Why doesn't the momentum exchange (or lack thereof) between photon and beam-splitter destroy the interference?

Question

I have a question (my very first here) related to 50/50 beam splitters as used in the Mach-Zehnder interferometers (see for example the Wikipedia page).

Let's concentrate on the input beam splitter: A continuous light beam (the input) is split 50/50, one in the 90° direction and one in the forward direction of the input beam. Momentum/energy conservation teaches us that deflected part of the beam must exert a (tiny) force on the beam splitter. The forward part of the beam will not.

In the single photon (quantum) case, the beam-splitter should get a slight "kick" if the photon is deflected, while the beam splitter is left in its original state if the photon goes straight through.

If I got this right so far, we can now increase the energy (and momentum) of the photon and decrease the mass of the beam splitter (now coupled to a sensitive piezo transducer or something) so that a photon "kick" can be recorded if deflected. Now "which way" information is available and interference should be destroyed as I understand this.

Would interference be restored if the transducer is disconnected? If yes, how do the photon "know" whether the transducer is connected or not? Maybe this is related to how much the photon/beam splitter system can become entangled to the environment?

If the answer is "No", how weak must the photon kick be in order for the interferometer to work? We already know that it works for ordinary light (very tiny photon kicks).

In other words: Why doesn't the momentum exchange (or lack thereof) between the photon and the beam-splitter (and the trace this leaves in the environment) destroy the interference?

Alternatively, couls someone just point me to relevant literature which discusses the transfer of momentum between between photons and optical elements?

Answer 1

It all boils down to how much information there is to, in principle, distinguish which way the photon went from the final state of the beam splitter, as encoded in the overlap between its two possible final states. The interference is destroyed because the photon gets entangled with the beam splitter, and the amount of entanglement depends on this overlap.

Say, then that if the photon goes straight through the beam splitter, to state $|{\to}\rangle$, the beam splitter stays put, at state $|0\rangle$, whereas if the photon gets deflected into state $|{\downarrow}\rangle$, the beam splitter gets some upwards momentum, $|{\Uparrow}\rangle$. If the result is a superposition, then, the total state of the system is entangled: $$|\Psi\rangle=|\to\rangle|0\rangle + |{\downarrow}\rangle|{\Uparrow}\rangle .$$

Regardless of what you do to the beam splitter - i.e. measure its state or just forget about it - in the absence of a measurement that introduces further interactions, the information you have available to produce an interference pattern on the photon side is given by the reduced density matrix obtained by taking the partial trace over the beam splitter.

Calculating this object is fairly simple. In the $\{|{\to}\rangle, |{\downarrow}\rangle\}$ basis, it is given by $$ \rm{Tr}_{\rm{BS}}(|\Psi\rangle\langle\Psi|) = \begin{pmatrix} 1 & \langle0|{\Uparrow}⟩ \\ \langle{\Uparrow}|0⟩&1 \end{pmatrix}. $$ If the beam splitter states are completely distinguishable, then they are orthogonal and what you get on the photon side is a completely mixed state, $|{\to}⟩⟨{\to}|+|{\downarrow}⟩⟨{\downarrow}|$, which is completely classical, and from which no interference can be extracted. Note that this happens regardless of whether you actually measure the beam splitter's momentum or not.

If there is no effect on the beam splitter, on the other hand, the states are the same, and the photon's density matrix corresponds to a pure state, $\left(|{\to}⟩+|{\downarrow}⟩\right)\left(⟨{\to}|+⟨{\downarrow}|\right)$. Then you will see complete interference, but you will have no "which way" information available, even in principle.

In any physical realization, of course, you're somewhere in the middle. Most realizations have very similar states for the beam splitters, which means that $\langle{\Uparrow}|0\rangle$ is very close to 1, and you get good interference, but as the states become more distinguishable, the contrast in the interference fringes is reduced.

I understand this can feel pretty thin. After all, how are we to know that we've eliminated all possible places where "which way" information may in principle be available? This is in fact how it goes down in the lab, and that's the reason observing things like Mandel dips is very, very touchy: if you want two photons to interfere, you need to make sure that they truly are indistinguishable - in spatial profile, displacement, spectrum, and timing - for otherwise there will be (possibly undetected) entanglement with some other mode, and that will reduce or destroy your interference contrast.

Answer 2

This is actually a very interesting question, along the lines of the Bohr-Einstein debates about quantum mechanics. Answers to these take a little time to work out, though. Another way to interpret the loss of coherence is that if the splitter is light enough (no pun intended), the velocity change in the splitter from the momentum kick, effectively measuring the which-way info, will give enough phase shift to the outgoing photon to destroying the interference.

As for the transducer, when connected, the compression of the transducer is translated into a voltage pulse across some load, damping out the transducer motion (transferring the information from the transducer to your measurement device). If you disconnect the transducer, the momentum kick will first compress the transducer, and now undamped by the measurement circuit, the transducer will oscillate until the motion is damped by friction. This results in a slight temperature rise that is effectively a measurement. The transducer still measures the which-way information.

BTW: if the splitter has very little friction, the splitter's motion will be entangled with the photon which-way information. I believe (having trouble finding the references at the moment) experiments with very small, high Q resonators in optical systems have displayed such effects. Optical cooling of a mechanical resonator [Nature 475, 359–363 (21 July 2011)] might be a place to start looking if you are interested in this.

[Edit: you can use the transducer in reverse, if you want. Devices called acousto-optic modulators (AOMs) use a transducer to induce pressure waves in a crystal. Light passing through this crystal can either absorb or emit photons into the acoustic wave. Energy and momentum conservation rule, so if a photon absorbs a phonon, then photon frequency increases and the photon gets a lateral kick. If the photon emits a phonon (stimulated emission via the existing phonons), the photon frequency drops and the photon gets a negative lateral kick. If you look at a laser beam going through an AOM, you will see two or more beams (depends on the alignment of the AOM) coming out at different angles corresponding to -1 phonon, 0 phonon, and +1 phonon changes (as well as weaker -2 and +2 phonon lines). The classical interpretation is that you have a moving index-of-refraction grating caused by the pressure wave that diffracts the +1 and -1 orders. Because the grating is moving with velocity of sound in the crystal, the +1 and -1 orders have slight energy shifts.]

Answer 3

A very interesting question. It looks like you are right, with a light enough beam splitter and a photon with enough momentum the interference effects will vanish as the beam splitter "measures" the photons.

Source: Quantum Processes Systems, and Information by B. Schumacher, et al. Section 10.4.