Physics 125cCourse Notes

Density Matrix Formalism040511 Frank Porter

1 Introduction

In this note we develop an elegant and powerful formulation of quantum me-chanics, the “density matrix” formalism. This formalism provides a structurein which we can address such matters as:

• We typically assume that it is permissible to work within an appropriatesubspace of the Hilbert space for the universe. Is this all right?

• In practice we often have situations involving statistical ensembles ofstates. We have not yet addressed how we might deal with this.

2 The Density Operator

Suppose that we have a state space, with a denumerable orthonormal basis{|un〉, n = 1, 2, . . .}. If the system is in state |ψ(t)〉 at time t, we have theexpansion in this basis:

|ψ(t)〉 =∑

n

an(t)|un〉. (1)

We’ll assume that |ψ(t)〉 is normalized, and hence:

〈ψ(t)|ψ(t)〉 = 1 =∑

n

∑

m

an(t)a∗m(t)〈um|un〉

=∑

n

|an(t)|2 (2)

Suppose that we have an observable (self-adjoint operator)Q. The matrixelements of Q in this basis are:

Qmn = 〈um|Qun〉 = 〈Qum|un〉 = 〈um|Q|un〉. (3)

The average (expectation) value of Q at time t, for the system in state |ψ(t)〉is:

〈Q〉 = 〈ψ(t)|Qψ(t)〉 =∑

n

∑

m

a∗m(t)an(t)Qmn. (4)

1

We see that 〈Q〉 is an expansion quadratic in the {an} coefficients.Consider the operator |ψ(t)〉〈ψ(t)|. It has matrix elements:

〈um|ψ(t)〉〈ψ(t)|un〉 = am(t)a∗n(t). (5)

These matrix elements appear in the calculation of 〈Q〉. Hence, define

ρ(t) ≡ |ψ(t)〉〈ψ(t)|. (6)

We call this the density operator. It is a Hermitian operator, with matrixelements

ρmn(t) = 〈um|ρ(t)un〉 = am(t)a∗n(t). (7)

Since ψ(t) is normalized, we also have that

1 =∑

n

|an(t)|2 =∑

n

ρnn(t) = Tr [ρ(t)] . (8)

We may now re-express the expectation value of observable Q using thedensity operator:

〈Q〉(t) =∑

m

∑

n

a∗m(t)a∗n(t)Qmn

=∑

m

∑

n

ρnm(t)Qmn

=∑

n

[ρ(t)Q]nn

= Tr [ρ(t)Q] . (9)

The time evolution of a state is given by the Schrodinger equation:

id

dt|ψ(t)〉 = H(t)|ψ(t)〉, (10)

where H(t) is the Hamiltonian. Thus, the time evolution of the densityoperator may be computed according to:

d

dtρ(t) =

d

dt[|ψ(t)〉〈ψ(t)|]

=1

iH(t)|ψ(t)〉〈ψ(t)| − 1

i|ψ(t)〉〈ψ(t)|H(t)

=1

i[H(t), ρ(t)] (11)

2

Suppose we wish to know the probability, P ({q}), that a measurement ofQ will yield a result in the set {q}. We compute this probability by projectingout of |ψ(t)〉 that protion which lies in the eigensubspace associated withobservables in the set {q}. Let P{q} be the projection operator. Then:

P ({q}) = 〈ψ(t)|P{q}ψ(t)〉= Tr

[P{q}ρ(t)

]. (12)

We note that the density operator, unlike the state vector, has no phaseambiquity. The same state is described by |ψ(t)〉 and |ψ′(t)〉 = eiθ|ψ(t)〉.Under this phase transformation, the density operator transforms as:

ρ(t) → ρ′(t) = eiθ|ψ(t)〉〈ψ(t)|e−iθ

= ρ(t). (13)

Furthermore, expectaion values are quadratic in |ψ(t)〉, but only linear inρ(t).

For the density operators we have been considering so far, we see that:

ρ2(t) = |ψ(t)〉〈ψ(t)||ψ(t)〉〈ψ(t)|= ρ(t). (14)

That is, ρ(t) is an idempotent operator. Hence,

Trρ2(t) = Trρ(t) = 1. (15)

Finally, notice that:

〈un|ρ(t)un〉 = ρnn(t) = |an(t)|2 ≥ 0 ∀n. (16)

Thus, for an arbitrary state |φ〉, 〈φ|ρ(t)φ〉 ≥ 0, as may be demonstrated byexpanding |φ〉 in the |u〉 basis. We conclude that ρ is a non-negative definiteoperator.

We postulate, in quantum mechanics, that the states of a system are inone-to-one correspondence with the non-negative definite density operatorsof trace 1 (defined on the Hilbert space).

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3 Statistical Mixtures

We may wish to consider cases where the system is in any of a number ofdifferent states, with various probabilities. The system may be in state |ψ1〉with probability p1, state |ψ2〉 with probability p2, and so forth (more gener-ally, we could consider states over some arbitrary, possibly non-denumerable,index set). We must have 1 ≥ pi ≥ 0 for i ∈ {index set}, and

∑i pi = 1.

Note that this situation is not the same thing as supposing that we are inthe state |ψ〉 = p1|ψ1〉+p2|ψ2〉+ · · · (or even with

√p1, etc.). Such statistical

mixtures might occur, for example, when we prepare a similar system (anatom, say) many times. In general, we will not be able to prepare the sameexact state every time, but will have some probability distribution of states.

