# Transformation of Jaynes-Cummings Hamiltonian

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.

• 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. Dec 1, 2015 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 $U=\begin{pmatrix} \exp(-i\delta t/2) &0 \\ 0 & \exp(i\delta t /2) \end{pmatrix}$ Dec 1, 2015 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. Dec 1, 2015 at 21:29
• In the post you give a candidate transformation, but you didn't show the result of applying that transformation... Dec 1, 2015 at 23:01
• 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. Dec 1, 2015 at 23:01

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.

• ok I tried to take this into account now. sorry for my late reply. Thanks again! Dec 15, 2015 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. Dec 16, 2015 at 17:33