Consider a two-level system described by the hamiltonian

$H = \hbar \omega_{eg} /2 \sigma_z \quad (1)$

The eigenenergies are $\pm \hbar \omega_{eg}/2$. Now, we add an interaction with an electromagnetic field, e.g.

$H = \hbar \omega_{eg} /2 \sigma_z - \vec{d} \cdot \vec{E}$.

The Hamiltonian can be transformed into the interaction picture and one gets

$H_{int} = \hbar /2 \begin{pmatrix} - \Delta & \Omega \\ \Omega^* & \Delta \end{pmatrix}$

with $\Omega $ the rabi frequency and $\Delta = \omega- \omega_{eg}$. Now, the eigenvalues are $\pm \hbar /2 \sqrt{\Omega^2 + \Delta ^2}$.

This raises several questions to me:

1.) If we now turn off the laser, e.g. set $\Omega = 0$, this means that the Eigenenergies are $\pm = \hbar/2 |\Delta|$. If the detuning is zero, this means that both levels have the same eigenergie. However, one should obtain the very same Eigenergies as in the case of equation (1). What am I not getting here? The rotating frame is oscillating at frequency $omega$, so the detuning should be zero independent from whether the laser is on or off. But this would mean that the resonance frequency of the atom has changed. That's impossible since changing from one frame to another cannot change the intrinsic properties of an atom.

2.) If one thinks about a pump-probe-experiment, the entire situation gets even more confusing for me. Think about a ladder scheme with ground stage $|g\rangle$, intermediate level $|e_1\rangle$ and the highest level $|e_2\rangle$. Now, if one pumps the transition $|g\rangle \leftrightarrow |e_1\rangle$ and probe the $|g\rangle \leftrightarrow |e_1\rangle$ transition, then one can observe the following:

  • with the pump off while scanning the probe, one observes a simple resonance at frequency $\omega{e_1,e_2}$

  • with the pump on at frequency $\Omega$, one observes two peaks with frequency $\omega{e_1,e_2} \pm \Omega/2$, so a total splitting of $\Omega$

This is confusing to me. Does this simply mean that the original eigenstates of the system $|g\rangle,|e_1\rangle, |e_1\rangle$ no longer exist and there is a complete new set of eigenstates with energies $\omega_{e2}, \omega_{e1}\pm \Omega/2$ and I complete have to forget about the ground state $|g\rangle$. What would happen if there was a 4th level $|e_3\rangle$ and which I use to probe to original transition $\omega_{e_3,g}$? Would there be no no transition anymore?


I need to evolve what I precisely mean or where my precise questions comes from:

1.) Let's consider two level-system with eigenenergies $E_{\pm} =\hbar/2 \omega_{eg}$. This system would be represented by the Hamiltonian $H_0 = \hbar/2 \omega_{eg} \sigma_z$. This syste can be as well considered in some rotating frame which would give a Hamiltonian $H_{rot} = H_0$, if the unitary transormation is of the form $U = \exp(-i/\hbar \sigma_z t)$. So, if I calculate now the eigenenergies in the rotating frame, I still obtain $E_{\pm}$. However, the time evolution of the Eigenstates in this system would be described by

$i \hbar \frac{\partial}{\partial t} \psi_{rot}=\hbar /2 \begin{pmatrix} \Delta & 0 \\ 0 & -\Delta \end{pmatrix}$

with $\Delta = \omega_{eg}-\omega$ (maybe the minus sign should be the other way around). This means, that for $\Delta =0$, the eigenstates would be a constant and for $\Delta = 0$, they would be rotating at frequencies $\pm \Delta /2$. So far, so good.

But now, if I include the interaction part, the trouble starts.

The Hamiltonian in the Schrödinger picture is given by $H = \hbar /2 \omega_{eg} \sigma_z -\begin{pmatrix}0 & \Omega \exp(-i \omega t)\\ \Omega \exp(-i \omega t) & 0 \end{pmatrix}$

(I simply copied the interaction part from Vadim). Transformed to any rotating frame, e.g. $H_{rot} = \exp(i \omega/2\sigma_z t/2) H \exp(i \omega/2 \sigma_z t) = \hbar\omega_{eg} /2 \sigma_z + \hbar \Omega/2 \sigma_x $.

This Hamiltonian already gives different (not -time dependent Eigenvalues than $H$). How can this be different? On top of that, one instead takes the Hamiltonian that one uses in the schrödinger equation in the interaction picture to calculate the Eigenvalues, e.g.

$H_{rot}^{int} = \hbar \Omega/2 \sigma_x $.

