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Consider a photon with momentum p approaching a half-reflective mirror. After encountering the mirror it is in a superposition of momentum +p and -p and the mirror (we presume) is in a superposition of momentum 0p and +2p. A second photon would do the same, and the mirror would be in a superposition of 0, +2p and +4p (with +2p most likely). After n photons +np would be most likely but there would be very much a spread. Suppose we can set this up so none of the photons are later absorbed and the quantum superpositions are preserved. What then do we see when we look at the mirror, and what momentum would we measure?

It seems to me that we would see the mirror moving (and measure it as moving) at +np – which would be the radiation pressure from the reflection. In this scenario momentum would be conserved because the two ‘halves’ of each photon waveform would have different momenta (+p and –p) and so the ‘overall’ waveform would have 0p and the mirror would have received +p from the reflection.

But if this is correct, it leaves the question of what happens to the photon’s ‘overall’ momentum when it is absorbed. For if the half-reflected photon has zero overall momentum then upon absorption it will deliver no recoil to the absorbing atom. But is this correct, has it been tested? Do half-reflected photons deliver zero recoil to atoms when absorbed (this assumes that the half-reflected photon has not already lost its quantum superposition in some other way).

The alternative explanation of course is that the mirror itself enters a superposition until the half-reflected photon is absorbed, so that we don’t need to contend with the 'overall momentum' of the half-reflected photon at all, but simply find both superpositions (photon and mirror) resolved when the photon is absorbed. But this seems to me to be contradicted by the fact of radiation pressure from reflection, and thus the fact that the mirror (say) would be accelerated by reflection by a well-defined amount which wouldn’t need to wait for the reflected photons to be absorbed to observe what that is.

Really I want to know if this has been tested and what the outcome was?

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    $\begingroup$ Beam splitters provide two paths, and each photon randomly follows a single path. Thus beam splitters do NOT divide photons! You should reconstruct your question with this in mind. $\endgroup$ – Peter Diehr Dec 31 '17 at 13:30
  • $\begingroup$ @PeterDiehr. The OP obviously refers to quantum mechanics where a photon would be in a superposition of two paths until it is being detected. This seems like a duplicate of physics.stackexchange.com/q/368333/1 or physics.stackexchange.com/q/282410/109928 $\endgroup$ – Stéphane Rollandin Dec 31 '17 at 14:05
  • $\begingroup$ I've edited the question to make clear I am referring to the waveform of a single photon under the influence of a beam splitter. The question isn't a duplicate of those linked to as it refers specifically to photon recoil momentum and asks whether anyone is aware of experiments which investigate it using a beam splitter. $\endgroup$ – willjones1982 Dec 31 '17 at 16:09
  • $\begingroup$ Experimentally it is difficult to get both beams interact with an electron without refocusing them thereby changing the photon momentum. In theory if a photon in superposition of two momentum states gets absorbed by an electron at rest, the electron recoils into a superposition of these two momenta with an average momentum of zero in your setup. $\endgroup$ – Zhuoran He Dec 31 '17 at 16:13
  • $\begingroup$ In addition to Peter Diehr's comment that photons can not be divided (e.g. half reflected) nor are localized traveling entities, I can recommend this concise write down: quantum-field-theory.net/photon. To understand how the e.m. field interacts with matter needs a thorough study of QFT including Fermi's golden rule. $\endgroup$ – Jan Bos Jan 6 '18 at 3:17
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I'm adding another answer since you've substantially changed the question. I feel we're getting out of the realm of appropriate stackexchange etiquette here with so much discussion and changing thing around. Maybe a mod can suggest how best to proceed. Anyways, here's my answer.

I agree whole-heartedly with your first paragraph

I disagree whole-heartedly with your second paragraph. your first statement is:

It seems to me that we would see the mirror moving (and measure it as moving) at +np – which would be the radiation pressure from the reflection.

This is incorrect. If we measured the momentum of the mirror we could measure any momentum $0, +2p, +4p, \ldots, +np, \ldots, +2np$. As you've pointed out, it is mostly likely that we measure $+np$. However, since the mirror is in a superposition of many momentum states it is not possible for us to predict before-hand what momentum we will measure if we measure it. We can only ascribe probabilities to each possible momentum, where the probability is given by the weighting of that term in the description of the state of the mirror.

