# Explicit Matrix Form of the Time-Evolution Operator

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!

• 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. Commented Nov 24, 2020 at 18:42
• @Jbag1212 thank you for your hint, I have had a go at trying to answer the question myself :) Commented Nov 25, 2020 at 13:02

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 $$\alpha = \frac{\eta}{\hbar \omega} sin\omega t$$