# A mathematically rigorous treatment of adiabatic processes?

1. May 19, 2010

### petergreat

If a quantum system is subjected to a time dependent Hamiltonian with one parameter lambda, then its entropy does not change provided lambda is changed slowly enough between two values. How can this be proved rigorously? The http://en.wikipedia.org/wiki/Adiabatic_theorem" [Broken] is not enough, since this system is different in that it constantly re-thermalizes itself to a Boltzmann distribution in the process.

I know this is a standard topic in any statistical physics course, but I haven't seen one which is mathematically fully convincing to me.

Last edited by a moderator: May 4, 2017
2. May 19, 2010

### vaibhav1803

Well lets give it a try,i dont know a damn about adiabatic theorem so lets try not to think about any theorem already quoted,but get to standard thermodynamic adiabatic process there your "parameter lambda",i expect it to show up as somehow related to "gamma, the ratio of specific heats" We know that the system has ZERO change in external interaction energy(Q) thus giving us some lead to follow, you must be knowing the thermodynamic treatment there, thenyou have to work your way backwards..
im most gratified if i am of any help here.

3. May 20, 2010

### DrDu

If the system is finite, than any time dependence will be a unitary transformation, whether adiabatic or not, so that S is constant. If the system is infinite, than its spectrum is continuous and it will be very hard to establish an equivalent of adiabaticity.

4. May 20, 2010

### petergreat

I'm not sure If I understand you. Even if the system is finite (composed of finitely many particles in a finite volume), an abrupt change (such as compression by a pistol) will disturb it and causes S to increase.

5. May 20, 2010

### DrDu

A perturbation just means that the hamiltonian changes with time. Nevertheless, even a time dependent hamiltonian will transform a pure state into another pure state, at least for a finite isolated system. Hence there is no increase of entropy. That's Liouvilles theorem.

6. May 20, 2010

### petergreat

Sure, every state goes to the corresponding state in the final configuration, if the Hamiltonian governs everything in our system, so the Gibbs entropy doesn't change. However, this is a thermal system, which constantly redistributes itself to a Boltzmann distribution. For example, if the initial energy levels have equal spacings, but the final energy levels are not equally spaced, then the initial occupation number for each energy level no longer satisfies the Boltzmann distribution for final energy levels. How can you show the entropy is constant even after this re-distribution?

7. May 21, 2010

### DrDu

Yes, but then you should somehow build in this decoherence into your rigorous treatment.

8. May 21, 2010

### vaibhav1803

can we not correlate with thermodynamic model..would be simpler to understand although a bit more tiring to start off...??