# Understanding - how universe/reality plays dice

1. Apr 7, 2012

### San K

According to the more popular interpretations of (parts of) QM:

photon (properties) are in an indeterminate state prior to measurement.

the outcome of the photon properties, upon measurement, is totally random

however after measurement the photon has definite/fixed/deterministic properties.

Question:

Now does this (recently measured) photon stay deterministic for ever? (because it has been measure via a polarizer or some other thing/material/apparatus).

how does this photon go back to being totally random?

which experiments/actions put a photon into
a) determinate state
b) into a totally random state?

While typing this another thought occurred (as a possible answer, not sure if it correct) -

when we measure one property, the other property becomes totally random thus not all the properties of a photon can be totally deterministic at the same time.

Definition:
Dice = a metaphor for causeless, totally/inherently random outcome. Thus this metaphorical dice is, of course, different from the real dice. The outcome of the real dice is predictable (as they say -- at least in theory) if you knew all the initial conditions and the in-process conditions (such as angular moment, velocity, air conditions etc)

Last edited: Apr 8, 2012
2. Apr 7, 2012

### StevieTNZ

Photon is destroyed first time it is measured.

I would say so.

3. Apr 7, 2012

### San K

the photons that pass through the polarizer are still existing (not destroyed) and have a "fixed" spin....

does this photon stay deterministic forever ?

how can we put this photon back into a totally random state?

i guess the answer would (roughly) be to measure a property that is a conjugate (of the property we can to make indeterminate).

4. Apr 7, 2012

### StevieTNZ

Ah - only the subsequent detection of the photon after the polariser makes the polarisation definite. Until then it is still in superposition. E.g. 45 polarised -> H filter: 1/2 probability. Detection: Yes! I'm H polarised, destroyed. No detection, V polarised, destroyed.

5. Apr 7, 2012

### San K

what is destroyed?

50% of the photons are passing through. the H-photons are passing through, they continue to exist

6. Apr 8, 2012

### StevieTNZ

But to know that they're H polarised, you must DETECT THEM. Otherwise it's a plain passing/failing superposition, which continues as indeterminate.

7. Apr 10, 2012

### wilmor51

Perhaps you cannot destroy a photon
Maybe when a photon is deterministic it would exist only in our universe and so can be observed
Maybe when the SAME photon is random it has passed to another universe or another NOW so cannot be observed

8. Apr 10, 2012

### San K

not sure if i got what you are saying. so let's go over the events

event 1: entangled photons created

event 2: sent through a polarizer

event 3: sent through a second polarizer

event 4: emerges from the second polarizer

where is the photon being destroyed? for/during each event -- what are the entities that are entangled (per popular/common interpretations)?

Last edited: Apr 10, 2012
9. Apr 11, 2012

### mathal

Photons that pass through a horizontal polarizer and then 'try' to pass through a vertical polarizer fail-i.e. to the degree that the polarizers are 'perfect' in their construction, no photons pass through.
However if the photons path is through a horizontal, then a diagonal and then a vertical polarizer some photons will pass through, and can be detected -an event. I know this is hard to accept but until the 'path' of the photon in finished nothing can be said about it. Nothing definitive can be said about any quantum particle 'in transit'-i.e. not definitively localized at a measureable event.
To the degree that physicists can collate data from experiments, they have extrapolated mathematical principles of what events will happen and in what proportions these events will occur. What an individual 'photon' does to satisfy this collective collated data is unknowable.

Back to entanglement.
It doesn't matter what you do with polarizers on one of the two entangled photons. As long as the final measurement on each photon is the same- and you don't even need to use polarization in this measurement- it just has to be the same form of measurement- if you do this the result will be identical for both entangled photons. If you don't use the identical form of measurement, you are measuring two distinct unentangled photons (and it's your 'fault').

But, what do I know, I'm not a 'mentor'.

mathal

quoation marks are vital to understanding the written word

Last edited: Apr 11, 2012
10. Apr 11, 2012

### StevieTNZ

Either (1) still in superposition, or (2) you detect it with a detector and it is destroyed (polarisation state becomes definite).

Starting from (1) "still in superposition"
Still in superposition, or you detect it with a detector and it is destroyed.

11. Apr 11, 2012

### Delta Kilo

Huh? When the photon goes through the polarizing filter it gets measured and its wavefunction collapses into state of definite polarization.

Instead of filter, consider polarizing beam splitter with detector in one arm, which is basically the same thing. If you know the photon went in and you also know it did not hit the detector, then you can be fairly certain it went the other way and in state of definite polarization.

Interestingly it is the interaction of the photon with detector that causes wavefunction collapse even if the photon decides to go the other way in the end. Whether it gets absorbed by the detector or not does not matter, it is sufficient that it has non-zero amplitude to do so.

