Time-Dependent Statistical Mechanics
13. Quantum theory of linear response
H.C.A.
November 5, 2009
1
The quantum mechanical response function
(detailed discussion)
Suppose we have a system that is in equilibrium before any field is applied.
Let t0 be some time before the field is applied. We will ultimately use t0 as
the time for constructing the Heisenberg representation. But for the moment
we are going to use a Schrodinger representation.
We apply a field that couples to B and we observe the property A as a
function of time. We want to calculate the linear response of the measured
value of the property A to the field that couples to B.
We have
HS (t) = Ĥ0 − g(t)B̂
H0 is the (Schrodinger representation) time-independent Hamiltonian of the
system when it is isolated from the surroundings, and there is an external
field, with amplitude g(t) that couples to the dynamical variable B. Then
we have the
|ψS (t) = Û (t, t0 )|ψS (t0 )
where Û (t, t0 ) is an operator that satisfies the Schrodinger equation
i
∂U (t, t0 )
= − ĤS (t)Û (t, t0 )
∂t
h̄
1
Continuation
of Lecture 12
10/29/09
not discussed
in lecture
2009
with the initial condition
Û (t0 , t0 ) = Iˆ
where Iˆ is the identity operator.
For any one system, the quantum mechanical expectation value of A at time
t is
ψS (t)|Â|ψS (t) = ψS (t0 )|U (t, t0 )† ÂU (t, t0 )|ψS (t0 )
We want the ensemble average of this over the nonequilibrium ensemble.
ψS (t)|Â|ψS (t)
ne
= ψS (t0 )|U (t, t0 )† ÂU (t, t0 )|ψS (t0 )
ne
The interesting thing about the right side of this equation is that the only
thing in it that varies from one member of the ensemble to another is the
wave function ψS (t0 ). But the statistical properties of the wave function at
time t0 are those appropriate for an equilibrium canonical ensemble. The
right side then is equal to
Tr ρ̂Û (t, t0 )† ÂÛ (t, t0 ) =
ρii i|Û (t, t0 )† ÂÛ (t, t0 )|i = Û (t, t0 )† ÂÛ (t, t0 )
i
eq
Thus we have
average
value of A at time
t in the nonequilibrium ensemble
= ψS (t)|Â|ψS (t)
=
ne
†
Û (t, t0 ) ÂÛ (t, t0 )
eq
= equilibrium average value of the operator Û (t, t0 )† ÂÛ (t, t0 )
When we evaluate this, we will get an expression for the response function
that is appropriate for quantum mechanical systems. To evaluate this, we
need to obtain an expression for the operator U (t, t0 ) to low order in perturbation theory in f .
To zeroth order in the applied field, the time evolution operator is just
e−iĤ0 (t−t0 )/h̄
2
Let’s look for a solution for U that is of the form1
Û (t, t0 ) = e−iĤ0 (t−t0 )/h̄ Û (t, t0 )
Then Û (t, t0 ), in the absence of the external field would be just the identity
operator. In other words, we define Û as
Û (t, t0 ) = eiĤ0 (t−t0 )/h̄ Û (t, t0 )
Let’s derive an equation of motion for Û .
∂ Û (t, t0 )
∂t
dÛ (t, t0 )
i
Ĥ0 eiĤ0 (t−t0 )/h̄ Û (t, t0 ) + eiĤ0 (t−t0 )/h̄
=
h̄
dt
i
i
Ĥ(t)Û (t, t0 )
=
Ĥ0 eiĤ0 (t−t0 )/h̄ Û (t, t0 ) + eiĤ0 (t−t0 )/h̄ −
h̄
h̄
i
i =
Ĥ0 eiĤ0 (t−t0 )/h̄ Û (t, t0 ) + eiĤ0 (t−t0 )/h̄ −
Ĥ0 (t) − g(t)B̂ Û (t, t0 )
h̄
h̄
The Ĥ0 (t) terms cancel because Ĥ0 commutes with the exponential.
i
g(t)eiĤ0 (t−t0 )/h̄ B̂ Û (t, t0 )
h̄
i
=
g(t) eiĤ0 (t−t0 )/h̄ B̂e−iĤO (t−t0 )/h̄ Û (t, t0 )
h̄
i
=
g(t)B̂H (t)Û (t, t0 )
h̄
Note that the operator in parenthesis is the Heisenberg version of B for the
system in the absence of the external field.