We may ask, for such a system, for the probability P ({q}) that a mea-surement of Q will yield a result in the set {q}. For each state in our mixture,we have

Pn({q}) = 〈ψn|P{q}ψn〉= Tr

(ρnP{q}

), (17)

where ρn = |ψn〉〈ψn|. To determine the overall probability, we must sum overthe individual probabilities, weighted by pn:

P ({q}) =∑

n

pnPn({q})

=∑

n

pnTr(ρnP{q}

)

= Tr

(∑

n

pnρnP{q}

)

= Tr(ρP{q}

), (18)

whereρ ≡

∑

n

pnρn. (19)

Now ρ is the density operator of the system, and is a simple linear combina-tion of the individual density operators. Note that ρ is the “average” of theρn’s with respect to probability distribution pn.

Let us investigate this density operator:

• Since ρn are Hermitian, and pn are real, ρ is Hermitian.

4

• Trρ = Tr (∑

n pnρn) =∑

n pnTrρn =∑

n pn = 1.

• ρ is non-negative-definite: 〈φ|ρφ〉 =∑

n pn〈φ|ρnφ〉 ≥ 0.

• Let Q be an operator with eigenvalues qn. In the current situation, 〈Q〉refers to the average of Q over the statistical mixture. We have:

〈Q〉 =∑

n

qnP ({qn}) =∑

n

qnTr(ρP{qn}

)

= Tr

(ρ∑

n

qnP{qn}

)

= Tr(ρQ), since Q =∑

n

qnP{qn}. (20)

• We may determine the time evolution of ρ. For ρn(t) = |ψn(t)〉〈ψn(t)|we know (Eqn. 11) that

idρn(t)

dt= [H(t), ρn(t)] . (21)

Since ρ(t) is linear in the ρn, ρ(t) =∑

n pnρn(t), we have

idρ(t)

dt= [H(t), ρ(t)] . (22)

• Now look at

ρ2 =∑

m

∑

n

pmpnρmρn

=∑

m

∑

n

pmpn|ψm〉〈ψm|ψn〉〈ψn|

6= ρ, in general. (23)

What about the trace of ρ2? Let

|ψm〉 =∑

j

(am)j|uj〉. (24)

Then

ρ2 =∑

m

∑

n

pmpn|ψm〉〈ψm|ψn〉〈ψn|

=∑

m

∑

n

pmpn

∑

i

∑

j

(am)∗i (an)jδij

[∑

k

∑

`

(am)k(an)∗` |uk〉〈u`|]

=∑

m,n,i,k,`

pmpn(am)∗i (an)i(am)k(an)∗` |uk〉〈u`|. (25)

5

Let’s take the trace of this. Notice that Tr(|uk〉〈u`|) = δk`, so that

Tr(ρ2) =∑

m,n,i,k

pmpn(am)∗i (an)i(am)k(an)∗k. (26)

But 〈ψm|ψn〉 =∑

i(am)∗i (an)i, and thus:

Tr(ρ2) =∑

m

∑

n

pmpn|〈ψm|ψn〉|2

≤∑

m

∑

n

pmpn〈ψm|ψm〉〈ψn|ψn〉, (Schwarz inequality)

≤∑

m

pm

∑

n

pn

≤ 1. (27)

The reader is encouraged to check that equality holds if and only if thesystem can be in only one physical state (that is, all but one of the pn’scorresponding to independent states must be zero).

Note that, if Tr(ρ2) = 1, then ρ = |ψ〉〈ψ|, which is a projection operator.We encapsulate this observation into the definition:

Def: A state of a physical system is called a pure state if Tr(ρ2) = 1; thedensity operator is a projection. Otherwise, the system is said to be ina mixed state, or simply a mixture.

The diagonal matrix elements of ρ have a simple physical interpretation:

ρnn =∑

j

pj(ρj)nn

=∑

j

pj〈un|ψj〉〈ψj|un〉

=∑

j

pj|(aj)n|2. (28)

This is just the probability to find the system in state |un〉. Similarly, theoff-diagonal elements are

ρmn =∑

j

pj(aj)m(aj)∗n. (29)

The off-diagonal elements are called coherences. Note that it is possible tochoose a basis in which ρ is diagonal (since ρ is Hermitian). In such a basis,the coherences are all zero.

6

4 Measurements, Statistical Ensembles, and

Density Matrices

Having developed the basic density matrix formalism, let us now revisit it,filling in some motivational aspects. First, we consider the measurementprocess. It is useful here to regard an experiment as a two-stage process:

1. Preparation of the system.

2. Measurement of some physical aspect(s) of the system.

For example, we might prepare a system of atoms with the aid of spark gaps,magnetic fields, laser beams, etc., then make a measurement of the systemby looking at the radiation emitted. The distinction between preparationand measurement is not always clear, but we’ll use this notion to guide ourdiscussion.

We may further remark that we can imagine any measurement as a sortof “counter” experiment: First, consider an experiment as a repeated prepa-ration and measurement of a system, and refer to each measurement as an“event”. Think of the measuring device as an array of one or more “counters”that give a response (a “count”) if the variables of the system are within somerange. For example, we might be measuring the gamma ray energy spectrumin some nuclear process. We have a detector which absorbs a gamma rayand produces an electrical signal proportional to the absorbed energy. Thesignal is processed and ultimately sent to a multichannel analyzer (MCA)which increments the channel corresponding to the detected energy. In thiscase, the MCA is functioning as our array of counters.

The process is imagined to be repeated many times, and we are not con-cerned with issues of the statistics of finite counting here. The result of suchan experiment is expressed as the probability that the various counters willregister, given the appropriate preparation of the system. These probabilitiesmay include correlations.

Let us take this somewhat hazy notion and put it into more concretemathematical language: Associate with each counter a dichotomic vari-able, D, as follows:

If the counter registers in an event, D = 1.If the counter does not register in an event, D = 0.

We assert that we can, in principle, express all physical variables in terms ofdichotomic ones, so this appears to be a sufficiently general approach.

