# Quantum entanglement and Compton Effect

Suppose we generate 2 entangled photons (via a beam-splitter or similar apparatus). They have an arbitrary indeterminate initial frequency $\nu_i$. One of these photons is allowed to propagate through free space, the other is subjected to Compton scattering via a graphite scatterer apparatus.

The frequency of the scattered photon is detected by a detector to be $\nu_f$, and $\nu_i \gt \nu_f$ (By Law of Conservation of Energy). Since $E=hf$, initial energy of the photons $E_i \gt E_f$ , the final energy of the scattered photon.

Now, since the 2 photons were entangled, this change in state should be reflected by the photon propagating through free space, unhindered. If now the freely propagating photon is intercepted by a detector, the detector should measure its frequency as $\nu_f$ and its energy as $E_f$. The detector does not take more than a negligible amount of energy from the photon during the process of detection.

If the photon is so detected, how/where was the energy lost? If not, what should the detector detect and why?

Firstly, entanglement isn't magic. And it is next to meaningless to just say something is entangled. Entangled just means not factorizable. But is the lack of factorizability from the spatial degrees of freedom? From the polarization degrees of freedom? Is it super close to factorizable? Is it maximally entangled? Just saying it is entangled isn't really a description of anything.

Secondly, entanglement isn't magic. And it very much doesn't mean that a change in one thing causes a change in the other thing, it's much much different than that.

One way to entangle photons is to have a state like $$\frac{1}{\sqrt 2}\left|\nu_{i,low}+\right\rangle_A\otimes\left|\nu_{i,low}+\right\rangle_B+$$ $$\frac{1}{\sqrt 2}\left|\nu_{i,high}+\right\rangle_A\otimes\left|\nu_{i,high}+\right\rangle_B$$

Where all the photons have a $+$ denoting the same polarization and they are entangled to give the same frequency as each other, i.e. they both give $\nu_{i,low}$ or they both give $\nu_{i,high}$. And they are maximally entangled between these two possibilities. This is an example of entanglement.

This is exactly the kind of entanglement where the energies of the photons are correlated. But in order to have that correlation they do not start out with a definite energy. So what happens is that when the scattering happens you either get $\nu_{i,low}$ for both or they both give $\nu_{i,high}$. So after the scattering you get the whole subsystem is in $\left|\nu_{i,low}+\right\rangle_A\otimes\left|\nu_{f,low}+\right\rangle_B$ or the whole subsystem is in $\left|\nu_{i,high}+\right\rangle_A\otimes\left|\nu_{f,high}+\right\rangle_B$. Or, in other words, when the second photon acquires a definite energy, the other one gets a definite energy too, in a totally correlated fashion. And when they get definite energies, then they also become no longer entangled.

As an aside, it is sometimes possible to keep the entanglement in the sense that if no measurement has happened, like a scattering off a single electron and everything is isolated, you might get $$\frac{1}{\sqrt 2}\left|\nu_{i,low}+\right\rangle_A\otimes\left|\nu_{f,low}+\right\rangle_B\otimes\left|e_{low}\right\rangle+$$ $$\frac{1}{\sqrt 2}\left|\nu_{i,high}+\right\rangle_A\otimes\left|\nu_{f,high}+\right\rangle_B\otimes\left|e_{high}\right\rangle$$ where the $e$ is the state of the electron. But this is just a precursor to it breaking into the possibility $\left|\nu_{i,low}+\right\rangle_A\otimes\left|\nu_{f,low}+\right\rangle_B\otimes\left|e_{low}\right\rangle$ or the possibility $\left|\nu_{i,high}+\right\rangle_A\otimes\left|\nu_{f,high}+\right\rangle_B\otimes\left|e_{high}\right\rangle$.

So lets be clear what entanglement is. Rather than having a state where you have a definite energy you can have a state where you have a definite correlation between energies when later (if ever) they acquire a definite energy. And those possibilities of definite energy and definite correlation are just two possibilities, there are lots of possible entanglements and lots of possible degrees of entanglement.

If the photon is so detected, how/where was the energy lost?

This is no different than any superposition of two energies. You could ask in general where the energy comes from when it acquires a definite energy. And on the one hand some theories of quantum mechanics will explicitly answer that in detail and with the time dynamics of how it happens, most theories do not. And most theories will simply say it didn't have a definite energy beforehand. And so it didn't gain or lose energy, it acquires an energy and not that it changed from one definite energy to another definite energy.

If not, what should the detector detect and why?

Our setup was too vague. If they were entangled as I described then 50% of the time you see one of the possible initial energies and 50% of the time you see the other initial energy. But we could have made either energy be whatever we want. And we could have used any pair $(\alpha,\beta)$ with $|\alpha|^2+|\beta|^2=1$ rather than $(\frac{1}{\sqrt 2},\frac{1}{\sqrt 2})$ and then the probabilities would have been $|\alpha|^2$ and $|\beta|^2$ instead of 50-50 for the state $$\alpha\left|\nu_{i,low}+\right\rangle_A\otimes\left|\nu_{i,low}+\right\rangle_B+$$ $$\beta\left|\nu_{i,high}+\right\rangle_A\otimes\left|\nu_{i,high}+\right\rangle_B$$

But we do need the energies to be different to be entangled (because otherwise it isn't entangled) and we can't let $\alpha$ or $\beta$ equal zero (because otherwise it isn't entangled). But that still leaves lots of choices. Plus we can entangle the directions instead of the magnitudes. Or we can entangle the polarizations, and so forth.

• Timaeus: "One way to entangle photons is to have a state like $$\frac{1}{\sqrt 2}\left|\nu_{i,1}+\right\rangle_1\otimes\left|\nu_{i,1}+\right\rangle_2+\frac{1}{\sqrt 2}\left|\nu_{i,2}+\right\rangle_1\otimes\left|\nu_{i,2}+\right\rangle_2$$ [...]" -- Not to argue about style and taste, but your expression is even wrong by confusing indices which are distinct. Better write: $$\frac{1}{\sqrt 2}\left|\nu_{i,\text{low}} +\right\rangle_A\otimes\left|\nu_{i,\text{low}} +\right\rangle_B+\frac{1}{\sqrt 2}\left|\nu_{i,\text{high}} +\right\rangle_A\otimes\left|\nu_{i,\text{high}}+\right\rangle_B$$. – user12262 Oct 29 '15 at 6:41
• Timaeus: "Entangled just means not factorizable. [...] you have a definite correlation [...] there are lots of entanglements and degrees of entanglement [...] we do need the energies to be different to be entangled." -- These are fair, crisp descriptions; +1. I'd like your answer even more if you could address how this relates to calling only one pair "entangled" (cmp. OP: "Suppose we generate 2 entangled photons [...]"). p.s. If you decide to edit your answer there are also some typos to correct. – user12262 Oct 29 '15 at 6:43
• "then they also become longer entangled". I suppose you mean "no longer entangled". this should be corrected – anna v Oct 29 '15 at 15:43