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X (t) =
m n

m |ρ (t) X| m
=
m

= Tr [ρ (t) X] , (2.117)
½
States in quantum theory

where the trace operation is de¬ned by eqn (C.22). Each of the ket vectors |Ψe in
the ensemble evolves according to the Schr¨dinger equation, and the bra vectors Ψe |
o
obey the conjugate equation

’i Ψe (t)| = Ψe (t)| H , (2.118)
‚t
so the evolution equation for the density operator is

i ρ (t) = [H, ρ (t)] . (2.119)
‚t
By analogy with the Liouville equation for the classical distribution function (Huang,
1963, Sec. 4.3), this is called the quantum Liouville equation. The condition
(2.112), together with the normalization of the ensemble state vectors, means that
the density operator has unit trace,

Tr (ρ (t)) = 1 , (2.120)

and eqn (2.119) guarantees that this condition is valid at all times.
A pure state is described by an ensemble consisting of exactly one vector, so that
eqn (2.116) reduces to
ρ (t) = |Ψ (t) Ψ (t)| . (2.121)
This explicit statement can be replaced by the condition that ρ (t) is a projection
operator, i.e.
ρ2 (t) = ρ (t) . (2.122)
Thus for pure states
Tr ρ2 (t) = Tr (ρ (t)) = 1 , (2.123)
while for mixed states
Tr ρ2 (t) < 1 . (2.124)
For any observable X and any state ρ, either pure or mixed, an important statistical
property is given by the variance
2
V (X) = X 2 ’ X , (2.125)
2
where X = Tr (ρX). The easily veri¬ed identity V (X) = (X ’ X ) shows that
V (X) 0, and it also follows that V (X) = 0 when ρ is an eigenstate of X, i.e.
Xρ = ρX = »ρ. Conversely, every eigenstate of X satis¬es V (X) = 0. Since V (X)
is non-negative, the variance is often described in terms of the root mean square
(rms) deviation
2
X2 ’ X
∆X = V (X) = . (2.126)
¾ Quantization of cavity modes

B Mixed states arising from measurements
In quantum theory the act of measurement can produce a mixed state, even if the state
before the measurement is pure. For simplicity, we consider an observable X with a
discrete, nondegenerate spectrum. This means that the eigenvectors |xn , satisfying
X |xn = xn |xn , are unique (up to a phase). Suppose that we have complete infor-
mation about the initial state of the system, so that we can describe it by a pure state
|ψ . When a measurement of X is carried out, the Born interpretation tells us that
the eigenvalue xn will be found with probability pn = | xn |ψ |2 . The von Neumann
projection postulate further tells us that the system will be described by the pure state
|xn , if the measurement yields xn . This is the reduction of the wave packet. Now con-
sider the following situation. We know that a measurement of X has been performed,
but we do not know which eigenvalue of X was actually observed. In this case there is
no way to pick out one eigenstate from the rest. Thus we have an ensemble consisting
of all the eigenstates of X, and the density operator for this ensemble is

pn |xn xn | .
ρmeas = (2.127)
n

Thus a measurement will change the original pure state into a mixed state, if the
knowledge of which eigenvalue was obtained is not available.

2.3.3 General properties of the density operator
So far we have only considered observables with nondegenerate eigenvalues, but in
general some of the eigenvalues xξ of X are degenerate, i.e. there are several linearly
independent solutions of the eigenvalue problem X |Ψ = xξ |Ψ . The number of solu-
tions is the degree of degeneracy, denoted by dξ (X). A familiar example is X = J 2 ,
where J is the angular momentum operator. The eigenvalue j (j + 1) 2 of J 2 has the
degeneracy 2j + 1 and the degenerate eigenstates can be labeled by the eigenvalues
m of Jz , with ’j m j. An example appropriate to the present context is the
operator
a† aks ,
Nk = (2.128)
ks
s

that counts the number of photons with wavevector k. If k has no vanishing com-
ponents, the eigenvalue problem Nk |Ψ = |Ψ has two independent solutions corre-
sponding to the two possible polarizations, so d1 (Nk ) = 2. In general, the common
eigenvectors for a given eigenvalue span a dξ (X)-dimensional subspace, called the
eigenspace Hξ (X). Let
|Ψξ1 , . . . , Ψξdξ (X) (2.129)