7

By repeatedly preparing the system and observing the counter D, we candetermine the probability that D registers: The average value of D, 〈D〉, isthe probability that D registers in the experiment. We refer to the particu-lar preparation of the system in the experiment as a statistical ensembleand call 〈D〉 the average of the dichotomic variable D with respect to thisensemble.

If we know the averages of all possible dichotomic variables, then the en-semble is completely known. The term “statistical ensemble” is synonymouswith a suitable set of averages of dichotomic variables (i.e., probabilities).Let us denote a statistical ensemble with the letter ρ. The use of the samesymbol as we used for the density matrix is not coincidental, as we shall see.The quantity 〈D〉ρ explicitly denotes the average of D for the ensemble ρ.Clearly:

0 ≤ 〈D〉ρ ≤ 1. (30)

D is precisely known for ensemble ρ if 〈D〉ρ = 0 or 〈D〉ρ = 1. Otherwise,variable D possesses a statistical spread. Note that we may prepare a system(for example, an atom) many times according to a given ensemble. However,this does not mean that the system is always in the same state.

We have the important concept of the superposition of two ensembles:Let ρ1 and ρ2 be two distinct ensembles. An ensemble ρ is said to be anincoherent superposition of ρ1 and ρ2 if there exists a number θ suchthat 0 < θ < 1, and for every dichotomic variable D we have:

〈D〉ρ = θ〈D〉ρ1 + (1 − θ)〈D〉ρ2. (31)

This is expressed symbolically as:

ρ = θρ1 + (1 − θ)ρ2, (32)

“ρ is a superposition of ρ1 and ρ2 with probabilities θ and 1 − θ.”We assume that if ρ1 and ρ2 are physically realizable, then any coherent

superposition of them is also physically realizable. For example, we mightprepare a beam of particles from two independent sources, each of whichmay hit our counter: ρ1 corresponds to source 1, ρ2 corresponds to source2. When both sources are on, the beam hitting the counter is an incoherentmixture of ρ1 and ρ2. We may compute the probability, P (1|hit), that aparticle hitting the counter is from beam 1. Using Bayes’ theorem:

P (1|hit) =P (hit|1)P (1)

P (hit)

8

=〈D〉ρ1θ

θ〈D〉ρ1 + (1 − θ)〈D〉ρ2

= θ〈D〉ρ1

〈D〉ρ. (33)

(34)

The generalization to an incoherent superposition of an arbitrary numberof ensembles is clear: Let ρ1, ρ2, . . . be a set of distinct statistical ensembles,and let θ1, θ2, . . . be a set of real numbers such that

θn > 0, and∑

n

θn = 1. (35)

The incoherent sum of these ensembles, with probabilities {θn} is denoted

ρ =∑

n

θnρn. (36)

This is to be interpreted as meaning that, for every dichotomic variable D:

〈D〉ρ =∑

n

θn〈D〉ρn. (37)

A particular prepared system is regarded as an element of the statisticalensemble. We have the intuitive notion that our level of information aboutan element from an ensemble ρ = θρ1 + (1 − θ)ρ2, which is an incoherentsuperposition of distinct ensembles ρ1 and ρ2, is less than our informationabout an element in either ρ1 or ρ2. For example, consider D a dichotomicvariable such that 〈D〉ρ1 6= 〈D〉ρ2. Such a variable must exist, since ρ1 6= ρ2.We have:

〈D〉ρ = θ〈D〉ρ1 + (1 − θ)〈D〉ρ2. (38)

Consider

〈D〉ρ −1

2= θ(〈D〉ρ1 −

1

2) + (1 − θ)(〈D〉ρ2 −

1

2). (39)

We find:∣∣∣∣〈D〉ρ −

1

2

∣∣∣∣ ≤ θ

∣∣∣∣〈D〉ρ1 −1

2

∣∣∣∣+ (1 − θ)

∣∣∣∣〈D〉ρ2 −1

2

∣∣∣∣

< max(∣∣∣∣〈D〉ρ1 −

1

2

∣∣∣∣ ,∣∣∣∣〈D〉ρ2 −

1

2

∣∣∣∣). (40)

9

What does this result tell us? The quantity |〈D〉ρ − 12| ∈ [0, 1

2] can be

regarded as a measure of the information we have about variable D for en-semble ρ. For example, if |〈D〉ρ − 1

2| = 1

2, then 〈D〉ρ = 1 or 0, and D is

precisely known for ensemble ρ. On the other hand, if |〈D〉ρ − 12| = 0, then

〈D〉ρ = 1/2, and each of the possibilities D = 0 and D = 1 is equally likely,corresponding to maximal ignorance about D for ensemble ρ. Thus, our in-equality says that, for at least one of ρ1 and ρ2, we know more about D thanfor the incoherent superposition ρ.

We may restate our definition of pure and mixed states:

Def: A pure ensemble (or pure state) is an ensemble which is not anincoherent superposition of any other two distinct ensembles. A mixedensemble (or mixed state) is an ensemble which is not pure.

Intuitively, a pure ensemble is a more carefully prepared ensemble – we havemore (in fact, maximal) information about the elements – than a mixedensemble.

The set of all physical statistical ensembles is a convex set,1 with an inco-herent superposition of two ensembles a convex combination of two elementsof the convex set. Pure states are the extreme points of the set – i.e., pointswhich are not convex combinations of other points.

So far, this discussion has been rather general, and we have not madeany quantum mechanical assumptions. In fact, let us think about classical

1Convex set: A subset K ⊂ Cn of n-dimensional complex Euclidean space is convexif, given any two points α, β ∈ K, the straight line segment joining α and β is entirelycontained in K:

. .

(a) (b) (c)

αβ

(a) Not a convex set. (b) A convex set. (c) A convex set: Any convex combination ofα, β, x = θα + (1 − θ)β, where 0 < θ < 1 is an element of the set.

