# 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.

• Notation: I suggest you use $J$ instead of $P$ for rotations, as $P$ refers commonly to translations. Jul 13, 2020 at 11:42

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 $$H = \omega_0 J_z - \omega J_z + \epsilon e^{i\omega t J_{z}} (J_{+}e^{-i\omega t} +J_{-} e^{i\omega t})e^{-i\omega t J_{z}} \\ = (\omega_0 - \omega ) J_z + \epsilon (J_{+} +J_{-} ) .$$

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)

• Thanks, how would applying Baker-Hausdorff look like? Jul 13, 2020 at 12:36
• Once you note that $[P_z, P_{\pm}] = \pm P_{\pm}$ then $U P_{\pm} U^{\dagger} = P_{\pm} \pm i\omega t P_{\pm} - \frac{(\omega t)^2}{2}P_{\pm} + \cdots$ where you get it from repeatedly commuting $P_z$ with the result of the previous commutator - which is always $P_{\pm}$. The series is $P_{\pm}e^{\pm i\omega t}$
– user245141
Jul 13, 2020 at 12:52