# When light reflects off a mirror, does the wave function collapse?

This question is specific to the Copenhagen interpretation, which states that the wave function collapses on interaction. If we have a beam of light reflected off a mirror, whether you see this light as a quantum field or a stream of photons, either way the reflected light is not the same as the original light. The original photons have been absorbed in the mirror and new photons reemitted in a different direction. The wave function also is different accordingly. So on one hand, it would seem that the original wave function collapses at the mirror.

However, a mirror reflection does not affect the results of interference experiments, such as the double slit experiment. Mirrors are used in lasers producing highly coherent light. And finally, mirrors are used in various quantum entanglement setups without affecting the state of entanglement. If this understanding is correct, then it would seem that a mirror does not collapse the wavefunction.

What is the correct understanding and the physical intuition behind it? If the reflected photon is not the original entangled photon, but is reemitted by the mirror, how can it possibly still be entangled with the counterpart of the original photon absorbed by the mirror?

• "which states that the wave function collapses on interaction." Er ... not quite. The collapse is associated with measurement. Keep in mind that the $V(\mathbf{x})$ in Schrödinger's equation is an interaction. Now, even giving a precise statement of what makes a "measurement" is non trivial, but a starting place is that a measurement leaves a record. Coherent reflection form a mirror generally does not leave a record. Commented Dec 3, 2017 at 19:26
• @dmckee Thanks for your insight, David! Essentially what you are saying is that a mirror reflection does not collapse the wavefunction. Right, but the question was why. Because it doesn't leave a record. Well, this only helps a little. How come an interaction with a detector leaves a record, but an interaction with a mirror does not? And how does "leaving a record" figure into the math? I hope you can expand to a full answer. What fun is it answering simple stuff? Take a real challenge here :) Commented Dec 3, 2017 at 20:12
• I asked a basically identical question here (except the mirror is replaced with a Stern-Gerlach apparatus, but everything in the answers applies to both). Hope this helps! Commented Dec 14, 2017 at 10:56
• You might be interested in evanescent waves. While they are demonstrated with a reflection off of a prism rather than a mirror, they demonstrate that reflection is much more complex process than it may appear at first glance. That might help with your understanding of mirrors as well! Commented Dec 31, 2017 at 16:15
• @CortAmmon Thank you, I'll check them out. They seem to be just standing waves with a fancy name, but I have no doubt this is way more complex than it superficially seems. Commented Jan 1, 2018 at 8:40

As I understand it, the question boils down to, "Why doesn't a mirror collapse the wavefunction of a photon that reflects off the mirror?". The answer is that the photon does not change the state of the mirror. After the photon has been reflected, the mirror is unchanged. There is no way to prove that the photon struck the mirror without also detecting the photon's path downstream. The only state-changing event that occurs to the photon is its detection where it strikes a screen (or camera sensor, or an observer's eye, etc.).

It is not correct that the wavefunction collapses upon interaction. It is "less incorrect" to say that the wavefunction collapses upon detection. "Less incorrect" requires a bit of explanation.

Detection is an interaction that results in an observer "knowing" that the wavefunction has collapsed. Yes, it's a vaguely circular definition. In the many worlds view of quantum mechanics, an observer "detects" the state of a particle and in doing so splits his world into as many different independent alternative worlds as there are possible values for the state of the particle.

In the case of the two-slit interferometer, the observer detects the position ("state") of each photon that strikes a screen. That "detection" (according to the MW view) is not really an observation of what the photon's state (i.e., location) is, but rather a projection of the observer's world onto one of the possible values of the photon's state. In a sense, the observer's world splits into all the possible worlds that would result immediately after the detection event, depending on the possible different state values that the photon might have.

The gedankenexperiment of Schroedinger's cat can be generalized to help explain this. Suppose we put the observer inside a box that is totally isolated from the rest of the universe, and the observer in the box detects a photon on a screen. We cannot know where on that screen the observer detected the photon, until we open the box and look at the observer's records. Moreover, according to Bell's theorem, from our perspective, the observer himself is a quantum object -- so the location of the photon's impact on the screen does not even have an actual value from our perspective until we open the box. The value is not hidden; it is indeterminate. The observer inside the box is sure to think he knows where the photon hit his screen, but from our perspective the observer is in a superposition of states until we "detect" what his state is: until we "open the box". As far as we are concerned, the observer co-exists in all possible states corresponding to different places the photon strikes his screen; and in each state he is sure he knows where the photon struck his screen -- but for each of his different states the photon landed a different place.

