# Collapse with and without measurement

1. Jul 4, 2015

### VALENCIANA

I read yesterday about a double-slit experiment in which the researchers set up the device ready to make the measurement, but didn´t make the measurement and the wave function collapsed. Then they did it by taking the measurement and it also collapsed.
What does it mean that just by setting up the device, but then not measuring, the
wave function anyway collapses??
Unfortunately I lost the website address. When I find it I will post it here

2. Jul 4, 2015

### atyy

We don't know whether the wave function is real, and we don't know whether collapse is real. We can think of the wave function and collapse as just a tool to calculate the probabilities of experimental outcomes, which are real.

The only time wave function collapse is absolutely needed is when the measurement is made and the result is recorded by the experimenter.

However, although it is optional, it is still correct to use wave function collapse even if the measurement is made but the result is not recorded by the experimenter.

3. Jul 4, 2015

### HallsofIvy

If there was no measurement how could they tell whether there was collapse or not?

4. Jul 4, 2015

### atyy

I don't know what the OP was actually talking about, but let me explain what I said.

If one makes a measurement and throws away the result, then one could say the wave function collapsed and we threw away the result.

Alternatively, since we made the measurement and threw away the result, we could say we never made the measurement, but the system interacted with the apparatus and decoherence occurred.

The two ways of thinking produce make exactly the same predictions.

So what they may have measured would be the effects of decoherence, which is "apparent collapse".

5. Jul 4, 2015

### VALENCIANA

Atyy: thanks for your explanations
In the experiment with delayed choice and erasure, you get a wave pattern in spite of the fact that the collapse of the wave function had already been measured by the detector Wouldn´t this mean that it is the existence or not of the information what defines the result?

6. Jul 4, 2015

### atyy

People say that, including very distinguished scientists like Feynman. Actually, I have trouble mapping it onto the standard quantum formalism, so I can't really answer your question since I don't understand that point of view. However, if you would like to see how the standard quantum formalism, including collapse, handles the delayed choice with erasure, a good reference is Bram Gaasbeek's Demystifying the Delayed Choice Experiments http://arxiv.org/abs/1007.3977.

7. Jul 4, 2015

### VALENCIANA

Atyy: what you said "We don't know whether the wave function is real, and we don't know whether collapse is real.", reminds me of those who think that waves are just mathematical probabilities, and that the detector forces the most probable one to become a particle.

8. Jul 4, 2015

### atyy

I certainly intended to leave that open as a possibility. In classical probability there is something called Bayesian updating that is very much like the collapse of the wave function. However, no one has found a way to map wave function collapse exactly onto Bayesian updating, so most leave it open as to whether the wave function is real or represents our knowledge similarly to a probability distribution. Of course a probability distribution can also represent real things, and the probability is due to our ignorance. Without hidden variables, quantum mechanics is different from classical probability in that the randomness represented by the quantum wave function cannot be interpreted as due to our ignorance of some reality. Technically, one says that in classical probability the state space is a simplex, whereas a quantum state space is not a simplex.

However, if one allows nonlocal hidden variables (eg. Bohmian Mechanics), we can embed quantum theory into classical probability, then we find that perhaps the wave function is real, and also a probability.

9. Jul 4, 2015

### VALENCIANA

That´s really interesting Atyy

Thanks for all your explanations.

10. Jul 4, 2015

### Derek Potter

That is almost a contradiction...

[MODE="Saturday night"]
That second statement hinges on the assumption that there is a single record of a single result. Apparent collapse works perfectly well. You can either mutter the Forbidden Words " Many Worlds Interpretation " or wax lyrical about proper and improper mixed states. But given the assumption that the wavefunction is ontic, and the assumption that everything has a state even if we can't measure it (a strong form of realism), decoherence obligingly presents the observer with an improper mixed state which appears, to the observer, exactly the same as a proper mixed state. But there is no actual wavefunction collapse in the established sense, since, like Mephistopheles striking a bargain with Faustus, decoherence exacts a high price for the observer's classical experience. The observer is entangled with the system he observes, and not only is he entangled but all possible outcomes are in superposition. Indeed all possible superpositions are equally valid. No wonder most people try to break free from their Faustian bargain in a last-minute repentence. "Bring back wavefunction collapse!" they wail as their quantum souls are dragged into the everlasting darkness of Copenhagen. [/MODE]

11. Jul 4, 2015

### VALENCIANA

Derek: Could you expand a little more on:
"The observer is entangled with the system he observes, and not only is he entangled but all possible outcomes are in superposition." ?