10

mechanics first. In classical physics, the pure states correspond to a completeabsence of any statistical spread in the dichotomic variables. If a preparationyields a pure state, then a repeated measurement of any variable will alwaysyield the same result, either 0 or 1. Experimentally, this does not seem to bethe case. Instead, no matter how carefully we prepare our ensemble, therewill always be at least one dichotomic variable D such that 〈D〉 = 1/2, corre-sponding to maximal statistical spread. Quantum mechanics (ignoring nowissues of superselection rules) also deals with the nature of dichotomic vari-ables and the set of ensembles, in a way which agrees so far with experiment.Let us restate some earlier postulates of quantum mechanics, modified andexpanded in this context:

1. To every physical system we associate a Hilbert space H. The pureensembles of the system are in 1:1 correspondence with the set of allone-dimensional projections in H. Such a projection, P , is an operatoron the Hilbert space satisfying (A):

(A)

P 2 = P idempotent,P † = P Hermitian,Tr(P ) = 1 “primitive”, or one-dimensional.

(41)

The set of all such projections is in one-to-one correspondence with theset of all rays2 in H. Alternatively, we say that there is a one-to-onecorrespondence between the rays and the pure states.

Given any ray R, we can pick a unit vector φ ∈ R, and the idempotentP associated with R is

(B) P = |φ〉〈φ|. (42)

Conversely, any idempotent with the properties (A) can also be writtenin the form (B).

Proof: We assume (see Exercises) that it has been demonstrated thatany linear operator in an n-dimensional Euclidean space may beexpressed as an n-term dyad, and that the extension of this ideato an infinite-dimensional separable space has been made. Hence,we may write:

P =∑

i

|ai〉〈bi|. (43)

2A ray is the set of all non-zero multiples of a given non-zero vector. Such a multipleis called an element of the ray.

11

Note that in some orthonormal basis {|ei〉}, the matrix elementsof P are Pij = 〈ei|P |ej〉, and hence,

P =∑

i,j

|ei〉Pij〈ej|. (44)

In the present case, P is Hermitian and therefore diagonalizable.Let {|ei〉} be a basis in which P is diagonal:

P =∑

i

|ei〉Pii〈ei|. (45)

Since P † = P , the Pii are all real. Calculate:

P 2 =∑

i,j

|ei〉Pii〈ei|ej〉Pjj〈ej|

=∑

i

|ei〉P 2ii〈ei|

= P, (46)

where the latter equality can be true if and only if P 2ii = Pii for

all i. That is, for each i we must either have Pii = 1 or Pii = 0.But we must also have Tr(P ) =

∑i Pii = 1, which holds if exactly

one Pii 6= 0, say Paa. In this basis,

P = |ea〉〈ea| (47)

The ray R associated with P is then {c|ea〉; c 6= 0}.

2. To every dichotomic variable D there corresponds a projection on somesubspace of H. That is, such a variable is represented by an operatorD on H satisfying:

D† = D (48)

D2 = D 6= 0, (49)

the latter since the eigenvalues of D are 0 and 1.

3. The average of D in pure ensemble P (corresponding ot projection P )is:

〈D〉P = Tr(DP ) (50)

(if P = |φ〉〈φ|, then 〈D〉P = 〈φ|D|φ〉.

12

4. An arbitrary ensemble ρ is represented by a statistical operator,or density matrix, which we also denote by symbol ρ. This is aHermitian operator on H with spectral decomposition,

ρ =∑

i

riPi, (51)

where

PiPj = δij (52)∑

i

Pi = I (53)

ri ≥ 0 (54)∑

i

ri = 1. (55)

The set {ri} is the set of eigenvalues of the operator ρ. The propertiesof this density matrix are precisely as in our earlier discussion.

Our symbolic equation for the incoherent superposition of two ensem-bles, ρ = θρ1 +(1− θ)ρ2, can be interpreted as an equation for the cor-responding density matrices represented by the same symbols. Hence,the density matrix ρ describing the superposition of ρ1 and ρ2 withprobabilities θ and 1−θ is ρ = θρ1 +(1−θ)ρ2. Thus, if ρ is any densitymatrix, and D any dichotomic variable, then:

〈D〉ρ = Tr(Dρ). (56)

For example,

〈D〉ρ = 〈D〉θρ1+(1−θ)ρ2(57)

= θ〈D〉ρ1 + (1 − θ)〈D〉ρ2

= Tr(Dθρ1) + Tr [D(1 − θ)ρ2]

= Tr {D [θρ1 + (1 − θ)ρ2]}= Tr(Dρ)

5. We regard every projection as corresponding to an observable, i.e.,every primitive Hermitian idempotent P corresponds to an observable.If ρ is a density matrix, then

ρ = P ⇔ Tr(Pρ) = 1. (58)

13

Proof: Suppose ρ = P . Then Tr(P P ) = Tr(P ), since P 2 = P . ButTrP = 1, since P is primitive. Now suppose Tr(Pρ) = 1. Then

1 = Tr

(P∑

i

riPi

)

=∑

i

riTr(PPi). (59)

Expand the one-dimensional projection operator in the basis inwhich Pi = |ei〉〈ei|:

P =∑

j,k

|ej〉〈ej|P |ek〉〈ek|. (60)

Then:

1 =∑

i

riTr

∑

j,k

|ej〉〈ej|P |ek〉〈ek|ei〉〈ei|

=∑

i

ri

∑

j

〈ej|P |ei〉Tr (|ej〉〈ei|)

=∑

i

ri〈ei|P |ei〉. (61)

But we also have∑

i〈ei|P |ei〉 = 1 and∑

i ri = 1, with 0 ≤ ri ≤ 1.Thus,

∑i ri〈ei|P |ei〉 < 1, unless there is a k such that rk = 1,

and all of the other ri = 0, i 6= k. Hence, 〈ek|P |ek〉 = 1, orP = |ek〉〈ek| = ρ.

