Atom-photon interactions in the interaction picture (self answered)

[KDPhysics]

TL;DR Summary

How do we evaluate the exponential of a hamiltonian involving tensor products? Why do i get a zero result?

EDIT: I'M SO DUMB! I can't believe I can't multiply matrices together. Of course the result is not zero, the matrix on the left will be:
$$
\begin{pmatrix}
0 & e^{i\omega_at/2}\\
e^{-i\omega_at/2}&0
\end{pmatrix}
$$

So i was solving problem 3 from https://ocw.mit.edu/courses/physics...iii-spring-2018/assignments/MIT8_06S18ps5.pdf

Here is my working (i used properties 3 and 8 from https://en.wikipedia.org/wiki/Kronecker_product to simplify the exponential of a tensor product):

Clearly 0 does not make sense, but I don't understand where I could have made a mistake? It seems like from the way the perturbation is defined as having off-diagonal elements, there is no way multiplying it by the exponential of the unperturbed hamiltonian, which is diagonal, will yield a non-zero result.

 

[hutchphd]

Also how does $\sigma_z$ get pulled through the exponential?

 

[KDPhysics]

What do you mean by pulled through?

 

[KDPhysics]

The $\sigma_z$ is a diagonal matrix so to take its exponential we can simply take the exponential of the diagonal elements.

 

[Twigg]

Nice job spotting that! And don't beat yourself up over making this kind of error. It happens! I've definitely posted self-answered threads before

I think the way you're doing is the way the problem intended you to do this. If you wanted to get fancy and see a little deeper into the physics of this process, I encourage you to rewrite the atomic part of the 0th and 1st order parts of the Hamiltonian as a vector dotted with the standard pauli matrix vector, if that makes sense ($H = H_x \sigma_x + H_y \sigma_y + H_z \sigma_z$). The interaction picture result $\delta \tilde{H}$ will have a clear geometric interpretation that way.

 

[vanhees71]

Well, we learn most by recognizing our errors! Great job!

 

[KDPhysics]

that's the way I justify it XD