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Not literally when as in the moment it occurs, but if you're given a series of particles can you differentiate between the particles which have been collapsed and those that have not?

If you have a stream of particles, say 100 particles and you measure the spin etc on an arbitrary number say 7 of these particles. Then the stream is sent to another room, the person in the room doesn't know which 7 of these 100 have been collapsed. Is it possible for him to detect or otherwise figure out which 7 have been collapsed?

Dan
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  • Answer to your titles question is sure. Make you measurement device make a click when it has detected the particle - that is the sound of a wave function collapse. But answer to your body text question is no. A bit misleading title, I would say, if you ask "when...", but start your question with "not literally when...". – Mikael Kuisma Aug 21 '16 at 13:49
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    I answered a very similar question here. Basically, there's absolutely no difference between a "collapsed" state and a non-collapsed one. It's more like "before/after the measurement interaction". – knzhou Aug 21 '16 at 18:26
  • I must be missing something here. A collapsed state should act like a particle and a non-collapsed like a wave right? So if you had a single slit and a detector in the behind it and then also detector off to the sides.. the collapsed state should form a bar through the slit and the non-collapsed should spread out like a wave right? So then the detectors to the side should show higher activity for non-collapsed states then for collapsed states right? Let's assume background interference and internal noise are non-issues. – Dan Aug 23 '16 at 00:32

2 Answers2

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If the particles all happen to start out in a pure state, but could be oriented in any direction with uniform probability... that is indistinguishable from the completely mixed state.

Even if you know they are all oriented along a particular axis, but could be up or down with equal probability, that is still indistinguishable from the completely mixed state.

If a measurement occurs to something in the completely mixed state, but you don't know the result of the measurement... the end result is that the state of the particle is still in something indistinguishable from the completely mixed state.

So, there is no statistical difference at all between the measured and unmeasured particles.

Mathematically, the state space of a qubit is the Bloch sphere -- the pure states are the surface of the sphere, and the mixed states are the interior of the sphere.

The mixture of (it doesn't matter if the mixture is for physical or ignorance reasons) states is the weighted average of their positions in 3-space. So our premise says the state of the qubit is at the center of the sphere.

When a 'measurement' is performed on the particle, you can express it as the following procedure:

  • You project the state orthogonally onto the axis you are doing the measurement around
  • The result is either 'up' or 'down' along that axis, and the state of the particle shifts to the corresponding point on the surface of the sphere

However, the second step is only seen in systems that have access to the result of the 'measurement'. Otherwise, be it due to ignorance or decoherence or whatever explanation we like to give to things, the result is that only the first step happens.

Anyways, the center of the sphere already lies on every axis of measurement, so the projection has no effect.

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C Cowan and R Tumulka, in their paper "Can One Detect Whether a Wave Function Has Collapsed?", show that with prior knowledge of the wave function, collapse can be detected, see https://arxiv.org/abs/1307.0810

user156531
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  • is it possible to know if one of the two entangled particles has collapsed first / second ? – Vlad May 06 '19 at 08:19