All that is background for saying that as long as there is no possible way to know that the photon reflected off a given mirror in an interferometer, the photon wavefunction takes all available paths including those in which it does not reflect of the mirror -- and it will thus form the interference pattern we observe.

If we were to make the mirror so tiny and thin that its recoil could in principle be detected , there would be no interference pattern. I know some folks will ponder what might happen in the gray area between using an extremely tiny & thin mirror and using a normal several-to-many grams mirror. I don't know if anyone has done the experiment, but I'll bet what happens is that the interference pattern's contrast is reduced as the mirror's recoil approaches detectability. It would be an experiment worth doing.

• A meaningful answer, very insightful and thought provoking. Thanks so much! Commented Aug 19, 2018 at 6:15
• @S.Mcgrew you are saying that "The answer is that the photon does not change the state of the mirror. After the photon has been reflected, the mirror is unchanged." What happens, is that some of the photons in the EM wave exert pressure on the mirror. You are right, that the elastically scattered photons that build up the mirror image, will not affect (in case of a perfect mirror) the mirror. But since it is not a perfect mirror, there will be some photons inelastically scattered, and some really (not virtually) absorbed. These photons should change the wavefunction of the mirror. Commented Aug 19, 2018 at 16:35
• Tht's right. The possibility that a photon may change the state of the mirror results in a reduction of the contrast in the interference pattern observed from recording many photon impacts on the screen. Commented Aug 19, 2018 at 17:01

This question is specific to the Copenhagen interpretation, which states that the wave function collapses on interaction.

No more photon after absorption, no more wave function. Obvious, even without Copenhagen interpretation.

If we have a beam of light reflected off a mirror, ... the original photons have been absorbed in the mirror and new photons reemitted in a different direction. The wave function also is different accordingly. So on one hand, it would seem that the original wave function collapses at the mirror.

Right, the new wave function is different with the frequency of the photon (slightly red shifted) and with the direction of propagation.

However, a mirror reflection does not affect the results of interference experiments, such as the double slit experiment.

Could not agree. Shifting one of the mirrors the intensity distribution on the observers screen changes. And for a Gaussian beam and well adjusted mirrors the spot of the beam could be dislocated to the reflecting part of the half transparent mirror (marked in the image below with a double arrow) or to the transparent part of the mirror. The intensity distribution from the incoming beam stays unchanged in this case.

See this video, the important point starts at 5:00).

That is very different from the intensity distribution behind a double slit where the intensity doubles at the crests and is zero between. Here for some mirror positions the intensity at the crest is only have the possible intensity and the other half energy is distributed through the half transparent mirror. And for a special setup the light goes 100% back through the half transparent mirror and for another setup it gets deflected 100% by this mirror.

Mirrors are used in lasers producing highly coherent light. And finally, mirrors are used in various quantum entanglement setups without affecting the state of entanglement. If this understanding is correct, then it would seem that a mirror does not collapse the wavefunction. What is the correct understanding and the physical intuition behind it?

The counterargument could be that the mirrors rearrange the photons state all in the same manner.

If the reflected photon is not the original entangled photon, but is reemitted by the mirror, how can it possibly still be entangled with the counterpart of the original photon absorbed by the mirror?

After realizing that the wave function of the photons of an interferometer setup are not additive but only re-distributed in different ways dependent from the mirror positions there should be another interpretation. For example it has to be investigated how the photons produce phonons in the mirrors and how this retarded the transmissive and reflective part of the mirrors.

Reflection doesn't collapse the wave function because it occurs as a logical consequence of the photon waveform undergoing refraction at a boundary under the Fresnel equations (which determine the proportion of the wave refracted and reflected at any given angle). It is therefore a property of the photon waveform and has nothing to do with wave function collapse. In other words, reflection does not involve photons being absorbed and emitted but each photon as a waveform being subject to refraction and reflection under constraints internal to its nature as a waveform as expressed in the Fresnel equations.

• Your answer seems to be a mix of quantum and classical concepts that don't normally work together. For example, the photon does not have a classical waveform. Classical and quantum waves are not the same. A classical wave consists of alternating real electric and magnetic fields. A quantum wave is a single complex wave of probability density of detecting a particle. Commented Jan 1, 2018 at 8:37
• The EM wave is the quantum waveform of the photon - it doesn't have any other. And the wave amplitude gives the probability density of the photon being detected i.e. interacting as a particle e.g absorption. Commented Jan 1, 2018 at 20:44
• The relation there is a bit more complex: physics.stackexchange.com/questions/93430/… Commented Jan 1, 2018 at 21:15