# Is the collapse of a wave function deterministic or random?

1. Dec 7, 2015

### Happiness

Suppose I measure the position of a particle, and I find it to be at point C.

By deterministic, I mean if we know the wave function of the measuring instrument (and of course also the wave function of the particle before measurement) then we can, in principle, know that the particle is going to be at point C during the measurement.

And by random, I mean there is no way we can, in principle, know that the particle is going to be at point C.

2. Dec 7, 2015

It's random. Even if you know the exact wave function, it only tells you how likely it is for the particle to be measured at point C.

3. Dec 7, 2015

### .Scott

Without the measurement being made, there is no position of the particle. However, there is one wrinkle to this. If the particle is one of an entangled pair of particles, and the position of its twin has been measured, then the position of this particle is known and the result of its position measurement is very predictable.

So in some cases we have enough information available to make a prediction.

But more generally, we do not have enough information collected - and in most cases we need to suspect that it would be impossible to assemble that information - making it effectively (if not also in principle) impossible.

4. Dec 7, 2015

### Happiness

Can I say that since the measuring instrument is the cause of the collapse of the particle's wave function, the measuring instrument should contain information that tells us which state the particle's wave function is going to collapse to during measurement?

Last edited: Dec 7, 2015
5. Dec 7, 2015

### Staff: Mentor

You cannot. That's one of the implications of Bell's theorem (google for it, or try this website maintained by our own @DrChinese).

6. Dec 7, 2015

### Staff: Mentor

As far as the mathematics of quantum mechanics are concerned, it is random. There's no way it cannot be, as the theory is explicitly about the probability of a measurement having a given result - you take the wave function, you crank it through the math, and you end up with a probability (or an expectation value if you integrate the probabilities across all the possible results)... And then QM has told you all it's ever going to tell you. If you want more, you have to perform the measurement and find out if the particle is there or not.

7. Dec 7, 2015

### .Scott

Here's another experiment that illustrates what needs to be considered when determining where a particle lands.

A Mach-Zehnder interferometer is a fairly simple device that splits a laser beam and recombines it to form an interference pattern.
It's described here: https://web.phys.ksu.edu/vqmorig/tutorials/online/wave_part/
And shown here:

Once it's set up and running, it looks like this:

Actually, with a red laser, the projection looks like this:

The first thing to consider is this, not all of the photons coming out of the laser land in the same place - even though the target (the screen) isn't changing. So clearly something more than the target is involved.

Notice in particular that there are dark areas in the pattern where photons never go. Each photon interferes with itself so that when it lands on the screen, it can be anywhere except the darkest places.

In that first diagram, you will see that there is a path A and a path B. If I block path A, the result will be a solid red blur being projected onto the screen.
Now consider the photons that are landing where those dark bands used to be. The reason they are getting there is because path A is blocked. But they obviously didn't take path A, because if they did, they would have hit the block and would never have reached the screen. Instead they took path B. They reached their location on the screen because they could have hit the block but didn't. So even things that didn't happen can affect where the particle is detected.

8. Dec 7, 2015

### Happiness

Bell's theorem shows that a particle did not have a determinate position before the measurement, but it does not rule out the possibility that the measuring instrument causes the collapse of a particle's wave function in a deterministic way.

Last edited: Dec 7, 2015
9. Dec 7, 2015

### Heinera

In what way would you distinguish "have a determinate position before the measurement" from "collapse of a particle's wave function in a deterministic way"?

10. Dec 7, 2015

### Happiness

If a particle has a determinate position before the measurement, then it was somewhere even though we do not know where it was.

If the wave function of a particle collapses in a deterministic way, then it is possible to have a mechanism describing the collapse.

11. Dec 7, 2015

### Heinera

But if the wave function is known, and the mechanism describing the collapse is known (and determinate), what prevents me from deducing the position of the particle without actually doing the measurement?

12. Dec 7, 2015

### Happiness

Because the particle did not have a determinate position before the measurement, provided it's pre-measurement wave function is not sharply peaked at any point. The particle exists more like a ripple on a water surface, but the measurement process collects all the energy of the ripple and places it at a determinate position, for example.

13. Dec 7, 2015

### Staff: Mentor

But what determines whether the collapse is to one of the possible results or another? There has to be something that that we can use, if we just knew what it was, to calculate the determined result of your hypothetical deterministic process. Where does that something fit in Bell's $\lambda$ variables that specify the state?

(Actually, I do believe that there is an intellectually coherent way for collapse to be deterministic.... but it takes you all the way to superdeterminism. Bell hasn't left a lot room in between).

