# How does the Hong-Ou-Mandel (HOM) effect conserve photon momentum?

HOM is a two-photon interference effect where temporally overlapped identical photons coming perpendicular to a beam splitter must leave it in the same direction.

How is momentum conserved in this process? Initially, the total momentum of the photons is only on the x-axis, but after the beam splitter, it has a y-axis component.

Does some momentum pass to the beam splitter itself? • Works the same as with a mirror. Some of the momentum is transferred to the mirror or beamsplitter. Sep 7 at 11:54
• physics.stackexchange.com/questions/119743/… Sep 7 at 15:35

The only easy way to see generic momentum conservation in quantum experiments is to not look at just one possible outcome, but rather to look at expectation values (weighted averages, over all possibilities). This viewpoint is guaranteed to conseve momentum because of Ehrenfest's Theorem which says that any classical mechanics equation is also true if you substitute every classical quantity for its quantum expectation value. Therefore $$\frac{d}{dt}

=-\left< \frac{d}{dx} V(x) \right>.$$

And without a gradient in the potential, the expectation value of the momentum will be conserved.

That is evidently true in the HOM experiment; since each of the two possible outcomes have a 50% probability, the expectation value is easy to calculate, and momentum will be conserved on average.

Now, what about in any particular run of the experiment? As you point out, it looks like momentum is not conserved. This is true in just about any quantum experiment with either photons or massive particles; even though momentum is conserved on average, any one particular outcome doesn't seem to obey this rule. So is momentum conservation fundamental or just emergent on larger scales?

It's impossible to definitively answer this last question without some particular resolution of the measurement problem in quantum foundations. On one hand you could argue that the "momentum" of an unmeasured quantum system is the expectation value of the momentum operator, with no more fine-grained way to define it, so it's already an emergent quantity. On the other hand, you could argue that at the deepest level a measurement is just an interaction, entangling the momentum correlations between the measured object and the measurement device, and always allowing some fundamental sort of momentum conservation. But both of these points will be debatable until we have an agreed-upon framework for the distinction, if any, between an interaction and a measurement -- as well as the even more important question of what is really going on when we're not looking.

• Thank you this was the answer I was searching for when I put a bounty on this question. I could see that on average momentum was conserved, but I couldn't understand the consequences of momentum not being conserved in one single measurement run. Sep 17 at 15:16
• This isn't correct. Momentum must be conserved in detailed in this case. The momentum is conveyed to the beamsplitter. This is similiar to a mirror reflecting a photon.
– JQK
Sep 17 at 15:59
• You mean, the classical beamsplitter, or the quantum state of the beamsplitter? :-). But yes, agreed, if you take a literal collapse viewpoint, then after the collapse, the momentum winds up in the classical beamsplitter. It has to be conserved back at the classical level. Sep 17 at 16:28
• Thanks for the answer. Could you expand it to include equations showing how the momentum is conserved in the non-collapsed system? Perhaps using a joint state for the two photons with the momentum operator eigenstates? Sep 25 at 15:35