Consider a mirror which is first hit by one photon, $\gamma_1$ and then a second photon, $\gamma_2$. The initial state of this system is

$$\lvert 0 \rangle_M \lvert +p\rangle_{\gamma_1} \lvert+p\rangle_{\gamma_2}$$

Where the subscripts refer to the state of the mirror, $M$, and the two photons, $\gamma_1$ and $\gamma_2$.

After the first photon hits the mirror the state is

$$\frac{1}{\sqrt{2}} \big( \lvert 0 \rangle_M \lvert +p\rangle_{\gamma_1} \lvert+p\rangle_{\gamma_2} +\lvert +2p \rangle_M \lvert -p \rangle_{\gamma_1} \lvert +p \rangle_{\gamma_2} \big) $$

After the second photon hits the mirror the quantum state is

$$ \frac{1}{\sqrt{2}}\Big(\frac{1}{\sqrt{2}}\big(\lvert 0 \rangle_M \lvert +p \rangle_{\gamma_1} \lvert +p \rangle_{\gamma_2} + \lvert +2p \rangle_M \lvert +p \rangle_{\gamma_1} \lvert -p \rangle_{\gamma_2}\big) + \frac{1}{\sqrt{2}}\big(\lvert +2p \rangle_M \lvert -p \rangle_{\gamma_1} \lvert +p \rangle_{\gamma_2} + \lvert +4p \rangle_M \lvert -p \rangle_{\gamma_1} \lvert -p \rangle_{\gamma_2}\big)\Big) $$

You see that upon each reflection each term splits into two terms. One where the mirror had no change in momentum and the photons momentum was not changed and one where the mirror got a kick of $+2p$ and the photon was reflected.

Say we now perform a measurement on the momentum of the mirror. The possible outcomes are $0$, $+2p$, or $+4p$.

If we measure $0$ then we know we have "collapsed" the quantum state into the first term or first "branch". This means that we know both $\gamma_1$ and $\gamma_2$ would reveal momentum $+p$ upon measurement of their momenta. The state has collapsed to $\lvert 0 \rangle_M \lvert +p \rangle_{\gamma_1} \lvert +p \rangle_{\gamma_2}$. Note that momentum is conserved.

If we measure the mirror to have momentum $+4p$ we know we are in the last branch and thus both $\gamma_1$ and $\gamma_2$ would reveal momentum $-p$ upon measurement of their momenta. The state has collapsed to $\lvert +4p \rangle_M \lvert -p \rangle_{\gamma_1} \lvert -p \rangle_{\gamma_2}$. Note that momentum is conserved.

Now, if we measure the momentum of the mirror to be $+2p$, then intuitively we know if we were to measure the momentum of the photons one of them would have been transmitted and one of them would have been reflected, but, just by measuring the momentum of the mirror we cannot determine which. This means that the state of the system after the measurement would be

$$ \frac{1}{\sqrt{2}}\big(\lvert +2p \rangle_M \lvert +p \rangle_{\gamma_1} \lvert -p \rangle_{\gamma_2} + \lvert +2p \rangle_M \lvert -p \rangle_{\gamma_1} \lvert +p \rangle_{\gamma_2} \big) $$

That is, even after measurement the system is still in a superposition. This is because the measurement didn't give us FULL information about the quantum state. You can see that the momentum is definite but the photon is still in a superposition state.

Perhaps this explication helps you already?

Anyways, back to your question and the second paragraph. It is not clear what you mean when you talk about 'halves' of the photon or the 'overall' photon momentum. I think what is confusing you is whatever you mean by 'overall' momentum. I'm pretty what you are referring to as 'overall' momentum is not a thing. Instead, you should think about the photons momentum as I have illustrated it above. The total state of the system a superposition of different terms in which different things happened. In each of these terms the photon has a well-defined momentum. Whenever an interaction happens each term can split into multiple other terms. These different branches split and split until a measurement is made. When a measurement is made the state "collapses" into the subspace of branches which are consistent with that measurement.

The language I am using here is borrowed from the many-worlds interpretation of quantum mechanics, but you need not adopt that interpretation for this simple description of superposition/entanglement to make sense.

