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.

  • 1
    $\begingroup$ Notation: I suggest you use $J$ instead of $P$ for rotations, as $P$ refers commonly to translations. $\endgroup$
    – devCharaf
    Jul 13, 2020 at 11:42

2 Answers 2


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)

  • $\begingroup$ Thanks, how would applying Baker-Hausdorff look like? $\endgroup$
    – MrDerDart
    Jul 13, 2020 at 12:36
  • $\begingroup$ 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 $P_{\pm}e^{\pm i\omega t}$ $\endgroup$
    – user245141
    Jul 13, 2020 at 12:52

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