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)