# A Adiabatic process - quantum mechanics and thermodynamics

1. Nov 17, 2016

### spaghetti3451

A diabatic process is defined as follows:

Rapidly changing conditions prevent the system from adapting its configuration during the process, hence the spatial probability density remains unchanged. Typically there is no eigenstate of the final Hamiltonian with the same functional form as the initial state. The system ends in a linear combination of states that sum to reproduce the initial probability density.

An adiabatic process is defined as follows:

Gradually changing conditions allow the system to adapt its configuration, hence the probability density is modified by the process. If the system starts in an eigenstate of the initial Hamiltonian, it will end in the corresponding eigenstate of the final Hamiltonian.

A diabatic process involves a sudden change of the Hamiltonian, while an adiabtic process involves a gradual change of the Hamiltonian.

The word adiabatic, in thermodynamics, is usually reserved for processes that do not involve the exchange of heat between the system and surroundings. Does this use of the word adiabatic have any relation to the use of the word adiabatic in quantum mechanics? How about the word diabatic?

2. Nov 17, 2016

### DrDu

Adiabatic is greek and means without transit. In QM it means that there are no transits between eigenstates. In thermodynamics, it means no transit of heat trough the boundaries.

3. Nov 17, 2016

### spaghetti3451

What does it mean for there to be no transits between eigenstates in the adiabatic limit?

So far as I understand, in the adiabatic limit, the Hamiltonian changes so slowly that, at each instant, the wavefunction is some eigenstate of the instantaneous Hamiltonian.

I don't see how the issue of 'no transits between eigenstates' arises.

4. Nov 17, 2016

### DrDu

Sorry, I don't get what you mean.

5. Nov 17, 2016

### spaghetti3451

You say that there are no transits between eigenstates in the adiabatic limit.

It would be helpful if you could explain this in more detail.

6. Nov 17, 2016

### Staff: Mentor

If the system is initially in an eigenstate of the Hamiltonian, in the adiabatic approximation its stays in an eigenstate of the Hamiltonian, even though that Hamiltonian is changing with time. Contrast that to the other extreme, the sudden approximation, where the initial state is mapped unchanged to the eigenstates of the modified Hamiltonian.

7. Nov 17, 2016

### DrDu

"Rapidly changing conditions prevent the system from adapting its configuration during the process, hence the spatial probability density remains unchanged. Typically there is no eigenstate of the final Hamiltonian with the same functional form as the initial state. The system ends in a linear combination of states that sum to reproduce the initial probability density."
So if the process is diabatic, it ends up in a linear combination of eigenstates. If you do a measurement, you will find the system exclulively in one eigenstate, which may not coincide with the initial one (or into what it has evolved continuously). In this sense you have made a transit from one eigenstate to the other.

8. Nov 17, 2016

### spaghetti3451

I get that, in the diabatic limit, the wavefunction remains unchanged, but upon measurement of the system (after the sudden shift of the Hamiltonian), the eigenstate measured is not the same as the initial eigenstate. (This is because the wavefunction (after the sudden shift of the Hamiltonian) is in a linear combination of eigenstate of the new Hamiltonian). So, there has been a transition between eigenstates.

But don't you also make a continuous transition from eigenstate to eigenstate (each instantaneous eigenstate belonging to the instantaneous Hamiltonian) when the Hamiltonian changes gradually in the adiabatic limit?

9. Nov 17, 2016

### DrDu

Yes, but the only eigenstates you have available are the instantaeous ones, especially once you stop modifying your hamiltonian.

10. Nov 17, 2016

### spaghetti3451

In the adiabatic limit where the Hamiltonian changes gradually, what determines the evolution of the eigenstates with time?

In other words, say you have two Hamiltonians $H(t_{0},x)$ and $H(t+\delta t,x)$. Say that the system is in an eigenstate $\psi(t_{0},x)$ of the Hamiltonian $H(t_{0},x)$.

Now, the Hamiltonian $H(t+\delta t,x)$ can have a multitude of eigenstates $\psi_{n}(t_{0}+\delta t,x)$.

How do you figure out which of the eigenstates $\psi_{n}(t_{0}+\delta t,x)$ the system has evolved into?

11. Nov 17, 2016

### DrDu

The main point in the derivation of the adiabatic theorem is that the initial eigenstate is not degenerate and does not become degenerate during time evolution. There is a nice discussion in the book "Quantum Mechanics" by Messiah. A monograph on how to make this as water tight as possible is the book by Stefan Teufel, Adiabatic Perturbation Theory in Quantum Dynamics.

12. Nov 17, 2016

### spaghetti3451

I am not talking about degeneracy here though.

I was wondering how Nature selects the eigenstate for the new Hamiltonian since the new Hamiltonian might have any number of eigenstates. (These new eigenstates may or may not have the same energy as the old eigenstate, since the Hamiltonian itself has changed with time, and so the energy must also change with time. Am I wrong?)

So, should there not be some criteria used to determine the evolution of the initial eigenstate with time?

13. Nov 17, 2016

### DrDu

Sure, you need some conditions on the hamiltonian. I think a sufficient condition is that H(t) is of the form $H_0+\lambda(t) V$ where V has to be relative compact with respect to $H_0$ and lambda being sufficiently smooth.

14. Nov 17, 2016

### spaghetti3451

What does it mean for $V$ to be relatively compact with respect to $H_0$?

Why do you think these conditions are sufficient to determine the new eigenstate?

15. Nov 17, 2016

### DrDu

Let's consider the easier case of V being absolutely compact. Then the absolute value of all the eigenvalues of V is bounded. Then, if $H_0$ has an isolated eigenvalue which is separated by a gap from the rest of the spectrum, $H_0+\epsilon V$ will have an eigenvalue which is nearer to the original eigenvalue than to the rest of the spectrum for sufficiently small $\epsilon$, or, if epsilon is a smooth function of time, for sufficiently short times.