# A loophole in Kubo formula

1. Nov 20, 2009

### hiyok

Hi, every one,

here i notice a loophole in the usual Kubo formalism for dealing with linear response. It assumes that the density operator evolve according to Heisenberg equation of motion. This implies that, the Boltzman factor is held time-independent, only the states evolving, which however cannot be exactly true. Because, the system under a perturbation will deviate from its present equilibrium state to reach, say, another equilibrium state after long time. In other words, in this case, the Boltzman factor must change.

BTW, as noted many years ago, the Kubo formulation has premised that, the system is initially in equilibrium, which may also be not true. This loophole seems filled with the Keldysh framework.

My question is, does Keldysh formaulation have anything to do with the first loophole ?

Thank you.

hiyok

2. Nov 20, 2009

### sokrates

Heisenberg Picture = States are constant, Operators are evolving

Schrodinger Picture = Operators are constant, states are evolving

Their difference = Semantic.

I don't think this is a serious problem either for Kubo or for Keldysh. As for the initial equilibrium assumption, doesn't hurt anything because these are steady state equations anyway. So you can take t--> inf. before you do anything.

Last edited: Nov 20, 2009
3. Nov 20, 2009

### xepma

The density matrix is not an observable, so in interaction picture it doesn't evolve like the other operators do (it's equation of motion is not the normal Heisenberg equation of motion).

4. Nov 20, 2009

### sokrates

Density matrix IS an observable. Its average value is the electron density inside the conductor.

It's spin up and spin down components can be (and is) measured in realistic experiments involving ferromagnetic contacts.

5. Nov 20, 2009

### hiyok

Hi, Socrates,

I agree that the difference between Heisenberg picture and Schrodinger picture is semantic. I also agree that, usually the equilibrium assumption does not cause great problems. But the question is not about this.

Let me put it more clearly. As is known, the density matrix is rho=\Sigma_i exp(-\beta E_i)|i><i|. In Kubo formulation, the Boltzman factor is assumed not varying with time, only the |i><i| evolving quantum mechanically. However, as I argued in the original post, the Boltzman factor should also change with time.

hiyok

6. Nov 20, 2009

### xepma

Maybe I've chosen my words a bit poor, but a density matrix is still a description of the state of the system, rather than an "independent" observable (independent as in: the observable operator doesn't depend on the state of the system). The density matrix contains information on the density of the system, but then again it contains all possible information of the system.

I mean, you wouldn't call the wavefunction an observable right? And isn't a pure state simply $$|\Psi(t)\rangle\langle\Psi(t)|$$? That doesn't really classify as an observable as far as I know..

By all means, correct me if I'm wrong..

7. Nov 20, 2009

### hiyok

I think it depends on how you define an observable. Obviously, the density matrix is real (Hermitian). So it bears real values. And these values can be determined experimentally.

Despite this, I don't want to call it an observable. Because, its average values are probabilities, which does not directly couple to any external field.

By 'observable', we mean at least two things:
(1) it is real;
(2) It couples to certain physical field.

E.g., charge current is an observable, because, first, it is real and second, it couples to magnetic field.

And in this sense, density matrix is NOT an observable.

Last edited: Nov 20, 2009
8. Nov 20, 2009

### f95toli

The density matrix is not an observable.
Consider a simple spin 1/2 system. It is true that the diagonal elements are related to "classical" probabilities and are therefore in some sense "observables" but the coherences (=the two off diagonal elements) are obviously not since they allow us to describe a system that is in a superposition of two states.
Hence, the density matrix for a spin 1/2 system contains exactly the same information as e.g. a Bloch sphere an you can quite easily calculate one from the other.

This can obviously also be generalized to systems with more degrees of freedom.

9. Nov 20, 2009

### sokrates

Not quite. Off-diagonal elements CAN be observed. For instance:

an x-directed spin-current incident to a z-directed magnet will observe the off-diagonal elements of the x-directed spin-current, because conversion to the z-basis will involve the off-diagonal elements of the initial matrix in diagonal entries.

So, having off-diagonal elements definitely does not mean they cannot be observed.
Just make a measurement in a different basis and you can observe this elements.

Last edited: Nov 20, 2009
10. Nov 20, 2009

### sokrates

I didn't know the second requirement. But even in that case, are you sure electron density is not coupled to any physical field?

Because charge-current and electron density can be DIRECTLY related in any quantum transport model.

11. Nov 20, 2009

### sokrates

The wavefunction is not an observable, but a quantity that is related to the "square" of the wavefunction, by definition, IS an observable.

Maybe I Am wrong, but this doesn't convince me, do you have a solid reference for this?

12. Nov 21, 2009

### hiyok

for the moment, I don't see what field density matrix |Psi><Psi| couples to.

13. Nov 21, 2009

### sokrates

According to your initial description, Kubo seemed to be OK with Heisenberg Picture- since Boltzmann factor is not varying with time.

14. Nov 23, 2009

### hiyok

Hi, Socrates, this is not what I intended. Let me restate my points.

As is known, the average of an observable $$\hat{A}$$ is given as $$A=Tr(\rho\hat{A})$$, where $$\rho$$ denotes the density matrix that reads $$\sum_i p_i|i\rangle\langle i|$$, $$p_i$$ being Boltzman factor under equilibrium. Suppose a system initially at equilibrium with a heat bath with temperature $$1/\beta$$, described by a Hamiltonian $$H_0$$. Now we switch on a perturbation, which shall grow and become constant after a duration. Under this perturbation, $$A$$ shall change. Within Kubo context, one calculates the change assuming that $$\rho$$ (and hence also $$A$$) abide by Heisenberg motion of equation, that is, $$p_i$$ has been assumed not varying with time. As we know, this equation describes unitary and coherent evolution of an isolated quantum system. On the other hand, a system in contact with a heat bath is surely an open system, i.e., its observable should not evolve unitarily, which then poses a problem to Kubo formalism. Actually, if the system evolved unitarily, it could not reach another equilibrium any more. But it should. A simple way to remedy this may be to let $$p_i$$ vary with time.

Last edited: Nov 23, 2009
15. Nov 24, 2009

### sokrates

You are the expert in Kubo so you have to tell us how the irreversibility is introduced there :) But doesn't it somewhere in the line relate conductance to the noise in the system (by fluctuation-dissipation theorem)?

I think your point is: If everything is Hermitian, and unitary, how do we have irreversible processes to restore equilibrium?

On the other hand, I believe that Keldysh or maybe the closely related but somewhat unique, updated version, NEGF-Landauer Formalism considers this very explicitly.

arxiv.org/pdf/0809.4460

explains this formalism in great detail.

But the essential physics related to your question is this:
The introduction of irreversibility is done by the non-Hermitian self energy matrices (sigmas), and this could even be seen in a simple one-particle Schrodinger equation:

$$(EI - H)\Psi = 0$$

becomes:

$$(EI - H - \Sigma)\Psi = 0$$

If you do not put any 'source' terms on the right hand side, the system evolves back to equilibrium because Sigma signifies a "resistance" or "leakage" term, and after a while the system goes to equilibrium with whatever process it is in contact with (modeled by Sigma). Without doing anything else, one could get the contact broadening, also. The stronger the system is coupled to its environment, the more the broadening is, in accordance with what one expects from HUP.

So I think in Keldysh, it is straightforward to identify the non-Hermitian processes that restore equilibrium.

Last edited: Nov 24, 2009