=
There are various ways to proceed to solve this equation. One of the most
straightforward involves converting it to an integral equation by integrating
over t starting at t0 . We get
Û (t, t0 ) = Iˆ +
it dt g(t )B̂H (t )Û (t , t0 )
h̄ t0
1
Note that the exponential factor appears on the left. It could have appeared on the
right or in a variety of other ways. We make this specific choice because it leads to an
expression for the operator we are trying to average, namely Û (t, t0 )† ÂÛ (t, t0 ) that is of
the form Û (t, t0 )† eiĤ0 (t−t0 )/h̄ Âe−iĤ0 (t−t0 )/h̄ Û (t, t0 ) = Û (t, t0 )† ÂH (t)Û (t, t0 ) and hence
contains the Heisenberg operator for A.
3
This is easy to solve by perturbation theory. Substitute the entire right side
for the U on the right and iterate. To lowest nontrivial order, we get
it ˆ
dt g(t )B̂H (t ) + O(g 2 )
Û (t, t0 ) = I +
h̄ t0
Also
†
Û (t, t0 )
it ˆ
= I−
dt g(t )B̂H (t )† + O(g 2 )
h̄ t0
it dt g(t )B̂H (t ) + O(g 2 )
= Iˆ −
h̄ t0
Here we used the fact that B̂S is a Hermitian operator, which implies that
B̂H (t) is also Hermitian. The proof of this latter statement is straightforward.
B̂H (t)† =
=
eiĤ0 (t−t0 )/h̄ B̂S e−iĤ0 (t−t0 )/h̄
e−iĤ0 (t−t0 )/h̄
†
†
B̂S† eiĤ0 (t−t0 )/h̄
†
= eiĤ0 (t−t0 )/h̄ B̂S† e−iĤ0 (t−t0 )/h̄
= B̂H (t)
The operator we want to calculate is
Û (t, t0 )† ÂÛ (t, t0 )
= Û (t, t0 )† eiĤ0 (t−t0 )/h̄ Âe−iĤ0 (t−t0 )/h̄ Û (t, t0 )
= Û (t, t0 )† ÂH (t)Û (t, t0 )
When we substitute our solution for Û (t, t0 ) and Û (t, t0 )† , we get a term
that is zeroth order in g and two terms that are of first order. The result
is
it it †
= ÂH (t) −
dt g(t )B̂H (t ) ÂH (t) +
dt g(t )ÂH (t)B̂H (t ) + O(g 2 )
h̄ t0
h̄ t0
i t dt ÂH (t), B̂H (t ) g(t ) + O(g 2 )
= ÂH (t) +
h̄ t0
We really want the equilibrium ensemble average of this. We get
average value of A at time t in the nonequilibrium ensemble
4
=
Û (t, t0 )† ÂÛ (t, t0 )
eq
i t = ÂH (t) +
dt ÂH (t), B̂H (t ) g(t ) + O(g 2 )
h̄ t0
t
i
= Aeq +
dt ÂH (t), B̂H (t ) g(t ) + O(t2 )
h̄ t0
eq
Comparison of this with the general linear response expression leads to an
expression for the quantum linear response function.
χAB (t − t ) =
i ÂH (t), B̂H (t ) Θ(t − t )
h̄
We expect that the left side depends only on the time interval, and we shall
show rather soon that the right side really does depend on the time interval,
not on the two separate times.
2
The quantum mechanical response function
(brief summary)
Consider the usual linear response situation.
• The system is in thermal equilibrium (but it obeys quantum mechanics
not classical mechanics).
• Its Hamiltonian has no explicit time dependence.
• We apply a weak time dependent field to the system starting some time
after t0 . This adds a term −g(t)B̂S to the Hamiltonian.
• We observe the response of some other property, whose Schrodinger
operator is ÂS . We calculate the average value of AH (t) in the nonequilibrium ensemble that results.
The calculation of this average using quantum mechanics is straightforward
but detailed.
5
The answer is of the form you would guess from linear response theory.
the average value of A at time t = ÂH (t) = Û (t, t0 )† ÂS Û (t, t0 )
= Aeq +
∞
−∞
eq
dt χAB (t − t )g(t ) + O(g )
where
χAB (t − t ) =
2
i ÂH (t), B̂H (t ) Θ(t − t )
h̄
Referring to our earlier definition of S, we see that
i
χAB (t − t ) = (SAB (t − t ) − SBA (t − t)) Θ(t − t )
h̄
This form makes it clear that χ depends only on the time interval. So we
have
i
(SAB (t) − SBA (−t)) Θ(t)
χAB (t) =
h̄
Let’s summarize a few definitions that we have accumulated in order to see
the relationships among various functions.
