# Thermal equilibration of a quantum system

1. Apr 20, 2007

### christianjb

OK, let's say we have solved Schrodinger's eqn. for a system composed of a large number of degrees of freedom.

We then start the wave-function off in an eigenstate of the nth energy level. It will never equilibrate- because the eigenstate is a stationary solution to S.E.

Even if we use an arbitrary wavefunction at time t=0, the wavefunction can always be expanded as a linear superposition of stationary eigenstates. Sure, the phase of each eigenstate will change as a function of time- but there can never be a transition between eigenstates.

So- is it possible at all for a closed quantum system to thermally equilibrate? If so- then how?

2. Apr 20, 2007

### christianjb

Oh- and I know that photon exchange will in practice give rise to thermal equilibration.

3. Apr 23, 2007

### Einstein Mcfly

I'm not sure I fully understand what you're asking. Let me try to restate it:
So have some large system that we can solve the SE for (and hence know the eigenfunctions), we "begin" with the system in the eigenstate of lowest energy and then put it in a box, closing the system. In order to do this you must ASSUME that your system is in a pure state and hence must be at 0K (please let me know if this is an incorrect or unclear statement). If you have something at 0K in a close system, in what sense does it need to "equilibrate"? There's nothing more equilibrated than a system in a pure state at 0K.

Also, you COULD expand such a wavefunction in a superposition of stationary states, but the coefficients of all terms other than the energetic ground state would be zero.

For this to be interesting, you'd have to begin with a system in thermal equilibrium at some temperature T, find the diagonal density matrix that describes such a state, perterb it somehow with a field or collision, and THEN put it in box and see if it equilibrates. I'm not sure I know exactly how to think about this; in the isolated molecule limit the models for dissipation aren't applicable and all you'd see is regular old wavepacket dephasing and rephasing, but no relaxation to the original (incoherent) thermal state.

4. Apr 23, 2007

### JPRitchie

This is not always a correct statement - witness absorption and emission spectra, especially for the hydrogen atom, explanations of the line spectra for which led to the postulation of energy quantization.

Also, remember there are electronic, vibrational, and rotational degrees of freedom.

Moreover, consider collisions between, say, molecules in a liquid or gas.

Lastly, molecules can chemically react.
-Jim

5. Apr 23, 2007

### christianjb

No- I'm talking about starting the system off in an arbitrary admixture of states, not the ground state.

I'm thinking of a computer simulation. Is it possible to obtain the correct thermal density matrix from one long trajectory- or does it really require running an ensemble of trajectories?

6. Apr 23, 2007

### christianjb

Yes, the system will eventually be thermalized by interactions with photons. However, I'm thinking of a computer simulation which doesn't include such effects.

7. Apr 23, 2007

### Einstein Mcfly

If you start the system off in some ARBITRARY mixed state and don't allow for collisions or decay by photon emission then your state would live on forever and in the basis of stationary states you'd see the normal coherent dephasing and rephasing of your wavepacket as time goes on. This seems vacuous to me; if you don't allow for dephasing and relaxation in your simulation, you won't see any dephasing and relaxation in your results. This makes me feel as if I still don't understand what you're asking.

I'd like to make a point about this idea of starting with a totally ARBITARTY mixture of states. In the absence of collisions (elastic or inelastic, it makes no difference) the only final states accessible for a system are those that have the same spectrum as the initial (thermal) state due to the rules of purely unitary evolution. If you're starting out with a system that has no symmetry breaking mechanism (ie, non-unitary evolution of the system) then you can't start out in any truly arbitrary state AND have that state be "physical". The eigenvalues of the density matrix in any non-relaxing environment must be the same at all times as the eigenvalues of the initial (in this case, Boltzman) matrix. This is not such a problem with vibrational and electronic degrees of freedom at normal temperatures, but is an important consideration for rotation and translation.

8. Apr 23, 2007

### christianjb

1) OK- but in the classical limit the system will eventually equilibrate.

2) Does this mean that a quantum dynamical simulation is useless if it doesn't include interaction with the photon bath?

3) Again- is it possible to prepare an ensemble of trajectories that reproduce the correct thermal density matrix? My guess is you can.

9. Apr 23, 2007

### Einstein Mcfly

1)What is the classical limit of such an isolated system with no relaxation dynamics? The reason that quantum systems ACT classical in certain limits is because of interaction with the environment (as I understand it). I feel like you're asking about the classical limit of a system that you're constraining to be inherently QM.

2) My response would be that it depends entirely on what system you're looking at, what properties you're looking at in the model and most importantly what the timescale of the relaxation would be. If it's much slower than the process you're looking at, you can ignore it, but the question you're asking seems to be a the very long term evolution of this state.

3) If you have something as purely QM as a system isolated from it's environment and put in some superposition state to evolve in time with no mechanism to interact with a bath or otherwise dissipate energy or alter phases it will not evolve into the thermal density matrix IMO. The thermal boltzman matrix is the state to which all trajectories must lead at finite temperature WHEN THERE IS A BATH that removes coherences between individual components of the wavefunction. I don't see how this happens in such an isolated system.

10. Apr 23, 2007

### JPRitchie

You're being too vague.

If you don't include all the real effects I mentioned that are possible in your simulation, your simulation won't be realistic. You'll have GIGO.

If you're trying to simulate a reasonably dense gas or liquid or other disordered state, symmetry considerations may not be signficant. If you're not, you might say so.

It worries me that you continually mention solving the SE, which normally would mean solving for the electronic degrees of freedom, because that doesn't include the vibrational and rotational wavefunctions that may play an important role in themalizing energy. And also that you aren't including photon re-emission as a mechanism, when you say you know "photon exchange will in practice give rise to thermal equilibration."
-Jim

11. Apr 23, 2007

### christianjb

Sorry for the confusion- I mean solving the S.E. for the nuclear degrees of freedom on the Born Oppenheimer surface.

12. Apr 23, 2007

### christianjb

1) A dynamical classical system on the Born Oppenheimer surface will reproduce a well-defined ensemble average (essentially by the ergodic theorem). I guess I'm asking about ergodicity using quantum dynamics, and why it doesn't appear to hold for quantum systems and yet works in the classical limit.

2) I'm interested in the long-time 'ergodic' limit.

3) I agree it won't happen for an isolated quantum system with no photon bath. Is it possible though to arrange an ensemble of systems that will reproduce trace averages over the density matrix?

13. Apr 23, 2007

### JPRitchie

That's still pretty vague. I hate guessing games.
-Jim