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. 
 A: 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.