SAB (t) =
ÂH (t)B̂H (0)
i
χAB (t) =
Θ(t) (SAB (t) − SBA (−t))
h̄
1
(SAB (t) + SBA (−t))
φAB (t) =
2
(In all these definitions, the average is performed for an equilibrium ensemble
and the system has a Hamiltonian that has no explicit time dependence.) SAB
is the most straightforwardly defined correlation function. It is simply the
equilibrium average of the product of two operators. φAB is a symmetrized
version of S. χAB contains an antisymmetrized version of S with additional
factors of the Heaviside function and i/h̄.
In a previous section, we asserted that an appropriate definition of a quantum
correlation function is our definition of S, which is
SAB (t) = ÂH (t)B̂H (0) = Tr ρ̂ÂH (t)B̂H (0)
This involves the trace of the operator product of the density operator and the
two Heisenberg operators. We now see why this particular type of quantity
is important. Quantities with this structure arise straightforwardly in the
quantum theory of linear response.
6
3
Absorption of energy from a single external
field
Let’s now consider the absorption of energy from an external field that couples to B. The initial stages of the calculation are identical to the corresponding classical calculation.
If we imagine that the field is initially off, then it is turned on and oscillates
at a particular frequency, and then it is turned off, then the energy gained by
this process is the difference between the average value of the Hamiltonian at
the end and the average value at the beginning. As in the classical case, the
rate of absorption is obtained by integrating the energy absorption over one
cycle. This involves calculating the interaction of the field with the induced
part expectation value of B, and the latter is given by the linear response
function. We find that only the out of phase part of the response contributes
to the energy absorption, and we get
g02 R(ω) = ω χ̂BB (ω)
2
where
χ̂BB (ω)
=
∞
−∞
dt χBB (t ) sin ωt
The quantum and classical calculations differ only in the form of the expression for the response function.
To proceed it will be helpful to manipulate the expression for χ̂ (ω).
χ̂ (ω) is defined as the sine Fourier transform of the response function. It is
a function of frequency.
It is the Fourier transform (i.e. the usual Fourier transform with complex
exponentials) of some function of time.
Let us call that function χ (t).
Let’s get an expression for that function of time.
7
We have
χ̂BB (ω)
=
∞
0
=
=
=
=
=
=
dt χBB (t ) sin ωt
1 ∞ dt χBB (t ) eiωt − e−iωt
2i −∞
∞
∞
1
iωt
−iωt
dt χBB (t )e −
dt χBB (t )e
2i −∞
−∞
∞
−∞
1
dt χBB (t )eiωt −
(−dt ) χBB (−t )eiωt
2i −∞
∞
∞
∞
1
iωt
iωt
dt χBB (t )e −
dt χBB (−t )e
2i −∞
−∞
∞
∞
1
iωt
iωt
dt χBB (t )e −
dt χBB (−t )e
2i −∞
−∞
∞
1
dt (χBB (t ) − χBB (−t )) eiωt
2i
−∞
Therefore
1
(χBB (t) − χBB (−t))
2i
1 i
=
(Θ(t) (SBB (t) − SBB (−t)) − Θ(−t) (SBB (−t) − SBB (t)))
2i h̄
1
(SBB (t) − SBB (−t))
=
2h̄
χBB (t) =
So add χBB (t) to the list of functions that are related to SBB (t). Note the
absence of Heaviside functions in this expression. Note also that this is an
odd function of time.
End of
Lecture
Recall that we need the Fourier transform of this. The result is straightfor12
ward.
10/29/09
8
χ̂BB (ω) =
1 ŜBB (ω) − ŜBB (−ω)
2h̄
To evaluate the energy absorption, it is clear that we need the Fourier transform of SBB (t).
Earlier we obtained a general expression for it.
ŜBB (ω) = 2π
ρii |Bij |2 δ(ω − ωji )
ij
where
ωji = (Ej − Ei )/h̄
is the quantum mechanical frequency associated with the energy difference
between the two states and
e−Ei /kB T
ρii =
Q
and Q is the quantum partition function.
Hence
π
ρii |Bij |2 (δ(ω − ωji ) − δ(−ω − ωji ))
h̄ ij
π
ρii |Bij |2 (δ(ω − ωji ) − δ(ω + ωji ))
=
h̄ ij
χ̂BB (ω) =
We get the following result for the rate of energy absorption.