Thus, P is the observable which tests whether an element of the sta-tistical ensemble is in the state corresponding to ray “P”.

6. In addition to the projection operators, we regard general self-adjointoperators as observables, and the laws of nature deal with these observ-ables. For example, we may consider operators with spectral resolutionsof the form:

Q =∑

i

qiPi =∑

i

qi|ei〉〈ei|, (62)

where PiPj = δijPi, and where the eigenvalues qi are real. We mayregard this as expressing the physical variable Q in terms of the di-chotomic variables Pi (noting that the eigenvalues of Pi are 0 and 1).

14

Hence it is natural to define the ensemble average of Q in an ensembleρ by:

〈Q〉ρ = 〈∑

i

qiPi〉ρ

=∑

i

qiTr(ρPi)

= Tr(ρQ). (63)

This completes our picture of the mathematical structure and postulatesof quantum mechanics in this somewhat new language. We see that weneed not discuss “state vectors” in quantum mechanics, we can talk about“ensembles” instead. In fact, the latter description has a more “physical”aspect, in the sense that experimentally we seem to be able to prepare systemsas statistical ensembles, but not so readily as pure states.

Of course, we have no proof that our experimental ensembles and di-chotomic variables must obey the above postulates. It may be that thereis some other theory which is more correct. However, there is so far no ex-perimental conflict with our orthodox theory, and we shall continue in thisvein.

5 Coherent Superpositions

Theorem: Let P1, P2 be two primitive Hermitian idempotents (i.e., rays, orpure states, with P † = P , P 2 = P , and TrP = 1). Then:

1 ≥ Tr(P1P2) ≥ 0. (64)

If Tr(P1P2) = 1, then P2 = P1. If Tr(P1P2) = 0, then P1P2 = 0 (vectorsin ray 1 are orthogonal to vectors in ray 2).

More generally, if ρ is a density matrix, and Q is any projection, then

1 ≥ Tr(Qρ) ≥ 0, (65)

Tr(Qρ) = 1 ⇔ Qρ = ρQ = ρ, (66)

Tr(Qρ) = 0 ⇔ Qρ = 0. (67)

Suppose we have orthogonal pure states, P1P2 = 0. There then exists aunique two parameter family of pure states {P} such that

Tr(PP1) + Tr(PP2) = 1. (68)

15

Any member P of this family is a ray corresponding to any vector in thetwo-dimensional subspace defined by the projection P1 + P2 = S. We saythat P is a coherent superposition of the pure states P1 and P2.

Let’s give an explicit construction of the operators P : Pick unit vector|e1〉 from ray P1 and |e2〉 from ray P2. Construct the following four operators:

S = P1 + P2 = |e1〉〈e1| + |e2〉〈e2| (69)

σ1 = |e1〉〈e2| + |e2〉〈e1| (70)

σ2 = i (|e2〉〈e1| − |e1〉〈e2|) (71)

σ3 = |e1〉〈e1| − |e2〉〈e2|. (72)

These operators satisfy the algebraic relations (noting the obvious similaritieswith the Pauli matrices):

S2 = S (73)

Sσi = σi (74)

σ2i = S (75)

[σi, σj] = iεijkσk. (76)

Let u = (u1, u2, u3) be a unit vector in three-dimensional Euclidean space.Define

P (u) ≡ 1

2(S + u · σσσ). (77)

The reader should demonstrate that P (u) is the most general coherent su-perposition of pure states P1 and P2. This set is parameterized by the two-parameter unit vector u. This, of course, is very characterstic of quantummechanics: If we have a “two-state” system we may form arbitrary super-positions |ψ〉 = α|ψ1〉 + β|ψ2〉 (assume 〈ψ1|ψ2〉 = 0). The overall phase isarbitrary, and the normalization constraint |α|2 + |β|2 = 1 uses another de-gree of freedom, hence two parameters are required to describe an arbitrarystate. Note that the coherent superposition of pure states is itself a purestate, unlike an incoherent superposition.

6 Density Matrices in a Finite-Dimensional

Hilbert Space

Consider a finite-dimensional Hilbert space H. The set of Hermitian opera-tors on H defines a real vector space (real, so that aQ is Hermitian if Q is

16

Hermitian). Call this vector space O (for vector space of Operators). Definea postive definite [(X,X) > 0 unless X = 0] symmetric [(X, Y ) = (Y,X)]scalar product on O by:

(X, Y ) ≡ Tr(XY ), (78)

for any two vectors (i.e., Hermitian operators) X, Y ∈ O. The set of alldensity matrices forms a convex subset of O, with norm ≤ 1.

Consider a complete orthonormal basis in O:

{B} = {B1, B2, . . .} ⊂ O such that Tr(BiBj) = δij. (79)

Expand any vector X ∈ O in this basis according to

X =∑

i

BiTr(BiX). (80)

For a density matrix ρ this expansion is

ρ =∑

i

BiTr(Biρ), (81)

but, as we have seen before, Tr(Biρ) = 〈Bi〉ρ is just the ensemble average ofobservable Bi in the ensemble ρ. Hence, the density matrix may be deter-mined through measurements, uniquely, if we measure the ensemble averagesof a complete set of operators.