be a basis for Hξ (X), then
|Ψξm Ψξm |
Pξ = (2.130)
m

is the projection operator onto Hξ (X).
¿
States in quantum theory

According to the standard rules of quantum theory (see eqns (C.26)“(C.28)) the
conditional probability that xξ is the result of a measurement of X, given that the
system is described by the pure state |Ψe , is

| Ψξm |Ψe |2 = Ψe |Pξ | Ψe .
p (xξ |Ψe ) = (2.131)
m

For the mixed state the overall probability of the result xξ is, therefore,

| Ψξm |Ψe |2 = Tr (ρPξ ) .
Pe
p (xξ ) = (2.132)
e m

Thus the general rule is that the probability for ¬nding a given value xξ is given by the
expectation value of the projection operator Pξ onto the corresponding eigenspace.
Other important mathematical properties of the density operator follow directly
from the de¬nition (2.116). For any state |Ψ , the expectation value of ρ is positive,

Pe | Ψe |Ψ |2
Ψ |ρ| Ψ = 0, (2.133)
e

so ρ is a positive-de¬nite operator. Combining this with the normalization condition
Ψ |ρ| Ψ 1 for any normalized state |Ψ . The Born interpretation
(2.120) implies 0
2
tells us that | Ψe |Ψ | is the probability that a measurement”say of the projection
operator |Ψ Ψ|”will leave the system in the state |Ψ , given that the system is
prepared in the pure state |Ψe ; therefore, eqn (2.133) tells us that Ψ |ρ| Ψ is the
probability that a measurement will lead to |Ψ , if the system is described by the
mixed state with density operator ρ.
In view of the importance of the superposition principle for pure states, it is natural
to ask if any similar principle applies to mixed states. The ¬rst thing to note is that
linear combinations of density operators are not generally physically acceptable density
operators. Thus if ρ1 and ρ2 are density operators, the combination ρ = Cρ1 +Dρ2 will
be hermitian only if C and D are both real. The condition Tr ρ = 1 further requires
C + D = 1. Finally, the positivity condition (2.133) must hold for all choices of |Ψ ,
and this can only be guaranteed by imposing C 0 and D 0. Therefore, only the
convex linear combinations

ρ = Cρ1 + (1 ’ C) ρ2 , 0 C 1 (2.134)

are guaranteed to be density matrices. This terminology is derived from the mathe-
matical notion of a convex set in the plane, i.e. a set that contains every straight line
joining any two of its points. The general form of eqn (2.134) is

ρ= Cn ρn , (2.135)
n

where each ρn is a density operator, and the coe¬cients satisfy the convexity condition

0 Cn 1 for all n and Cn = 1 . (2.136)
n
Quantization of cavity modes

The o¬-diagonal matrix elements of the density operator are also constrained by the
de¬nition (2.116). The normalization of the ensemble states |Ψe implies | Ψe |Ψ | 1,
so

| Ψ |ρ| ¦ | = Pe Ψ |Ψe Ψe |¦
e

Pe | Ψ |Ψe | | Ψe |¦ | 1, (2.137)
e

i.e. ρ is a bounded operator.
The arguments leading from the ensemble de¬nition of the density operator to its
properties can be reversed to yield the following statement. An operator ρ that is
(a) hermitian, (b) bounded, (c) positive, and (d) has unit trace is a possible density
operator. The associated ensemble can be de¬ned as the set of normalized eigenstates
of ρ corresponding to nonzero eigenvalues. Since every density operator has a complete
orthonormal set of eigenvectors, this last remark implies that it is always possible to
choose the ensemble to consist of mutually orthogonal states.

2.3.4 Degrees of mixing
So far the distinction between pure and mixed states is absolute, but ¬ner distinc-
tions are also useful. In other words, some states are more mixed than others. The
distinctions we will discuss arise most frequently for physical systems described by a
¬nite-dimensional Hilbert space, or equivalently, ensembles containing a ¬nite number
of pure states. This allows us to simplify the analysis by assuming that the Hilbert
space has dimension d < ∞. The inequality (2.124) suggests that the purity
P (ρ) = Tr ρ2 1 (2.138)
may be a useful measure of the degree of mixing associated with a density operator ρ.
By virtue of eqn (2.122), P (ρ) = 1 for a pure state; therefore, it is natural to say that
the state ρ2 is less pure (more mixed) than the state ρ1 if P (ρ2 ) < P (ρ1 ). Thus the
minimally pure (maximally mixed) state for an ensemble will be the one that achieves
the lower bound of P (ρ). In general the density operator can have the eigenvalue
0 with degeneracy (multiplicity) d0 < d, so the number of orthogonal states in the
ensemble is N = d ’ d0 . Using the eigenstates of ρ to evaluate the trace yields
N
P (ρ) = p2 , (2.139)
n
n=1