12. Jul 4, 2015

### Derek Potter

What aspect are you not clear about? You may just need to brush up on the basis concepts of QM but that would be rather a lot to deal with in a post!

13. Jul 4, 2015

### VALENCIANA

Do you mean that the observer is entangled with the system in a non local way in the sense that when the observer wants to measure there is a measurement in the way in which when one entangled particle is up, the other is instantly down?

14. Jul 5, 2015

### Staff: Mentor

Observation is actually a form of entanglement.

To understand what's going on you need to become acquainted with technicalities. I suggest the following:
https://www.amazon.com/Quantum-Mechanics-The-Theoretical-Minimum/dp/0465062903

The resolution of the issue you raised is in an assumption of the Copenhagen interpretation that at some point an observation appears here in the classical common-sense world. You find the earliest that occurs and that is where you put the quantum classical cut and treat everything classically from that point on. This means you don't worry about entanglement from that point on and usually you as an observer never becomes involved with it. That you do this is a bit of a blemish on Copenhagen ie you cant say exactly when that cut occurs. It doesn't disprove it or anything like that, but it would be best to not have to do that. A lot of progress has been made - but some issues still remain.

If you would like to pursue it further I suggest the following:
https://www.amazon.com/Where-Does-The-Weirdness-Mechanics/dp/0465067867

Thanks
Bill

15. Jul 5, 2015

### Derek Potter

No, not at all. The entanglement is nothing to do with non-locality. It simply means that the observer's state correlates with that of the particle.

A measurement is an interaction. If a particle in state |left> interacts with an observer, the observer enters a state of |having seen "left">. That is common-sense and pretty uncontroversial. In QM two states that occur together are a single state of the composite system. Mathematically it a product (written ⊗) of the two states (wavefunctions) - that of the particle and that of the observer. So if the possible outcomes are "left" and "right" we can say that the state of the system is
either
|left>(particle)⊗|having seen "left">(observer)
or
|right>(particle)⊗|having seen "right">(observer)
Obviously this is true whether the "particle" is an electron or a cat and if the observer is single photon or a human being surrounded by equipment.

Now it is easy enough to prepare particles in a state that is intermediate between any other two, a superposition. For instance a|left> + b|right>. Such states don't correspond to anything from familiar experience because our senses never see both outcomes at once. There is no such thing (for all practical purposes, FAPP) as an observer state
|having seen "left and right">(observer).

Instead the observer sees one thing or the other. So the question arises, if we prepare a particle in a superposition, what does the observer see? And the answer is perfectly logical, the |left>particle term interacts with the observer to create a |having seen "left"> term and ditto for "right" and a |having seen "right"> term. In fact the resultant state is a simple sum with no other terms - because the system is linear.

Thus the state of the composite system is
a|left>(particle)⊗|having seen "left">(observer) + b|right>(particle)⊗|having seen "right">(observer)

The previous either/or has been replaced by an addition sign!

What this implies is that the observer's state is entangled with that of the particle. You can't split them apart. Common sense, you might think, but remember that the wavefunction speaks of both outcomes, not just one. The big expression is a superposition of the two product states.

Hope that helps and that I have got all the details right. I am sure someone :) will correct me if I haven't.

Last edited: Jul 5, 2015
16. Jul 5, 2015

### Derek Potter

I guess that "someone" has to be me. Sorry, I was jumping ahead of myself. I was just talking about cross products denoted by a simple x, not ⊗. I only realized what I'd done when it was too late to correct it. It's only notation but I should have got it right.

Last edited: Jul 5, 2015
17. Jul 5, 2015

### VALENCIANA

Derek, I get it.
Fantastic explanation !!

18. Jul 7, 2015

### Swamp Thing

(Here $P{_{\lambda_i^A}}$ and $P{_{\lambda_j^B}}$ are the projectors on the eigenspaces corresponding to the eigenvalues ${\lambda_i^A}$ and ${\lambda_j^B}$).

Q: Is the term "product structure" referring to inner products? How does the "implied" statement follow?

19. Jul 7, 2015

### atyy

"Product structure" refers to tensor product. The "implied" statement is easier to see if you use Eq 10 and 11.

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