# Solving the Schrodinger equation with a time-dependent Hamiltonian

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.

• Which edition? What year? – Qmechanic Oct 11 '18 at 11:46
• The second edition international version published in 2011 by Pearson. – Matt0410 Oct 11 '18 at 12:30

## 1 Answer

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 $$K_{mn}(t) = \left\langle \psi_{m, t}, i \frac{ d \psi_{n,t} }{dt} \right \rangle$$.

• Okay this makes sense. So Sakurai has expanded the state $|\psi(t)\rangle$ in terms of simultaneous eigenstates as $| \psi(t) \rangle = \sum_n c_n(t) |\psi_{n,t} \rangle$. Does the $t$ in $|\psi(t)\rangle$ evolve by Schrodinger's equation still? If so, I find that $c(t) = c(0)\exp(-i \int_0^t E_{n,t} dt )$. I am confused between the usage of the parameter $t$, i.e. where to treat it as a simultaneous, non-evolving time, or where to treat it as an evolving time. – Matt0410 Oct 11 '18 at 12:54