where pn is the nth eigenvalue of ρ. In this notation, the trace condition (2.120) is just
N
pn = 1 , (2.140)
n=1

and the lower bound is found by minimizing P (ρ) subject to the constraint (2.140).
This can be done in several ways, e.g. by the method of Lagrange multipliers, with
the result that the maximally mixed state is de¬ned by
Mixed states of the electromagnetic ¬eld


, n = 1, . . . , N ,
1
N
pn = (2.141)
0 , n = N + 1, . . . , d .
In other words, the pure states in the ensemble de¬ning the maximally mixed state
occur with equal probability, and the purity is
1
P (ρ) =
. (2.142)
N
Another useful measure of the degree of mixing is provided by the von Neumann
entropy, which is de¬ned in general by
S (ρ) = ’ Tr (ρ ln ρ) . (2.143)
In the special case considered above, the von Neumann entropy is given by
N
S (ρ) = ’ pn ln pn , (2.144)
n=1

and maximizing this”subject to the constraint (2.140)”leads to the same de¬nition
of the maximally mixed state, with the value
S (ρ) = ln N (2.145)
of S (ρ). The von Neumann entropy plays an important role in the study of entangled
states in Chapter 6.

2.4 Mixed states of the electromagnetic ¬eld
2.4.1 Polarized light
As a concrete example of a mixed state, consider an experiment in which light from a
single atom is sent through a series of collimating pinholes. In each atomic transition,
exactly one photon with frequency ω = ∆E/ is emitted, where ∆E is the energy
di¬erence between the atomic states. The alignment of the pinholes determines the
unit vector k along the direction of propagation, so the experimental arrangement
determines the wavevector k = (ω/c) k. If the pinholes are perfectly circular, the
experimental preparation gives no information on the polarization of the transmitted
light. This means that the light observed on the far side of the collimator could be
described by either of the states
|Ψs = |1ks = a† |0 , (2.146)
ks

where s = ±1 labels right- and left-circularly-polarized light. Thus the relevant en-
semble is composed of the states |1k+ and |1k’ , with probabilities P+ and P’ , and
the density operator is
P+ 0
Ps |1ks 1ks | =
ρ= . (2.147)
P’
0
s

In the absence of any additional information equal probabilities are assigned to the
two polarizations, i.e. P+ = P’ = 1/2, and the light is said to be unpolarized. The
Quantization of cavity modes

opposite extreme occurs when the polarization is known with certainty, for example
P+ = 1, P’ = 0. This can be accomplished by inserting a polarization ¬lter after
the collimator. In this case, the light is said to be polarized, and the density operator
represents the pure state |1k+ . For the intermediate cases, a measure of the degree of
polarization is given by
P = |P+ ’ P’ | , (2.148)
which satis¬es 0 P 1, and has the values P = 0 for unpolarized light and P = 1
for polarized light.

A The second-order coherence matrix
The conclusions reached for the special case discussed above are also valid in a more
general setting (Mandel and Wolf, 1995, Sec. 6.2). We present here a simpli¬ed version
of the general discussion by de¬ning the second-order coherence matrix

Jss = Tr ρa† aks , (2.149)
ks

where the density operator ρ describes a monochromatic state, i.e. each state vector
|Ψe in the ensemble de¬ning ρ satis¬es ak s |Ψe = 0 for k = k. In this case we may
as well choose the z-axis along k, and set s = x, y, corresponding to linear polarization
vectors along the x- and y-axes respectively. The 2 — 2 matrix J is hermitian and posi-
tive de¬nite”see Appendix A.3.4”so the eigenvectors cp = (cpx , cpy ) and eigenvalues
np (p = 1, 2) de¬ned by
Jcp = np cp (2.150)
satisfy
c† cp = δpp and np 0. (2.151)
p
The eigenvectors of J de¬ne eigenpolarization vectors,
e p = c — e x + c— e y , (2.152)
px py

and corresponding creation and annihilation operators
a† = c— a† + c— a† , ap = cpx akx + cpy aky . (2.153)
px kx py ky
p

