# Why does the Eigen-Energy-Shift for the AC-Stark-Shift, calculated in the "rotating frame", actually matter?

As is neatly described in this answer, the AC-Stark Shift is the observation that the energy-eigenvalues of stationary states in the "rotating frame" of a two-level-system behave as (Adam Steck, 5.64) \begin{align} E_{\pm} = − \frac{\hbar \delta}{2} \pm \frac{\hbar \tilde{\Omega}}{2} \end{align} and thus they are split by $$\tilde{\Omega} = \hbar \sqrt{\delta^2 + \Omega^2}$$

However, we begin in the original frame with a hamiltonian \begin{align} H = \hbar ( \omega_0 | e \rangle \langle e | + \frac{\Omega}{2} (|e \rangle \langle g | e^{-i \omega t } + | g \rangle \langle e | e^{i\omega t }) \end{align}

We arrived in the rotating frame from the "usual frame" by the unitary transformation \begin{align} U = e^{i\omega t} | e \rangle \langle e | + |g \rangle \langle g | \end{align}

In the rotating frame, the time evolution is given by

\begin{align} \tilde{H}_n = U H U^{\dagger} + i \hbar (\partial_t U)U^{\dagger} = \hbar (\omega - \omega_0) | e \rangle \langle e| + \hbar \frac{\Omega}{2} ( | e \rangle \langle g | + | g \rangle \langle e | ) \end{align} Diagonalizing this hamiltonian gives the mentioned energy eigenvalues.

Now my question is: Why do this eigenvalues actually matter? I know that all (interaction) pictures give the same physics - that is, if I transform the state of the system as $$| \tilde{\Psi} \rangle = U | \Psi \rangle$$ and every operator as $$\tilde{O} = U O U^{\dagger}$$, then the matrix elements of the operators, the expectationvalues etc ... won't change. This answer marks that point, as does this one, but both don't adress the point that there is a difference between observables that will give the same eigenvalues after transformation, and a hamiltonian that still governs the schroedinger equation in the right way, but is not the same observable anymore after the transformation:

The eigenvalues we calculated are the eigenvales of $$\tilde{H}_n$$. This is the operator that governs the right time evolution in this frame of reference. But because our transformation into the rotating frame $$U$$ is time dependent, $$\tilde{H}_n$$ and $$H$$ are related by $$\tilde{H}_n = U H U^{\dagger} + i \hbar (\partial_t U)U^{\dagger}$$, and not just by $$\tilde{H}_n = U H U^{\dagger}$$. This is set up specifically to have the right time-evolution of the states. Note that this is not the way one would transform an observable to not change it's eigenvalues and vectors (modulo unitary transformation).

In other words: Yes, matrix elements and expectation values of $$H$$ can be frame-independently calculated, but in the rotating frame, the transformation of $$H$$ (that would have the same eigenvalues etc...) is $$\tilde{H} = U H U^{\dagger}$$, and not $$\tilde{H}_n = \tilde{H} + i \hbar (\partial_t U)U^{\dagger}$$.

Because of that, I don't see the link between the eigenvalues of $$\tilde{H}_n$$, to which one refers when one talks about the AC-Stark shift, and energy values or even expectation values of the original Hamiltonian $$H$$. How are they related? And in case they are not related - Why can I get this energy splitting, but calculate it only in one frame (the rotating frame), but not in the other (the frame that we usually do quantum physics in, and in which we motivated the shape of our operators)?

If one dares to ask, the question can be expanded into the broader question: Which is the right operator to measure the energy of a system? The answer to this question obviously can't be "the operator that generates the time evolution", because that operator does depend on the choice of a frame.

EDIT: I know that if one employs the more sophisticated Jaynes-Cummings model, and goes to the Heisenberg-picture, the Hamilton-Operator will be:

\begin{align} {\displaystyle H_{\mathrm {JC} }=\overbrace {\hbar \omega _{\mathrm {a} }\sigma ^{+}\sigma ^{-}} ^{H_{\mathrm {a} }}+\overbrace {\hbar \omega _{\mathrm {c} }a^{\dagger }a} ^{H_{\mathrm {c} }}+\overbrace {\hbar g\left(a^{\dagger }\sigma ^{-}+\sigma ^{+}a\right)} ^{H_{\mathrm {int} }}} \end{align}

Since the operator that generates the right time evolution (I didn't write "Hamilton operator" on purpose) in Schroedinger and in Heisenberg picture is the same (unlike in the rotating frame), there are no ambiguities here. Does that mean that in general, the AC-Stark shift as a shift of the measurable energy simply can't be explained in the semiclassical picture in a rigorous way?

• Jan 18, 2023 at 11:01
• @EmilioPisanty I linked to that answer ;) it was your answer given to my question, and this a kind of- follow up. Jan 18, 2023 at 11:08