We previously derived the time evolution operators for both Hamiltonian and Schrodinger equations.
A Heisenberg observable in classical or quantum mechanics will evolve by the time evolution operator
where the derivative is the Liouville operator, and the time evolution operator is the operator exponentiation of the integrated Liouville operator.
Now if we substitute back into the derivative
then integrate both sides
replacing with 1
we get the following relation.
And since it should hold for arbitrary observable, the operators on both sides of the equation must be equal.
If we keep substituting into itself, on iteration 0:
On iteration 1:
Then on iteration 2:
Then on iteration 3:
And so on. If we assume it converges the limit should be a solution
Just for syntax we will group all the higher order terms.
If we have a probability distribution over the phase space we can find the expected value of an observable by
And the probability distribution flows according the the continuity equation
where the state vector so we can split it up and say the below.
Use the chain rule
then substitue in Hamiltons equations.
Cancel those two terms.
It's just the Possion bracket
Define an operator that does this
If we have a density matrix over the phase space we can find the expected value of an observable by
where the density matrix is defined below.
If we take the derivative of the density matrix we use the chain rule
and substituting in Schrodingers equation we get
and the Hamiltonian is its on Hermetian so
It's just the commutator bracket
Define an operator that does this
Since the derivative of the probability is
Make a time evolution operator as the exponentiation
In the interaction picture we split the Hamiltonian into time independent and a time dependent part.
Define an interaction picture version.
The Possion bracket distributes over addition.
Then define 2 new operators for each term
Define an interaction picture Liouville operator.
The commutator bracket distributes over addition.
Then define 2 new operators for each term
and can prove with showing starting condition is the same and derivative is the same
We can do a Volterra series it as
and expand it
Then
Consider a closed system where the Hamiltonian is time independent, and is then perturbed by a force along some observable
Here
soHere is
So let's define
So that
Here is
So let's define
So that
So we get
And we find that
Assume that at time there was no perturbation and the system is in thermal equilibrium. For all the probability distribution is at some equilibrium distribution and there was no force .
If we have a probability distribution over the phase space we can find the expected value of an observable by
The difference in expected values is
Then put in the Volterra series
and move the integrals around
Call the factor with first order of the linear response function
put in the equilbrium distribution
And by definition the distribution shouldn't evolve under the unberturbed Hamiltonian (you can prove this).
so its like
where is a Liouville operator for only the equilibrium Hamiltonian. The time evolution operator distributes over multiplication.
And since , and operators commute with their own exponent:
Where is time shifted to measure in the future. And this is just the expected value at equilibrium
So
So the expected value is
And the difference is
Put the volterra series in there
Move some stuff around
Call this first order term the linear response function
Assume equilibrium at and put in the equilbium density matrix .
And you can check that it doesn't change in time from the equilbrium Hamiltonian
Then expand
and because
split
and remember the cycle property of traces, and cycle the second one
so
This is the Kubo formula
The time evolution operator
Next we consider imaginary time
Then cycle the trace again
fourier transform (linear response must be zero for negative times so the integral can start at zero)
And we can remove the phase shift in the first term
Now take the fourier transform
It shouldn't work for negative values so
integrate by parts
In the limit to infinity they should become uncorrelated
taking just the imaginary part
Taking just the real part of anything is
So doing that gives
And if then
And at equlibrium it's time invariant so you can shift both the offsets in the right integral by .
So we get a final formulat relating the spectral density to the linear respones function.
Taking just the imaginary part of anything is
so
And observables are Hermetian
Let A=B
And at equilibrium shifting the time shouldn't matter
Then combined makes
And we get
If we negative the frequency
And since is a real function the Imaginary component is an odd function
So in the quantum version unlike the classical version , the spectral density isn't the same when frequency is negated. To fix this we can average the two.
And thats just the formula for hyperbolic cotangent
for large temepratrues
so
And this recovers the classical version in the classical limit