Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

I was reading through the proof of the Adiabatic Theorem (in Sakurai) and I realised I'm not quite sure how Schrodinger Basis kets behave when we have a time-dependent Hamiltonian. I know that with a time-independent Hamiltonian the basis kets don't change in the Schrodinger Picture.

So if $|n;t\rangle$ are the energy eigenkets of $H(t)$ at time $t$ and $|\alpha;t\rangle$ is an arbitrary state at time $t$, is the following at all true? \begin{align*} |\alpha;t\rangle = \sum_n c_n(t)|n;t\rangle = \sum_n c_n(t) e^{i\theta_n(t)}|n,t_0\rangle \end{align*} where $\theta_n(t) = -\frac{1}{\hbar}\int_{t_0}^t H(t')\,dt'$ and $e^{i\theta_n(t)}$ is a time-evolution operator

Wikipedia and Sakurai both have (each in different notation): \begin{align*} |\alpha;t\rangle = \sum_n c_n(t) e^{i\theta_n(t)}|n;t\rangle \end{align*} I feel like I'm not understanding this properly at all

share|improve this question

1 Answer 1

up vote 2 down vote accepted

The basis of the Hilbert space in Schrödinger's picture is assumed to be time-independent regardless of any properties of the Hamiltonian. The Hamiltonian is just another operator. If the Hamiltonian is time-dependent, its eigenstates and eigenvalues are obviously time-dependent, too.

Both equations you write down only express the fact that the basis of eigenstates of $H(t)$ is still a basis, so a general ket vector, including the actual state vector of the system, may be expanded as a linear superposition of these basic vectors with some general complex coefficients $c_n(t)$. The two expansions only differ by the phase one includes into the coefficients $c_n(t)$ or into the basis vectors $|n;t\rangle$. One convention includes the phase $\exp(i\theta_n(t))$, another one doesn't, and so on. Obviously, there is no "universally mandatory" rule that would dictate the right phase of these vectors so there's some freedom about the notation. Note that a phase factor times an eigenstate is still an eigenstate.

Whatever your convention for the phases is, if you carefully follow the maths and remember what the symbols mean – the defining equations – you will be able to derive the invariant claims about the adiabatic theorem. The Wikipedia-Sakurai conventions treat the phases wisely and naturally, to speed up the derivations.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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