# Does measurement necessarily change the state?

1. Jan 17, 2015

### Larry Pendarvis

Suppose one scientist performs a series of experiments, and another scientist does the same thing. But unbeknown to either of them, a highly evolved alien measures all of the particles the first scientist is going to test, just before he does his measurements.
Question: Would the two scientists find different results from their experiments?

2. Jan 17, 2015

### Staff: Mentor

It depends on the nature of the measurement that the alien performs.

In the formalism, everything that happens before a particular measurement is considered part of the preparation procedure, and the theory predicts the statistical distribution of the results of measurements made on a system after a particular preparation procedure. Thus, the alien's measurement (and any other measurement or interaction) is part of the preparation procedure for the first scientist's measurements.

Two examples:
1) Particles pass through a vertically oriented Stern-Gerlach device, we discard all the spin-down ones, our scientists measure the spin on the vertical orientation and expect to see 100% spin-up. The alien performs a measurement along the horizontal axis; now the first scientist will get a 50/50 distribution of spin-up and spin-down while the second will get 100% spin-up.
2) Same conditions, but alien measures along the vertical axis; now both scientists get 100% spin-up.

3. Jan 17, 2015

### morrobay

For clarification could you reply in this format:
t0 Alien measurement on what axis with what result
t1 First scientist measurement on what axis with what result
t2 Second scientist measurement on what axis with what result

4. Jan 17, 2015

### Larry Pendarvis

Your SG device has already made a measurement. The particles have a definite (and known) spin before the scientists' and alien's measurements. How about we leave that out and prepare particles whose spin is undetermined? Since we are doing an experiment designed to measure spin. And measuring a known quantity is uninteresting.

5. Jan 17, 2015

### Larry Pendarvis

t0 vertical, 50/50
t1 any, 50/50
t2 same as t1, 50/50

Actually, none of the axes matter, we will get 50/50 as long as we don't discard any.

Last edited: Jan 17, 2015
6. Jan 17, 2015

### StevieTNZ

If you are measuring the same axis of the system that you did at t1, at t2, you would see the same result as t1.

7. Jan 17, 2015

### Larry Pendarvis

Consider two cases:
(A) The scientist measures spin along whatever axes he chooses. He has prepared the particles and they have no definite spins.
(B) The scientist knows for a fact that someone has already measured his particles on the vertical axis. He even knows that the previous measurements' results are known to the previous measurer. Thus every particle has an exact, known spin, but the scientist does not know the correlation - he does not know WHICH particle has an up spin and which has a down spin, he only knows that half of them were measured up and half were measured down.

It seems that there is no physical experiment the scientist can do that can distinguish between (A) and (B). Therefore we can, if we choose, always speak of particles having an actual measured spin known to someone in some galaxy. Whether they do or do not makes no difference to physics.
I imagine Bell might disagree, but that seems to be the conclusion here so far.

Last edited: Jan 17, 2015
8. Jan 17, 2015

### StevieTNZ

I asked the following question to Caslav Brukner back in 2012 (when I was still developing my knowledge around QM):
Caslav replied:
EDIT: I trust you know enough about the spin basis (x, y and z axises) that will confirm that you cannot differentiate between your situation (A) and situation (B).

Last edited: Jan 17, 2015
9. Jan 17, 2015

### Larry Pendarvis

The particles that have been measured are not in a superposition. Reduction has occurred and the wave function has "collapsed". Each particle is either up or down spin. The observables have been observed. It is simply a set of such known-spin particles. But such a set of particles gives the very same measurements as a set of particles in superposition.
Or... is it possible to separate the two types of particles of known spin? Then the scientist could do experiments with that knowledge. Is such a separation a "measurement"? Does it change any states?

Last edited: Jan 18, 2015
10. Jan 18, 2015

### Larry Pendarvis

BTW, the title of this thread is misleading. That is not my question. The answer to the title question might be answered by remembering that there are experimentally proven interaction-free measurements. There is no need to ask the question.
http://en.wikipedia.org/wiki/Elitzur–Vaidman_bomb_tester

11. Jan 18, 2015

### Staff: Mentor

What title would you suggest?