[Edit: If you're willing to accept the state of the other (distant, spacelike-separated) as part of the "measuring apparatus" you can construct a consistent but non-local deterministic theory. Without a specific candidate theory though, it's not going to be experimentally distinguishable from the randomness that's inherent in the mathematical formulation of QM]

Last edited: Dec 7, 2015
14. Dec 7, 2015

### Heinera

And if this measurement process that "collects all the energy" is deterministic, then it is equivalent to the particle having a specific property before the measurement, as far as Bell's theorem is concerned.

15. Dec 7, 2015

### Happiness

Bell's theorem shows that local realism is incompatible with quantum mechanics, but this is not what I'm asking.

A clearer phrasing of my question would be as follows:
We know that the wave function of a particle evolves in a deterministic way according to the Schrodinger equation when the particle is not being measured. Suppose I measure the position of a particle, and I find it to be at point C. We know that its wave function evolves to one that is sharply peaked at point C during the measurement. Is such evolution of the wave function deterministic or random?

16. Dec 7, 2015

### .Scott

So far, that is a matter of personal opinion. QM, at its current development, does not address this. To my knowledge, there is no experiment that has been proposed that would differentiate between a truly random result and one that is determined by the non-local environment in a deterministic way.

I have argued in other threads that if the result is truly random, then the decision is being made by adding information to the universe. Something that is inconsistent with QM. However, I have not been persuasive.

17. Dec 7, 2015

### Staff: Mentor

Collapse is not part of the QM formalism, only some interpretations.

The answers to the questions you are asking depends entirely on interpretation.

Pick an interpretation, any interpretation:
https://en.wikipedia.org/wiki/Interpretations_of_quantum_mechanics

Once you have done that then answering your questions will be much easier.

Thanks
Bill

18. Dec 7, 2015

### TESL@

Also, please don't choose Bohm or the consciousness thing.

19. Dec 7, 2015

### Staff: Mentor

That's an interpretation you are putting on it.

The QM formalism is silent on what's going on when not observed.

As I said in a previous post, pick an interpretation then answering your questions may be possible. The QM formalism is silent on many things so you must pick an actual interpretation to ask meaningful questions.

Thanks
Bill

20. Dec 7, 2015

### Staff: Mentor

Bohm's OK - but this conciousness thing really annoys me. Still its a legit interpretation and can be chosen for illustrative purposes if that's the OP's thing.

Thanks
Bill

21. Dec 8, 2015

### Happiness

What does it mean by the wave function never collapses in some interpretations? It seems like all interpretations have to agree that the wave function collapses but may disagree on the interpretation and the ontology of the collapse. By collapse, I mean the evolution of the wave function from one that is spread out in space to one that is sharply peaked at a point during a measurement. Some interpretations may argue that the collapse of the wave function is not due to a physical change in the system but only due to a change in the information of the system available to an observer, but they still cannot deny that the wave function collapses. Hence, the collapse still needs to be addressed regardless of the interpretation one takes.

22. Dec 8, 2015

### Staff: Mentor

It means that a measurement that does not destroy what's being measured is, in some interpretations, a state preparation procedure. This does away with the concept of collapse that is supposed to instantaneously change the state. State and state preparation procedure are, in those interpretations, synonymous. It also means that collapse never happens in MW. There are probably other ways to do away with it but cant think of them off the top of my head.

Thanks
Bill

23. Dec 8, 2015

### Happiness

A measurement that does not destroy what's being measured would be a measurement done on a particle whose wave function is already peaked at a point, so naturally there isn't a collapse (or there is but a trivial one; it just evolves to the same wave function). But what about a measurement done on a particle whose wave function is spread out. Shouldn't there be a collapse?

Does MW mean many-worlds interpretation? How does it do away with the collapse? Even if all the possible outcomes are realised but in different parallel worlds, we still observe only one outcome and hence the wave function collapses.

Wikipedia: "In many-worlds, the subjective appearance of wavefunction collapse is explained by ..." So even many-worlds interpretation does not deny that the wave function collapses.

24. Dec 8, 2015

### Staff: Mentor

That's incorrect. In the double slit experiment the slit does a position measurement on a particle without an actual position.

It interprets the elements of a mixed state after decoherence as separate worlds so no change in the wave function happens.

You do understand the difference between actual and appearance don't you?

Thanks
Bill

Last edited by a moderator: Dec 8, 2015
25. Dec 8, 2015

### Happiness

Could you explain more?