1 Introduction 2 The Density Operator

1 Physics 125cCourse NotesDensity matrix Formalism040511 Frank Porter1 IntroductionIn 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 The Density OperatorSuppose that we have a state space, with a denumerable orthonormal basis{|un ,n=1,2,..}. If the system is in state| (t) at timet, we have theexpansion in this basis:| (t) = nan(t)|un .(1)We ll assume that| (t) is normalized, and hence: (t)| (t) =1 = n man(t)a m(t) um|un = n|an(t)|2(2)Suppose that we have an observable (self-adjoint Operator )Q. The matrixelements ofQin this basis are:Qmn= um|Qun = Qum|un = um|Q|un.

2 (3)The average (expectation) value ofQat timet, for the system in state| (t) is: Q = (t)|Q (t) = n ma m(t)an(t)Qmn.(4)1We see that Q is an expansion quadratic in the{an} the Operator | (t) (t)|. It has matrix elements: um| (t) (t)|un =am(t)a n(t).(5)These matrix elements appear in the calculation of Q . Hence, define (t) | (t) (t)|.(6)We call this thedensity Operator . It is a Hermitian Operator , with matrixelements mn(t)= um| (t)un =am(t)a n(t).(7)Since (t) is normalized, we also have that1= n|an(t)|2= n nn(t)=Tr[ (t)].(8)We may now re-express the expectation value of observableQusing thedensity Operator : Q (t)= m na m(t)a n(t)Qmn= m n nm(t)Qmn= n[ (t)Q]nn=Tr[ (t)Q].(9)The time evolution of a state is given by the Schr odinger equation:iddt| (t) =H(t)| (t) ,(10)whereH(t) is the Hamiltonian. Thus, the time evolution of the densityoperator may be computed according to:ddt (t)=ddt[| (t) (t)|]=1iH(t)| (t) (t)| 1i| (t) (t)|H(t)=1i[H(t), (t)](11)2 Suppose we wish to know the probability,P({q}), that a measurement ofQwill yield a result in the set{q}.

3 We compute this probability by projectingout of| (t) that protion which lies in the eigensubspace associated withobservables in the set{q}. LetP{q}be the projection Operator . Then:P({q})= (t)|P{q} (t) =Tr[P{q} (t)].(12)We note that the Density Operator , unlike the state vector, has no phaseambiquity. The same state is described by| (t) and| (t) =ei | (t) .Under this phase transformation, the Density Operator transforms as: (t) (t)=ei | (t) (t)|e i = (t).(13)Furthermore, expectaion values are quadratic in| (t) , but only linear in (t).For the Density operators we have been considering so far, we see that: 2(t)=| (t) (t)|| (t) (t)|= (t).(14)That is, (t) is anidempotentoperator. Hence,Tr 2(t)=Tr (t)=1.(15)Finally, notice that: un| (t)un = nn(t)=|an(t)|2 0 n.(16)Thus, for an arbitrary state| , | (t) 0, as may be demonstrated byexpanding| in the|u basis. We conclude that is a non-negative 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).

4 33 Statistical MixturesWe 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| 1 with probabilityp1, state| 2 with probabilityp2, and so forth (more gener-ally, we could consider states over some arbitrary, possibly non-denumerable,index set). We must have 1 pi 0 fori {index set}, and ipi= that this situation isnotthe same thing as supposing that we are inthe state| =p1| 1 +p2| 2 + (or even with p1,etc.). Such statisticalmixtures 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 may ask, for such a system, for the probabilityP({q}) that a mea-surement ofQwill yield a result in the set{q}. For each state in our mixture,we havePn({q})= n|P{q} n =Tr( nP{q}),(17)where n=| n n|. To determine the overall probability, we must sum overthe individual probabilities, weighted bypn:P({q})= npnPn({q})= npnTr( nP{q})=Tr( npn nP{q})=Tr( P{q}),(18)where npn n.

