3
$\begingroup$

Adiabatic theorem states that: A physical system remains in its instantaneous eigenstate if a given perturbation is acting on it slowly enough and if there is a gap between the eigenvalue and the rest of the Hamiltonian's spectrum.

That said, slowly enough adiabatic evolution ensures that it remains in the same instantaneous eigenstate. What happen if adiabatic evolution fails, for example, not slow enough? Assume the system starts as a pure instantaneous eigenstate. Will it go to:

  1. a non-eigenstate.
  2. a combination of instantaneous eigenstates with the same eigenvalue (in case of degeneracy) or
  3. one another instantaneous eigenstate?
$\endgroup$
3
  • 2
    $\begingroup$ This is weirdly phrased. A combination of eigenstates (with different eigenvalues) is a non-eigenstate. But yes, 1/2 is what happens. $\endgroup$
    – knzhou
    Jul 11, 2016 at 4:41
  • $\begingroup$ The canonical example of a non-adiabatic process is a Landau-Zener transition: en.wikipedia.org/wiki/Landau%E2%80%93Zener_formula $\endgroup$
    – Rococo
    Jul 11, 2016 at 5:04
  • $\begingroup$ @knzhou yes, but why 3 is not possible? $\endgroup$
    – diff
    Jul 11, 2016 at 6:05

1 Answer 1

3
$\begingroup$

Let's begin with knzho's comment:

A combination of eigenstates (with different eigenvalues) is a non-eigenstate. But yes, 1/2 is what happens.

and take up your further question in comment:

yes, but why 3 is not possible? –

The answer is that a quantum system's unitary evolution is defined by the operator $\psi(t) = \exp\left(-\frac{i}{\hbar}\,\hat{H}\right)\,\psi(0)$ and this evolution is always continuous and differentiable by dint of the Stone theorem that asserts a one-to-one correspondence between self adjoint operators on a Hilbert space (like the energy observable $\hat{H}$) and strongly continuous one parameter unitary groups.

It is not necessary that the self adjoint operator should be bounded (equivalent to the notion of continuous in Hilbert spaces) for this theorem to hold.

Perturbation of a system is equivalent to altering the perturbed system's eigenstates, so any general evolution wrought by such perturbation can be thought of as imparting a sequence of operators of the form $\exp\left(-\frac{i}{\hbar}\,\hat{H}_j\right)$ to the initial state, which was an eigenstate of the original system.

Therefore, in the light of the Stone theorem, if the evolution fails to be adiabatic, the superposition weights for the combined eigenstates (of the original, unperturbed system) cannot jump discontinuously, so the system state must evolve continuously and differentiably through a superposition of eigenstates before it reaches another pure eigenstate.

Possibility 3 cannot therefore happen instantaneously.

However, you can, with enough knowledge of the system, choose the perturbation to steer the system state from one eigenstate of the original unperturbed system to another eigenstate over a nonzero time.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.