Download Lecture 34: The `Density Operator

Lecture 34:
The `Density Operator’
Phy851 Fall 2009
The QM `density operator’
• HAS NOTHING TO DO WITH MASS PER
UNIT VOLUME
• The density operator formalism is a
generalization of the Pure State QM we
have 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 `coherent
superposition’
– Statistical mixture: it is either a or b,
but we don’t know which one
• No interference effects
– Coherent superposition: it is both a
and b at the same time
• Quantum interference effects appear
Pure State quantum Mechanics
• The goal of quantum mechanics is to
make predictions regarding the
outcomes of measurements
• Using the formalism we have developed
so far, the procedure is as follows:
– Take an initial state vector
– Evolve it according to Schrödinger's
equation until the time the measurement
takes place
– Use the projector onto eigenstates of the
observable to predict the probabilities for
different results
– To confirm the prediction, one would
prepare a system in a known initial state,
make the measurement, then re-prepare
the same initial state and make the same
measurement after the same evolution
time. With enough repetitions, the results
should show statistical agreement with
the results of quantum theory
Expectation Value
• The expectation value of an operator is
defined (with respect to state |ψ〉) as:
A ≡ ψ Aψ
• The interpretation is the average of the
results of many measurements of the
observable A on a system prepared in
state |ψ〉.
– Proof:
A ≡ ψ Aψ
= ∑ ψ an an A ψ
n
= ∑ ψ an an an ψ
n
= ∑ an ψ ψ an an
n
2
= ∑ ψ an an
n
= ∑ p ( an ) an
n
This is clearly the weighted average of
all possible outcomes
Statistical mixture of states
• What if we cannot know the exact initial
quantum 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 with
probability 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 the
probability of obtaining result an of a
measurement of observable A?
– Clearly, the probability would be
〈ψ1|an〉〈an|ψ1〉 with probability p1, and
〈ψ2|an〉〈an|ψ2〉 with probability p2.
P(an ) = P(an | ψ 1 ) P(ψ 1 ) + P(an | ψ 2 ) P(ψ 2 )
• Thus the frequency with which an would
be obtained over many repetitions would
be
2
2
p ( an ) = an ψ 1
P1 + an ψ 2
P2
The Density `Operator’
• For the previous example, Let us define
a `density operator’ for the system as:
ρ = ψ 1 ψ 1 P1 + ψ 2 ψ 2 P2
This will describe the
state of the system, in
place of a wavefunction
• The probability to obtain result an could
then obtained in the following manner:
P(an ) = Tr{ρ I (an )}
I ( an ) = an an
• Proof:
P(an ) = Tr{ρ I (an )}
= ∑ m ρ an an m
{|m〉} is a
complete basis
m
= ∑ m (P1 ψ 1 ψ 1 + P2 ψ 2 ψ 2 )an an m
m
= ∑ an m m (P1 ψ 1 ψ 1 + P2 ψ 2 ψ 2 )an
m
= an (P1 ψ 1 ψ 1 + P2 ψ 2 ψ 2 )an
= P1 an ψ 1 ψ 1 an + P2 an ψ 2 ψ 2 an
= an ψ 1
2
P1 + an ψ 2
2
P2
Generic Density Operator
• For a ‘statistical mixture’ of the states
{|ψj〉} with respective probabilities {Pj},
the density operator is thus:
ρ = ∑ Pj ψ j ψ j
j
• The sum of the Pj’s is Unity:
∑P
j
=1
j
• The |ψj〉’s are required to be normalized
to one, but are not necessarily
orthogonal
– For example, we could say that with 50%
probability, an electron is in state |↑〉, and
the other 50% of the time it is in state
(|↑〉+|↓〉)/√2
1
1 (↑ + ↓ ) ( ↑ + ↓ )
ρ= ↑ ↑+
2
2
2
2
3
1
1
1
= ↑ ↑+ ↑ ↓+ ↓ ↑+ ↓ ↓
4
4
4
4
€
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 matrix
description:
ρ=ψ ψ
• Every density matrix does not have a
pure 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:
ρ=ψ ψ
• Start from:
• Pick any orthonormal basis that spans the
Hilbert space, for which |ψ〉 is the first basis
vector
• In any such basis, we will have the matrix
elements
m ρ n = δ m ,1δ n ,1
1

0
ρ =
0

M

0
0
0
M
0
0
0
M
L

L
L

O
Testing for purity cont.
• Test #2:
– In any basis, the pure state will satisfy for
every m,n:
ρ mn ρ nm = ρ mm ρ nn
– A partially mixed state will satisfy for at
least one pair of m,n values:
0 < ρ mn ρ nm < ρ mm ρ nn
– And a totally mixed state will satisfy for at
least one pair of m, n values:
ρ mn = ρ nm = 0 and
ρ mm ρ nm ≠ 0
• Examples in spin-1/2 system:
ρ= ↑ ↑
ρ=
(↑ + ↓ )(↑ + ↓ )
2
2
3
1
1
1
ρ == ↑ ↑ + ↑ ↓ + ↓ ↑ + ↓ ↓
4
4
4
4
3
1
ρ == ↑ ↑ + ↓ ↓
4
4
 12
 1
2
1
2
1
2



