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Imagine a source of photons at the center of a spherical shell of detectors at radius $R$.

Assume the photons are emitted one at a time.

Now if photons are particles that are highly likely to travel on straight paths at velocity $c$ then one would expect the following behavior:

At time $t=0$ as the photon is emitted in a particular direction the source recoils in the opposite direction. Later at time $t=R/c$ the photon is absorbed by one of the detectors which recoils as it absorbs the photon.

But quantum mechanics says that the photon is actually emitted at time $t=0$ as a spherical wave that expands out to the detectors at the velocity $c$.

While the spherical wave is in transit from the source to the detectors the source cannot recoil in any particular direction as no direction has been picked out yet by the photon detection.

So does the source only recoil when the photon is absorbed at time $t=R/c$ or is its recoil somehow backdated to $t=0$ to be consistent with the particle picture?

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  • $\begingroup$ The state is a superposition $|\psi\rangle = \int_\vec{p} |\text{photon momentum }\vec{p},\text{ source momentum }-\vec{p}\rangle$, i.e. the photon is entangled with the recoiling source so that momentum conservation is satisfied at all times. EDIT: just saw John Rennie's answer go up. $\endgroup$
    – Michael
    Commented Oct 22, 2013 at 15:05

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I'm assuming that you've set up the experiment so emission is equally probable in all directions. If so then what you have is a variant of the EPR paradox. After emission of the photon the source and the photon form an entangled system. When you measure the momentum of the photon this collapses the system (other interpretations are available) and simultaneously determines the momentum of the source and vice versa. Until you make a measurement on the system the photon and source do not have any well defined momenta or indeed position.

The apparent superluminal communication between the photon and source presents no problems because it is impossible to use it to transmit any information.

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  • $\begingroup$ So when I measure the momentum of the photon at $t=R/c$ the momentum of the source is determined simultaneously. Does this retrospectively determine the momentum of the source and photon at $t=0$ as well? $\endgroup$ Commented Oct 22, 2013 at 16:01
  • $\begingroup$ yes, but consider carefully the semantics involved. you don't actually change anything retrospectively that you didn't set up to measure.. I would rather phrase it like this: no matter how you chose to measure either the center recoil or which detector fires, both measurements will always be consistent. this is the core of what QM actually predicts. $\endgroup$
    – BjornW
    Commented Oct 22, 2013 at 16:38
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    $\begingroup$ @JohnEastmond: the answer depends on what interpretation of QM you favour. If you are a Copenhagenist the momentum of neither the source nor the photon was determined at $t = 0$, or indeed at any time prior to your measurement. If you are a decoherence + MWI fan then your measurement selected the world in which the momentum has the result you measured and in that world the momentum always had the result you measured (so your measurement didn't change anything). $\endgroup$ Commented Oct 22, 2013 at 16:42
  • $\begingroup$ Has recoil ever been measured when a photon was emitted? Or is it just assumed? $\endgroup$ Commented May 21, 2023 at 7:25
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The question at the heart is whether the source will recoil even if none of the photons are detected. Copenhagenists argue that the question is not meaningful until one measures the momentum of the atom. But this does not explain the phenomenon of lateral spreading of a well collimated beam of atoms induced by spontaneous emission. To determine the spreading effect it is not necessary to know the momentum of any of the atoms, you can just stick a small detector somewhere on-axis downstream to see if the flux of the atoms is reduced by the spreading, i.e. you only need to measure the position of the atoms.

In quantum optics the photon concept is never enlisted until the moment of detection, only then will the raising and lowering operators of 2nd quantization' be involved. So for all intents and purposes one is now tempted to describe the emission of undetected photons by means of classical outgoing spherical waves. It is well known that such waves lead to no recoil of the source, however, as there is no preferred direction of recoil. The fact that the aforementioned beam of atoms spreads out would seem to contradict this model. In other words, even if one does not observe any of the photons, each is still emitted into some specific (though random) direction, i.e. the emitted photons are linear photons and not spherical ones, and a small detector tucked along some particular direction would not see at time t=R/c manyYoung slit' style spherical waves from all the atoms that emitted at t=0 interfering each other. Rather, it will only catch one packet of plane waves, the photon that happened to be emitted by some atom in the source along the right direction.

The emission of linear photons does not in fact contradict Copenhagen, because without detecting any of the photons all one knows is that the beam of atoms has spread, and there is no way of telling which atom recoiled in what direction.

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