I have been given a time-dependent Hamiltonian $H = \eta$ cos $\omega t$ $\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$ and asked to calculate explicitly in matrix form the time-evolution operator $U(0, t)$ associated to $H$.

I am completely stuck on how to do this. Do I use $U(0,t) = e^\frac{-iHt}{\hbar}$ ? Although I believe this only holds if H is time-independent.

And if so, how do I write this explicitly in matrix form?

Thanks for any help!

  • $\begingroup$ Yes, you'll need to use the exponential of the matrix. The exponential of a matrix is defined by the "power series" expansion. Work out the various powers of $H$ and then write the infinite sum. $\endgroup$
    – Jbag1212
    Commented Nov 24, 2020 at 18:42
  • $\begingroup$ @Jbag1212 thank you for your hint, I have had a go at trying to answer the question myself :) $\endgroup$
    – B Delamera
    Commented Nov 25, 2020 at 13:02

1 Answer 1


Use $U(t,0) = exp [ -\frac{i}{\hbar} \int_{0}^{t} H(t') dt' ] = exp [ - \frac{i}{\hbar} \begin{pmatrix}0 & \frac{\eta}{\omega}sin\omega t\\ \frac{\eta}{\omega}sin \omega t & 0 \end{pmatrix}]$

Then let $A = - \frac{i}{\hbar} \begin{pmatrix}0 & \frac{\eta}{\omega}sin\omega t\\ \frac{\eta}{\omega}sin \omega t & 0 \end{pmatrix}$

Work out $e^A$ by diagonalizing A (writing $A=SDS^{-1}$) and using the power series expansion $e^{A} = \sum_{n=0}^{\infty} \frac{A^n}{n!}$

And you should get $U(t,o) = e^{A} = \begin{pmatrix}cos\alpha & -isin\alpha \\ -isin\alpha & cos\alpha \end{pmatrix}$ where $\alpha = \frac{\eta}{\hbar \omega} sin\omega t$


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