the Schrödinger equation as
ih̄
∂ψ
= Hψ.
∂t
8
2
h̄
When there is no potential H = − 2m
∇2 and we get the equation for a free particle
ih̄
∂ψ
h̄2 2
=−
∇ ψ.
∂t
2m
We see that the existence of deBroglie waves implies the Schrödinger equation. We can also
go the other way. If we assume a wave of the form ψ = ψ0 eı(k·r−ωt) then the Schrödinger
equation implies h̄ω = h̄2 k 2/2m = p2 /2m and we get deBroglie waves. It is important
to recognize that the Schrödinger equation is linear. If ψ1 and ψ2 are solutions then ψ =
c1 ψ1 + c2 ψ2 is also a solution for arbitrary constants c1 , c2 .
1.
Conservation of probability
The Schrödinger equation conserves the total probability for the particle to be somewhere.
Normalized solutions satisfy
Z
3
dP =
Z
d3 r |ψ|2 = 1.
This equation has to be interpreted in a different way for free particles, since in that case
the integral of |ψ|2 diverges and is δ function normalized. We will clarify this point later on.
Nevertheless, we can always calculate the time rate of change of the probability integral.
We have
Z
Z
dψ ∗ dψ ∗
d
3
2
3
d r |ψ| = d r ψ
+
ψ
dt
dt
dt
2
Z
i −h̄2 2 ∗
−i
−h̄ 2
3
∗ ∗
∗
= dr ψ
∇ ψ +V ψ +ψ
∇ ψ+Vψ
h̄ 2m
h̄
2m
Z
Z
i
−ih̄
3
∗
2
=
d r (V − V )|ψ| +
d3 r ψ∇2ψ ∗ − ψ ∗ ∇2ψ .
h̄
2m
Assume V is real so V ∗ − V = 0. We are left with
Z
d3 r ψ∇2ψ ∗ − ψ ∗∇2 ψ = (ψ∇ · ψ ∗ − ψ ∗∇ · ψ) |boundary
Z
− d3 r (∇ · ψ∇ · ψ ∗ − ∇ · ψ ∗ ∇ · ψ)
= (ψ∇ · ψ ∗ − ψ ∗∇ · ψ) |boundary.
Assuming that the wavefunction vanishes on a boundary at infinity we get zero for the the
integral so probability is conserved.
Dirac Notation and rules of Quantum Mechanics
9
Dirac notation provides an abstract representation theory of quantum mechanics. The
notation is elegant and powerful and widely used. The authoritative reference is Dirac’s
book on quantum mechanics2. The theory employs three types of objects: states, operators,
and c-numbers (complex numbers). In the following we give a brief introduction to the main
notational rules and ideas.
II.
STATES AND OPERATORS
A quantum state is represented by the ket |ψi. The ket should be thought of as an abstract symbol for a quantum state. It is not the same as ψ(x) which is the position space
representation of the state. We will soon see how to calculate ψ(x) from |ψi.
The Hermitian conjugate is the bra hψ|. The inner product called a bra-ket or braket is
hφ|ψi = c (a number).
(1)
If c = hφ|ψi then the complex conjugate is c∗ = hφ|ψi∗ = hψ|φi. Kets and bras exist in
a Hilbert space which is a generalization of the three dimensional linear vector space of
Euclidean geometry to a complex valued space with possibly infinitely many dimensions.
The inner product is linear
hφ| (a1|ψ1i + a2|ψ2i) = hφ|a1 |ψ1i + hφ|a2|ψ2i = a1 hφ|ψ1i + a2 hφ|ψ2i.
(2)
Operators are denoted by a hat Â.
 (c1 |ψ1i + c2 |ψ2i) = c1 Â|ψ1 i + c2 Â|ψ2 i.
The matrix element of an operator is
hφ|Â|ψi = hφ| (Â|ψi) = (hφ|Â) |ψi = c (a number).
(3)
The complex conjugate of the matrix element is
hφ|Â|ψi∗ = hψ|†|φi = c∗
(4)
where † is the Hermitian conjugate of Â. When  is represented by a matrix the Hermitian
conjugate is found by transposing the matrix and then taking the complex conjugate of each
2
P. A. M. Dirac, The Principles of Quantum Mechanics, 4th Ed., Oxford University Press, Oxford (1958).
10
matrix element. The operation of taking the Hermitian conjugate of a combination of
numbers, states, and operators involves changing c → c∗ , |ψi → hψ|, hψ| → |ψi,  → †
and reversing the order of all elements. For example
c1† hφ|B̂|ψihξ|
†
= c∗1 |ξihψ|B̂ † |φiÂ.
The combination of a ket and a bra is an operator
|φihψ|.
When dealing with a finite dimensional Hilbert space we can represent kets as column
vectors, bras as row vectors and matrix elements as inner products:
 
a1
 
a 
 2
 
.
 
|ψi =   , hψ| = a∗1 a∗2 . . . a∗n ,
.
 
 
.
 
an
 
a1
 
a 
 2
 
 . 
 
hψ|ψi = a∗1 a∗2 . . . a∗n   = |a1|2 + |a2|2 + ... + |an |2 .
.
 
 
.
 
an
The outer product is therefore a matrix
 


a1
2
∗
∗
 
|
a
a
a
|a
.
.
a
1
1
1
2
n
a 


 2

∗
2
∗
 
a
a
|a
|
.
.
a
a
2
2 n
1 2
 . 



  ∗ ∗
∗

|ψihψ| =   a1 a2 . . . an =  .
. . . . 
.
.


 
 .

 
.
.
.
.


.
 
