5.73 Lecture #19
19 - 1
Density Matrices I
(See CTDL pp. 252-263, 295-307**, 153-163, 199-202, 290-294)
Last time: Variational Method

˜˜
Linear variation: 0 = H − ε S ⇒ 0 = H − ε

dε

ψ = ∑ cnφ n
=0

dc n
n

1
[Variation vs. pert. theory
TODAY
ψ phase ambiguity – but for all observables each state always appears as a bra
and a ket.
what is needed to encode motion in the probability density? A superposition of
eigenstates belonging to several different values of E.
coherent superposition vs. statistical mixture: think about polarized light
ρ
no phase ambiguity
“coherences” in off-diagonal position
“populations” along diagonal
A = Tr (ρ A) = Tr ( Aρ )
Quantum Beats
prepared state → ρ
detection → D (detect or destroy coherences)
ρ (t )
Α
t
∂Α
d
i
Α =
Η , Α] +
[
h
∂t
dt
dρ
= [Η (t ), ρ ]
ih
* state:
dt
ρ
* evolution: H
* detection: D
modified 10/9/02 10:20 AM
5.73 Lecture #19
19 - 2
Let us define a quantity called “Density Matrix”
ρ≡ ψ ψ
a pure state
ψ can be any sort of QM wavefunction
* eigenstate of H
* coherent superposition of several eigenstates of H
but ψ cannot represent a statistical (i.e. incoherent) mixture of several different ψ’s
However, ρ can represent a statistical (i.e. equilibrium) mixture of states
ρ ≡ ∑ p k ψ k ψ k = ∑ p k ρk
k
k
∑pk =1
Example
*
one beam of linarly polarized light, with the polarization axis
(ε-field)
ê
y
*
(
ê = 2 −1/2 ê x + ê y
45°
)
x
two beams of linearly polarized light, 50% along êx, 50% along êy.
These 2 cases seem to be identical if you make 2 measurements with
analyzer polarizers along êx then êy. But they are different with
respect to 2 measurements with analyzer polarizers along 2 −1/2 ê x + ê y
then 2 −1/ 2 ê x − ê y .
In the statistical mixture, it does not matter how the analyzer is oriented.
(
)
(
)
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5.73 Lecture #19
19 - 3
What are the properties of ρ?
1. ρ for a pure state is Hermitian with positive elements along diagonal and
other elements off-diagonal.
ρnm = n ψ ψ m
c*m
cn
ψ = ∑ cn n
ρnm = c n c*m
( )nm
but ρ†
*
= ρmn
=[ mψ ψ n
can expand |ψ⟩
in any basis set
but H eigenbasis
is most useful.
]
*
= ψ m n ψ = n ψ ψ m = ρnm
∴ ρ† = ρ
2
ρnn = n ψ ψ n = c n c*n = c n ≥ 0
positive along diagonal
2. 2 × 2 Example
Coherent Superposition vs. Statistical Mixture
1
ψ = 2 −1/ 2  ±1
ρcs =
1 1
1  1 ±1
=
1
±
1
(
)
2  ±1
2  ±1 1 
Trace ρ = 1
Now consider a statistical mixture state.
modified 10/9/02 10:20 AM
5.73 Lecture #19
19 - 4
1  1
1
(1 0) +  10 (0 1)
0


2
2
1  1 0
trace ρ = 1
=  0 1
2
ρsm =
The difference is in the off-diagonal positions of ρ
diagonal elements
→ “populations”
off-diagonal elements → “coherences”
(statistical mixture states have
strictly diagonal ρ)
Expectation values of  in terms of ρ
A = ψ A ψ = ∑ ψ |k k A j j| ψ
j,k
= ∑ j ψ ψ k A kj
j,k
ρ jk
= ∑ (ρ A) jj ≡ Trace(ρ A)
j


ρ) 
Could have arranged the factors ∑ A kj j ψ ψ k = ∑ (Aρ )kk = Trace(Aρ


j,k
k
A = Trace( Aρ ) = Trace(ρ A)
So ρ describes state of system, A describes a measurement to be
made on the system
simple prescription for calculating ⟨A⟩
The separation between initial preparation, evolution, and
measurement of a specific observable is very convenient and
instructive.
modified 10/9/02 10:20 AM
5.73 Lecture #19
19 - 5
Example: Quantum Beats
Preparation, evolution, detection
magically prepare some coherent superposition state Ψ(t)
Ψ( t ) = N ∑ a n ψ n e
n
−iE n t h
Several eigenstates of H.
Evolve freely without
any time-dependent
intervention

2
N = ∑ a n 
n

−1/2
normalization
ρ ( t ) = Ψ(t) Ψ(t)
Case (1): Detection: only one of the eigenstates, ψ1, in the superposition is
capable of giving fluorescence that our detector can “see”.
Thus
 1 0 L
D = ψ1 ψ1 =  0 0 0 
M 0 0
a projection operator
(designed to project out only |ψ1⟩
part of state vector or ρ11 part of ρ.
 a 2 a a* e −( E1 −E 2 )t h
L
1
1 2

