# B Superposition of measurement apparatus

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1. Apr 26, 2016

### entropy1

Does measuring a polarized photon after it passed a polarization filter put the measurement apparatus in a superposition of detected/not-deteced (the photon)? Does this depend on whether the photon is part of an entangled pair?

2. Apr 26, 2016

### StevieTNZ

According to QM formalism, the photon is in a superposition of passing and not passing the filter. Therefore the apparatus is in a superposition of detected/not-detected.

With entangled photons, what would be used would be wave plates and polarizing beam splitters to measure the photons. I did ask Bruce Rosenblum, one of the authors of 'Quantum Enigma', that when the photons hit the measuring apparatus, because the photons are absorbed is the entanglement between the two photons now transferred to entanglement between the two apparatus. He said yes.

3. Apr 26, 2016

### entropy1

I was wondering, suppose we have an entangled pair of particles, individually measured at perfectly isolated labs (with the exception of the incomming particles). Now both researchers are in superposition of having seen their particle detected or not detected or property being up or down or something like that. Now say that they have a classical phone to ring each other. If and when they ring each other and convey the results of their measurements to each other, they now know each other's measurement results and will agree on them. They may send along photo's of their apparatuses' readout. So, are the researches now no longer in superposition? (since the results are in?)

Last edited: Apr 26, 2016
4. Apr 26, 2016

### Strilanc

The final state in the described situation is $|HH \text{Alice}_H \text{Bob}_H\rangle + |VV \text{Alice}_V \text{Bob}_V\rangle$.

Whether or not it makes sense to think of the experimenters as really in the superposition or as having collapsed it into the two cases depends on your interpretation. Suffice it to say that the final states differ so much that they are never going to interfere. Decoherence has occurred.

5. Apr 26, 2016

### entropy1

I know a little of product states, but I can't interpret the ones you gave. Of what are they a product?

Does $|HH \text{Alice}_H \text{Bob}_H\rangle$ mean: both particles in state H and both measurents results are H?

6. Apr 26, 2016

### Strilanc

Yes. Also it's supposed to imply including the experimenters having seen the result, and all of the other minute differences in the environment that that would entail.

7. Apr 26, 2016

### entropy1

So you say "decoherence has occurred". However, when the researches haven't rung each other yet, the measurement results are decohered in the seperate individual labs already, aren't they? Yet, in that stage the labs are still in superposition of two results (detected/not deteced/up/down). What is special about the latter decoherence, when the researchers ring each other, that make the results as it were to collapse into a definite shared result?

Last edited: Apr 26, 2016
8. Apr 26, 2016

### StevieTNZ

They won't know the result of their measurement because the researchers are still in superposition.

Last edited: Apr 26, 2016
9. Apr 26, 2016

### Strilanc

Decoherence will have occurred as soon as either of the measurements has been done. There's no way to keep the labs isolated in the sense you want.

It's as if you were asking for the experimenters to be wearing socks with identical color and also asking for the colors to be totally unrelated until they called each other on the phone. It makes no sense.

10. Apr 27, 2016

### entropy1

Do you mean that decoherence destroys superposition?

11. Apr 27, 2016

### Strilanc

No, "is decohered" is a property that any quantum state can have. It basically means that subsets of it have become thermodynamically separated: they differ so much that they won't meet again to interfere. Whether or not this triggers collapse depends on your favorite interpretation. In many-worlds it doesn't, in Copenhagen it does.

12. Apr 27, 2016

### Staff: Mentor

Yes.

Its transformed into a mixed state. The concept of superposition does not apply to mixed states - only to pure states.

This is very technical and cant be explained at the lay level.

Thanks
Bill