It is not di¬cult to show that
np = Tr ρa† ap , (2.154)
p

i.e. the eigenvalue np is the average number of photons with eigenpolarization ep . If ρ
describes an unpolarized state, then di¬erent polarizations must be uncorrelated and
the number of photons in either polarization must be equal, i.e.
n1 0
J= , (2.155)
20 1
where
n = Tr ρ a† akx + a† aky (2.156)
kx ky

is the average total number of photons. If ρ describes complete polarization, then
the occupation number for one of the eigenpolarizations must vanish, e.g. n2 = 0.
Mixed states of the electromagnetic ¬eld

Since det J = n1 n2 , this means that completely polarized states are characterized by
det J = 0. In this general setting, the degree of polarization is de¬ned by
|n1 ’ n2 |
P= , (2.157)
n1 + n2
where P = 0 and P = 1 respectively correspond to unpolarized and completely polar-
ized light.

B The Stokes parameters
Since J is a 2 — 2 matrix, we can exploit the well known fact”see Appendix C.3.1”
that any 2 — 2 matrix can be expressed as a linear combination of the Pauli matrices.
For this application, we write the expansion as
1 1 1 1
S0 σ0 + S1 σ3 + S2 σ1 ’ S3 σ2 ,
J= (2.158)
2 2 2 2
where σ0 is the 2 — 2 identity matrix and σ1 , σ2 , and σ3 are the Pauli matrices given
by the standard representation (C.30). This awkward formulation guarantees that the
c-number coe¬cients Sµ are the traditional Stokes parameters. According to eqn
(C.40) they are given by

S0 = Tr (Jσ0 ) , S1 = Tr (Jσ3 ) , S2 = Tr (Jσ1 ) , S3 = ’ Tr (Jσ2 ) . (2.159)

The Stokes parameters yield a useful geometrical picture of the coherence matrix, since
the necessary condition det (J) 0 translates to
2 2 2 2
S 1 + S2 + S3 S0 . (2.160)

If we interpret (S1 , S2 , S3 ) as a point in a three-dimensional space, then for a ¬xed
value of S0 the states of the ¬eld occupy a sphere”called the Poincar´ sphere”of
e
radius S0 . The origin, S1 = S2 = S3 = 0, corresponds to unpolarized light, since this
is the only case for which J is proportional to the identity. The condition det J = 0
for completely polarized light is
2 2 2 2
S 1 + S2 + S3 = S0 , (2.161)

which describes points on the surface of the sphere. Intermediate states of polarization
correspond to points in the interior of the sphere.
The Poincar´ sphere is often used to describe the pure states of a single photon,
e
e.g.
Cs a† |0 .
|ψ = (2.162)
ks
s

In this case S0 = 1, and the points on the surface of the Poincar´ sphere can be labeled
e
by the standard spherical coordinates (θ, φ). The north pole, θ = 0, and the south pole,
θ = π, respectively describe right- and left-circular polarizations. Linear polarizations
are represented by points on the equator, and elliptical polarizations by points in the
northern and southern hemispheres.
Quantization of cavity modes

2.4.2 Thermal light
A very important example of a mixed state arises when the ¬eld is treated as a thermo-
dynamic system in contact with a thermal reservoir at temperature T , e.g. the walls
of the cavity. Under these circumstances, any complete set of states can be chosen
for the ensemble, since we have no information that allows the exclusion of any pure
state of the ¬eld. Exchange of energy with the walls is the mechanism for attaining
thermal equilibrium, so it is natural to use the energy eigenstates”i.e. the number
states |n ”for this purpose.
The general rules of statistical mechanics (Chandler, 1987, Sec. 3.7) tell us
that the probability for a given energy E is proportional to exp (’βE), where
β = 1/kB T and kB is Boltzmann™s constant. Thus the probability distribution is
Pn = Z ’1 exp ’βEn , where Z ’1 is the normalization constant required to satisfy
eqn (2.112), and
En = ω κ nκ . (2.163)
κ

Substituting this probability distribution into eqn (2.116) gives the density operator

1 1
e’βEn |n n| =
ρ= exp (’βHem ) . (2.164)
Z Z
n

The normalization constant Z, which is called the partition function, is determined
by imposing Tr (ρ) = 1 to get

Z = Tr [exp (’βHem )] . (2.165)

Evaluating the trace in the number-state basis yields


exp ’β
Z= nκ ω κ = Zκ , (2.166)

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