7 Entropy, Mixing, Correlations

For this discussion, we need to first define the concept of a function of anoperator. Consider a self-adjoint operator Q, with a pure point spectrumconsisting of (real) eigenvalues {qi; i = 1, 2, . . .} and no finite point of accu-mulation.3 Let |k〉 denote the eigenvector corresponding to eigenvalue qk, andassume it has been normalized. Then {|k〉} forms a complete orthonormalset, i.e.:

〈k|j〉 = δkj; I =∑

k

|k〉〈k|. (82)

3Abstractly, a point of accumulation (or a limit point) is a point x ∈ S ⊂ T , whereT is a topological space, if every neighborhood N(x) contains a point of S distinct fromx.

17

The spectral resolution of Q is given by:

Q =∑

k

qk|k〉〈k|. (83)

Let Σ(Q) denote the spectrum {q} of Q. If f(q) is any function definedon Σ(Q), we define the operator f(Q) by:

f(Q) ≡∑

k

f(qk)|k〉〈k|. (84)

For example,Q2 =

∑

k

q2k|k〉〈k|, (85)

which may be compared with

Q2 =∑

k,j

qkqj|k〉〈k|j〉〈j| (86)

=∑

k

q2k|k〉〈k|,

which is what we hope should happen. In particular, we may perform Taylorseries expansions of functions of operators.

We wish to define a measure of the amount of (or lack of) informationconcerning the elements of a statistical ensemble ρ. Thus, define the entropys = s(ρ) by:

s ≡ −Tr(ρ ln ρ) (= −〈ln ρ〉ρ). (87)

Note that, with an expansion (spectral decomposition) of ρ according to

ρ =∑

i

riPi =∑

i

ri|ei〉〈ei|, (88)

we haveln ρ =

∑

i

(ln ri)Pi, (89)

and hence

s = −Tr

[∑

i

(ln ri)ρPi

]

= −∑

i

ln riTr(ρPi)

= −∑

i

ln riTr(∑

j

rjPjPi)

= −∑

i

ri ln ri. (90)

18

Since 0 ≤ ri ≤ 1, we always have s ≥ 0, and also s = 0 if and only if theensemble is a pure state. Roughly speaking, the more non-zero ri’s there are,that is the more the number of pure states involved, the greater the entropy.

Consistent with our classical thermodynamic notion that entropy in-creases with “mixing”, we have the “von Neumann mixing theorem”:

Theorem: If 0 < θ < 1, and ρ1 6= ρ2, then:

s [θρ1 + (1 − θ)ρ2] > θs(ρ1) + (1 − θ)s(ρ2). (91)

8 Combination of Systems

Consider the situation where the system of interest may be regarded as the“combination” of two subsystems, 1 and 2. For example, perhaps the systemconsists of two atoms. For simplicity of illustration, assume that the statesof system 1 alone form a finite-dimensional Hilbert space H1, and the statesof system 2 alone form another finite-dimensional Hilbert space H2. Thecombined system is then associated with Hilbert space H = H1 ⊗ H2. Forexample, we may have a two-dimensional space H1 and a three-dimensionalspace H2, with sets of vectors:

{(ab

)}and

αβγ

, (92)

respectively. Then the product space consists of direct product vectors ofthe form:

aαbαaβbβaγbγ

. (93)

The operators on H which refer only to subsystem 1 are of the form X⊗I,and the operators on H which refer only to subsystem 2 are of the form I⊗Y

19

(X is an operator on H1 and Y is an operator on H2). For example:

X ⊗ I =(x1 x2

x3 x4

)⊗

1 0 00 1 00 0 1

=

x1 x2

x3 x40 0

0x1 x2

x3 x40

0 0x1 x2

x3 x4

. (94)

We see that this operator does not mix up the components α, β, γ vectors inH2.

Consider now an operator on H of the special form Z = X ⊗ Y . Define“partial traces” for such an operator according to the mappings:

Tr1(Z) = Tr1(X ⊗ Y ) ≡ Y Tr(X) (95)

Tr2(Z) = Tr2(X ⊗ Y ) ≡ XTr(Y ) (96)

For our example:

Z = X ⊗ Y =(x1 x2

x3 x4

)⊗

y1 y2 y3

y4 y5 y6

y7 y8 y9

(97)

=

x1y1 x2y1 x1y2 x2y2 x1y3 x2y3

x3y1 x4y1 x3y2 x4y2 x3y3 x4y3

x1y4 x2y4 x1y5 x2y5 x1y6 x2y6

x3y4 x4y4 x3y5 x4y5 x3y6 x4y6

x1y7 x2y7 x1y8 x2y8 x1y9 x2y9

x3y7 x4y7 x3y8 x4y8 x3y9 x4y9

, (98)

and thus, for example,

Tr1(Z) = (x1 + x4)

y1 y2 y3

y4 y5 y6

y7 y8 y9

, (99)

and alsoTr [Tr1(Z)] = (x1 + x4)(y1 + y5 + y9) = Tr(Z). (100)

These mappings thus map operators on H of this from into operators on H1

or on H2.An arbitray linear operator on H may be expressed as a linear combina-

tion of operators of this form, and we extend the definition of Tr1 and Tr2 by

20

linearity to all operators on H. For example, suppose Z = X1⊗Y1 +X2⊗Y2.Then

Tr1(Z) = Tr1(X1 ⊗ Y1 +X2 ⊗ Y2)

= Tr1(X1 ⊗ Y1) + Tr1(X2 ⊗ Y2)

= Y1Tr(X1) + Y2Tr(X2), (101)

and the result is an operator on H2.Now let ρ be a density matrix on H, describing a statistical ensemble of

the combined system. Define “reduced density matrices” for subsystems 1and 2:

ρ1 ≡ Tr2(ρ), ρ2 ≡ Tr1(ρ). (102)

The interpretation is that ρ1 summarizes all of the information contained in ρabout the variables of subsystem 1 alone, and similarly for ρ2. For example,if X is any operator on system 1 alone:

〈X〉ρ = Tr [ρ(X ⊗ I)]

= 〈X〉ρ1 = Tr(Xρ1). (103)

From the reduced density matrices ρ1 and ρ2 we can form a new densitymatrix on H:

ρ12 = ρ1 ⊗ ρ2. (104)

It contains the same information which ρ1 and ρ2 contain together — ρ12

describes a statistical ensemble for which the variables of subsystem 1 arecompletely uncorrelated with the variables of subsystem 2. If ρ is not ofthis form (ρ 6= ρ12), then ρ describes an ensemble for which there is somecorrelation between the variables of the two subsystems.