However, from what I understand this Hamiltonian is only used to calculate the time-evolution in the Schrödinger picture and not the calculate the Eigenvalue, since the actual Hamiltonien $H_{int} = U^\dagger H_{schrödinger} U = H_0 + H_{int}^{I}$ should be used to calculate the Eigenenergies (with $H_{int}^{I}$ I mean the interaction part of the Hamiltonian in the interaction picture).

So, therefore I have the following question: Do I have to include $H_0$ (so in my example the atomic part $\hbar/2 \omega_{eg} \sigma_z $) to calculate the energies in any rotating frame? So are the actual energies $\pm \hbar \omega/2 \pm \sqrt{\Omega^2 + \Delta^2}$

  • $\begingroup$ If there is no laser, isn’t $\Delta=\omega_{eg}$? $\endgroup$ Oct 23, 2020 at 9:12
  • $\begingroup$ This is where Im not sure. I would have thought that Im writing down the Hamiltonian in a different reference frame. This should not change anything $\endgroup$
    – anonymous
    Oct 23, 2020 at 9:13
  • $\begingroup$ Could you spell in more details the transformation to the rotating reference frame? I think your misunderstanding stems from not following through all the steps, some of which are tricky. $\endgroup$
    – Roger V.
    Oct 27, 2020 at 10:02

2 Answers 2


The Hamiltonian here is transformed to a rotating reference frame, which rotates with frequency $\omega$. This transformation is independent on whether the laser (i.e. the interaction term in the Hamiltonian) is actually present. As long as the frequency $\omega$ of this rotating reference frame is different from $\omega_{eg}$, the diagonal elements of the Hamiltonian cannot vanish.

One can use different formalisms for deriving the transformed Hamiltonian, so I merely try to sketch a few steps to explain the difference. Let us start with Hamiltonian $$ \hat{H} = \frac{\hbar}{2} \begin{bmatrix} \omega_{eg} & \Omega e^{-i\omega t}\\ \Omega e^{+i\omega t} & -\omega_{eg} \end{bmatrix}, $$ where the rotating wave approximation is already made. The Schrödinger equation is then $$ i\hbar\dot{\psi}_e = \frac{1}{2}\left(\hbar\omega_{eg}\psi_e + \hbar\Omega e^{-i\omega t}\psi_g\right),\\ i\hbar\dot{\psi}_g = \frac{1}{2}\left(\hbar\Omega e^{+i\omega t}\psi_e - \hbar\omega_{eg}\psi_g\right). $$ Transforming to the interaction picture could now be done by introducing new variables $$ \psi_e = \phi_e e^{-\frac{i}{2}\omega_{eg}t}, \psi_g = \phi_e e^{+\frac{i}{2}\omega_{eg}t}. $$ This would eliminate the diagonal part of the Hamiltonian, whereas the non-diagonal terms would oscillate with the frequency of detuning $\omega - \omega_{eg}$: $$ i\hbar\dot{\phi}_e = \frac{1}{2}\hbar\Omega e^{-i(\omega - \omega_{eg}) t}\phi_g,\\ i\hbar\dot{\phi}_g = \frac{1}{2}\hbar\Omega e^{+i(\omega - \omega_{eg}) t}\phi_e. $$

The Rabi Hamiltonian is however obtained by a different transformation: $$ \psi_e = \varphi_e e^{-\frac{i}{2}\omega t}, \psi_g = \varphi_e e^{+\frac{i}{2}\omega t}, $$ which results in $$ i\hbar\dot{\varphi}_e = \frac{1}{2}\left(\hbar(\omega_{eg}-\omega)\varphi_e + \hbar\Omega \varphi_g\right),\\ i\hbar\dot{\varphi}_g = \frac{1}{2}\left(\hbar\Omega \varphi_e - \hbar(\omega_{eg}-\omega)\varphi_g\right), $$ which is then attributed to an effective time-independent Hamitonian $$ \hat{H} = \frac{\hbar}{2} \begin{bmatrix} \omega_{eg}-\omega & \Omega\\ \Omega & -(\omega_{eg}-\omega) \end{bmatrix}. $$