R(ω) =
πg02 ρii |Bij |2 (δ(ω − ωji ) − δ(ω + ωji ))
ω
2h̄ ij
There are several things to note.
1. It is an even function of ω. For simplicity, let’s restrict our attention to
positive frequencies.
2. For positive frequency, the ω factor in front is positive. The ρii and |Bij |2
factors are clearly positive. However, some of the delta functions are positive
in magnitude and some are negative.
9
Lecture
13
11/5/09
• In the positive terms, the ωij is must be positive if the term is to be
nonzero, so Ej > Ei . This corresponds to the absorption of a quantum
of energy from the external field, with the system making a transition
from state i to state j. Note that the ρii factor makes sense; the term
that corresponds to such a transition should be proportional to the
probability that the system is in state i to begin with.
• In the negative terms, the ωji must be negative if the term is to be
nonzero, so Ej < Ei . This corresponds to the emission of a quantum
of energy into the external field, with the system making a transition
from state i to a state j of lower energy.
If we were considering an electromagnetic radiation field as the external field,
then these two terms would correspond to absorption and stimulated emission
of photons. The theory of the radiation field that we are using is essentially
a classical theory (even though its interaction with the system is quantum
mechanical and the system itself is quantum mechanical). Thus this theory
does not include the physics of spontaneous emission.2
3. The Second Law of Thermodynamics still holds, so the overall result had
better be positive, even though it has negative contributions. We will now
demonstrate this.
2
It is reasonable that a theory that is calculating the response of a system to an applied
field would not describe spontaneous processes that take place even in the absence of the
applied field. However, no description of radiation that does not take into account the
quantum nature of the radiation field can provide a theory of the spontaneous emission
process.
10
slide
R(ω) =
πg02 ρii |Bij |2 (δ(ω − ωji ) − δ(ω + ωji ))
ω
2h̄ ij
It is an even function of ω. (Restrict attention to positive frequencies.)
The ω factor in front is positive.
The ρii and |Bij |2 factors are clearly positive.
Some of the delta functions have a + sign and some have a − sign.
• + sign: ωji must be positive if the term is nonzero, so Ej > Ei .
describes the absorption of energy from the external field, with the system
making a transition from state i to state j of higher energy.
• − sign: ωji must be negative if the term is nonzero, so Ej < Ei .
describes the emission of a quantum of energy into the external field, with
the system making a transition from state i to a state j of lower energy.
end of slide
Let us break the sum into three types of terms:
• those with Ej > Ei ;
• those with Ej = Ei . These terms are zero, as is easily seen, because of
the factor of ω out front and because the two delta functions cancel.
• those with Ej < Ei .
11
R(ω) =
πg02
ω
ρii |Bij |2 (δ(ω − ωji ) − δ(ω + ωji ))
2h̄ ij(Ei <Ej )
+
=
πg02
ρii |Bij |2 (δ(ω − ωji ) − δ(ω + ωji ))
ω
2h̄ ij(Ei <Ej )
+
=
πg02
ρjj |Bji |2 (δ(ω − ωij ) − δ(ω + ωij ))
ω
2h̄ ij(Ej >Ei )
πg02
ρii |Bij |2 (δ(ω − ωji ) − δ(ω + ωji ))
ω
2h̄ ij(Ei <Ej )
+
=
πg02
ρii |Bij |2 (δ(ω − ωji ) − δ(ω + ωji ))
ω
2h̄ ij(Ei >Ej )
πg02
ω
ρjj |Bji |2 (δ(ω + ωji ) − δ(ω − ωji ))
2h̄ ij(Ej >Ei )
πg02
(ρii − ρjj ) |Bij |2 (δ(ω − ωji ) − δ(ω + ωji ))
ω
2h̄ ij(Ei <Ej )
(In getting the second equality, we merely interchanged the dummy summation variables i and j in the second sum. In getting the third equality, we
interchanged the subscripts on the ω quantities and changed their signs in
the second term. This is valid, given the definition of the ωij above. At this
point, the two sums are very similar and can be combined to tive the last
line.) For positive frequency, this is clearly positive, since ρii > ρjj if Ei < Ej .
