Lecture 34: The `Density Operator’

1 Lecture 34:The ` density Operator Phy851 Fall 2009 The QM ` density operator HAS NOTHING TO DO WITH MASS PERUNIT VOLUME The density operator formalism is ageneralization of the Pure State QM wehave used so far. New concept: Mixed state Used for: Describing open quantum systems Incorporating our ignorance into ourquantum theory Main idea: We need to distinguish between a`statistical mixture and a `coherentsuperposition Statistical mixture: it is either a or b,but we don t know which one No interference effects Coherent superposition: it is both aand b at the same time Quantum interference effects appearPure State quantum Mechanics The goal of quantum mechanics is tomake predictions regarding theoutcomes of measurements Using the formalism we have developedso far, the procedure is as follows.

2 Take an initial state vector Evolve it according to Schr dinger'sequation until the time the measurementtakes place Use the projector onto eigenstates of theobservable to predict the probabilities fordifferent results To confirm the prediction, one wouldprepare a system in a known initial state,make the measurement, then re-preparethe same initial state and make the samemeasurement after the same evolutiontime. With enough repetitions, the resultsshould show statistical agreement withthe results of quantum theoryExpectation Value The expectation value of an operator isdefined (with respect to state | ) as: The interpretation is the average of theresults of many measurements of theobservable A on a system prepared instate | . Proof: AA AA Aaannn = nnnnaaa =nnnnaaa = nnnaa =2 nnnaap =)(This is clearly the weighted average ofall possible outcomesStatistical mixture of states What if we cannot know the exact initialquantum state of our system?

3 For example, suppose we only know thetemperature, T, of our system? Suppose I know that with probability P1,the system is in state | 1 , while withprobability P2, the system is in state | 2 . This is called a statistical mixture of thestates | 1 and | 2 . In this case, what would be theprobability of obtaining result an of ameasurement of observable A? Clearly, the probability would be 1|an an| 1 with probability p1, and 2|an an| 2 with probability p2. Thus the frequency with which an wouldbe obtained over many repetitions wouldbe222121)(PaPaapnnn +=)()|()()|()(2211 PaPPaPaPnnn+=The density `Operator For the previous example, Let us definea ` density operator for the system as: The probability to obtain result an couldthen obtained in the following manner: Proof:222111PP +=)}({)(nnaITraP =nnnaaaI=)()}({)(nnaITraP = =mnnmaam {|m } is acomplete basis() +=mnnmaaPPm222111 () +=mnnaPPmma222111 ()nnaPPa222111 +=nnnnaaPaaP222111 +=222121 PaPann +=This will describe thestate of the system, inplace of a wavefunctionGeneric density Operator For a statistical mixture of the states{| j } with respective probabilities {Pj},the density operator is thus: The sum of the Pj s is Unity.

4 The | j s are required to be normalizedto one, but are not necessarilyorthogonal For example, we could say that with 50%probability, an electron is in state | , andthe other 50% of the time it is in state(| +| )/ 2 =jjjjP =jjP1 =12 +12 + ()2 + ()2=34 +14 +14 +14 This state is only `partially mixed ,meaning interference effects arereduced, but not eliminatedDensity matrix of a pure state Every pure state has a density matrixdescription: Every density matrix does not have apure state description Any density matrix can be tested to see ifit corresponds to a pure state or not: Test #1: If it is a pure state, it will have exactlyone non-zero eigenvalue equal to unity Proof: Start from: Pick any orthonormal basis that spans theHilbert space, for which | is the first basisvector In any such basis, we will have the matrixelements = =1,1,nmnm = =OMMMLLL000000001 Testing for purity cont.

5 Test #2: In any basis, the pure state will satisfy forevery m,n: A partially mixed state will satisfy for atleast one pair of m,n values: And a totally mixed state will satisfy for atleast one pair of m, n values: Examples in spin-1/2 system:nnmmnmmn =nnmmnmmn <<000 ==nmmmnmmnand + + + ==41414143 ()()22 + + = = + ==4143 0001 21212121 41414143 414300 Probabilities and`Coherence In a given basis, the diagonal elementsare always the probabilities to be in thecorresponding states: The off diagonals are a measure of the coherence between any two of the basisstates. Coherence is maximized when: 0001 21212121 41414143 414300nnmmnmmn =Rule 1: Normalization Consider the trace of the densityoperator =jjjjP jjjjPTr =}{ =jjP1}{= TrSince the Pj s are probabilities, theymust sum to unityRule 2: Expectation Values The expectation value of any operator Ais defined as: For a pure state this gives the usualresult: For a mixed state, it gives: A=Tr A{}= A }{ATrA =jjjjjjjjApApTrA = =Rule 3: Equation of motion For a closed system: Pure state will remain pure underHamiltonian evolution For an open system, will have additionalterms: Called master equation Example: 2 level atom interacting withquantized electric field.

6 Master equation describes state of systemonly, not the `environment , but includeseffects of coupling to environment Pure state can evolve into mixed state = + = dtddtddtdHiHi hh+ =[] ,Hih& = = ihH, [] 2ee + ee()+ ge egExample: Interference fringes Consider a system which is in either acoherent, or incoherent (mixture)superposition of two momentum states k,and k: Coherent superposition: Incoherent mixture: =12k+ k()kkkkkkkk + + +=21212121 P(x)=Tr xx{}=x x)2cos(1)(kxxP+=Fringes!kkkk +=2121 NA= 1)(=xPNo fringes!Entanglement Gives theIllusion of decoherence Consider a small system in a pure state. Itis initially decoupled from theenvironment: Then turn on coupling to the environment: Let the interaction be non-dissipative System states do not decay to lower energystates Strong interaction: assume that different|s states drive | into orthogonal states)()(),(esssessc = ),(),(),()0(esesesU = )()()()(),(essesesssU = ssess =,)( The `reduced system densityoperator Suppose we want to make predictionsfor system observables only Definition of system observable : Take expectation value: Define the `reduced system densityoperator : Physical predictions regarding systemobservables depend only on (s).

7 ()(essIAA = As=Tr (s,e)A(s) I(e){})()()()(),()()(,esesesesnmnmIAnm = )()()(),()()(ssneesesmmAnnm = (s)=n(e) (s,e)n(e)n =Tre (s,e){})()()()(ssssmsmAmA = As=Trs (s)A(s){}Entanglement mimics`collapse Return to our entangled state of thesystem + environment: Compute density matrix : Compute the reduced system densityoperator:)()(),(essssessc = ),(es =)()()()(,essesssssssscc = =csc s s, s Tres(s) s(e) s (s) s (e){} (s)=Tre (s,e){}ssssssssscc = )(,)(2sssssc = Collapse of the state Conclusion: Any subsequent measurementon the system, will give results as if thesystem were in only one of the |s , chosen atrandom, with probability Ps = |cs|2 This is also how we would describe the`collapse of the wavefunction Yet, the true state of the whole system is not`collapsed : We see that the entanglement betweensystem and env.)

8 Mimics `collapse Is collapse during measurement real orillusion? Pointer States: for a measuring device towork properly, the assumption, s | s = s,s will only be true if the system basis states,{|s }, are the eigenstates of the observablebeing measured)(2)(ssssssc = )()(),(essssessc =