∗
∗
2
a1 an a2an . . |an |
an
Note that |ψihψ| has real, nonnegative entries on the diagonal. This is not the case for the
more general operator |φihψ|.
11
It is important to understand that c-numbers obey the usual associative and commutative
rules of multiplication so that
ha|bihc|di = hc|diha|bi
and
ha|bi|cihd| = |cihd|ha|bi
but operators do not satisfy the commutative rule so in general
(|aihb|)(|cihd|) 6= (|cihd|)(|aihb|).
Operators do satisfy the associative rule so that
(|aihb|)(|cihd|) = |ai(hb||ci)hd| = |aihb|cihd| = hb|ci|aihd|.
A.
Observables
Observables are represented by Hermitian operators which satisfy † = Â. Hermitian operators are also referred to as self-adjoint. The expectation value of a Hermitian operator is
real:
a∗ = hÂi∗ = hψ|Â|ψi∗ = hψ|†|ψi = hψ|Â|ψi = a.
(5)
Denote the eigenstates of a Hermitian operator by |ni. They satisfy
Â|ni = an |ni.
The eigenvalues are real since
hm|Â|mi = hm|am|mi = am hm|mi = am
and
a∗m = hm|Â|mi∗ = hm|† |mi = hm|Â|mi = am .
States corresponding to different eigenvalues are orthogonal. We assume the states are
normalized so that hm|ni = δmn . To prove orthogonality we calculate
hm|Â|ni = hm|an |ni = an hm|ni
and
†
hm|Â|ni = hm| |ni = † |mi |ni = (am |mi)† |ni = a∗m hm|ni = am hm|ni.
12
Thus
(am − an )hm|ni = 0
so hm|ni = 0 if m 6= n. We can write this as hm|ni = δmn when the kets are normalized.
The eigenstates |ni of a Hermitian operator form a complete set. Therefore for an arbitrary
ket |ψi
|ψi =
where
hn|ψi = hn|
Thus
|ψi =
∞
X
n=0
cn |ni =
∞
X
j=0
∞
X
n=0
∞
X
n=0
∞
X
cj |ji =
j=0
hn|ψi|ni =
cn |ni
(6)
cj hn|ji =
∞
X
n=0
∞
X
cj δnj = cn .
j=0
|nihn|ψi =
∞
X
n=0
!
|nihn| |ψi.
Since this is true for arbitrary |ψi we can write the identity operator as
Iˆ =
∞
X
n=0
|nihn|.
(7)
A component of |ψi can be found by operating with the projection operator P̂n = |nihn|.
We have
P̂n |ψi = |nihn|
∞
X
(8)
(P̂n )2 = P̂n P̂n = (|nihn|) (|nihn|) = |ni (hn|ni) hn| = |nihn| = P̂n .
(9)
The projection operator is idempotent:
j=0
cj hn|ji = |ni
∞
X
cj δnj = cn |ni.
j=0
cj |ji = |ni
∞
X
j=0
The inner product of two states can be expressed in terms of the coefficients of their decomP
P
position. We write |ψi = n cn |ni, |φi = n bn |ni. Then
hφ|ψi =
X
m
b∗m hm|
X
n
cn |ni =
XX
m
b∗m cn δmn =
n
X
b∗n cn .
n
The spectral representation of an operator is found from
!
!
X
X
XX
 = IˆÂIˆ =
|mihm| Â
|nihn| =
|mihm|an |nihn|
m
=
XX
m
n
an |miδmn hn| =
n
X
n
an |nihn|.
13
m
n
(10)
Thus
 =
X
an |nihn|.
n
(11)
The representation (11) is diagonal since we have expressed  in a basis of the eigenvectors
of Â. If we choose some other set of basis vectors {|mi} (not the eigenvectors of Â) then the
representation will not be diagonal. Thus
 =
X
an |nihn|
X
X
n
=
an
n
=
m
X
n,m,m0
=
X
m,m0
|mihm| |nihn|
an |micmn cnm0 hm0|
X
m0
|m0ihm0 |
!
umm0 |mihm0 |.
where cmn = hm|ni and umm0 =
B.
!
P
n
an cmn cnm0 .
Degeneracy
Each eigenvalue aα may be associated with a subspace of dimension nα > 1. The nα degenerate eigenvectors can be orthonormalized to span the subspace. In this case we label the eigenvectors with an additional parameter r as |α, ri where r = 1, 2...nα and hβ, s|α, ri = δαβ δrs .
The eigenvalue relation is then
Â|α, ri = aα |α, ri for r = 1, ...nα.
(12)
The identity operator can be written as
Iˆ =
∞ X
nα
X
|α, rihα, r|.
(13)
aα|α, rihα, r|.
(14)
α=0 r=1
The spectral decomposition of the operator is
 =
∞ X
nα
X
α=0 r=1
14
III.
CONTINUOUS BASIS
A Dirac ket |ψi should be thought of as an abstract symbol for a quantum state. It is not
tied to any particular representation. By taking inner products we can find a representation
in terms of a discrete set of basis states as in Eq. (6). This can also be generalized to a
continuous basis |ξi using the orthogonality condition
hξ 0 |ξi = δ(ξ − ξ 0 )
and the representation of the unit operator
Z
ˆ
I = dξ |ξihξ|.
An arbitrary state |ψi can be expanded as
Z
Z
Z
ˆ
|ψi = I|ψi =
dξ |ξihξ| |ψi = dξ hξ||ψi|ξi = dξ ψ(ξ)|ξi
where we have introduced the wavefunction ψ(ξ) = hξ|ψi. The wavefunction ψ(ξ) gives the
amplitude of the decomposition of the state |ψi into the basis ket |ξi.
Using a continuous basis we can calculate the matrix elements of operators as follows.
ˆ ∂ ) that is some function of ξˆ and ∂ . For arbitrary states
Consider a general operator Â(ξ,
∂ ξ̂
∂ ξ̂
|ψi, |φi we have
†
Z
∂
0 0
0
ˆ
hφ|Â|ψi =
dξ |ξ ihξ |φi Â(ξ, )
dξ |ξ ihξ |ψi
∂ ξˆ
Z
ˆ ∂ )|ξ 0 ihξ 0 |ψi.
= dξ 00 dξ 0 hφ|ξ 00 ihξ 00 |Â(ξ,
∂ ξˆ
Z
00
00
00
(15)
Now the matrix element in the middle of the last expression is
ˆ ∂ )|ξ 0 i = A(ξ, ∂ )hξ 00 ||ξ 0 i = A(ξ, ∂ )δ(ξ 00 − ξ 0 ) = A(ξ 0 , ∂ )δ(ξ 00 − ξ 0 ).
hξ 00 |Â(ξ,
∂ξ
∂ξ
∂ξ 0
∂ ξˆ
Thus (15) becomes
hφ|Â|ψi =
Z
dξ 00 dξ 0 hφ|ξ 00 iδ(ξ 00 − ξ 0 )A(ξ 0,
∂
)hξ 0 |ψi
∂ξ 0
Z
∂
= dξ φ∗(ξ)A(ξ, )ψ(ξ).
∂ξ
=
Z
∂
)hξ 0 |ψi
0
∂ξ
dξ 0 hφ|ξ 0 iA(ξ 0 ,
(16)
Equation (16) gives the general formula for evaluating a matrix element in terms of an
expansion in a continuous basis.
15
A.
Representation of Derivatives
Given a ket |ψi we can define another ket |dψ/dξi whose representation is the derivative
of the original one. This new ket is the result of transforming the original one with an
operator and we write the transforming operator as
d
dξˆ
so
dψ
d
|ψi = | i.
dξˆ
dξˆ
The matrix element of the differential operator is
Z
Z
Z
d
d 0 0
d 0
dψ(ξ 0 )
0
0
0
hφ| |ψi = dξ hφ| |ξ ihξ |ψi = dξ hφ| |ξ iψ(ξ ) = dξ 0 φ∗ (ξ 0 )
.
dξ 0
dξˆ
dξˆ
dξˆ
Assuming the wavefunctions vanish at infinity an integration by parts gives
Z
Z
0
∗ 0
0 ∗ 0 dψ(ξ )
0 dφ (ξ )
dξ φ (ξ )
=
−
dξ
ψ(ξ 0 ).
0
0
dξ
dξ
(17)
(18)
Comparing (17) and (18) we get
hφ|
dφ∗(ξ 0 )
dφ
d 0
= −h |ξ 0 i
|ξ i = −
0
dξ
dξˆ
dξˆ
so
hφ|
d
dφ
= −h |
dξˆ
dξˆ
(19)
and
†
ˆ
d/dξˆ = −d/dξ.
The minus sign in this equation is important. It guarantees that the position representation
†
of the momentum is Hermitian, and a valid observable since id/dξˆ = id/dξˆ and e.g. the
momentum operator −ih̄d/dx̂ is a Hermitian operator.
If you find the steps from (17) - (19) unnecessarily complicated an alternative is to simply
note that
d
d
hφ| |ψi = hφ|
|ψi
dξˆ
dξˆ
Z
dψ
= dξ φ∗
dξ
Z
dφ∗
= − dξ
ψ
dξ
dφ
= −h |ψi
dξ
d
= hφ|
|ψi.
dξˆ
16
Thus
hφ|
B.
d
dφ
= −h |.
dξˆ
dξˆ
Position and momentum representations
We can use the above results for a continuous basis to find the relation between the
position and momentum wavefunctions ψ(x) and φ(p). Consider position and momentum
eigenkets: x̂|xi = x|xi and p̂|pi = p|pi. The matrix element of p̂ between different basis kets
is
hx|p̂|pi = hx|p|pi = phx|pi.
Using the position representation of the momentum operator we can also write this as
∂
∂
hx|p̂|pi = hx| −ih̄
|pi = −ih̄ hx|pi.
(20)
∂ x̂
∂x
This follows also from the general expression for matrix elements Eq. (16) 3 .
Comparing we see that
ip
∂
hx|pi = hx|pi
∂x
h̄
which has solution
hx|pi = Neıpx/h̄
with N a normalization constant.
We then observe that
ψ(x) = hx|ψi = hx|
and
Z
Z