Let's extend this example a little more. Imagine we DON'T measure the momentum of the mirror or the first photon, but instead measure the momentum of photon 2, $\gamma_2$. Imagine we measure $+p$. Then the quantum state collapses to

$$ \frac{1}{\sqrt{2}}\big(\lvert 0 \rangle_M \lvert +p \rangle_{\gamma_1} \lvert +p \rangle_{\gamma_2} + \lvert +2p \rangle_M \lvert -p \rangle_{\gamma_1} \lvert +p \rangle_{\gamma_2} \big) $$

We know the momentum of the second photon, but the first photon and the mirror remain in a superposition state and the whole system remains in an entangled state.

You keep asking if this has been tested. I'm not sure exactly what experiment you are imagining, but I can tell you that if you shine light on a beamsplitter and then use one output of the beamsplitter to illuminate atoms the light will certainly impart the expected momentum onto the atoms. I have performed this experiment.


In the comments you ask about an experiment in which a single photon hits a beamsplitter and confirmation that beamsplitter is only seen to be in the $\lvert 0 \rangle_M$ or $\lvert +2p \rangle_M$ and never $\lvert +p \rangle_M$. I can't think off of the top of my head of an experiment that does PRECISELY this. The first reason is that it is very difficult to measure the recoil of a massive mirror due to a single photon. I think many would say it is impossible. However, I have worked in the field of optomechanics where people regularly see single-photon, single-phonon interactions between an optical field (photons) and some mechanical object such as a mirror. Perhaps you can look up experiments on optomechanics to see if there is a specific experiment which satisfies your question.

What I can say is that the concepts of superposition, entanglement, and radiation pressure have been thoroughly studied and the theory underlying countless experiments relies on these concepts. The measurement of the mirror in state $\lvert +p \rangle_M$ would contradict all of these experimental results so I can say with certainty that if this experiment was able to be performed with the required precision you would not measure the mirror to be in state $\lvert +p \rangle_M$.

What I can also say is that the the single photon interaction is very similar to an EPR experiment for example. Notice that that photon and mirror form an EPR entangled state after the interaction. Many EPR pair experiments have been performed to test Bell's inequality for example, and these are all consistent with the usual results of quantum mechanics. These EPR experiments also demonstrate a kind of conservation law. if $\lvert \uparrow \rangle$ and $\lvert \downarrow \rangle$ represent angular momentum states then the EPR state

$$ \frac{1}{\sqrt{2}}\big(\lvert \uparrow \rangle_1 \lvert \downarrow \rangle_2 + \lvert \downarrow \rangle_1 \lvert \uparrow \rangle_2 \big) $$

exhibits conservation of momentum in each "branch" just like the photons and beamsplitter. That is, if one of the particles is measured in state $\lvert \uparrow \rangle$ then we KNOW we could not measure the other particle to be in state $\lvert \rightarrow \rangle$, for example, because that would violate conservation of momentum. That is, unless the particles interact with something else which can carry away momentum.

Anyways, the point is these are basic results in superposition/entanglement upon which a lot of quantum theory and quantum experiments rely so I am certain of these results. There is probably a specific experiment out there in the field of single photon single phonon optomechanics or atom interferometry but I can't point to it now. EPR/Bell's inequality measurements may be of interest to you as well.

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  • $\begingroup$ Ok so the mirror can have definite momentum while the photons remain in superposition through the photons becoming entangled via their interaction with the mirror. In the two photon example this means that if the mirror momentum is observed as +2p and one photon is later observed as +p then the other must turn out to be -p. Are you able to confirm that in the one photon example the mirror momentum is only ever measured as either 0p or +2p (and never p) or is this measurement too tricky to obtain accurately? $\endgroup$ – willjones1982 Jan 6 '18 at 18:09
  • $\begingroup$ Nice, I agree with everything you said here. The part about the mirror entangling the photons is very key. In quantum optics we sometimes encounter interactions which cause entanglement of this kind and we sometimes call these interactions "beamsplitter interactions" even if they aren't literally implemented by a beamsplitter. $\endgroup$ – jgerber Jan 6 '18 at 18:27
  • $\begingroup$ I'm adding a section in the answer regarding experimental realizations now. $\endgroup$ – jgerber Jan 6 '18 at 18:30
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This question is about conservation of momentum in the presence of a superposition. Imagine that a photon is travelling to the right with photon momentum $+\hbar k$. It then hits a 50% reflective, 50% transmissive mirror (this is the beamsplitter you are imagining). The photon is now in a superposition of travelling to the right with momentum $+\hbar k$ as before and travelling to the left with momentum $-\hbar k$.