PS: I'm not a mentor either :)

12. Apr 11, 2012

### San K

well said delta ton......

however I am not sure at what point/event does the wave-function collapse

13. Apr 11, 2012

### StevieTNZ

Completely contrary to what Anton Zeilinger says in his book "Dance of the Photons".

14. Apr 11, 2012

### Ken G

The way to tell the difference between a "collapsed" state and a "superposition" state is whether or not you can still get interference between the different possibilities. Since you can send a photon through a polarizing beamsplitter, and later rejoin the branches, and still get interference between them, the state is still a superposition rather than a collapsed state. This is like a two-slit experiment-- the photon remains in a superposition of passing through each slit, even if you have perpendicular polarizers in the two slits, because you can "erase" the polarization information by passing both beams through a polarizer at a 45 degree angle to the first set. After doing that, you can recover a two-slit diffraction pattern, even though the polarization at some "intermediate stages" clearly connects with which slit the photon went through. If that information is no longer available at the final screen, it's still a superposition, and you can still get a two-slit pattern.

15. Apr 11, 2012

### DrChinese

So there is a little truth to both sides. You need to look at the final complete context. Then you should be able to make comments about what kinds of things are happening within.

16. Apr 11, 2012

### Delta Kilo

Yes, it's a tricky subject. In Ken G example, each polarizer does collapse the wavefunction into a particular base with regards to spin. But it does not affect position nor the overall phase (which is unimportant for individual photons but matters a lot for the entangled pair). Specifically, observable operator of the polarizer (or rather its tensor product with identity operator, as applied to the combined wavefunction of the pair) is degenerate with respect to these, so it lets them through unaffected.

17. Apr 11, 2012

### ThomasT

What others said about photons, etc., plus I'll add my two cents.

It's the measurement results that are random. (But not always. Eg., the predictable correlation between θ and rate of coincidental detection in Bell tests is nonrandom and is thus compatible with the assumption of an underlying determinism.)

Whether the deep(er) reality is deterministic or indeterministic is unknown and unknowable. But insofar as physical science involves the search for a fundamental dynamical mechanism, then it's assuming that our universe is evolving deterministically. In fact, the assumption of a deterministic evolution is part and parcel of our daily navigations through the world of our experience.

It might be a good idea to just disregard some, maybe most, of the ordinary language surrounding QM.

18. Apr 11, 2012

### mathal

Photons can only get 'measured' once. Filters that differentiate polarization in an either/or way do not measure the photon. Half-silvered mirrors do not measure the photon. In an experiment, they offer options that must 'all' be considered, not just the 'real' ones that would result from measuring (detecting-ending the existence of) photons immediately after the device.

As Ken G pointed out a polarizing beam splitter can be incorporated to act in a similar way to the polarized filter. That is not the way that you used it.

With the polarizing beam splitter you don't even need a detector. All you need to do is throw the option of polarized light opposite to that that would 'pass' through the polarized filter -away -out of the experiment, and you have a different experiment.

My uneducated guess is that after passing through this polarizing beam splitter with the vertical polarization option necessarily eliminated (-a measureable, knowable effect), no light subsequently would pass through the diagonal and vertical polarized filters. -this is in reference to my first post in this thread -number 9 -where photons got through -no intervening 'knowledge' to gum up the works.

You didn't address the issue of photons managing to pass through a horizantal and a vertical polarized filter because of the diagonal polarized filter in between. This result is in complete opposition to the notion that each filter blocks it's complement from going any further along a particular path.

mathal

19. Apr 12, 2012

### Delta Kilo

That is exactly how I used it. With detector blocking one output, it will have exactly the same effect as polarizing filter. Also, the absorption of the photon by the polarizing filter is in theory detectable, so it is a measurement. Eg. polarizer heats up when it blocks the light and does not when it lets it pass through. My example with separate splitter/detector just makes it explicit.

Note that while splitter can be configured as a filter, the reverse is not true. In the filter the measurement takes place, wavefunction collapses and the original photon polarization is lost forever.

This is incorrect.

It is such an archetypal example, I thought it was obvious.
This happens precisely because photon's polarization is measured by the filter. At the first filter, whatever photons manage to go through have their wavefunctions collapsed into |H> state. Since |H> and |V> are orthogonal (<H|V>=0), they would never pass through the vertical filter. Diagonal filter measures and collapses the wavefunction again, this time into |45°> state. Since |45°> has some |V> component (<45°|V>=1/√2), half of the photons do pass though vertical filter.

DK

20. Apr 12, 2012

### mathal

Thank you for the explanation. I still have a lot to learn.
mathal