dp |pihp| |ψi = N
Z
∗
dp eıpx/h̄ ψ(p)
Z
ψ(p) = hp|ψi = hp|
dx |xihx| |ψi = N
dp e−ıpx/h̄ ψ(x).
R
R
Requiring that dx |ψ(x)|2 = dp |ψ(p)|2 gives |N|2 = 1/h. We choose N to be real and
obtain
ψ(x) = √
1
Z
2πh̄ Z
1
dp eıpx/h̄ ψ(p)
dx e−ıpx/h̄ψ(x),
2πh̄
which demonstrates that ψ(x) and ψ(p) are related by a Fourier transform. For clarity we
ψ(p) = √
sometimes write φ(p) instead of ψ(p).
3
To see this use |xi =
R
dξ |ξihξ|xi =
∂
∂
x) ∂ξ
hξ|pi = −ih̄ ∂x
hx|pi.
R
dξ δ(ξ − x)|ξi, so (16) gives hx|p̂|pi = −ih̄hx| ∂∂x̂ |pi = −ih̄
17
R
dξ δ(ξ −
IV.
COMMUTING OBSERVABLES
If a state |ψi is an eigenstate of two observables  and B̂ with eigenvalues α and β then
ÂB̂|ψi = αβ|ψi = βα|ψi = B̂ Â|ψi.
(21)
A necessary and sufficient condition for  and B̂ to have a common complete set of eigenstates is that  and B̂ commute:
[Â, B̂] = ÂB̂ − B̂ Â = 0.
(22)
The uncertainty (variance) of an operator is
h(∆Â)2 i = hÂ2i − hÂi2 .
(23)
The generalized uncertainty relation for two noncommuting observables is
1
(∆A)(∆B) ≥ |h[Â, B̂]i|
2
q
q
2
where ∆A = h(∆Â) i, ∆B = h(∆B̂)2 i.
(24)
There are many ways to prove this, e.g. with the Schwarz inequality. Here is an alternative
approach. Define Â0 = Â − hÂi so hÂ0i = 0,
h(∆Â0 )2i = hÂ02i = (∆A)2
and similarly for B̂ 0 = B̂ − hB̂i. We note that [Â0, B̂ 0] = [Â, B̂].
Consider an arbitrary ket |ψi and the quantity (Â0 + iλB̂ 0)|ψi with λ real. The square
of the norm is
||(Â0 + iλB̂ 0)|ψi||2 = hψ|(Â0 − iλB̂ 0)(Â0 + iλB̂ 0)|ψi
= hψ|Â02|ψi + λ2 hψ|B̂ 02|ψi + iλhψ|[Â0, B̂ 0]|ψi
= λ2 (∆B)2 + iλhψ|[Â, B̂]|ψi + (∆A)2
≥ 0.
The last term is the commutator of two Hermitian operators which is anti-Hermitian (it
has imaginary eigenvalues), so i times the commutator is a Hermitian operator. The above
expression is therefore a quadratic function of λ with real coefficients, and is therefore strictly
18
real. It is non-negative for all values of λ, and therefore must not have two different real
roots, since if that were the case the expression would have to change sign and be somewhere
negative. For this to be so the discriminant must be negative or zero which corresponds to
1
(∆A)(∆B) ≥ |hψ|[Â, B̂]|ψi|.
2
Since this is true for any |ψi we can write
1
(∆A)(∆B) ≥ |h[Â, B̂]i|
2
as desired.
As an example of application of the commutation relation (24) consider x̂ and p̂. Their
commutator can be evaluated in the position representation as
hψ|[x̂, p̂]|ψi = hψ|(x̂p̂ − p̂x̂)|ψi
Z
∂
∂
∗
−
x ψ
= −ih̄ dx ψ x
∂x ∂x
Z
∂ψ
∂ψ
∗
= −ih̄ dx ψ x
−ψ−x
∂x
∂x
Z
= ih̄ dx ψ ∗ψ
= hψ|ih̄|ψi.
Since |ψi is an arbitrary ket we have [x̂, p̂] = ih̄ and using (24)
(∆x)(∆p) ≥
h̄
2
which is the usual position-momentum form of the uncertainty relation.
V.
EVOLUTION IN TIME
Time evolution is governed by the Hamiltonian Ĥ and
ih̄
d
|ψ(t)i = Ĥ|ψ(t)i.
dt
(25)
2
h̄
For a particle of mass m in the coordinate representation Ĥ = − 2m
∇2 + V (r) and we get
the Schrödinger equation
ih̄
∂ψ(r, t)
h̄2 2
=−
∇ ψ(r, t) + V (r)ψ(r, t).
∂t
2m
19
(26)
The formal solution for time evolution is
|ψ(t)i = Û(t, t0)|ψ(t0)i
where the time evolution operator is
i
Û(t, t0) = exp −
h̄
Z
t
0
0
dt Ĥ(t )
t0
(27)
+
and [....]+ indicates time ordering of operator products:
when t1 ≥ t2 ≥ ... ≥ tn .
h
Â(t1 )...Â(tn )
i
+
= Â(t1 )...Â(tn )
When Ĥ is not explicitly time dependent this simplifies to
i
Û (t, t0) = exp − Ĥ(t − t0) .
h̄
(28)
The time evolution operator is unitary which means Û † = Û −1 so
ˆ
Û † Û = Û Û † = I.
(29)
The time evolution of an initial state |ψ(t0)i can be determined explicitly by expanding in
energy eigenfunctions Ĥ|ni = En |ni. Assume the initial state
|ψ(t0)i =
X
n
hn|ψ(t0)i|ni =
X
n
cn0 |ni
and the time dependent state
|ψ(t)i =
X
n
hn|ψ(t)i|ni =
X
cn (t)|ni.
n
Plugging in to the Schrödinger equation gives
ih̄
X dcn
n
dt
|ni =
X
n
Ĥcn |ni =
X
n
En cn |ni.
Projecting out the mth component by operating on both sides with hm| gives
ih̄
dcm
= Em cm
dt
which is solved by
cm (t) = cm0 e−ıEm (t−t0 )/h̄ .
Thus
|ψ(t)i =
X
cn (t)|ni =
n
X
n
20
cn0 e−ıEn (t−t0 )/h̄ |ni.
A.
Schrödinger and Heisenberg pictures
There are several different ways of working with time evolution.
In the Schrödinger picture the state |ψ(t)iS is a function of time while all observables
S
=
ÂS are constant. The state vector evolves according to the Schrödinger equation ih̄ d|ψi
dt
ĤS |ψiS and the expectation value of an operator at time t is
a(t) = hÂS i
= S hψ(t)|ÂS |ψ(t)iS .
Here the subscripts S refer to the Schrödinger picture.
We can cast this in a different form as follows. The expectation value of an operator at
time t is
a(t) = hÂS i
= S hψ(t)|ÂS |ψ(t)iS
= hψ(0)|Û † (t)ÂS Û(t)|ψ(0)i.
We can define a time dependent operator by ÂH = Û † (t)ÂS Û(t), and time independent
states by |ψiH = |ψ(0)i such that
a(t) = H hψ|ÂH |ψiH .
This is the Heisenberg picture in which the kets are stationary but the operators evolve in
time. The equation of motion for the operator is found from
∂ Â
d †
dÂH (t)
H
=
Û (t)ÂS Û(t) +
dt
dt
∂t
dÛ † (t)
dÛ(t) ∂ ÂH
=
ÂS Û (t) + Û † (t)ÂS
+
dt
dt
∂t
i †
−i
∂ ÂH
†
=
Û Ĥ ÂS Û(t) + Û (t)ÂS
Ĥ Û(t) +
h̄
h̄
∂t
=
i
∂ ÂH
[Ĥ, ÂH ] +
h̄
∂t
where the last lines follows from [Û, Ĥ] = 0, and for completeness we have included the
possibility of an explicit time dependence (not due to Û(t)) in ÂH . Thus
dÂH
i
∂ ÂH
= − [ÂH , Ĥ] +
,
dt
h̄
∂t
21
(30)
which is referred to as the Heisenberg equation (although it was first written down by Dirac).
This equation is usually more difficult to solve than the Schrödinger equation. However, it
provides a clear picture of the correspondence between quantum and classical mechanics.
B.
Quantum - Classical correspondence and Ehrenfest’s theorem
In classical mechanics we can describe a dynamical system by coordinates qj and momenta
pj . The Hamiltonian of the system is a function H(qj , pj ) of the coordinates and momenta
which satisfy Hamilton’s equations
∂H
∂qj
=
,
∂t
∂pj
∂pj
∂H
=−
.
∂t
∂qj
For a particle in a potential V (r), H = p2 /2m + V (r) so Hamilton’s equations give dr/dt =
p/m and dp/dt = −∇V (r).
For any two dynamical quantities A, B we define the Poisson bracket {A, B} by
X ∂A ∂B
∂A ∂B
−
.
{A, B} =
∂qj ∂pj
∂pj ∂qj
j
Clearly {A, B} = −{B, A}.
The time derivative of the quantity A is
dA X ∂A ∂qj
∂A ∂pj
∂A
=
+
+
dt
∂qj ∂t
∂pj ∂t
∂t
j
X ∂A ∂H
∂A ∂H
∂A
=
−
+
∂qj ∂pj
∂pj ∂qj
∂t
j
= {A, H} +
∂A
.
∂t
(31)
The similarity between the quantum equation (30) and the classical equation (31) is remarkable. We see that the quantum equation of motion for the operator  can be found
by writing down the classical equation for the dynamical variable A in terms of Poisson
brackets and making the substitution
i
{A, H} → − [ÂH , Ĥ].
h̄
(32)
Alternatively, we may think of quantum mechanics as the more fundamental theorem
which classical mechanics is an approximation to. In the limit of h̄ → 0 the transformation
i
− [ÂH , Ĥ] → {A, H}
h̄
22
(33)
reveals the limiting classical equation of motion. Note that while (32) should always lead to
