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We interpret the electron's wave function as a probabilistic wave function. During a measurement, it has the probability to collapse to any of the eigenstates of the measurement operator based on the square of the amplitudes of the wave function expressed in terms of the eigenstates.

On the other hand, photon is the result of oscillation of electromagnetic field. We also know that photon exhibits quantum behavior like electron, such as photon can also undergo quantum entanglement.

Questions:

  1. Does photon undergoes collapse of wave function during a measurement?
  2. Does it follow the interpretation that the probability for the photon to collapse to any eigenstate equals to the square of the amplitudes of the electromagnetic wave?
  3. What happen to the electromagnetic wave when we perform measurement on the photon? Does the electromagnetic wave undergoes drastic change like the collapse of electron wave function?
  4. In the double slit experiment for photon, are we able to measure which slit the photon passes through? And if we make the measurement, does the interference pattern changes?
Qmechanic
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JNL
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2 Answers2

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Photons as necessarily relativistic particles do not have wavefunctions in the usual sense because there are no good relativistic position operators, so the question as written collapses already at the first step. See this question and this question for more discussion of photon wavefunctions.

Classical electromagnetic waves correspond not to states of definite photon number, but to coherent states with indefinite photon number. The electric and magnetic fields do not have definite values in a state of definite photon number, so in particular not for the state of a single photon, see also this question and its answers.

And in any case, in order to ask what happens to the EM fields of a photon after a position measurement, you first would have to explain how you're even measuring the position of a photon without destroying it - the usual methods are to put some kind of screen into its path, and then the photon is absorbed by the screen at some position, which is a method of measurement that destroys the photon - you don't have a photon with "collapsed wavefunction" after that, you just don't have a photon.

The situation is not much better for any other observables - non-destructive photon measurements are hard, and most non-destructive ways of observing photons are merely counting them instead of measuring any specific observable.

While many non-specialist texts tend to present the idea of a photon as something simple and uncontroversial, the opposite is true - see e.g. Ján Lalinský's answer and Bosoneando's answer to the question of what a "photon" actually is.

ACuriousMind
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    I was thinking about the measurement of the polarization direction in the quantum entanglement thought experiment where two entangled photons are sent far apart. The measurement of one photon, as destructive as it may be, will collapse the wave function of the other instantaneously, is that correct? How would an EM field, which is supposedly not a probabilistic wave function, do that?

    The other question is, as the simultaneity is not absolute in theory of relativity, in which reference frame does the wave function collapse instantaneously?

     – JNL Apr 29 '23 at 23:14
  • @JNL maybe you should ann an edit to the question mentioning the polarization experiment you have in mind. Anyway, this answer is really good for the question as it is now. – Quillo May 01 '23 at 10:40
  • @JNL Whether or not the wave function of the other particle "collapses instantaneously" is interpretation-dependent and in any case the measurement of one part of an entangled pair cannot cause any observable effect on the other end (Bob can never tell whether Alice has done her measurement or not) due to the no-communication theorem. We already have countless questions on this site about that. – ACuriousMind May 01 '23 at 10:47
  • @ACuriousMind I understand that there is no information transferred beyond speed of light. But that is not what I was asking. My question is if EM wave collapsible? If so, why? All the interpretations are for probabilistic wave functions, not EM wave. Your answer to that is "non-destructive photon measurements are hard", so it is meaningless to ask if EM wave collapses or not. But this is not the case in quantum entanglement experiment, where measurement maybe destructive, but photon wave function still collapses. – JNL May 01 '23 at 15:40
  • @JNL I don't know what you mean by "the EM wave" "collapsing". As my answer says, for definite number of photons the EM fields do not have definite values. Measuring anything about the photon - destructive or not - doesn't do anything to change that. – ACuriousMind May 01 '23 at 16:58
  • Hmm.. Ok, "for definite number of photons the EM fields do not have definite values", that is because E and B operators do not commute with number operator, correct? If we start with a system with definite number of photons, and measure E, then we will have different probability to get different value of E, right? So we can say the wave function collapse? By collapse I meant the wave function becomes the eigenstate of an observable operator after measurement. The operator might not be position operator, since as you said, position operator is not defined for photon. – JNL May 01 '23 at 23:40
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Let's say we have one EM-wave sending antenna and two identical receiving antennas.

When it happens that for one millisecond one antenna receives no energy and the other antenna receives double the normal energy, then some observers might say that the EM-wave collapsed.

Quantum mechanics and the law of large numbers tells us that these kind of events are normally very rare, but become less rare when energy sent per second is decreased. I mean, when the number of energy packets sent per millisecond is not a very large number, then the law of large numbers does not work so well.

stuffu
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