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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 our
quantum 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 appear
Pure 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:
– Take an initial state vector– Evolve it according to Schrödinger's
equation 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 theory
Expectation Value
• The expectation value of an operator isdefined (with respect to state |ψ〉) as:
• The interpretation is the average of theresults of many measurements of theobservable A on a system prepared instate |ψ〉.– Proof:
ψψ AA ≡
ψψ AA ≡
ψψ Aaa nn
n∑=
ψψ nnn
n aaa∑=
nn
nn aaa∑= ψψ
nn
n aa∑=2
ψ
nn
n aap∑= )(
This is clearly the weighted average ofall possible outcomes
Statistical mixture of states
• What if we cannot know the exact initialquantum state of our system?– For example, suppose we only know the
temperature, T, of our system?
• Suppose I know that with probability P1,the system is in state |ψ1〉, while withprobability P2, the system is in state |ψ2〉.– This is called a statistical mixture of the
states |ψ1〉 and |ψ2〉.
• In this case, what would be theprobability of obtaining result an of ameasurement of observable A?– Clearly, the probability would be
〈ψ1|an〉〈an|ψ1〉 with probability p1, and〈ψ2|an〉〈an|ψ2〉 with probability p2.
• Thus the frequency with which an wouldbe obtained over many repetitions wouldbe
2
2
21
2
1)( PaPaap nnn ψψ +=
)()|()()|()( 2211 ψψψψ PaPPaPaP nnn +=
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:
222111 PP ψψψψρ +=
)}({)( nn aITraP ρ=nnn aaaI =)(
)}({)( nn aITraP ρ=
∑=m
nn maam ρ
{|m〉} is acomplete basis
( )∑ +=m
nn maaPPm 222111 ψψψψ
( )∑ +=m
nn aPPmma 222111 ψψψψ
( ) nn aPPa 222111 ψψψψ +=
nnnn aaPaaP 222111 ψψψψ +=
2
2
21
2
1 PaPa nn ψψ +=
This will describe thestate of the system, inplace of a wavefunction
Generic Density Operator
• For a ‘statistical mixture’ of the states{|ψj〉} with respective probabilities {Pj},the density operator is thus:
• The sum of the Pj’s is Unity:
• The |ψj〉’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
∑=j
jjjP ψψρ
∑ =j
jP 1
€
ρ =12↑ ↑ +
12↑ + ↓( )2
↑ + ↓( )2
=34↑ ↑ +
14↑ ↓ +
14↓ ↑ +
14↓ ↓
This state is only `partially mixed’,meaning interference effects are
reduced, but not eliminated
Density 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 if
it corresponds to a pure state or not:
• Test #1:– If it is a pure state, it will have exactly
one non-zero eigenvalue equal to unity– Proof:
• Start from:• Pick any orthonormal basis that spans the
Hilbert space, for which |ψ〉 is the first basisvector
• In any such basis, we will have the matrixelements
ψψρ =
ψψρ =
1,1, nmnm δδρ =
=
OMMM
L
L
L
000
000
001
ρ
Testing for purity cont.
• Test #2:– In any basis, the pure state will satisfy for
every 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 ρρρρ =
nnmmnmmn ρρρρ <<0
00 ≠== nmmmnmmn and ρρρρ
↓↓+↑↓+↓↑+↑↑==4
1
4
1
4
1
4
3ρ
( )( )22
↓+↑↓+↑=ρ
↑↑=ρ
↓↓+↑↑==4
1
4
3ρ
00
01
21
21
21
21
41
41
41
43
41
43
0
0
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:
00
01
21
21
21
21
41
41
41
43
41
43
0
0
nnmmnmmn ρρρρ =
Rule 1: Normalization
• Consider the trace of the densityoperator
∑=j
jjjP ψψρ
jjj
jPTr ψψρ ∑=}{
∑=j
jP
1}{ =ρTr
Since the Pj’s are probabilities, theymust sum to unity
Rule 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 ψ ψ A{ }= ψ Aψ
}{ ATrA ρ=
jjj
j
jjjj
Ap
ApTrA
ψψ
ψψ
∑
∑
=
=
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 with
quantized electric field.
– Master equation describes state of systemonly, not the `environment’, but includeseffects of coupling to environment
– Pure state can evolve into mixed state
ψψρ =
+
= ψψψψρdt
d
dt
d
dt
d
Hi
Hi
ψψψψhh
+−=
[ ]ρρ ,Hih& −=
€
˙ ρ = −ihH,ρ[ ] − Γ
2e e ρ + ρ e e( ) + Γ g e ρ e g
Example: 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:
€
ψ =12
k + −k( )
kkkkkkkk −−+−+−+=2
1
2
1
2
1
2
1ρ
€
P(x) = Tr ρ x x{ } = x ρ x
)2cos(1)( kxxP += Fringes!
kkkk −−+=2
1
2
1ρ
NA=ψ
1)( =xP No 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 energy
states
• Strong interaction: assume that different|s〉 states drive |φ〉 into orthogonal states
)()(),( es
ss
essc φψ ⊗
= ∑
),(),(),()0(
esesesU ψψ =′
)()()()(),( e
s
seses ssU φφ ⊗=⊗
ss
e
ss ′′ = ,
)(δφφ
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 ρ(s):
)()( ess IAA ⊗=
€
As = Tr ρ(s,e ) A(s) ⊗ I(e ){ })()()()(),()()(
,
eseseses
nm
nmIAnm ⊗⊗⊗=∑ ρ
)()()(),()()( ss
n
eeses
m
mAnnm ∑∑= ρ
€
ρ(s) = n (e )ρ(s,e ) n (e )
n∑ = Tre ρ
(s,e ){ }
)()()()( ssss
ms mAmA ρ∑=
€
As = Trs ρ(s)A(s){ }
Entanglement mimics`collapse’
• Return to our entangled state of thesystem + environment:
• Compute density matrix:
• Compute the reduced system densityoperator:
)()(),( e
s
s
ss
essc φψ ⊗=∑
),( esψψρ =
)()()()(
,
e
s
se
s
s
ssss sscc ′
′
∗′ ⊗′⊗=∑ φφ
€
= csc ′ s ∗
s, ′ s ∑ Tre s (s)
⊗ φs(e ) ′ s (s) ⊗ φ ′ s
(e ){ }
€
ρ(s) = Tre ψ ψ(s,e ){ }
ss
s
ssss sscc φφ ′
′
∗′ ′=∑ )(
,)(2 s
ss ssc∑=
‘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. mimics `collapse’– Is collapse during measurement real or
illusion?
• Pointer States: for a measuring device towork properly, the assumption, 〈φs |φs’ 〉 = δs,s’will only be true if the system basis states,{|s 〉}, are the eigenstates of the observablebeing measured
)(2)( s
ss
s ssc∑=ρ
)()(),( e
s
s
ss
essc φψ ⊗=∑