a valid quantum equation, the substitution (33) may be meaningless since there are quantum
problems for which no classical analog exists. A prime example is the spin of an electron.
Finally, we note that the time evolution of the expectation value of a quantum variable
has a close analogy with the time evolution of the corresponding classical quantity. This
analogy is already apparent in equations (30,31), which when written in terms of expectation
values is referred to as Ehrenfest’s theorem.
To see this note that for any differentiable operator function F̂ = F (q̂j , p̂j ) we have the
following commutators
[q̂j , F̂ ] = ih̄
∂ F̂
,
∂ p̂j
[p̂j , F̂ ] = −ih̄
From (30) with F̂ = Ĥ we obtain
*
+
d
∂ Ĥ
hq̂j i =
,
dt
∂ p̂j
d
hp̂j i = −
dt
Thus for a particle in potential V (r̂) with Ĥ =
hp̂i
d
hr̂i =
,
dt
m
p̂2
2m
∂ F̂
.
∂ q̂j
*
∂ Ĥ
∂ q̂j
+
.
+ V (r̂) we get
d
hp̂i = −h∇V (r̂)i.
dt
(34)
The first of Eqs. (34) is identical to the corresponding classical equation obtained by putting
hr̂i → r, hp̂i → p. However, the second equation differs from the corresponding classical
equation
d
hp̂i = ∇V (r)|r=hr̂i
dt
since, in general,
h∇V (r̂)i =
6 ∇V (r)|r=hr̂i .
The difference betweens these two expressions is negligible provided the potential is close to
constant over the extent of the deBroglie wave packet of the particle. When the potential is
constant over the extent of the deBroglie wave packet the expectation values of the quantum
variables agree with the results of a classical description.
VI.
MEASUREMENTS
Consider an observable  with eigenspectrum |ni satisfying Â|ni = an |ni. An arbitrary state
P∞
|ψi can be expanded as |ψi = n=0 cn |ni. The result of a measurement of the observable A
23
is
hÂi = hψ|Â|ψi =
X
m
c∗m hm|Â
X
n
cn |ni =
XX
m
n
c∗m cn an hm|ni =
X
n
|cn |2an .
(35)
The probability of the measurement result being the eigenvalue an is
P (an ) = |cn |2 = |hn|ψi|2 = ||P̂n |ψi||2 = hψ|P̂n |ψi.
(36)
After  has been measured and given the result an the new state of the system is
|ψ 0i =
P̂n |ψi
||P̂n |ψi||
=
cn
|ni.
|cn |
(37)
A subsequent measurement of hÂi will return an with unit probability.
The uncertainty in the measurement of an operator depends on the state being measured.
Recall the definition of the variance
h(∆Â)2 i = hÂ2i − hÂi2 .
Let us assume |ψi = a|0i + b|1i describes a two-state particle and that  has eigenvalues
Â|0i = λ0 |0i, Â|1i = λ1 |1i. The variance of  is
h(∆Â)2 i = λ20 |a|2 + λ21 |b|2 − (λ0 |a|2 + λ1 |b|2 )2
= λ20 (|a|2 − |a|4) + λ21 (|b|2 − |b|4) − 2λ0 λ1 |a|2|b|2.
√
If a = 0 or b = 0 we are in an eigenstate of  and h(∆Â)2i = 0. If a = b = 1/ 2 then
h(∆Â)2i = λ20 /4 + λ21 /4 − 2λ0 λ1 /4. This variance is often referred to as quantum projection
noise. Variation of the measurement variance as the state changes is a signature of a quantum
limited measurement.
24
VII.
PRINCIPLES OF QUANTUM MECHANICS
We can summarize the formalism given above in a small set of principles.
1. A physical system is associated with a Hilbert space EH containing ket vectors. At time
t the physical state is completely described by a ket |ψ(t)i residing in EH .
2. Any physical quantity A is associated with a Hermitian operator  that acts on kets
in EH . The result of a measurement of A is always one of the eigenvalues an of Â. The
probability of measuring an is P (an ) = ||P̂n |ψi||2 where P̂n = |nihn| is the projector onto
the ket |ni. After the measurement the system will be in the new state |ψ 0i =
P̂n |ψi
.
||P̂n |ψi||
3. Time evolution is governed by the Hamiltonian according to ih̄ d|ψ(t)i
= Ĥ|ψ(t)i. The
dt
solution can be written in terms of a time evolution operator Û as |ψ(t)i = Û(t, t0)|ψ(t0)i
with Û a unitary operator.
How to translate these simple rules into explicit procedures is often far from obvious, and
learning how to do so constitutes a course in Quantum Mechanics.
25
VIII.
DENSITY MATRIX THEORY
When we have a statistical mixture of states the state vector |ψi is equal to one of the
set {|ψii} with probability Pi . If the state is described by any one of the |ψi i the quantum
mechanical expectation value of an operator Ô is simply hÔii = hψi |Ô|ψi i. When we have a
statistical mixture we define the expectation value of an operator as
hÔi =
X
i
Pi hÔii =
X
i
Pi hψi |Ô|ψii.
(38)
It should be emphasized that the probabilistic nature of the expectation value hÔi is not
quantum mechanical in origin but arises from our imperfect knowledge of the state |ψi.
This could for example be due to imperfect preparation of the state. In addition there is
the quantum mechanical uncertainty due to the probabilistic interpretation of hÔii . The
P
P
probabilities satisfy 0 ≤ Pi ≤ 1, i Pi = 1, and i Pi2 ≤ 1. A pure state refers to the
situation where only one Pi = 1 and all the other Pi vanish. In this case hÔi = hÔii and we
recover our usual quantum mechanical result. If this is not the case we refer to the state as
a mixed state.
In order to deal with situations where we only have statistical knowledge of the wave
function we introduce the density matrix defined by
ρ̂ =
X
i
Pi |ψiihψi |.
For a pure state this reduces to ρ̂ = |ψihψ|. We now introduce a complete set of orthonormal
basis states |ni using which we can write
ˆ ii =
|ψi i = I|ψ
X
n
!
|nihn| |ψi i =
X
n
hn|ψi i|ni =
X
n
cin |ni
with cin = hn|ψi i. The trace of the density matrix is the sum of the diagonal components
26
which is
Tr[ρ̂] =
X
n
=
X
n
=
X
hn|ρ̂|ni
hn|
Pi
=
Pi
=
X
X
cin0 c∗in00 hn|n0 ihn00 |ni
cin c∗in
n
i
X
i
Pi |ψi ihψi |ni
n,n0 ,n00
i
X
X
Pi = 1.
(39)
i
which corresponds to conservation of probability. Some important properties of density
matrices are ρ̂ = ρ̂†, and hj|ρ̂2 |ii ≤ hj|ρ̂|ii. The equality holds for pure states with ρ̂2 = ρ̂ in
which case the density matrix is referred to as idempotent.
The expectation value of an arbitrary operator Ô is given by
X
hÔi =
Pi hψi |Ô|ψii
i
=
X
Pi
n0 ,n00
i
=
X
Pi
=
i
=
X
n00
=
X
n00
=
X
n00
X
n0 ,n00
i
X
X
Pi
X
n0 ,n00
hn00 |
hn00 |ρ̂
hψi |(|n0ihn0 |)Ô(|n00ihn00 |)|ψi i
hψi |n0ihn0 |Ô|n00ihn00 |ψi i
hn00 |ψi ihψi |n0 ihn0 |Ô|n00i
X
i
X
n0
Pi |ψi ihψi |
!
X
n0
|n0 ihn0 |Ô|n00i
|n0 ihn0 |Ô|n00i
hn00 |ρ̂Ô|n00i
h i
= Tr ρ̂Ô .
It can be shown that the trace is unchanged with cyclic reordering of the operators so for
any operator
h i
h i
hÔi = Tr ρ̂Ô = Tr Ôρ̂ .
(40)
Equation (40) defines the expectation value of an operator when the state is only known
statistically. The equation of motion for the density matrix is
i
dρ̂
= [ρ̂, Ĥ].
dt
h̄
27
(41)
This can be derived by plugging the definition of ρ̂ into the Schrödinger equation. Comparing
with (30) for the time evolution of a Heisenberg frame operator we see that Eq. (41) is of
the same form but differs by a minus sign.
Let’s now look at a few examples. Consider a single atom with two levels in a pure state.
The density matrix is ρ̂ = |ψihψ| and using |ψi = cg |gi + ce |ei gives