 34 0 


1
0
4

1 0


0
0


 34
 1
4
1
4
1
4



Probabilities and`Coherence’
• In a given basis, the diagonal elements
are always the probabilities to be in the
corresponding states:
• The off diagonals are a measure of the
‘coherence’ between any two of the basis
states.
 12
 1
2
1
2
1
2



1 0


0 0
 34
 1
4
1
4
1
4



 34 0 


1
0 4 
– Coherence is maximized when:
ρ mn ρ nm = ρ mm ρ nn
Rule 1: Normalization
• Consider the trace of the density
operator
ρ = ∑ Pj ψ j ψ j
j
Tr{ρ} = ∑ Pj ψ j ψ j
j
= ∑ Pj
j
Tr{ρ} = 1
Since the Pj’s are probabilities, they
must sum to unity
Rule 2: Expectation Values
• The expectation value of any operator A
is defined as:
A = Tr{ρA}
• For a pure state this gives the usual
result:
A = Tr{ ψ ψ A}
= ψ Aψ
•€For a mixed state, it gives:


A = Tr ∑ p j ψ j ψ j A
 j

= ∑ p j ψ j Aψ j
j
Rule 3: Equation of motion
• For a closed system:
ρ=ψ ψ
d
d

d

ρ = ψ ψ + ψ  ψ 
dt
 dt

 dt

i
i
=− Hψ ψ + ψ ψ H
h
h
ρ& = −ih[H , ρ ]
– Pure state will remain pure under
Hamiltonian evolution
• For an open system, will have additional
terms:
– Called ‘master equation’
– Example: 2 –level atom interacting with
quantized electric field.
ρ˙ = −
€
i
Γ
[H, ρ] − ( e e ρ + ρ e e ) + Γ g e ρ e g
h
2
– Master equation describes state of system
only, not the `environment’, but includes
effects of coupling to environment
– Pure state can evolve into mixed state
Example: Interference fringes
• Consider a system which is in either a
coherent, or incoherent (mixture)
superposition of two momentum states k,
and –k:
– Coherent superposition:
ψ =
ρ=
1
k + −k
(
2
)
1
1
1
1
k k + k −k + −k k + −k −k
2
2
2
2
€
P(x) = Tr{ρ x x } = x ρ x
P( x) = 1 + cos(2kx)
Fringes!
€ – Incoherent mixture:
ψ = NA
1
1
ρ = k k + −k −k
2
2
P( x) = 1
No fringes!
Entanglement Gives the
Illusion of decoherence
• Consider a small system in a pure state. It
is initially decoupled from the
environment:
( s ,e )
ψ

=  ∑ cs s
 s
(s)

⊗ φ

(e)
• Then turn on coupling to the environment:
ψ′
( s ,e )
=U
( s ,e )
ψ (0)
( s ,e )
• Let the interaction be non-dissipative
– System states do not decay to lower energy
states
U
( s ,e )
s
(s)
⊗φ
(e)
= s
(s)
⊗ φs
(e)
• Strong interaction: assume that different
|s〉 states drive |φ〉 into orthogonal states
φs φs′
(e)
= δ s , s′
The `reduced system density
operator’
• Suppose we want to make predictions
for system observables only
– Definition of ‘system observable’:
As = A( s ) ⊗ I ( e )
• Take expectation value:
As = Tr{ρ (s,e ) A(s) ⊗ I (e )}
=∑ m
(s)
⊗ n
(e)
ρ
( s ,e )
(s)
A ⊗I
(e)
(e)
(s)
m
(s)
m
(s)
m,n
=∑ m
(s)
m
∑
n
(e)
ρ
( s ,e )
n
A
n
• Define the `reduced system density
operator’:
ρ (s) = ∑ n
(e )
ρ (s,e ) n
(e )
= Tre {ρ (s,e )}
n
• Physical predictions regarding system
observables depend only on ρ(s):
€
As = ∑ m
(s)
(s)
ρ A
(s)
m
As = Trs{ρ (s) A(s)}
m
(s)
⊗ n
(e)
€
Entanglement mimics
`collapse’
• Return to our entangled state of the
system + environment:
ψ
( s ,e )
= ∑ cs s
(s)
⊗ φs
(e)
s
• Compute density matrix:
ρ=ψ ψ
( s ,e )
= ∑c c s
∗
s s′
(s)
⊗ φs
(e)
s′
(s)
⊗ φs′
(e)
s , s′
• Compute the reduced system density
operator:
{
(s)
ρ = Tre ψ ψ
(s,e )
{
= ∑ c c Tre s
∗
s s′
s, s′
= ∑ c c s s′
∗
s s′
}
(s)
(s)
s , s′
2
= ∑ cs s s
s
(s)
⊗ φs
φs′ φs
(e )
s′
(s)
⊗ φ s′
(e )
}
‘Collapse’ of the state
ρ
(s)
2
= ∑ cs s s
(s)
s
• Conclusion: Any subsequent measurement
on the system, will give results as if the
system were in only one of the |s〉, chosen at
random, 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’:
ψ
( s ,e )
= ∑ cs s
(s)
⊗ φs
(e)
s
• We see that the entanglement between
system and env. mimics `collapse’
– Is collapse during measurement real or
illusion?
• Pointer States: for a measuring device to
work properly, the assumption, 〈φs |φs’ 〉 = δs,s’
will only be true if the system basis states,
{|s 〉}, are the eigenstates of the observable
being measured