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

• I
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.

Last edited:
• Kolmo, vanhees71 and Twigg

Homework Helper
2022 Award
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.

• vanhees71 and hutchphd
Gold Member
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 and KDPhysics
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