ρ̂ = 
|cg |2
cg ce ∗
(cg c∗e )∗ |ce |2

.
√
The maximum possible value of the off-diagonal entries occurs for |cg | = |ce | = 1/ 2 giving
an off-diagonal entry with magnitude 1/2. If the atom is in a coherent superposition of
√
ground and excited states |ψi = (1/ 2)(|gi + |ei), and the density matrix is
ρ̂ =
X
i

Pi |ψi ihψi | = 
1
2
1
2
1
2
1
2

.
The nonzero off-diagonal elements are referred to as coherences. They show that the expectation value of for example the dipole operator d̂ = er̂ will be nonzero. On the other hand
an atom that is prepared in an incoherent mixture of ground and excited states has density
matrix
ρ̂ =
X
i

1
2
1
1
Pi |ψi ihψi | = |gihg| + |eihe| = 
2
2
0
0
1
2

.
The off-diagonal elements are zero and there is no coherence. Note that measurements of
the probabilities of the ground and excited states do not distinguish between the two cases.
As a second example of a mixed state consider a two-level atom in thermal equilibrium
at temperature T. The probability of occupation of state |ji is proportional to he−Ĥ/kB T i =
e−Ej /kB T . Therefore the correctly normalized density operator is
ρ̂ =
e−Ĥ/kB T
h
i
−
Ĥ/k
T
B
Tr e


1
0
1


=
−h̄ω
/k
T
a
B
−h̄ωa /kB T
1+e
0 e


h̄ωa /2kB T
e
0
1

.
=
−h̄ωa /2kB T
2 cosh(h̄ωa /2kB T )
0
e
28
In the case of a sample of N atoms the wavefunction can be written as
|ψi = c0|1g 2g ...Ng i + (c1 |1e 2g ...Ng i + .....) + CN +2 |1e 2e ...Ng i + ..... + C2N |1e 2e ...Nei.
When N is large this is a very complicated wavefunction with 2N coefficients. When the
atoms are uncorrelated, that is to say the state of any atom is not dependent on the states
of the other atoms, the total density matrix is just the product of the density matrices of
the individual atoms,
ρ=
Y
ρ(1) ⊗ ρ(2) ⊗ .... ⊗ ρ(N ).
A system of N two-level atoms is described by a composite density matrix of dimensions
2N × 2N . For N large we have approximately 22N /2 independent complex quantities which
highlights the difficulty of solving quantum dynamical many body problems.
A.
Composite systems
Density matrices are important for describing composite quantum systems. Consider two
subsystems A, B in the product state
|ψi = |φiA ⊗ |χiB .
Here ⊗ denotes the tensor product. We will often omit the ⊗ symbol and write
|ψi = |φiA |χiB or |ψi = |φA χB i or |ψi = |φχi.
In this last version the subsystems are identified by their ordering in the ket. The corresponding bra is
hψ| = |ψi† = hφ|A hχ|B or hψ| = hφA , χB | or hψ| = hφχ|.
Unfortunately the other convention is also in use for which
hψ| = hχ|B hφ|A = hχφ|.
In many cases the ordering will be clear from the context. If not, then it is a good idea to
be explicit by using subscripts to indicate the subsystems.
Using the partial trace the density matrix of the joint state ρAB can be reduced to give
ρA = TrB (ρAB ) or ρB = TrA (ρAB ).
29
The reduced density matrices ρA or ρB encapsulate what we know about the subsystems
individually. In many applications of density matrices the composite system may consist of
a material object A such as an electron or an atom or a piece of condensed matter containing
many atoms. The second half of the system B may be the set of oscillator modes describing
a radiation field that is couped to the matter. We are often interested in the state and the
dynamics of the matter, but have only statistical knowledge of the radiation field B. In
this case the result of measurements performed on the matter are found using the reduced
density matrix ρ̂A .
As an example of a basic composite system consider the two-particle Bell state
|ψi =
|01i + |10i
√
.
2
(42)
If we measure one of the particles individually we will get the answer 0 or 1 with 50%
probability. However, we will also know that a measurement of the other particle would
reveal the opposite state. This type of behavior, where there is no local certainty, but strong
two-particle correlations is characteristic of entanglement, and is revealed by contrasting the
full density matrix with the reduced density matrices. We find
ρAB = |ψihψ|
|01i + |10i h01| + h10|
√
√
=
2
2
|01ih01| + |01ih10| + |10ih01| + |10ih10|
=
2