12. Jan 18, 2015

### Larry Pendarvis

Beats me. The measurements I am talking about do change the state. So maybe something like "Can we distinguish a mixed population from superposition?"

13. Jan 18, 2015

### StevieTNZ

Spin up along vertical axis is a superposition in the other two axises. My example is also where the ensembles of photons are in a definite state (which is why Caslav calls them a mixed state).

14. Jan 18, 2015

### Larry Pendarvis

Sorry, I guess I should have said "the particles that have been measured are not in a superposition in the vertical axis."
But measuring a spin-up particle on the vertical always gives the same result.
A mixture of known up and down is not conceptually the same as a population of superpositions on the vertical axis. And it ought to be distinguishable physically (otherwise who needs QM), but I don't see how.

15. Jan 18, 2015

### StevieTNZ

Maybe utilise a Mach-Zender interferometer? Not sure... hopefully someone else can come along and help us both out!

16. Jan 18, 2015

### Larry Pendarvis

A Stern-Gerlach apparatus will separate the up from the down, in the mixed set of ups and downs. And there will be no change in spin.
But it will give the very same result with a population of superpositions, and in that case the state IS changed to definitely up or down.

17. Jan 18, 2015

### Larry Pendarvis

18. Jan 18, 2015

### morrobay

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 ?

Last edited: Jan 18, 2015
19. Jan 18, 2015

### Larry Pendarvis

Strange how we keep coming back to Bell's Theorem even with something this straightforward, not involving EPR or entanglement. There are not pairs of entangled particles, only single particles in superposition.

20. Jan 19, 2015

### Larry Pendarvis

This can be carried much further. Suppose you measure the mixed population not just along the vertical axis but say π/6.; you would still get a 50/50 result, I believe. And any other angle will give the same result, would it not? And if the mixed population had been measured on some other axis, you would STILL always get the same 50/50 mix no matter what axis you measured.
So as long as you have a random mixed population such that every particle has been measured along some axis, then no matter which axes you measure you will always get a 50/50 result. Thus how can you distinguish between a mixed population of arbitrary previously-measured spins (fixed states, no superposition and no entanglement), and a population of arbitrary spin superpositions?
I did not do the math just now; perhaps the π/6 measurement would NOT give a 50/50 result, but I can't imagine how it could favor one direction over the opposite direction. Measuring an up would not give 50/50 when measured on the π/6 axis, but that would be exactly compensated for by how much the down would also not give a 50/50.

21. Jan 19, 2015

### DrChinese

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.

22. Jan 19, 2015

### Staff: Mentor

Not so strange if you consider that the bolded text above is basically a restatement of the EPR claim that "I know that if were to measure property X I would get value Y" is equivalent to "property X as the definite value Y". Thus, this single-particle case has everything to do with EPR; and even though we cannot do an Aspect-style experiment without entanglement, the essential role of Bell's theorem in rejecting EPR local reality is relevant.

23. Jan 19, 2015

### Staff: Mentor

That is a different question than the the one that started this thread (which involved an alien introducing an extra step in the state preparation procedure before measurement) but the answer turns out to be the same: It depends on the specifics of how the states are prepared and the measurement being made.

For an example with spin-1/2 particles, using Stern-Gerlach devices for the preparation and measurement, and representing spin in the up, down, left, and right directions as $|U\rangle$, $|D\rangle$, $|L\rangle$, $|R\rangle$:

I prepare ensemble A in a a statistical mixture of $|L\rangle$ and $|R\rangle$ and ensemble B in the superposition $\frac{\sqrt{2}}{2}(|L\rangle + |R\rangle)$. These will be indistinguishable if I measure spin along the left-right axis but will produce very different results if I measure along the vertical axis.

24. Jan 19, 2015

### Larry Pendarvis

The results will be 50/50 up and down in the first case.
What do you expect the ratio to be in the second case for the statistical mixture? Roughly. More up than down, or more down than up?
What do you expect the ratio to be in the second case for the superposition ensemble? Roughly. More up than down, or more down than up?

25. Jan 19, 2015

### Staff: Mentor

Ensemble A will produce 50/50 up and down. Ensemble B will produce 100% up. Don't take my word for it, calculate it.