Does measurement necessarily change the state?

[Larry Pendarvis]

It is true that an individual particle in a superposition (per A) does not appear different to an observer than a single one from a group of mixed H> and V> (per B). But I am not so sure that a group of such particles cannot be differentiated, assuming they are all from either a stream of A's or a stream of B's. Not sure if that is what you are talking about, but if it is, I think I would disagree.

I am thinking specifically of photon polarization tests and things that you can do with entangled pairs. Specifically: you can provide a light source of indefinite H/V polarization as input to a PDC crystal, and you WILL get some pairs of polarization entangled photons out. If you provide a light source of DEFINITE H/V polarization as input to a PDC crystal, you do NOT get any pairs of polarization entangled photons out. Entangled pairs produce different statistics than ones in a product state. The difference in statistics may not be much though, and might not be enough to measure reliably. Not sure. But that's the concept.

That is intriguing, but my description of (B) did not involve any H>, only V>:
"(B) The scientist knows for a fact that someone has already measured his particles on the vertical axis."
So (B) involves only up and down, no left no right.
Still, it would be interesting to prepare two such ensembles as you describe and see if a PDC crystal can be used to distinguish between an ensemble of definite H/V polarization and an ensemble of indefinite H/V polarization... creating entanglement where there was none, in order to make a measurement. If you can physically distinguish between the two ensembles in this way, it might demonstrate nonlocality without relying on Bell's Theorem, closing all loopholes.

 

[DrChinese]

That is intriguing, but my description of (B) did not involve any H>, only V>:
"(B) The scientist knows for a fact that someone has already measured his particles on the vertical axis."
So (B) involves only up and down, no left no right.

And my example was photons and their polarization. V> and H> are paired for photons the way Up> and Down> relate for electrons.

 

[Larry Pendarvis]

And my example was photons and their polarization. V> and H> are paired for photons the way Up> and Down> relate for electrons.

When you measure a million electrons in the vertical axis, you get 1/2 million U and 1/2 million D. What happens when you measure the polarization of a million photons in a given axis?

 

[Larry Pendarvis]

The scientist without the alien has input particles in superposition, and the one with the alien deals with a statistical ensemble of definite spins.

If you cannot even in principle distinguish between the ensemble of pure states and the ensemble of mixed states, then there is no Black Hole Information Paradox - there was no information to begin with.
From John Baez's web site:
http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/info_loss.html
"Take a quantum system in a pure state and throw it into a black hole. Wait for some amount of time until the hole has evaporated enough to return to its mass previous to throwing anything in. What we start with is a pure state and a black hole of mass M. What we end up with is a thermal state and a black hole of mass M. We have found a process (apparently) that converts a pure state into a thermal state. But, and here's the kicker, a thermal state is a MIXED state (described quantum mechanically by a density matrix rather than a wave function). In transforming between a mixed state and a pure state, one must throw away information. For instance, in our example we took a state described by a set of eigenvalues and coefficients, a large set of numbers, and transformed it into a state described by temperature, one number. All the other structure of the state was lost in the transformation."

 

[Larry Pendarvis]

How about setting up a CHSH type inequality experiment for the mixed population in one case,
( if it is possible to sample and send these particles to detectors A and B in accord with experiment).
And for the entangled/superposition population in the other case. The measurement results will differ, right ?

I have been told that entanglement doesn't even make sense in this context.

 

[DrChinese]

When you measure a million electrons in the vertical axis, you get 1/2 million U and 1/2 million D. What happens when you measure the polarization of a million photons in a given axis?

1/2 million H and 1/2 million V, assuming they are polarization entangled.

 

[Larry Pendarvis]

1/2 million H and 1/2 million V, assuming they are polarization entangled.

okay then they are not

 

[Nugatory]

PeterDonis's advice in your other thread (https://www.physicsforums.com/threa...e-and-mixed-states.794432/page-2#post-4989278) applies here as well. This thread is closed.