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

In summary, the problem asks for the product of two matrices, one on the left and one on the right, and it's not possible to get a result that is not zero. The solution is to take the exponential of the diagonal elements of the matrices.
  • #1
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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):

CamScanner 05-20-2021 17.19.45_1.jpg

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.
 
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  • #2
Also how does ##\sigma_z## get pulled through the exponential?
 
  • #3
What do you mean by pulled through?
 
  • #4
The ##\sigma_z## is a diagonal matrix so to take its exponential we can simply take the exponential of the diagonal elements.
 
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  • #5
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 :oldbiggrin:

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.
 
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  • #6
Well, we learn most by recognizing our errors! Great job!
 
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  • #7
that's the way I justify it XD
 

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

1. What is the interaction picture in atom-photon interactions?

The interaction picture is a mathematical tool used to describe the time evolution of a quantum system, such as an atom interacting with a photon. It separates the Hamiltonian (energy operator) of the system into two parts: the free Hamiltonian, which describes the system in the absence of any external interactions, and the interaction Hamiltonian, which accounts for the interactions between the atom and photon.

2. How do atom-photon interactions occur in the interaction picture?

In the interaction picture, the state of the system (e.g. the atom) is represented by a time-dependent wavefunction, which evolves according to the free Hamiltonian. The interaction Hamiltonian then acts on this wavefunction to account for the interactions with the photon. This allows us to study the effects of the interaction separately from the free evolution of the system.

3. What is the significance of the interaction picture in studying atom-photon interactions?

The interaction picture allows us to simplify the mathematical description of atom-photon interactions by separating the free evolution of the system from the effects of the interactions. This makes it easier to analyze and understand the dynamics of the system and make predictions about the behavior of the atom and photon.

4. How does the interaction picture differ from the Schrödinger picture in atom-photon interactions?

In the Schrödinger picture, the state of the system is represented by a time-dependent wavefunction that evolves according to the total Hamiltonian (including both the free and interaction Hamiltonians). In the interaction picture, the state of the system is represented by a time-dependent wavefunction that evolves according to the free Hamiltonian, while the interaction Hamiltonian acts on the wavefunction to account for the interactions. In other words, the interaction picture separates the effects of the interactions from the free evolution of the system.

5. What are some applications of the interaction picture in atom-photon interactions?

The interaction picture is commonly used in quantum optics and quantum information processing, where the interactions between atoms and photons play a crucial role. It allows us to study various phenomena, such as photon absorption and emission, Rabi oscillations, and quantum entanglement, and make predictions about the behavior of the system. It is also used in the development of quantum technologies, such as quantum computing and quantum communication.

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