# A two-level system absorbs a detuned photon. Where does the extra energy go?

Let's consider simple two-level system with frequency gap of $\omega_0$ between ground and excited state. Now, when we turn on external electromagnetic field with frequency $\omega < \omega_0$, there is a non-zero probability that, starting from the ground state, the system will become excited.

The energy of excited state is $\hbar \omega_0$, but each photon of the field contributes only $\hbar \omega < \hbar \omega_0$. So my question is - what's with the missing $\hbar (\omega_0 - \omega)$?

I heard the explanation that it is measuring equipment that fills the energy gap, but I'm not quite convinced. Could you help?

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Other Q can help physics.stackexchange.com/questions/4047/… – user299 Jun 2 '12 at 19:45
Thanks, but I find linked problem a tiny bit different. Before any measurement process the ground state $|0\rangle$ mixes with excited one giving $a_0 |0\rangle + a_1 |1 \rangle$. I understand that such a mixing during measurement process can result in gaining energy from aparatus. But here theory predicts some gain in energy before this act. – qoqosz Jun 2 '12 at 20:29
Are you asking how you can excite an atom when the frequency is only slightly detuned? – Ron Maimon Jun 2 '12 at 21:17
More or less that's the point. – qoqosz Jun 2 '12 at 21:28

Rabi oscillations occur, in their simplest form, when a two-level system interacts with a quasi-resonant classical electric field; the total hamiltonian is then $$H_\textrm{s.c.}=\frac{\hbar\omega_0}{2}\hat{\sigma}_3+\frac{\hbar\Omega_0}{2}\cos(\omega t) \hat{\sigma}_1.$$ (For the atom in the ground state at $t=0$, the excitation probability then shows Rabi oscillations of the form $P=\frac{\Omega_0^2}{\Omega_0^2+\delta^2} \sin^2\left( \Omega t\right)$ for $\delta=\omega-\omega_0$ and $\Omega=\sqrt{\Omega_0^2+\delta^2}$.)

The important thing to realize is that this is the total energy of the system. Since the hamiltonian is not time-independent, there is no need for the atomic system to conserve energy. Of course, if the laser pulse stops and the atom is left in the excited state, the extra energy is taken from the light field.

To account for energy conservation, then, you need to include the light field into the reckoning. This is done by the Dicke model, for which the total hamiltonian is $$H_\textrm{D}=\frac{\hbar\omega_0}{2}\hat{\sigma}_3+\hbar\omega \hat{a}^\dagger \hat{a} +\frac{\hbar\Omega_0}{2}\left(\hat{a}+\hat{a}^\dagger\right) \hat{\sigma}_1.$$ This hamiltonian is barely integrable and no closed-form time-dependent evolutions are known, so it is usually simplified to its rotating-wave approximation counterpart, the Jaynes-Cummings hamiltonian $$H_\textrm{J.C.}=\frac{\hbar\omega_0}{2}\hat{\sigma}_3+\hbar\omega \hat{a}^\dagger \hat{a} +\frac{\hbar\Omega_0}{2}\left(\hat{a}\hat{\sigma}_+ +\hat{a}^\dagger \hat{\sigma}_-\right).$$ This hamiltonian leaves invariant the subspaces spanned by $\{|g,n\rangle, |e, n-1 \rangle\}$, and the probability of excitation shows Rabi oscillations of the form $$P=\frac{n \Omega_0^2}{n\Omega_0^2+\delta^2} \sin^2\left( \Omega t\right)$$ with $\Omega=\sqrt{n\Omega_0^2+\delta^2}$.

Thus, there is still an oscillation between levels of "different energy", since in this process a photon of energy $\hbar\omega$ is absorbed (and subsequently re-emitted, of course) to excite the atom up by $\hbar\omega_0$, and these two may not match, as you point out.

So what's going on? The hamiltonian is time-independent so that energy must be conserved. The resolution is that only the total hamiltonian need be conserved, and the notions of photon and atomic excitation are only approximate ones because they represent eigenstates of only part of the total hamiltonian. The true eigenstates are linear combinations of these and one can speak of excitations of the whole of the system, which are indeed conserved.

So where does that leave the problem then? Say the atom is moving across the laser beam profile, so that the coupling $\Omega_0$ increases from zero to some maximum and returns to zero. Rabi oscillations will show up as a function of the interaction time, which one can control for example by changing the atom's speed (this is actually rather common!). Say, then, that the system starts up with the atom in the ground state and the field in a number state, with the field mode slightly red-detuned as in your set-up. Then with some probability the atom will emerge excited and the field will have lost a photon, which are (now!) valid notions since the coupling has been turned off. Where did the energy go?

You can understand this as a system being adiabatically brought near a level anti-crossing similar to the one below. (image credit)

As the coupling is increased, the system (initially in a non-perturbed eigenstate) is brought impulsively to the middle of the anti-crossing, which puts it in a superposition of the true Jaynes-Cummings eigenstates. These two eigenstates then evolve at their own frequencies (which are respectively slightly higher and lower than the original) and by their interference make the Rabi oscillations. When the atom exits the beam, the hamiltonian impulsively returns to the uncoupled hamiltonian, and the atom is frozen into a superposition of the two uncoupled eigenstates, even if they do have different energies.

Where did the extra energy come from in this case? Well, the beam entry and exit were impulsive, so that again energy need not really be conserved (if the entry and exit are adiabatic then the system will remain in a total-hamiltonian eigenstate and will not show oscillations). In this particular case, then, it is probably taken from the atomic motion, which drives these impulsive transitions, and is affected by a light-induced potential. You would then need to add an atomic-position hamiltonian and account for the beam profile... and there's no ending to the amount of extra frills you can add.

In general, though, the driving principles are that you can only "conserve energy" when you've got a closed, time-independent system, and that energy conservation only makes sense when you consider the total energy of the system, including interaction energies, which are not simply transition-driving terms but do represent additional energy. Finally, one must be careful when speaking of "photons" when dealing with interacting systems - it is only total-hamiltonian excitations that are truly physical (though of course the free-hamiltonian excitations are very much useful notions).

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Thank you for very detailed explanation. – qoqosz Jun 3 '12 at 7:25
I trust the explanation, but fail to follow at first line of math. Do you have a "handwaving" 2-liner answer for mere mortals like me ? Its OK if it is not practical to abandon math for the sake of human language, but I hoped to have a chance of understanding the paradox – user299 Jun 3 '12 at 22:33
Basically, the red-shifted absorption is usually associated with a time-dependent process that does not account for the field energy, so there isn't a paradox since there are no "photons". Even if you account for the field energy, the presence of an interaction hamiltonian means that "photons" are not the true physical excitations of the system, which are eigenstates of the total hamiltonian. – Emilio Pisanty Jun 4 '12 at 0:40
The math details are not important. What matters is that the semiclassical hamiltonian is time-dependent, so Noether's theorem for time-energy does not apply. The second and third hamiltonians remove this time dependence by considering the field, but the only important detail is that the total eigenstates are not the uncoupled ones. – Emilio Pisanty Jun 4 '12 at 0:42