Time-independent case
Let is look first at a problem where the Hamiltonian does not contain explicit time dependence. Such a problem is described by Schrödinegr equation $$ i\hbar\partial_t |\psi(t)\rangle = \hat{H}|\psi(t)\rangle, $$ where as the observables are given by $$ O = \langle \psi(t)|\hat{O}|\psi(t)\rangle. $$ The general solution of the time-dependent Schrödinger equation is $$ |\psi(t)\rangle = e^{-i\hat{H}t/\hbar}|\psi(0)\rangle, $$ and the observable of the Hamiltonian (i.e. the observed energy) is $$ E = \langle \psi(t)|\hat{H}|\psi(t)\rangle = \langle \psi(0)|\hat{H}|\psi(0)\rangle. $$ Rather than using the formal solution, one often uses the decomposition of the wave function into the eigenfunctions of the Hamiltonian operator, which makes the time-dependent Schrödinger equation trivially solvable: $$ |\psi(t)\rangle = \sum_n |n,t\rangle = \sum_ne^{-iE_nt/\hbar}|n\rangle, \text{ where } \hat{H}|n\rangle = E_n|n\rangle . $$

Time-dependent case
Let us now look at the time-dependent Schrödinger equation with $$ \hat{H} = \hat{H}_0 + \hat{\Omega},\\ \hat{H}_0 =\frac{\hbar\omega_{eg}}{2}\hat{\sigma}_z,\\ \hat{\Omega} = \frac{\hbar\Omega}{2}\begin{bmatrix}0 & e^{-i\omega t}\\e^{+i\omega t}&0\end{bmatrix} = \frac{\hbar\Omega}{2}\left[\hat{\sigma}_x\cos(\omega t) + \hat{\sigma}_y\sin(\omega t)\right] $$ We can transform the problem to a rotating wave representation $$ |\psi(t)\rangle = U(t)|\varphi(t)\rangle,\\ i\hbar\partial_t |\psi(t)\rangle = i\hbar \partial_t U(t)|\varphi(t)\rangle + U(t)i\hbar\partial_t|\varphi(t)\rangle = \hat{H}U(t)|\varphi(t), $$ so that $$ i\hbar\partial_t|\varphi(t)\rangle = \left[U^\dagger(t)\hat{H}U(t) - i\hbar U^\dagger(t)\partial_t U(t)\right] |\varphi(t)\rangle = \hat{H}_{int}|\varphi(t)\rangle $$ (remonder: $U^\dagger = U^{-1}\Leftrightarrow U^\dagger U = 1$). The operators are transformed in such a way that the observables remain unchanged: $$ O(t) = \langle\psi(t)|\hat{O}|\psi(t)\rangle = \langle\varphi(t)|U^\dagger(t)\hat{O}U(t)|\varphi(t)\rangle =\langle\varphi(t)|\hat{\tilde{O}}|\varphi(t)\rangle $$ If we explicitly choose the transformation as $$ U(t) = e^{-i\omega t \sigma_z/2} = \begin{bmatrix}e^{-i\omega t/2}&0\\0&e^{i\omega t/2}\end{bmatrix}, $$ we obtain $$ \tilde{H}_0 = H_0, \hat{\tilde{\Omega}} = \frac{\hbar\Omega}{2}\sigma_x, U^\dagger(t)\partial_tU(t) = -\frac{\hbar\omega}{2}\sigma_z,\\ H_{int} = \frac{\hbar}{2}\left[(\omega_{eg}-\omega)\sigma_z + \Omega\sigma_x\right] =\frac{\hbar}{2} \begin{bmatrix} \omega_{eg}-\omega &\Omega\\ \Omega & \omega_{eg}-\omega \end{bmatrix} . $$ An important point here is that the the operator that determines the evolution in time, $$H_{int} = U^\dagger(t)\hat{H}U(t) - i\hbar U^\dagger(t)\partial_t U(t),$$ is not the same as the transformed energy operator $$\tilde{H} = U^\dagger(t)\hat{H}U(t).$$ One can now determine the eigenfunctions by solving the Schrödinger equation in the interaction picture, $$ i\hbar\partial_t|\varphi(t)\rangle = |\varphi(t)\rangle = \hat{H}_{int}|\varphi(t)\rangle, $$ and use them to calculate the expectation of energy $H_0$.

Update 2
Now, when solving the full problem, the evolution of the wave function is determined by $$|\psi(t)\rangle = U(t)|\varphi(t)\rangle = e^{-i\omega\sigma_z t/2}|\varphi(t)\rangle,\\ \partial_t |\varphi(t)\rangle = H_{int}|\varphi(t)\rangle = \frac{\hbar}{2}\left[(\omega_{eg}-\omega)\sigma_z + \Omega\sigma_x\right]|\varphi(t)\rangle $$ The last equation can be solved just as I outlined in the description of the time-independent case: either in operator terms or by calculating the eigenvalues and eigenvectors of $H_{int}$. One can then use this solution to calculate the expectation values of $H_0$ or $H$, interpreting them as time-dependent energies of the system. In discussions of Rabi oscillations $H_{int}$ is often treated as if it were the true Hamiltonian of the system, by discussing its eigenvalues and eigenfunctions. This terminology is however based only on the analogy with eigenvalue problems. However this analogy should not be taken too literally, as we are dealing with a time-dependent problem here.

