In the standard derivation of the adiabatic approximation (see Sakurai in Modern Quantum Mechanics, Wikipedia) a differential equation for the coefficients is reached as $$ i\hbar \dot{c}_m(t) + i\hbar \sum_n c_n(t) \langle m(t)|\dot{n}(t)\rangle = c_m(t)E_m(t) \tag{1} $$

Now in order to find the inner product $\langle{m(t)|\dot{n}(t)}$, usually the method is to differentiate the time-independent Schrodinger equation using the product rule and use orthogonality of basis states to find

$$ \langle m(t)|\dot{n}(t)\rangle = -\frac{\langle m(t)|\dot H (t)|n(t)\rangle}{E_m(t)-E_n(t)} \tag{2} $$

But why can't the time-dependent Schrodinger equation be used again on the basis state $|n(t)\rangle$?

$$ i\hbar |\dot n(t)\rangle = H(t)|n(t)\rangle = E_n(t)|n(t)\rangle \tag{3} $$

Clearly this is wrong since substituting into $(1)$ gives that $\dot{c}_m(t) = 0$, but where is the flaw?

  • $\begingroup$ You are missing some $\langle$ and $|$ $\endgroup$
    – Mauricio
    Commented Apr 14, 2023 at 21:57
  • $\begingroup$ It was done by second editor. I have no idea why he did such a ridiculous thing… $\endgroup$
    – Siam
    Commented Apr 14, 2023 at 22:02

1 Answer 1


It’s a simple question of definition. Your $|n,t\rangle$ are the instantaneous eigenvectors of the Hamiltonian $H(t)$. There is a priori no reason why they should obey the Schrödinger equation. They are defined independent from it after all.

A mathematical investigation confirms this hunch. By definition: $$ H(t)|n,t\rangle=E_n(t) |n,t\rangle $$ with $E_n(t)$ the n-th eigenvalue of $H(t)$ (assuming the Hamiltonian stays nondegenerate). You could try to take the time derivative of the defining eigenvalue problem. However, you don’t get the SE, but rather equation (2) by projecting on the eigenbasis.

You can check on specific examples for confirmation. Try a spin 1/2 particle in a rotating magnetic field which can be solved analytically. The instantaneous energy eigenstates are not solution of the Schrödinger equation.

Hope this helps.


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.