Thus, taking into account both the absorption and stimulated emission, the
net result is always an absorption of energy from the field when the system
is in equilibrium.
slide
The Second Law of Thermodynamics still holds, so the overall result had
better be positive. Straightforward manipulations give:
R(ω) =
πg02
ω
(ρii − ρjj ) |Bij |2 (δ(ω − ωji ) − δ(ω + ωji ))
2h̄ ij(Ei <Ej )
which is manifestly nonnegative.
end of slide
12
4
Quantum mechanical fluctuation-dissipation
theorem
slide
Let’s summarize a few definitions that we have accumulated in order to see
the relationships among various functions.
SAB (t) =
ÂH (t)B̂H (0)
i
χAB (t) =
Θ(t) (SAB (t) − SBA (−t))
h̄
1
(SAB (t) + SBA (−t))
φAB (t) =
2
1
1
(χBB (t) − χBB (−t)) =
(SBB (t) − SBB (−t))
χBB (t) =
2i
2h̄
(In all these definitions, the average is performed for an equilibrium ensemble and the system has a Hamiltonian that has no explicit time dependence.)
end of slide
There are various versions of the quantum mechanical fluctuation-dissipation
theorem. The qualitative idea is that the absorption of energy from an external field is related to the behavior of fluctuations in the absence of the
external field. The result we got above relating χ̂BB (ω) to ŜBB (ω) was one
example of a statement of this idea. However, we can relate χ , which is
the quantity most closely connected to dissipation, to the some of the other
quantities that are related to fluctuations.
To do this, let’s first derive a useful property of Ŝ(ω).
13
We showed earlier that
ŜBB (ω) = 2π
ρii |Bij |2 δ(ω − ωji )
ij
Then
SBB (−ω) = 2π
ρii |Bij |2 δ(−ω − ωji ) = 2π
ij
= 2π
ρii |Bij | δ(ω − ωij ) = 2π
ρjj
jj
= 2π
ρii |Bij |2 δ(ω + ωji )
ij
2
ij
= 2π
ρii
ρjj |Bij |2 δ(ω − ωji )
ji
ρii |Bij |2 δ(ω − ωji )
e−(Ej −Ei )/kB T ρii |Bij |2 δ(ω − ωji )
jj
= 2π
e−h̄ωji /kB T ρii |Bij |2 δ(ω − ωji )
jj
= 2π
e−h̄ω/kB T ρii |Bij |2 δ(ω − ωji ) = e−h̄ω/kB T SBB (ω)
jj
The result is
SBB (−ω) = e−h̄ω/kB T SBB (ω)
Thus
χ̂BB (ω) =
1 1
ŜBB (ω) 1 − e−h̄ω/kB T
ŜBB (ω) − ŜBB (−ω) =
2h̄
2h̄
Similarly
φ̂BB (ω) =
1
1
ŜBB (ω) + ŜBB (−ω) = ŜBB (ω) 1 + e−h̄ω/kB T
2
2
Hence
φ̂BB (ω)
= h̄ coth(h̄ω/2kB T )
χ̂ (ω)
φ̂BB (ω) = h̄ coth(h̄ω/2kB T )χBB (ω)
This is another statement of the fluctuation-dissipation theorem.
We can use this to get an expression for the rate of absorption of energy in
terms of φ̂BB . We get
R(ω) =
g2
g02 ωχBB (ω) = 0 ω tanh(h̄ω/2kB T )φ̂BB (ω)
2
2h̄
14
This is another statement of the fluctuation-dissipation theorem, with the
left side representing dissipation and the right side representing fluctuations.
Let’s consider the classical limit of this. The classical limit corresponds to
h̄ → 0, or h̄ω/2kB T → 0. For small x,
tanh x = x + O(x3 )
So
g02 2
ω φ̂BB (ω)
4kB T
This is the same form as the classical result, if we make the reasonable
guess that the quantum correlation function φ̂BB (t) approaches the classical
correlation function CAA (t) in the classical limit.
R(ω) →
slide
φ̂BB (ω) = h̄ coth(h̄ω/2kB T )χBB (ω)
Get an expression for the rate of absorption of energy in terms of φ̂BB .
R(ω) =
g02 g2
ωχBB (ω) = 0 ω tanh(h̄ω/2kB T )φ̂BB (ω)
2
2h̄
Take the classical limit.
The classical limit corresponds to h̄ → 0, or h̄ω/2kB T → 0.
For small x,
tanh x = x + O(x3 )
So
g02 2
ω φ̂BB (ω)
4kB T
The same form as the classical result, if we make the reasonable guess
R(ω) →
φ̂BB (t) → CAA (t)
in the classical limit.
end of slide
15
Lecture
13
11/5/09
continued in
N14