Half wave plate and angular momentum

Summary: the energy change is negligible and if it is not, the energy difference comes from a frequency change of the photon.

One must realize that unless the plate was already quickly rotating before the experiment, the energy stored in the rotation of the plate at the end is negligible relatively to the energy of the photon for the same reason why the electron carries most of the kinetic energy in the hydrogen atom even though both electron and proton are orbiting around the shared center-of-mass. In the latter case, the energy of a particle goes like $p^2/2m$, so for a fixed value of $p^2$ - and indeed, the proton's and electron's $p$ only differ by a sign - the lighter particle carries much more kinetic energy.

This is totally true for the rotational motion as well. The kinetic energy of rotation is $J^2/2I$ where $J$ is the angular momentum and $I$ is the moment of inertia. For a single photon, the angular momentum relative to the direction of the photon's motion, $J_1=-\hbar$, is changed to $J_2=+\hbar$. The overall change is $J=+2\hbar$ which becomes the angular momentum of the plate. When you square it, you clearly get a negligible number - that is divided by an $O(1)$ value of the moment of inertia.

So the energy obtained in the form of the increased rotation, caused by a single photon's change of the polarization, is tiny. If you study where this small amount comes from, you will find out that the photon's frequency is decreased by a tiny fraction so that its reduced energy exactly compensates the increase of the kinetic energy of the plate. But we can only see this change as nonzero if we allow the plate to have an arbitrary angular frequency before the experiment.

It's not too difficult to explicitly check that this statement works: if $J$ jumps by $2\hbar$, the energy $J^2/2I$ jumps by $J\times \Delta J/I=2J\hbar/I$. Because $E=\hbar\omega_\gamma$ for photons, the energy conservation law requires that the frequency of the photon decreases by $\delta \omega = -2J/I$ where $J$ is the plate's angular momentum before it changed the photon's polarization. But indeed, $-2J/I=-2\omega$ where $\omega$ is the angular frequency of the plate before the experiment. That's exactly the frequency change that the photon experiences. Why? Because the periodicity of the photon is conserved if measured relatively to the rotating plate - so its angular frequency changes from $|-\omega_\gamma|$ to $|+\omega_\gamma|$. But relatively to the inertial system, it's changed from $|-\omega_\gamma-\omega|$ to $|+\omega_\gamma-\omega|$ i.e. it drops by $2\omega$. It's just like comparing sidereal days and solar days.

What would happen with a single photon? I was working with a single-photon case from the beginning because it's a reasonable approach. A large electromagnetic wave may always be represented as a coherent state of many photons, so one just multiplies the results for a single photon. Of course, for a single photon, the "electromagnetic wave" must actually be represented as a wave function determining the probability amplitudes. At any rate, it is true that such wave plates may guarantee that every single photon that enters with a particular polarization leaves with another polarization.

If you measure the "exact" polarizations of the final photon, you will get the right one with probability of 100 percent. If you measure a different one, you may get odds between 0 and 100 percent. Every photon will be ultimately detected at a single point; the wave function knows about the odds.

Lubos is right: the (tiny) energy comes from the frequency shift of the final photon. That's one of the forms of the rotational Doppler effect.