2
a2
ρ = N2 
2

a
3

O

ρ12 = 1 Ψ Ψ 2
ρ12 = N 2a1e −E1 t ha*2 e +iE 2 t h
a 2
1

2
D t = Trace(Dρ) = N Trace 0
 M

= N2 a 1
2
D picks out only 1st
row of ρ.
a 1a *2 e − iω12 t
0
M
stuff L

0
0
M
M 
no time dependence!
modified 10/9/02 10:20 AM
5.73 Lecture #19
19 - 6
case (2): a particular linear combination of eigenstates is bright: the initial
state 2–1/2(ψ1 + ψ2) has ⟨D⟩ = 1.
a projection operator.
1
How much of the original
D=
ψ1 + ψ 2 ψ1 + ψ 2
state is present in the
2
evolved state?
(
)(
)
1
ψ1 ψ1 + ψ 2 ψ 2 + ψ1 ψ 2
2
 1 0 0 L  0 0 0 L  0
1  0 0 0 0   0 1 0 0   0
=  0 0 0 0 + 0 0 0 0 + 0
2 
 
 
 M 0 0 0   M 0 0 0   M
 1 1 0 L
1 1 1 0 0 
=
2 0 0 0 0 
M 0 0 0
=
[
+ ψ 2 ψ1
1
0
0
0
0
0
0
0
]
L  0 0
0  1 0
0  + 0 0
0 M 0
0
0
0
0
L 
0
0
0  

 1 -1 0
1

-1/ 2
 -1 1 0
−
if
the
bright
state
had
been
2
ψ
ψ
then
=
,
D
(
)
1
2

2 0 0 0

0 0 0


1 2
why do we need to look at
Trace(Dρ ) = N Trace

only the 1,2 block of ρ?
2


1
(Dρ )11 = N 2  a1 2 + a1*a 2 e +i( E1 −E 2 )t h 


2
1
(Dρ )22 = N 2  a 2 2 + a1a*2 e −i( E1 −E 2 )t h 


2
1
2
2
Trace(Dρ ) = N 2 a1 + a 2 + 2 Re a1*a 2 e +iω12 t
2
[
[
0 
0 
0 
0 
]]
beat note at ω12
ρ ) would
[if the bright state had been 2–1/2(ψ1–ψ2), then Tr(Dρ
be the same except for –2Re[
] ]
2
If a1 = a2
2
(and a1, a2 real), Trace(Dρ) = N 2 a1 2 [1 ± cos ω12t ]
(N2 = 1/2)
QUANTUM BEAT! 100% modulation!
modified 10/9/02 10:20 AM
5.73 Lecture #19
19 - 7
So we see that the same Ψ(x,t) or ρ (t) can look simple or complicated depending on
the nature of the measurement operator! The measurement operator is designed to
be sensitive only to specific coherences (i.e. locations in ρ) which oscillate at ωij.
THIS IS THE REASON WHY WE SEPARATE PREPARATION AND
OBSERVATION SO CLEANLY.
Time evolution of ρ nm and A
Start with the time-dependent Schrödinger equation:
HΨ = ih
∂

H Ψ = ih ∂t Ψ

 Ψ H = − ih ∂ Ψ

∂t
∂Ψ
∂t
for time-independent H we know Ψ( t ) = ∑ a n ψ n e −iE n t h
n
1. ρ(t)
ρ (t) = Ψ(t) Ψ(t)
ρnn (t) = n Ψ(t) Ψ(t) n = a n
2
a time independent
“population” in state n.
−i E −E t h
ρnm (t) = a n a*m e ( n m ) = a n a*m e −iω nm t a “coherence” which
oscillates at ωnm (eigenstate
energy differences /h)
Α⟩t
2. ⟨Α
Recall ih
∂Ψ
= HΨ
∂t
∂
∂

∂A
∂

A =
Ψ A Ψ + Ψ
Ψ + Ψ A Ψ 
∂t
∂t
 ∂t

 ∂t

−1
∂A
1

=
+ Ψ A H Ψ 
Ψ H AΨ +
∂t
ih
 ih

i
∂A
Heisenberg Equation
= [ H , A] +
of Motion
h
∂t
[
]
Note that nothing has been assumed about the time-dependence of H.
modified 10/9/02 10:20 AM
5.73 Lecture #19
19 - 8
If A commutes with H (regardless of whether H is time-dependent), there
is no dynamics as far as observable A is concerned. However, if A does not
commute with H, there can be dynamics of ⟨A⟩ even if both A and H are
time-independent.
∂ρ
Similarly can derive ih = [H( t), ρ] evolution of ρ under H(t). This is a matrix equation.
∂t
It specifies the time
dependence of each
If H is
element of ρ.
time
dependent
Summarize
ρA) = Tr(Aρ
ρ)
⟨A⟩ = Tr(ρ
info about quantity
being measured
ih
info about state on which
measurement is to be made
∂ρ
= [H, ρ ]
∂t
time
evolution
state
initial state: ρ

 each expressed independently in
time evolution of ρ : H  the form of matrices which can be
observable quantity: A  easily read (or designed!).
NMR pulse gymnastics
statistical mixture states - use same machinery BUT add the independent ρk
matrices with weights pk that correspond to their fractional populations.
ρ is Hermitian so can be diagonalized by T†ρT. However, if ρ is
time-dependent, T would have to be time-dependent. This
transformation gives a representation without any coherences in ρ̂ρ
even if we started with a coherent superposition state. No problem
because this transformation will undiagonalize H, thereby
reintroducing time dependencies.
modified 10/9/02 10:20 AM