5 (19)Now is the Density Operator of the system, and is a simple linear combina-tion of the individual Density operators. Note that is the average of the n s with respect to probability us investigate this Density Operator : Since nare Hermitian, andpnare real, is Tr =Tr( npn n)= npnTr n= npn=1. is non-negative-definite: | = npn | n 0. LetQbe an Operator with eigenvaluesqn. In the current situation, Q refers to the average ofQover the statistical mixture. We have: Q = nqnP({qn})= nqnTr( P{qn})=Tr( nqnP{qn})=Tr( Q),sinceQ= nqnP{qn}.(20) We may determine the time evolution of .For n(t)=| n(t) n(t)|we know (Eqn. 11) thatid n(t)dt=[H(t), n(t)].(21)Since (t) is linear in the n, (t)= npn n(t), we haveid (t)dt=[H(t), (t)].(22) Now look at 2= m npmpn m n= m npmpn| m m| n n|6= ,in general.(23)What about the trace of 2? Let| m = j(am)j|uj .(24)Then 2= m npmpn| m m| n n|= m npmpn i j(am) i(an)j ij [ k `(am)k(an) `|uk u`|]= m,n,i,k,`pmpn(am) i(an)i(am)k(an) `|uk u`|.

6 (25)5 Let s take the trace of this. Notice that Tr(|uk u`|)= k`, so thatTr( 2)= m,n,i,kpmpn(am) i(an)i(am)k(an) k.(26)But m| n = i(am) i(an)i, and thus:Tr( 2)= m npmpn| m| n |2 m npmpn m| m n| n ,(Schwarz inequality) mpm npn 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 thepn scorresponding to independent states must be zero).Note that, if Tr( 2) = 1, then =| |, which is a projection encapsulate this observation into the definition:Def:A state of a physical system is called apure stateif Tr( 2) = 1; thedensity Operator is a projection. Otherwise, the system is said to be inamixed state, or simply a diagonal matrix elements of have a simple physical interpretation: nn= jpj( j)nn= jpj un| j j|un = jpj|(aj)n|2.(28)This is just the probability to find the system in state|un . Similarly, theoff-diagonal elements are mn= jpj(aj)m(aj) n.(29)The off-diagonal elements are calledcoherences.

7 Note that it is possible tochoose a basis in which is diagonal (since is Hermitian). In such a basis,the coherences are all Measurements, Statistical Ensembles, andDensity MatricesHaving 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 Measurement of some physical aspect(s) of the 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 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.

8 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 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 us take this somewhat hazy notion and put it into more concretemathematical language: Associate with each counter adichotomic vari-able,D, as follows:If the counter registers in an event,D= the counter does not register in an event,D= assert that we can, in principle, express all physical variables in terms ofdichotomic ones, so this appears to be a sufficiently general repeatedly preparing the system and observing the counterD, we candetermine the probability thatDregisters: The average value ofD, D ,isthe probability thatDregisters in the experiment.

9 We refer to the particu-lar preparation of the system in the experiment as astatistical ensembleand call D the average of the dichotomic variableDwith respect to 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 ( , probabilities).Let us denote a statistical ensemble with the letter . The use of the samesymbol as we used for the Density matrix is not coincidental, as we shall quantity D explicitly denotes the average ofDfor the ensemble .Clearly:0 D 1.(30)Dis precisely known for ensemble if D =0or D = 1. Otherwise,variableDpossesses 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 have the important concept of the superposition of two ensembles:Let 1and 2be two distinct ensembles.

10 An ensemble is said to be anincoherent superpositionof 1and 2if there exists a number suchthat 0< <1, and for every dichotomic variableDwe have: D = D 1+(1 ) D 2.(31)This is expressed symbolically as: = 1+(1 ) 2,(32) is a superposition of 1and 2with probabilities and 1 . We assume that if 1and 2are physically realizable, then any coherentsuperposition of them is also physically realizable. For example, we mightprepare a beam of particles from two independent sources, each of whichmay hit our counter: 1corresponds to source 1, 2corresponds to source2. When both sources are on, the beam hitting the counter is an incoherentmixture of 1and 2. 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 1 D 1+(1 ) D 2= D 1 D .(33)(34)The generalization to an incoherent superposition of an arbitrary numberof ensembles is clear: Let 1, 2,..be a set of distinct statistical ensembles,and let 1, 2.