For the entropy in particular, we have

s(ρ12) = s(ρ1 ⊗ ρ2) = s(ρ1) + s(ρ2). (105)

Proof: We can choose a basis in which ρ1 and ρ2 are diagonal, and in thisbasis ρ12 = ρ1 ⊗ ρ2 is also diagonal. Denote the diagonal elements ofρ1 as di, i.e., di ≡ (ρ1)ii, and the diagonal elements of ρ2 as δi. Thenthe diagonal elements of ρ12 are given by all products of the form diδj,where i = 1, 2, . . . , n1, and j = 1, 2, . . . , n2, and where n1 and n2 are

21

the dimensions of H1 and H2, respectively. Thus,

s(ρ12) = −Tr(ρ12 ln ρ12)

= −n1∑

i=1

n2∑

j=1

(diδj) ln(diδj). (106)

We compare this with (noting that Trρ1 = Trρ2 = 1):

s(ρ1) + s(ρ2) = −

n1∑

i=1

di ln di +n2∑

j=1

δj ln δj

= −

n1∑

i=1

n2∑

j=1

δjdi ln di +n1∑

i=1

n2∑

j=1

diδj ln δj

= −n1∑

i=1

n2∑

j=1

diδj(ln di + ln δj)

= s(ρ12). (107)

Thus, the entropy for an ensemble (ρ12) for which the subsystems areuncorrelated is just equal to the sum of the entropies of the reduced ensem-bles for the subsystems. When there are correlations, we should expect aninequality instead, since in this case ρ contains additional information con-cerning the correlations, which is not present in ρ1 and ρ2 (ρ12 = ρ1⊗ρ2 6= ρ).Then:

s(ρ12) = s(ρ1) + s(ρ2) ≥ s(ρ), (108)

where equality holds if and only if ρ = ρ12, that is, if there are no correlations.It is interesting that this inequality is specific for −x ln x, in the following

sense: Let s(ρ) = Tr [f(ρ)]. If this inequality, including the condition forequality, holds for all finite-dimensional Hilbert spaces H1 and H2, and alldensity matrices ρ on H = H1 ⊗ H2, then f(x) = −kx ln x, where k > 0(and we may take k = 1). Since this inequality appears to be determinedby physical considerations, this becomes a strong argument for the forms(ρ) = −Tr(ρ ln ρ) for the entropy.

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9 Some Statistical Mechanics

Consider a Hamiltonian H with point spectrum {ωi; i = 1, 2, . . .}, boundedbelow. The partition function, Z(T ), for temperature T > 0 is defined by:

Z(T ) ≡∞∑

k=1

e−ωk/T . (109)

We are assuming that this sum converges. The density matrix (or statis-tical operator) for the canonical distribution is given by:

ρ = ρ(T ) =e−H/T

Z(T )(110)

=1

Z(T )

∞∑

k=1

|k〉〈k|e−ωk/T . (111)

This makes intuitive sense – our canonical, thermodynamic distribution con-sists of a mixture of states, with each state receiving a “weight” of exp(−ωk/T ).Note that

Z(T ) =∞∑

k=1

e−ωk/T = Tr

( ∞∑

k=1

|k〉〈k|e−ωk/T

)

= Tr(e−H/T

). (112)

(113)

Hence, Tr [ρ(T )] = 1.The ensemble average of any observable (self-adjoint operator), Q, in the

canonical ensemble is:〈Q〉ρ = Tr [Qρ(T )] . (114)

For example, the mean energy is:

U = 〈H〉ρ

=1

Z(T )Tr(He−H/T

)

=T 2

Z(T )∂T Tr

(e−H/T

)

=T 2

Z(T )∂T TrZ(T ) (115)

= T 2∂T ln [Z(T )] . (116)

23

The entropy is:

S = −Tr(ρ ln ρ)

= −Tr

{e−H/T

Z(T )

[−HT

− lnZ(T )]}

=U

T+ ln [Z(T )] . (117)

If we define the Helmholtz free energy, F = −T lnZ, then S = −∂TF .Alternatively, U = TS + F .

10 Exercises

1. Show that any linear operator in an n-dimensional Euclidean spacemay be expressed as an n-term dyad. Show that this may be extendedto an infinite-dimensional Euclidean space.

2. Suppose we have a system with total angular momentum 1. Pick abasis corresponding to the three eigenvectors of the z-component ofangular momentum, Jz, with eigenvalues +1, 0,−1, respectively. Weare given an ensemble described by density matrix:

ρ =1

4

2 1 11 1 01 0 1

.

(a) Is ρ a permissible density matrix? Give your reasoning. For theremainder of this problem, assume that it is permissible. Does itdescribe a pure or mixed state? Give your reasoning.

(b) Given the ensemble described by ρ, what is the average value ofJz?

(c) What is the spread (standard deviation) in measured values of Jz?

3. Prove the first theorem in section 5.

4. Prove the von Neumann mixing theorem.

5. Show that an arbitrary linear operator on a product space H = H1⊗H2

may be expressed as a linear combination of operators of the formZ = X ⊗ Y .

