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This is the Jaynes-Cummings Hamiltonian in the interaction picture: $H_i^n = \hbar g \sqrt{n+1}\begin{pmatrix} 0 & \exp(-i\delta t)\\ \exp(i\delta t) & 0 \end{pmatrix}$

I want to transform it into another basis, so that it looks like this:

$H_{i2}^n = \hbar \begin{pmatrix} -\delta/2 & g \sqrt{n+1}\\ g \sqrt{n+1} & \delta/2 \end{pmatrix}$

I present the solution that I found with help from DanielSank and march below.

Mechanix
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  • Can you think physically about what you need to do? Try writing the original $H$ in terms of Pauli operators and then think about what it means to get rid of the time dependence. If you can show some effort I will help you the rest of the way. On this site, we require questions to ask some specific and conceptual. Broad appeals to solve problems like this are actually considered off topic and reason to close a question. – DanielSank Dec 01 '15 at 21:15
  • Hi DanielSank, as far as I understand, getting rid of the time dependence means in this case to transform our reference frame so that it rotates exactly with this detuning frequency $\delta$, so that the observed frame is stationary. At the moment we are looking at a system that rotates with the frequency of the driving field. However, that doesn't seem to be everything that is necessary to do here. When I transform this hamiltonian with a matrix $U=\begin{pmatrix} \exp(-i\delta t/2) &0 \ 0 & \exp(i\delta t /2) \end{pmatrix}$ – Mechanix Dec 01 '15 at 21:29
  • ... then i can kill all time dependence (i'm left with constant off-diagonal elements), but i can't make diagonal elements appear like this. – Mechanix Dec 01 '15 at 21:29
  • It's not clear to me what to do after writing it in terms of pauli operators: $H_i^n=\hbar g \sqrt{n+1}\frac{1}{2}\left((\sigma_x+i\sigma_y)\exp(-i\delta t) + (\sigma_x-i\sigma_y)\exp(i\delta t) \right)$ ... for the other hamiltonian i get: $-\hbar \sigma_z \delta/2 + \hbar g \sqrt{n+1}\sigma_x$ – Mechanix Dec 01 '15 at 21:48
  • Could you put your work into the main post? – DanielSank Dec 01 '15 at 21:55
  • Sure! No problem! – Mechanix Dec 01 '15 at 22:38
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    In the post you give a candidate transformation, but you didn't show the result of applying that transformation... – DanielSank Dec 01 '15 at 23:01
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    Note that when transforming the Hamiltonian, you actually have to transform the left-hand (time-dependent) side of the Schrodinger equation, too, and since the transformation is time-dependent, you get more terms when applying the product rule to the product of the transformation matrix and the state vector. That's what will bring in those $\delta$'s on the diagonal. – march Dec 01 '15 at 23:01
  • @march Yeah, I feel like that is the thing that so preciously few people know about in problems like this. – DanielSank Dec 01 '15 at 23:02
  • @DanielSank. It helps having grown up in a theoretical quantum optics group. – march Dec 01 '15 at 23:04
  • @march The weird part is that although I learned about the "interaction picture" in school, we never did it the way you and I are talking about here. In other words, I learned how to move time dependence back and forth between the state and the operators, but I never learned that you can actually get rid of it entirely by going into a rotating frame. I had to figure that out myself. It's amazing how un-optimized our pedagogical programs can be some times! – DanielSank Dec 01 '15 at 23:18
  • @DanielSank. I do remember some homework problem where we brute-forced getting rid of the time-dependence in the Hamiltonian by guessing the form of the solution, but yes, I later had to generalize the procedure for myself. – march Dec 01 '15 at 23:26
  • Why don't you move the solution you've put in the post into an answer? Self-answered questions are perfectly fine! – DanielSank Dec 02 '15 at 00:02

1 Answers1

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Following DanielSank's suggestion, I tried writing both Hamiltonians in terms of Pauli operators: $$H_i^n=\hbar g \sqrt{n+1}\frac{1}{2}\left[(\sigma_x+i\sigma_y)e^{-i\delta t} + (\sigma_x-i\sigma_y)e^{i\delta t} \right],$$ and $$H_{i2}^n = -\hbar \sigma_z \delta/2 + \hbar g \sqrt{n+1}\sigma_x.$$

So I'm basically looking for a transformation that converts the content of the brackets into $\sigma_x$ and also makes this other term appear.

If I want to get rid of the time dependence, I can transform everything back into the Schroedinger picture by using $$U=\begin{pmatrix} \exp(i\delta t/2) &0 \\ 0 & \exp(-i\delta t /2) \end{pmatrix}.$$

This gives: $$UH_i^n U^+ = \hbar \sqrt{n+1}\sigma_x=H^*$$

I try to apply this $U$ to the Schrodinger equation from the left side:

$$UH_i^n|\Psi\rangle = i\hbar U \partial_t |\Psi\rangle $$

Then I insert $U^{-1}U$ before the state $|\Psi\rangle$ on both sides:

$$U H_i^n U^{-1}U |\Psi\rangle = i\hbar U\partial_t U^{-1}U |\Psi\rangle $$

Then I rename $$U|\Psi\rangle \equiv |\Psi'\rangle$$

Now product rule, as march suggested:

$$UHU^{-1} |\Psi'\rangle = i\hbar UU^{-1}\partial_t |\Psi'\rangle + i\hbar U \partial_t( U^{-1}) |\Psi'\rangle$$ which leads to $$\left(H^* - \hbar \begin{pmatrix} \delta/2 &0 \\ 0 &-\delta/2 \end{pmatrix}\right) |\Psi'\rangle =i\hbar\partial_t | \Psi' \rangle$$

The part of the left side is now exactly what I was looking for!

Update

A follow-up question generalizes this special case.

Community
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Mechanix
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  • ok I tried to take this into account now. sorry for my late reply. Thanks again! – Mechanix Dec 15 '15 at 21:49
  • Looks good! I edited the formatting a bit and added a link to @DanielSank's follow-up-ish question. And upvoted both, which somehow I had neglected to do before. – march Dec 16 '15 at 17:33