Your misunderstanding is to think that this means that the photons total momentum is $0$. But this is incorrect. The correct way to think about this is that the photon is in a superposition of both having momentum $+\hbar k$ and having momentum $-\hbar k$.

The next part of your question is about the subsequent absorption of this photon after the beam-splitting. Well here is what you will find. If you find that photon is absorbed by an absorber (atom) to the RIGHT of the beamsplitter then that means the photon was observed with momentum $+\hbar k$ so this atoms will absorb momentum $+\hbar k$ and start travelling to the right. In this case no photon will be observed travelling to the left.

If instead you find the photon to be absorbed by an atom to the LEFT of the beamsplitter then that photo is observed to have momentum $-\hbar k$ so that atom will absorb momentum $-\hbar k$ and start travelling to the left.

The obvious question at this point is the case when the photon has momentum $-\hbar k$ because it seems like conservation of momentum has been broken. Well, the answer is that when the photon becomes a superposition of travelling left and right the beamsplitter ALSO becomes in a superposition of having recoiled and not having recoiled. That is:

IF the photon is found to be travelling right THEN the mirror will be found to not be moving, with momentum $0$.

IF the photon is travelling left THEN the mirror will be found to have momentum $+2 \hbar k$, to compensate the momentum of the photon which has reversed direction.

Another way to say this is that the photon and mirror become entangled after this experiment.

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  • $\begingroup$ The problem with this answer is it implies all macro bodies are put in a quantum state of superposition following all reflection until some 'detection' occurs to resolve the indeterminacy one way or the other. But radiation pressure can occur as a result of reflection and that causes a macro change in momentum regardless of whether the reflected particles have yet been absorbed or detected. So the photon waveform must be capable of conveying a particular (not superposition) of momentum on bodies as it is reflected. So it won't do to argue the mirror, a macro body, is placed in a quantum state. $\endgroup$ – willjones1982 Jan 1 '18 at 21:04
  • $\begingroup$ In this comment you are considering a different situation. In the OP and my response we consider an optic which has 50% transmission, 50% reflection. When the photon hits this optic it goes into a 50-50 superposition of going left and right. The (macroscopic) optic also goes into a 50-50 superposition of moving and not moving. In this comment you consider an optic which is 0% transmissive, 100% reflective. As a result, when the photon hits it the photon is in a 100-0 superposition of reflecting/transmitting, similarly the mirror is in a 100-0 superposition of recoiling/not recoiling. $\endgroup$ – jgerber Jan 2 '18 at 19:38
  • $\begingroup$ In other words, in the case of a 100% reflective optic nothing ever enters a superposition. You claim that "The problem with this answer is it implies all macro bodies are put in a quantum state of superposition following all reflection until some 'detection' occurs to resolve the indeterminacy one way or the other.". This statement is incorrect. My answer does NOT imply that "all macro bodies are put in a quantum state of superposition". The optic is only put in a superposition if the photon which hits it goes into a superposition. $\endgroup$ – jgerber Jan 2 '18 at 19:42
  • $\begingroup$ You say 'in the case of a 100% reflective optic nothing ever enters a superposition'. But that is obviously untrue: the photon waveform is itself a superposition, with the probability of 'detecting' the photon given by the amplitude of the wave - that's what the double slits experiment shows. The half-reflective mirror is just a special case which would be particularly easy to test (has it been tested?). The point is that photons generally exist in superposition as they travel as waves and diffract etc. and yet they still cause radiation pressure i.e. convey definite momentum. $\endgroup$ – willjones1982 Jan 3 '18 at 21:36
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    $\begingroup$ It seems you are misunderstanding what a photon is and what is meant by the photon waveform. A photon is an excitation of a particular spatial mode of the electromagnetic field. For example that spatial mode could be a plane wave travelling to the right. The "waveform" of this photon is the plane wave. So it is true that the photon is spatially delocalized. However, the momentum of the photon is well defined and travelling to the right. It is not in a momentum superposition. If this photon hits a 50-50 beam splitter it now splits into a superposition of having positive and negative momentum. $\endgroup$ – jgerber Jan 4 '18 at 4:58

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