# Thought Experiment Confusion

1. Jan 28, 2012

### StevieTNZ

Anyone who follows some of my posts knows I've fallen in love with GianCarlo Ghirardi's thought experiment.

(apologises for the equations - I don't know how to work LaTeX).

The experimental set-up is a 45 degree polarised photon goes through a birefringent crystal. On the other side of the crystal is an apparatus which detects whether a photon comes from the ordinary ray (vertical polarisation – apparatus shows A+1) or extraordinary ray (horizontal polarisation – apparatus shows A-1). This apparatus does not absorb the photon, so allows us to perform a further test on the photon.

The whole situation is now described by the following equation:
=1/2 |45>(1/2 |A+1>+|A-1>) + |135> (1/2 |A+1>-|A-1>)
The photon is in a superposition of V and H polarisations, and when going through a subsequent polariser, orientated at 45 degrees, has 1/2 probability of passing or failing.

On the apparatus that measures which way the photon exits the crystal, is another observable called Z, which takes on the value X if the apparatus is in the state |A+1>+|A-1> or Y if in the state |A+1>-|A-1>. Observable Z takes on X if the photon passes the 45 degree test, and Y if not.

All of this seems to be in accord with QM. As no collapse is meant to occur after the crystal, we should describe the photon as being in a superposition of V and H, as well as 45 and 135.

The issues I'm going to raise:
If we observe the apparatus in either A+1 or A-1, then the photon is in V or H polarisation. Observable Z takes on X or Y (A+1: 1/2 X, Y; A-1: 1/2 X, Y [as a consequence of the inverse above]). So, if observable Z is showing X, the photon can fail the test.

What I'm wondering is if we've set up the experiment and prior to sending the photon through the subsequent polariser see a definite pointer, how does QM take that into account when its saying it shouldn't occur (and obviously affects subsequent results)?

In Bohm Mechanics, after the photon exists the crystal and is detected by the apparatus, is it in a definite polarisation? Then if we look at the apparatus, we'll see A+1 or A-1, and subsequently send the photon through the polariser, we would still see quantum mechanical results (X correlated to passing, etc.) If Bohm Mechanics is correct, and QM is describing the situation, looking at the apparatus shouldn't make a difference because even not looking at it doesn't change the fact that it is in a definite A+1 or A-1. But if we do see it, and if we do see X being correlated to passing (due to hidden variables), then we might need to conclude Bohm Mechanics is correct.

But I'm seeing an issue as because we CAN look at the apparatus, and because it SHOULD be in a superposition before the photon reaches the polariser, how to account this if QM is correct (obviously we should be saying the apparatus is in a superposition, even though we're seeing otherwise)?

Whether we update the wave function after we look, I'm not sure. Because Brian Cox seems to make it clear that if we want to compute the probability of future events, we need to take into account everyway those future states can arise from. From what I gather, he is saying everytime we see a definite state, we can't strictly say coherence is lost.

I'm not too sure, though. Hoping to get some views on this.

2. Jan 28, 2012

### questionpost

The term weak measurement comes to mind when reading over some of this http://physicsworld.com/cws/article/news/48126

When you say we "look into it", what your actually saying I think is that we would need to bounce a photon off of the photon and have it reach our retinas to observe that it's in a specific location, which I don't think happens.
The only way this can work is we observe the photon, which means it's absorbed by our retinas or an apparatus and thus it does not make it to the crystal at all, or we don't measure it and it passes through the crystal. You can't measure an individual photon and have it keep going on as if you didn't measure it.

From what I gather, polarizing a photon adds to its certainty but does not make it completely certain, which is why you can work backwards to build a wave function in that weak-measurement article and not a defined path composed of straight lines.

Last edited: Jan 28, 2012
3. Jan 28, 2012

### StevieTNZ

I don't think you understand entirely the experiment. It is possible in principle to have photon detectors that don't absorb it. And I'm talking nothing about measuring its location prior to the photon reaching the crystal and apparatus, nor looking at the photon. I'm looking at the macroscopic apparatus.

4. Jan 29, 2012

### questionpost

If you have photon detectors that don't actually absorb it, they aren't completely measuring it they are just making it more certain one way or another, but not actually determining location, like with the article I linked, so the photon just ends up having a different probability wave that is probability a little more narrow but can still polarize in both directions.

5. Jan 31, 2012

### StevieTNZ

If observable Z takes on either X or Y, does that mean that the pointer takes on A+1 or A-1? That would seem to imply the photon is V or H polarised, and with 1/2 probability of passing or failing the 45 degree test, X could be correlated to failing, etc.

6. Jan 31, 2012

### lugita15

StevieTNZ, you have to understand that when two photons are in the singlet (entangled) state, it means that for any angle θ you happen to align the polarizing filter with, if you put a photon through the filter it has a 50-50 chance of being measured as polarized along θ or polarized along the angle perpendicular to θ. In other words, there is NO angle θ such that the photon is always guaranteed to be polarized along θ. And yet, the amazing fact of quantum mechanics is that if the photons are in the singlet state and both polarizers are set at the same angle θ, then (to use your terminology) both will be measured "ordinary" or both measured "extraordinary", and again it's 50-50 which will happen.

So no, the mere fact that a photon has a 50-50 chance of being measured polarized along 45° or 135° does not mean that it has a 100% chance of being measured polarized along 0° or 90°. (On the other hand, the converse is true: if a photon has definite polarization along 0° or 90°, then it does have a 50-50 chance of being measured polarized along 45° or 135°. But definite polarization is very different from a singlet or entangled state of two photons)

Last edited: Jan 31, 2012
7. Feb 1, 2012

### StevieTNZ

I'm not entirely sure your comment has anything to do with this thought experiment. I haven't been saying being measured in 45 or 135 degrees means it is 100% V or H.

8. Feb 1, 2012

9. Feb 1, 2012

### StevieTNZ

Ah ok. Makes sense now!

10. Feb 2, 2012

### Karl Coryat

I don't get how in his experiment, information about the photon gets into the apparatus (as it would have to if the photon's presence is verified) without the photon/apparatus system undergoing decoherence -- regardless of what other tests may be going on. At least with respect to polarization. Is the apparatus in fact performing a weak measurement, as another person mentioned?

If the experiment is just a way to show that a macroscopic pointer needle does not exist in a cat state of both A+1 and A-1, then I don't get why it's so interesting.