0 0 0 0



1
0 1 1 0
= 

2 0 1 1 0


0 0 0 0
which displays the maximum possible coherence. Indeed the Bell state (42) is the maximally
entnagled state of two, two-level objects.
30
On the other hand
ρA = TrB (ρAB )
X
|01ih01| + |01ih10| + |10ih01| + |10ih10|
=
|xiB
B hx|
2
x
=
=
|0ih0| + |1ih1|
 2
1 1 0 
2 0 1
is a mixed state with no coherence.
For reference let’s calculate the partial trace for an arbitrary mixed state of two spin 1/2
objects. The most general 4 × 4 matrix is

ρAB
a b c d



 e f g h


=
.
 i j k l


m n o p
Of course the coefficients have to satisfy several conditions for this to be a valid density
matrix. The reduced density matrices are


a+f c+h

ρA = TrB (ρAB ) = 
i+n k+p
and
IX.


a+k b+l
.
ρB = TrA (ρAB ) = 
e+o f +p
TIME EVOLUTION OF OPEN QUANTUM SYSTEMS
The unitary evolution of the density operator is described by
i
dρ̂
= [ρ̂, Ĥ]
dt
h̄
(43)
with Ĥ the Hamiltonian. Since Ĥ is a hermitian operator the solution for ρ̂(t) is described
by a unitary transformation
ρ̂(t) = U(t, t0)ρ̂(t0 )U † (t, t0)
31
FIG. 2. Quantum system and environment. The total Hamiltonian Ĥ and the combined density
operator ρ̂SE consist of parts describing the system, the environment, and their interaction.
with U the unitary time evolution operator. Unitary time evolution is characteristic of
“closed” quantum systems that are isolated from the environment.
On the other hand we often deal with “open” quantum systems that interact with the
environment. This is depicted qualitatively in Fig. 2. In such cases we may observe nonunitary evolution. A familiar example is the decay of an excited atomic level. The atom
decays to a lower level and emits a photon, but is never observed to spontaneously transition
from a lower to a higher level in the absence of some source of energy in the environment.
The dynamics is therefore non-unitary since independent of the initial conditions the atom
is always observed in the lower level at long times.
This apparent departure from unitary evolution can be understood from different points
of view. If we believe that quantum mechanics applies to everything then the non-unitary
dynamics is simple an expression of our lack of knowledge of the environment. If we had
a complete description of the environment, i.e. knew the Hamiltonian, there should be a
unitary description of the combined dynamics of the system and the environment. In the
example of the decaying atom the emitted photon, after rattling around in the environment,
should eventually return to the atom and have the possibility of exciting it to a higher level,
thereby restoring unitary dynamics.
An alternative viewpoint is that quantum mechanics only applies to systems of a finite
size, and that at some point we have to give up the quantum description and use classical
mechanics. Otherwise we would predict quantum behavior in macroscopic objects, which is
32
never observed. This is the famous Schrödinger’s cat paradox. A problem with this point of
view is that there is no obvious scale at which to insert a boundary between quantum and
classical behavior. Indeed experimental advances of the last few decades have succeeded in
observing quantum behavior in larger and larger objects.
Whatever our point of view about these questions, it is true that we often wish to predict
the evolution of a quantum system without full knowledge of the environment. This leads
to a genaralization of Eq. (43) to describe effectively non-unitary evolution. Let’s now
make these ideas more precise4. Consider a quantum system S and an environment E. The
density matrix describing system and environment is ρ̂SE and the Hamiltonian is Ĥ = ĤS +
ĤE + ĤSE . Here ĤS is the system Hamiltonian, ĤE is the Hamiltonian of the environment,
and ĤSE describes the coupling between the system and the environment.
Let’s assume that the system and the environment are initially in a separable state,
ρ̂SE (t0 ) = ρ̂S (t0) ⊗ ρ̂E (t0). At a later time t we have
ρ̂SE (t) = U ρ̂SE (t0)U †
where the time evolution operator U is defined by the total Hamiltonian Ĥ. The state of
the system alone at a later time is
ρ̂S (t) = TrE [ρ̂SE ] = TrE [U ρ̂SE (t0 )U † ] = TrE [U ρ̂S (t0 ) ⊗ ρ̂E (t0 )U † ].
The evolution of the system density operator can be formally described as due to the action
of a superoperator E,
ρ̂S (t) = E[ρ̂S (t0 )].
The map E is referred to as a quantum process, or as a superoperator, since it maps operators
to operators.
Define a basis of environment states |Ej i then
ρ̂S (t) = TrE [U ρ̂S (t0) ⊗ ρ̂E (t0)U † ]
X
=
hEj |U ρ̂S (t0) ⊗ ρ̂E (t0)U † |Ej i.
j
4
We primarily draw on the treatment in B. Schumacher and M. Westmoreland, Quantum processes systems,
& information, Cambridge University Press, Cambridge (2010).
33
Let the initial state of the environment be ρE (t0) = |E0 ihE0 | and the initial system state be
a pure state ρS (t0) = |S0ihS0 | then
ρ̂S (t) =
X
j
=
X
j
=
X
j
hEj |U|S0 ihS0 | ⊗ |E0 ihE0 |U † |Ej i
hEj |U|S0 i|E0 i ⊗ hS0 |hE0 |U † |Ej i
hEj |U|E0 i|S0 i ⊗ hS0 |hE0 |U † |Ej i.
If we now define an operator Âj by
Âj |S0i = hEj |U|E0 i|S0 i
then
ρ̂S (t) =
X
j
Âj |S0ihS0 |†j .
Since any density operator can be written as a sum of pure state density operators we arrive
at
ρ̂S (t) = E[ρ̂S (t0)] =
X
Âj ρ̂S (t0)†j .
(44)
j
The result of these manipulation is that we have an expression for the evolution of the
system called the operator sum representation of the process E. The Âj are often called
Kraus operators. There are two important things to note about Eq. (44). First, we have
found a representation for the time evolution in terms of operators Âj that act only on the
system. Second, the operators are not unitary and the temporal dynamics of the system
density matrix is therefore not unitary. The act of tracing over the environment effectively
