Solving the Schrodinger equation with a time-dependent Hamiltonian

Question

I am trying to find the general solution to the Schrodinger equation with a time-dependent Hamiltonian:

$$ i \frac{\partial}{\partial t}| \psi(t) \rangle = H(t) | \psi(t) \rangle.$$

My Hamiltonian evolves over time but it remains Hermitian, so at any given time $t$ I have an orthonormal basis of eigenstates $ \{ |\psi_n(t) \rangle \}$ where

$$H(t)|\psi_n (t) \rangle= E_n(t) |\psi_n(t) \rangle.$$

My starting point to solve the Schrodinger equation would be to expand my state $|\psi(t) \rangle$ in an orthonormal basis. I will choose my basis to be the eigenstates of $H(0)$, i.e. the set $\{ |\psi_n(0) \rangle \}$, so I have

$$ |\psi(t) \rangle = \sum_n c_n(t) |\psi_n(0) \rangle $$

What if I wanted to expand $| \psi(t) \rangle$ in terms of the time-dependent eigenstates $\{ |\psi_n(t) \rangle \}$ at some arbitary non-zero time? Well I know how these states evolve so I can relate them to $\{ |\psi_n(0) \rangle \} $: the states $\{ | \psi_n(t) \rangle \} $ solve the Schrodinger equation as

$$ i \frac{\partial}{\partial t} |\psi_n(t) \rangle = H(t) |\psi_n(t) \rangle \\ = E_n(t) |\psi_n(t)\rangle \\ \Rightarrow |\psi_n(t)\rangle=\exp \bigg(-i\int_0^tE_n(t')dt' \bigg)|\psi_n(0)\rangle $$

Which gives me the relationship between $\{ |\psi_n(0) \}$ and $\{ |\psi_n(t) \rangle \}$. I can invert this to find

$$ |\psi_n(0)\rangle = \exp \bigg(i\int_0^tE_n(t')dt' \bigg)|\psi_n(t)\rangle. $$

Substuting this into my expression for $|\psi(t)\rangle$, I have

$$ |\psi(t) \rangle = \sum_n c_n(t)\exp \bigg(i\int_0^tE_n(t')dt' \bigg)|\psi_n(t)\rangle $$

On page 346 of Sakurai's Modern Quantum Mechanics, eq. (5.6.5), he has expanded the general solution $|\psi(t)\rangle$ just like this but the integral phase has a minus sign out the front. I do not see where I have gone wrong in my reasoning above. The following analysis in Sakurai's book to prove the Adiabatic theorem requires the minus sign to work so I would very much like some hints! Thank you in advance.

Answer 1

You assume that

$i \frac{d}{d t} \psi_n(t) = H(t) \psi_n(t) \ , $

which is not true. I think the confusion arises because of the notation for $\psi_n(t)$, which are often called the instantaneous eigenstates. Let's choose the notation $\psi_{n,\tau}$, solving:

$H(\tau) \psi_{n,\tau} = E_{n,\tau} \psi_{n,\tau} \ .$

Now of course we may find solutions $\psi_{n,\tau}(s)$ to the time-dependent Schrödinger problem

$i \frac{d}{d t} \psi_{n,\tau}(t) = H(s) \psi_{n,\tau}(t) \ \ \ \ , \ \psi_{n,\tau}(0) = \psi_{n,\tau} \ ,$

But the Schrödinger equation does not govern the dependence of $\psi_{n,\tau}$ on the parameter $\tau$ ( which i have chosen to be a greek letter, to stress that it's role is very different from the time $t$).

The set $\{\psi_{n,\tau}\}_n$ is a basis like any other basis, so one may expand the vector at time $t$ in this basis using time-dependent coefficients $\tilde{c}_n(t)$:

$\psi(t) = \sum_n \tilde{c}_n(t) \psi_{n,\tau} \ . $

But this doesn't really bring an advantage in solving the Schrödinger problem, because the $\psi_{n\tau}$ are not eigenvectors of $H(t)$ for $t \neq \tau$.

What you did is expanding $\psi(t)$ in the instantaneous eigenstates $\psi_{n,t}$ as

$\psi(t) = \sum_n c_n(t) \psi_{n,t}$.

Now acting with $H(t)$ gives

$H(t) \psi(t) = \sum_n E_{n,t} c_n(t) \psi_{n,t} \ .$

However, now the time-derivative acting on $\psi(t)$ is more complicated:

$i \frac{ d\psi(t)}{dt}(t) = \sum_{n} \left[ i \frac{ d c_n}{dt}(t) \psi_{n,t} + c_n(t) i \frac{ d \psi_{n,t} }{dt} \right] \ .$

You may know project to $\psi_{m,t}$, but the equation you get is

$ i \frac{ d c_m}{dt}(t) = E_{m,t} c_m(t) - \sum_{n} c_n(t) \left\langle \psi_{m, t}, i \frac{ d \psi_{n,t} }{dt} \right \rangle \ ;$

And this is no longer easy to solve. You might be interested in the topic of adiabatic quantum mechanics, which is basically a pertubation theory neglecting off-diagonal elements of the matrix $ K_{mn}(t) = \left\langle \psi_{m, t}, i \frac{ d \psi_{n,t} }{dt} \right \rangle$.