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QM says that if we have many particles they have a common wavefunction. Also QM says that when you measure a particle or observe it, you collapse its wavefunction. That must be a logical mistake. Now lets look at a laser beam as an example. The laser produces a plane wave which is comprised by many photons. Now if we observe (annihilate) one photon a part of the wavefunction must collapse. But then this means that the wavefunction must be comprised by many individual wavefunctions of any separate photon. So it can not be common but a sum of individual wavefunctions (by the way sphearical in form which overlap to build a plane wave). What is your opinion about this issue?

Qmechanic
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Mercury
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1 Answers1

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QM says that if we have many particles they have a common wavefunction.

It is not compulsory. Many particles may have incoherent wavefunctions and be described by a classical density matrix,

Also QM says that when you measure a particle or observe it, you collapse its wavefunction.

Collapse is an anthropomorphic way of talking about mathematics.The wavefunction $Ψ$ of a particle is a solution of a quantum mechanical equation with given boundary conditions, a complex function. The $Ψ^*Ψ$ is the predicted probability distribution. An interaction, measurement is an interaction too, will pick up one instance of a probability distribution that has to be accumulated to validate the prediction. Many instances are necessary to build up a probability distribution, as in this double slit single electron experiment.

That must be a logical mistake. Now lets look at a laser beam as an example. The laser produces a plane wave which is comprised by many photons.

The laser is a good example of a collective coherent wavefunction which is defined by the whole system, including the lasing source. This MIT video gives a good explanation of the superposition of two coherent laser beams. Superposition is not interaction. It is the addition of the two beams collective wavefunctions.

Now if we observe (annihilate) one photon a part of the wavefunction must collapse.

We cannot observe one photon out of the zillions in the laser $Ψ$ .

One can make the laser beam very weak so that individual photons can be recorded, and their wave function retains the coherence to show the probability distribution as seen in this double slit experiment single photon at a time. The collective wave function and its complex conjugate squared are there, even for one photon.

But then this means that the wavefunction must be comprised by many individual wavefunctions of any separate photon. So it can not be common but a sum of individual wavefunctions (by the way sphearical in form which overlap to build a plane wave).

Note, it is wavefunctions and they have phases and can be coherent building the mathematics of one collective coherent wavefunction , two or many they are complex functions. BUT the probability is the total $Ψ^*Ψ$, a real distribution .

What is your opinion about this issue?

Once the boundary conditions are set , the measurement picks up a probability instant.

anna v
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  • Thanks Anna V, a very well structured answer. Though I don't cleared my case. What is exactly what you say? So I put an example. Suppose we have diminished the beam to two photons. The wavefunction is still a plane wave, isn't it? Now we register one of the photons. Does the wavefunction become a spherical wave now (instantly)? That is the question! – Mercury Nov 13 '18 at 19:59
  • Another issue I considered recently is some counter relativity case. They are certainly many particles systems with a common WF (wavefunction). The way particles move is governed by the WF. When a particle (or part of them) interacts or is annihilated (or observed) the WF must instantly change. But this must have an instant effect on the other particles as they are governed by the new WF. And this happens superluminally in my opinion. And by the way there is no interaction whatsoever. This is all very disturbing? What do you think about? – Mercury Nov 13 '18 at 20:20
  • it will depend on the method of detection , i.e. the further boundary condition problem including the screen or photomultiplier. For a screen the photon is absorbed and a new photon or more will come from deexcitation probably of different energy and probably spherically symmetric to the original direction. – anna v Nov 14 '18 at 04:52
  • Look at this series of videos of how photons behave in the coherent wavefunction in which they are generated by the laser https://www.youtube.com/watch?v=J4Ecq7hIzYU . It gives a feeling for how superposition operates. In this particular case the same function describes quantum mechanically individual photons, and with classical electrodynamics the coherent wave of the emergent beam. Superposition gives interference patterns from single photon at a time – anna v Nov 14 '18 at 04:59
  • https://www.google.gr/search?q=swiss+double+slit+single+photon&ie=utf-8&oe=utf-8&client=firefox-b&gfe_rd=cr&ei=sGFmWYrFCO_v8AfrtLGADQ to the one seen in the video where there are zillions of photons.Superposition is addition of the complex wavefunctions, not interaction. The total wavefunction squared, then gives the probability distribution which is the observed interference pattern. Photons are not bricks in building up the classical wave. – anna v Nov 14 '18 at 05:03
  • Rather , an analogy, the wavefunction is a "map" for the boundary conditions which forces single photons and their superposition to follow the "flow".. If one photon leaves, the wavefunction does not change for all the rest. It is the boundary conditions that build up the whole or the single. – anna v Nov 14 '18 at 05:03
  • So you basically say that nothing changes in the WF when the photons are absorbed (on by one). - The boundary condition don't change. But then this undermines the principal of collapse (and the Bohr complementarity)? How do you keep it? – Mercury Nov 15 '18 at 13:47
  • Also the rest photons are governed by other photons which are long gone (just by the superposition imposed before?). This is more than interaction (a ghost interaction)!!?? – Mercury Nov 15 '18 at 14:07
  • This reminds me that I haven't seen an explanation anywhere why all laser photons are in the same direction, but I will ask this as a separate question. In a question titled why all laser photons are in the same direction why all laser photons are in the same direction "Why all laser photons are in the same direction?". – Mercury Nov 15 '18 at 14:10
  • The surperposition is in the mathematical wavefunction which does not need any single real photon, it is a mathematical, model that works. When one calculates the trajectory of a rocket, given the velocity, angle etc, is the rocket there? – anna v Nov 15 '18 at 14:17
  • The newtonian mathematical model describes successfully the trajectory whether you shoot the rocket or not. It is the same with quantum mechanical models, the boundary conditions describe successfully the probability distribution for the given boundary conditions. A measured photon is an instance in the probability distribution as throwing a 6 is an instance in the probability distribution of throwing a dice.. Collapse is an anthropomorphic way to say : "throw the dice". – anna v Nov 15 '18 at 14:18
  • for the direction see Motl's answer here https://physics.stackexchange.com/questions/65711/why-is-the-photon-emitted-in-the-same-direction-as-incoming-radiation-in-laser – anna v Nov 15 '18 at 14:27