leads to non-unitary system evolution.
The Kraus operators satisfy a normalization condition since
1 = Tr[ρ̂S (t)] = Tr[
X
j
Âj |S0ihS0 |†j ] = Tr[hS0|
X
must hold for any pure system state |S0 i. Therefore
X
ˆ
†j Âj = I.
j
34
j
†j Âj |S0 i] = hS0 |
X
j
†j Âj |S0i
We can now use the operator sum representation to find a general form for the time
evolution of the system density operator when coupled to an environment. Consider an
infinitesimal time evolution
E[ρ̂] = ρ̂ + δ ρ̂ =
X
Âj ρ̂†j
(45)
j
with δ ρ̂ ∼ δt. Here we have dropped the subscript S with the understanding that ρ̂ is the
reduced density operator of the system. Assume one of the Kraus operators Â0 is dominant
and is given by
Â0 = Î + L̂0 δt
and the other operators are
√
Âj = L̂j δt,
j 6= 0.
Then
Â0ρ̂†0 = ρ̂ + (L̂0 ρ̂ + ρ̂L̂†0 )δt + O(δt2)
and
Âj ρ̂†j = L̂j ρ̂L̂†j δt.
To order δt (45) becomes
ρ̂ + δ ρ̂ = ρ̂ +
L̂0 ρ̂ + ρ̂L̂†0 +
X
L̂j ρ̂L̂†j
j6=0
!
δt.
This implies the differential equation
X
dρ̂
= L̂0 ρ̂ + ρ̂L̂†0 +
L̂j ρ̂L̂†j .
dt
j6=0
If we compare with (43) we see that in order to recover the Hamiltonian part of the dynamics
we must put L̂0 = −iĤ/h̄ + L00 so that
X
†
dρ̂
i
= [ρ̂, Ĥ] + L̂00ρ̂ + ρ̂L̂00 +
L̂j ρ̂L̂†j .
dt
h̄
j6=0
Since Tr[ρ̂] = 1 at all times we require Tr[dρ̂/dt] = 0 which results in the condition L̂00 =
P
− 21 j L̂†j L̂j . Plugging in we get
X
dρ̂
i
1
1
= [ρ̂, Ĥ] +
L̂j ρ̂L̂†j − L̂†j L̂j ρ̂ − ρ̂L̂†j L̂j .
dt
h̄
2
2
j
35
This is known as the Lindblad equation5 and is widely used to study open quantum system
dynamics. The specific form of the Lindblad operators L̂j depends on the problem being
treated. We will see an example for the interaction of atomic energy eigenstates with a
radiation field.
5
G. Lindblad, On the Generators of Quantum Dynamical Semigroups, Commun. Math. Phys. 48, 119
(1976).
36
X.
ONE-DIMENSIONAL HARMONIC OSCILLATOR
The time independent Schrödinger equation for a one-dimensional harmonic oscillator is
−
h̄2 d2 ψ 1
+ mω 2 x2ψ = Eψ.
2
2m dx
2
It is convenient to introduce a characteristic energy E0 = h̄ω, length x0 =
p
h̄/(mω), and
dimensionless position coordinate ξ = x/x0 . In terms of these new variables we have
d2 ψ
2E
2
+
− ξ ψ = 0.
dξ 2
E0
(46)
To find solutions to this equation consider the limit ξ → ∞ for which ψ 00 − ξ 2 ψ = 0,
where prime denotes the derivative with respect to ξ. Asymptotic solutions are ψ ∼ e±ξ
2 /2
.
To avoid a singularity at infinity we choose the decaying exponential and therefore seek a
solution of (46) in the form ψ = f(ξ)e−ξ
2 /2
. The function f then satisfies
E
1
00
0
−
f = 0.
f − 2ξf + 2
E0 2
The solutions to this equation are the Hermite polynomials Hn (ξ) with integer n = E/E0 −
1/2. The energy eigenvalues are therefore En = E0 (n + 1/2) = h̄ω(n + 1/2). The normalized
R∞
eigenfunctions satisfying −∞ dξ |ψ|2 = 1 are
un =
r
1
π 1/2n!2n
Hn (ξ)e−ξ
2 /2
.
The first few Hermite polynomials are
A.
H0 = 1
(47)
H1 = 2x
(48)
H2 = 4x2 − 2.
(49)
One-dimensional Quartic Oscillator
The quartic oscillator with potential U(x) = (aEho /2)(x/x0 )4 satisfies the Schrödinger
equation
−
h̄2 d2 ψ
+ (aEho /2)(x/x0 )4ψ = Eψ
2m dx2
37
where the constant a is dimensionless. Introducing ξ = x/x0 gives
E
a 4
00
ψ +2
− ξ ψ=0
Eho
2
where Eho = h̄ω. There are no closed form analytical solutions to this equation. We proceed
by seeking an approximate solution in the form
ψ(ξ) =
N
X
cn un (ξ)
n=0
where the un are normalized solutions to a harmonic oscillator and the cn are constants to
be determined. Using the differential equation for the harmonic oscillator we find
X
2E
4
2
cn un = 0.
−aξ + ξ − (2n + 1) +
Eho
n
(50)
We can convert this into a set of linear equations for the coefficients cn using the following
integrals
Z
Z
∞
2
dξ um ξ un =
−∞
∞
4
dξ um ξ un =
−∞
1
+ d0 δm,n + d2+ δm,n+2 + d2− δm,n−2
2
3
+ d̃0 δm,n + d˜2+ δm,n+2 + d̃2− δm,n−2
4
+ d˜4+ δm,n+4 + d˜4− δm,n−4
(51)
(52)
where
p
p
(n + 1)(n + 2)
n(n − 1)
d0 = n,
d2+ =
, d2− =
2
2
p
p
(2n
−
1)
n(n − 1)
1
3 2
d̃0 = (n + n), d˜2+ = (2n + 3) (n + 2)(n + 1), d˜2− =
2p
2
2
p
(n + 4)(n + 3)(n + 2)(n + 1)
n(n − 1)(n − 2)(n − 3)
d̃4+ =
, d˜4− =
.
4
4
Multiplying by um and integrating over all space transforms Eq. (50) into a set of algebraic
equations
Mc = λc
(53)
where λ = (−2E + 34 aEho )/Eho + 1/2 and c = (c0, c1 , ..., cN ) is the vector of coefficients. The
energy eigenvalues are therefore given by
Eho
En =
2
1 3
+ a − λn
2 4
38
ξ
ξ
0.5
0.4
0.3
0.2
0.1
0
-3 -2 -1 0 1 2 3
|ψ|2
0.5
0.4
0.3
0.2
0.1
0
-3 -2 -1 0 1 2 3
|ψ|2
|ψ|2
|ψ|2
0.6
0.5
0.4
0.3
0.2
0.1
0
-3 -2 -1 0 1 2 3
0.4
0.3
0.2
0.1
0
-3 -2 -1 0 1 2 3
ξ
ξ
FIG. 3. Probability distributions of the first four eigenmodes for the quartic oscillator (solid lines)
and the harmonic oscillator (dashed lines). In all cases the eigenfunctions have been normalized
R∞
such that −∞ dξ |ψ|2 = 1.
where λn is the nth eigenvalue of Eq. (53). The elements of the matrix M are
Mmn = −2n + d0 − ad˜0 δm,n
˜
+ d2+ − ad̃2+ δm,n+2 + d2− − ad2− δm,n−2
−ad̃4+ δm,n+4 − ad˜4− δm,n−4 .
As an example we can take a = 1, Eho = 2. Using N = 20 I find the first few eigenvalues to be .1896, −2.550, −6.206 and the corresponding energies are E0 = 1.06037, E1 =
3.7997, E2 = 7.4558, E3 = 11.646. These agree well with values published in the literature.
As shown in Fig. 3 the eigenfunctions look quite similar to those of the harmonic oscillator,
but are localized more strongly near the origin.
XI.
TWO DIMENSIONAL HARMONIC OSCILLATOR
In two dimensions the harmonic oscillator potential can be written as U = 12 mω 2 r2 =
1
mω 2(x2
2
+ y 2). The Schrödinger equation is separable in Cartesian coordinates so the 2-D
solutions can be written as products of 1-D solutions. The eigenfunctions and eigenvalues
are labeled by two quantum numbers m, n and the energies are Emn = h̄ω(m + n + 1) with
degeneracy m + n + 1.