24

6. Let us try to improve our understanding of the discussions on the den-sity matrix formalism, and the connections with “information” or “en-tropy” that we have made. Thus, we consider a simple “two-state”system. Let ρ be any general density matrix operating on the two-dimensional Hilbert space of this system.

(a) Calculate the entropy, s = −Tr(ρ ln ρ) corresponding to this den-sity matrix. Express your result in terms of a single real para-meter. Make sure the interpretation of this parameter is clear, aswell as its range.

(b) Make a graph of the entropy as a function of the parameter. Whatis the entropy for a pure state? Interpret your graph in terms ofknowledge about a system taken from an ensemble with densitymatrix ρ.

(c) Consider a system with ensemble ρ a mixture of two ensemblesρ1, ρ2:

ρ = θρ1 + (1 − θ)ρ2, 0 ≤ θ ≤ 1 (118)

As an example, suppose

ρ1 =1

2

(1 00 1

), and ρ2 =

1

2

(1 11 1

), (119)

in some basis. Prove that VonNeuman’s mixing theorem holds forthis example:

s(ρ) ≥ θs(ρ1) + (1 − θ)s(ρ2), (120)

with equality iff θ = 0 or θ = 1.

7. Consider an N -dimensional Hilbert space. We define the real vectorspace, O of Hermitian operators on this Hilbert space. We define ascalar product on this vector space according to:

(x, y) = Tr(xy), ∀x, y ∈ O. (121)

Consider a basis {B} of orthonormal operators in O. The set of den-sity operators is a subset of this vector space, and we may expand anarbitrary density matrix as:

ρ =∑

i

BiTr(Biρ) =∑

i

Bi〈Bi〉ρ. (122)

25

By measuring the average values for the basis operators, we can thusdetermine the expansion coefficients for ρ.

(a) How many such measurements are required to completely deter-mine ρ?

(b) If ρ is known to be a pure state, how many measurements arerequired?

8. Two scientists (they happen to be twins, named “Oivil” and “Livio”,but never mind. . . ) decide to do the following experiment: They setup a light source, which emits two photons at a time, back-to-back inthe laboratory frame. The ensemble is given by:

ρ =1

2(|LL〉〈LL| + |RR〉〈RR|), (123)

where “L” refers to left-handed polarization, and “R” refers to right-handed polarization. Thus, |LR〉 would refer to a state in which photonnumber 1 (defined as the photon which is aimed at scientist Oivil, say)is left-handed, and photon number 2 (the photon aimed at scientistLivio) is right-handed.

These scientists (one of whom is of a diabolical bent) decide to play agame with Nature: Oivil (of course) stays in the lab, while Livio treksto a point a light-year away. The light source is turned on and emitstwo photons, one directed toward each scientist. Oivil soon measuresthe polarization of his photon; it is left-handed. He quickly makes anote that his brother is going to see a left-handed photon, sometimeafter next Christmas.

Christmas has come and gone, and finally Livio sees his photon, andmeasures its polarization. He sends a message back to his brother Oivil,who learns in yet another year what he knew all along: Livio’s photonwas left-handed.

Oivil then has a sneaky idea. He secretly changes the apparatus, with-out telling his forlorn brother. Now the ensemble is:

ρ =1

2(|LL〉 + |RR〉)(〈LL| + 〈RR|). (124)

He causes another pair of photons to be emitted with this new appa-ratus, and repeats the experiment. The result is identical to the firstexperiment.

26

(a) Was Oivil just lucky, or will he get the right answer every time,for each apparatus? Demonstrate your answer explicitly, in thedensity matrix formalism.

(b) What is the probability that Livio will observe a left-handed pho-ton, or a right-handed photon, for each apparatus? Is there aproblem with causality here? How can Oivil know what Livio isgoing to see, long before he sees it? Discuss! Feel free to modifythe experiment to illustrate any points you wish to make.

9. Let us consider the application of the density matrix formalism to theproblem of a spin-1/2 particle (such as an electron) in a static externalmagnetic field. In general, a particle with spin may carry a magneticmoment, oriented along the spin direction (by symmetry). For spin-1/2, we have that the magnetic moment (operator) is thus of the form:

µµµ =1

2γσσσ, (125)

where σσσ are the Pauli matrices, the 12

is by convention, and γ is aconstant, giving the strength of the moment, called the gyromagneticratio. The term in the Hamiltonian for such a magnetic moment in anexternal magnetic field, BBB is just:

H = −µµµ ·BBB. (126)

Our spin-1/2 particle may have some spin-orientation, or “polarizationvector”, given by:

PPP = 〈σσσ〉. (127)

Drawing from our classical intuition, we might expect that in the ex-ternal magnetic field the polarization vector will exhibit a precessionabout the field direction. Let us investigate this.

Recall that the expectation value of an operator may be computed fromthe density matrix according to:

〈A〉 = Tr(ρA). (128)

Furthermore, recall that the time evolution of the density matrix isgiven by:

i∂ρ

∂t= [H(t), ρ(t)]. (129)

27

What is the time evolution, dPPP/dt, of the polarization vector? Expressyour answer as simply as you can (more credit will be given for rightanswers that are more physically transparent than for right answerswhich are not). Note that we make no assumption concerning thepurity of the state.

10. Let us consider a system of N spin-1/2 particles (see the previous prob-lem) per unit volume in thermal equilibrium, in our external magneticfield BBB. Recall that the canonical distribution is:

ρ =e−H/T

Z, (130)

with partition function:

Z = Tr(e−H/T

). (131)

Such a system of particles will tend to orient along the magnetic field,resulting in a bulk magnetization (having units of magnetic momentper unit volume), MMM .

(a) Give an expression for this magnetization (don’t work too hard toevaluate).

(b) What is the magnetization in the high-temperature limit, to lowestnon-trivial order (this I want you to evaluate as completely as youcan!)?

28