Transformation to rotating frame I want to apply a transformation to the rotating frame of a two level system such that a state in the transformed frame is $ |\hat{\phi} \rangle = U |\phi \rangle$, where U is the generator of rotations $ U = e^{i\omega J_{z}t}$ with its Hermitian conjugate $U^\dagger $.
Given the Hamiltonian in the stationary frame $H = \omega_0 J_z + \epsilon (J_{+}e^{-i\omega t} +J_{-} e^{i\omega t})$, I want to derive the expression for the Hamiltonian in the rotating frame, which should be $ \hat{H} = (\omega_0 -\omega)J_z + \epsilon(J_{+} +J_{-})$.
So far, I proceeded as follows:
Demanding that both$|\phi\rangle$ and $|\hat{\phi} \rangle$ satisfy the time-dependent Schroedinger equation, we may write (let $\hbar =1$) :
$i \frac{d}{dt}|\hat{\phi} \rangle = \hat{H}|\hat{\phi} \rangle$,
where the LHS evaluates to $i \frac{d}{dt}|\hat{\phi} \rangle = i \frac{dU}{dt}|\phi\rangle + U i \frac{d}{dt}|\phi\rangle = - \omega J_{z} e^{i\omega J_z t}|\phi\rangle + UH|\phi\rangle$.
Since U is unitary, we can also write $|\phi\rangle = U^\dagger|\hat{\phi}\rangle $ and substitute this into the expression to above to find
$i \frac{d}{dt}|\hat{\phi} \rangle  = (- \omega J_{z} + UHU^{\dagger})|\hat{\phi} \rangle $.
This would imply that $\hat{H} = (- \omega J_{z} + UHU^{\dagger}) $
I would highly appreciate any help how to get to the desired result.
 A: You got
$$ i\frac{d}{dt}|\hat{\phi}\rangle = -\omega P_z U |\phi\rangle + UH|\phi\rangle$$
but you want the equation for $|\hat{\phi}\rangle$. That's no problem - you just plug in $U^{\dagger}U=1$ to get
$$ i\frac{d}{dt}|\hat{\phi}\rangle = -\omega P_z U |\phi\rangle + UHU^{\dagger}U|\phi\rangle = \left[UHU^{\dagger}-\omega P_z\right]|\hat{\phi}\rangle$$
and you can identify
$$ \hat{H} = UHU^{\dagger}-\omega P_z$$
as $UP_{\pm}U^{\dagger} = e^{\pm i\omega t}P_{\pm} $ (you can get that from Baker-Hausdorff or just by checking for the two possible values of $P_z$ by hand) you get the desired Hamiltonian.
(p.s. here I assumed that $P_z$ has eigenvalues $\pm 1/2$, if that's not the case then you have to adjust the transformation accordingly)
A: I suppose this is not a homework problem I'm denying you anymore. There is no CBH rearrangement involved.
You wish to evaluate
$$\hat{H} = - \omega J_{z} + e^{i\omega t J_{z}} He^{-i\omega t J_{z}} , $$
given
$$
[J_z,J_{\pm}]= \pm J_{\pm},
$$
whence you can easily prove
$$
J_{\pm}~ f(J_z) = f(J_z\mp 1) ~J_{\pm},
$$
for any function f.
This expression, then readily reduces to
$$
H = \omega_0 J_z - \omega J_z + \epsilon e^{i\omega t J_{z}} (J_{+}e^{-i\omega t} +J_{-} e^{i\omega t})e^{-i\omega t J_{z}} \\ =
(\omega_0 - \omega ) J_z + \epsilon   (J_{+}  +J_{-}  ) .
$$