  • $\begingroup$ "As long as the frequency ω of this rotating reference frame is different from ωeg, the diagonal elements of the Hamiltonian cannot vanish." But that's exactly the case that I ask about. In that cas ethe diagonal terms are zero and the eigenergies are suddenly the same?! $\endgroup$
    – anonymous
    Oct 27, 2020 at 17:36
  • $\begingroup$ I don't quite understand what case you are talking about: $\Omega=0, \omega\neq 0$? Your Hamiltonians are the same as mine, but the initial Hamiltonian in the one in the rotating frame are not the same, of course. $\endgroup$
    – Roger V.
    Oct 27, 2020 at 18:03
  • $\begingroup$ I edited my question and phrased more clearly what I am asking about $\endgroup$
    – anonymous
    Oct 28, 2020 at 20:25
  • $\begingroup$ @anonymous when performing canonical transformation, you transform both operators and the states - you are moving time dependence from ones to the others. Eigenvalue problem is only a step in solving time-dependent Schrodinger equation, when the time dependence is removed. So it is not surprising that your eigenvalue ptoblems are different. What should remain unchanged is the observables. $\endgroup$
    – Roger V.
    Oct 28, 2020 at 21:33
  • 1
    $\begingroup$ Thanks a Lot! I think I got it ;) $\endgroup$
    – anonymous
    Oct 31, 2020 at 8:04

Lets think about a picture

As a more "pictorial" answer (agreeing perfectly with Vadim's mathematics).

Think of the simplest system, a harmonic oscillator. An energy eigenstate "whirls" around the phase space (x, p plane) over time at the oscillator frequency. This is what the oscillator frequency means.



Now we move to a frame that rotates in phase space. IE our "camera" is whirling around the x, p plane as well. (Often people say "moving to the rotating frame" or similar without reference to phase space, in my view this is needlessly confusing as it can make people think that some kind of frame that actually rotates in real space is being invoked).

By spinning our point of view in phase space at the oscillator frequency we can choose for the state that is actually orbiting around phase space to appear stationary. This means that, in the phase-space rotating frame this state has zero energy. (Energy IS, up to $\hbar$ how quickly things whirl around phase space).

The energy being different in the rotating frame is the entire point of the rotating frame. We are getting all the boring fiddly stuff to do with the systems free evolution and sweeping it under the carpet so that we can focus on the interesting bits (the interaction). If you are disturbed by the energy changing when we do the Unitary transform to the rotating frame then recall how only energy differences ever really matter. We can add "+10 energy" to every state with no change in anything at all. ($H_{new} = H_{old} + 10$), no difference to anything.

Your System

When you moved to the interaction picture you applied a Unitary that evolved with time, $U(t)$. In order to see what a particular state, $|\psi>$ found in the interaction picture "really looks like" in the lab you need to apply the dagger of that unitary from earlier, $|\psi_{lab}> = U^{\dagger}(t) | \psi_{rotating} >$.

In the example system you give (for zero detuning) one of the two (degenerate) eigenstates you found in the rotating frame will (in the lab frame) be:

$|\psi (t)>$ = cos($\omega t$) |up> + sin($\omega t$) |down>

While the other will be similar, but phase shifted, eg. (I maybe missing a "-"):

$|\psi (t)>$ = sin($\omega t$) |up> - cos($\omega t$) |down>

Notice that the two states were degenerate and stationary in the rotating frame. In the lab frame they are not stationary (they evolve in time), which makes sense as something standing still from a rotating point of view must be spinning in actuality. Note also that in the lab frame they are (on average) at the same energy over time, which is how the degeneracy in the rotating frame carries over.

  • $\begingroup$ This "We can add "+10 energy" to every state with no change in anything at all. (Hnew=Hold+10), no difference to anything" is not correct in this case, is it? What we do is that we can chage our referecne system such that it is rotating at the same speed as one of both eigenstates. However, the other Eigenstate is then rotating at frequency omega and the difference in energy is still $\hbar \omega$ $\endgroup$
    – anonymous
    Oct 28, 2020 at 20:27
  • $\begingroup$ @anonymous. You are correct. We are not adding "+10" energy, we are instead rotating our reference frame. At no point do we add "+" any energy. I only mentioned that because in the original question the poster seemed worried about the energy changing: I was trying to highlight that the energy changing when you re-arrange the Hamiltonian is not something that you necessarily need to worry about. $\endgroup$
    – Dast
    Oct 29, 2020 at 9:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.