Altwernatively the 2-D harmonic oscillator can be solved using polar coordinates (r, φ)
with x = r cos φ, y = r sin φ. This will be advantageous for studying anharmonic extensions
since ρ4 and higher potentials are not separable in Cartesian coordinates. In a polar representation we label the solutions with quantum numbers n, l where n refers to the radial
dependence and l determines the azimuthal behavior. The requirement of a single valued
39
wavefunction will result in only integer values of l being admissible. We will find that for a
given n there are 2n + 1 degenerate solutions corresponding to 0 ≤ |l| ≤ n.
The Schrödinger equation in polar coordinates is
1 ∂ 2ψ
1
h̄2 ∂ 2 ψ 1 ∂ψ
−
+
+ 2 2 + mω 2 r2 ψ = Eψ.
2
2m ∂r
r ∂r
r ∂φ
2
We seek solutions in the form ψ = f(r)Φ(φ). Requiring that the angular dependence is
single valued we use Φ(φ) =
h̄2
−
2m
√1 eılφ
2π
with l integer. The radial equation is then
d2 R 1 dR
l2
1
+
− 2 R + mω 2 r2 R = ER.
2
dr
r dr
r
2
As in the one-dimensional case we introduce characteristic energy E0 = h̄ω, length r0 =
p
h̄/(mω), and dimensionless radial coordinate ρ = r/r0 . Using the new variables the radial
equation can be written as
2
d R 1 dR
l2
2E
2
+
− 2R +
− ρ R = 0.
dρ2
ρ dρ
ρ
E0
For small ρ , ρ2 R00 + ρR0 − l2R ∼ 0. This is solved by R ∼ ρl . To prevent divergence at the
2 /2
origin we use R ∼ ρ|l| . For large ρ, R00 − ρ2R ∼ 0. This is solved by R ∼ e±ρ
2 /2
seek a solution in the form R = f(ρ)ρ|l| e−ρ
. We therefore
. The function f satisfies
ρf 00 + (1 + 2|l| − 2ρ2 )f 0 + 2(
E
− |l| − 1)ρf = 0.
E0
To remove the terms in ρ2 we make a change of variables x = ρ2 and g(x) = f(ρ). Then
√
f 0 = g 02 x, f 00 = g 004x + 2g 0 and
xg 00 + (1 + |l| − x)g 0 + (E/2E0 − |l|/2 − 1/2)g = 0.
|l|
The solutions of this equation are the associated Laguerre polynomials Ln (x) where n =
E/2E0 − l/2 − 1/2 is integer. The energy is En = E0 (2n + |l| + 1).
The solutions of the Schrödinger equation can therefore be written
unl (ρ, φ) = Rnl (ρ)Φl (φ)
1
|l| 2 |l| −ρ2 /2
ılφ
√ e
= Ln (ρ )ρ e
.
2π
(54)
Although the wavefunctions written above are correct they do not provide the most
convenient labeling and ordering of the states. We can order the energy eigenvalues in a
40
v states in polar basis (v, l)
degeneracy states in Cartesian basis (m, n) degeneracy
0 (0,0)
1
(0,0)
1
1 (1,1), (1,-1)
2
(1,0), (0,1)
2
2 (2,2), (2,-2), (2,0)
3
(2,0), (0,2), (1,1)
3
3 (3,3), (3,-3), (3,1), (3,-1)
4
(2,1), (1,2), (3,0), (0,3)
4
4 (4,4), (4,-4), (4,2), (4,-2), (4,0)
5
(4,0), (0,4), (3,1), (1,3), (2,2)
5
TABLE I. Labeling of 2-D harmonic oscillator states in a polar (v, l) basis or Cartesian (m, n)
basis.
manner that makes the analogy with the 1-D case more apparent by introducing an effective
vibrational quantum v = 2n + |l| so that
Ev = Eho (v + 1).
The last term in parentheses is 1 as opposed to 1/2 in the 1-D case as we now have two
units of ground state energy, one from the x motion and one from the y motion. We can
therefore write the wavefunctions as
1 ılφ
|l|
2 |l| −ρ2 /2
√ e
uvl (ρ, φ) = Rvl (ρ)Φl (φ) = L v−|l| (ρ )ρ e
.
2
2π
(55)
where v = 0, 1, 2, ... and l = v, v − 2, v − 4, ...1 or 0. States differing only in the sign of
l are degenerate so the degeneracy of the eigenvalue Ev is v + 1. We can verify that the
same degeneracy is obtained with a Cartesian representation of the 2-D harmonic oscillator
solutions as shown in Table XI.
A.
Two dimensional quartic oscillator
In two dimensions the quartic oscillator potential can be written as U =
aEho
(r/r0 )4
2
with
a a dimensionless number proportional to the energy scale which we take as the ground state
p
of the harmonic oscillator Eho = h̄ω, and r0 = h̄/mω is the characteristic length scale.
The frequency ω sets the scale for the energy and the spatial dependence. The Schrödinger
equation is
h̄2
−
2m
∂ 2ψ 1 ∂ψ
1 ∂ 2ψ
+
+
∂r2
r ∂r
r2 ∂φ2
41
+
aEho 4
r ψ = Eψ.
2r04
Introducing ρ = r/r0 we get
2
1 ∂ 2ψ
2E
∂ ψ 1 ∂ψ
4
+
+
+
− aρ ψ = 0.
∂ρ2
ρ ∂ρ ρ2 ∂φ2
Eho
The quartic term is diagonal in the azimuthal variable so l remains a good quantum
number and we can seek approximate solutions in the form
ψl (ρ, φ) = Φl (φ)
N
X
cv Rvl (ρ).
v=0
This leads to the equation
N X
2E
4
2
−aρ + ρ − 2(v + 1) +
cv Rvl = 0.
Eho
v=0
(56)
To convert this into a set of linear equations for the coefficients cv we use the integrals
Z ∞
(n + k)!
dx Lkm (x)Lkn (x)xk e−x =
δmn ,
n!
0
Z ∞
2
(n + k)!
dx Lkn (x) xk+1 e−x =
(2n + k + 1),
(57)
n!
0
together with the recurrence relation
(n + 1)Lkn+1 (x) = (2n + k + 1 − x)Lkn (x) − (n + k)Lkn−1 (x).
It is then readily shown that
Z ∞
dρ ρ Rml Rnl = d0 δm,n ,
0
Z ∞
dρ ρ Rml ρ2 Rnl = d0 d00 δm,n + d0 d01+ δm,n+1 + d0 d01− δm,n−1
Z0 ∞
dρ ρ Rml ρ4 Rnl = d0 d000 δm,n + d0 d001+ δm,n+1 + d0 d001− δm,n−1
(58)
0
+ d0 d002+ δm,n+2 + d0 d002− δm,n−2
(59)
where
1 (n + |l|)!
,
2
n!
d00 = (2n + |l| + 1),
d0 =
d01+ = −(n + 1),
d000 = [(k + 2)(k + 1) + 6n(n + k + 1)],
d001− = −2(2n + |l|)(n + |l|),
d01− = −(n + |l|)
d001+ = −2(2n + |l| + 2)(n + 1),
d002+ = (n + 2)(n + 1),
d002− = (n + |l|)(n + |l| − 1).
(60)
42
40
35
Enl/Eho
30
25
20
15
10
2
1
5
l=0
0
1
0
2
3
4
n
FIG. 4. Energy eigenvalues for the 2d quartic oscillator with a = 1, scaled to Eho = 2.
With these results we transform Eq. (56) into the algebraic equations
Mc = λc
(61)
where λ = 2(|l| + 1) − 2E/Eho and c = (c0 , c1, ..., cN ) is the vector of coefficients. The energy
eigenvalues are therefore given by
Enl = Eho
λnl
|l| + 1 −
2
where λnl is the nth eigenvalue of Eq. (61) evaluated with azimuthal quantum number l.
The elements of the matrix M are
Mmn = (−4n + d00 − ad000 ) δm,n
+ d01+ − ad001+ δm,n+1 + d01− − ad001− δm,n−1
−ad002+ δm,n+2 − ad002− δm,n−2 .
The first few energy values shown in Fig. 4 for Eho = 2 and a = 1 are E00 = 2.345, E10 =
9.530, E20 = 18.74, E01 = 5.394, E11 = 13.81, E21 = 23.78, E02 = 8.928, E12 = 18.31, E22 =
28.96.
43
44