# Questions about measurement of a particle in superposition

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1. Jan 26, 2016

### entropy1

I have a very basic question about superposition.

Suppose two experimenters do a measurement on a particle in superposition. Does de measurement apparatus become in superposition? Do the experimenters become in superposition?

What is superposition?

2. Jan 26, 2016

### Staff: Mentor

Answering in reverse order: superposition is the mathematical fact that the state of a quantum particle can always be written as the sum of other states. Note that "always" - every quantum state is a superposition of other states, in the same way that every vector can always be written as the sum of other vectors. In fact, the mathematical formalism of quantum mechanics starts with the axiom that the possible states of a quantum particle form a vector space.

I'm not sure why you are specifying two experimenters instead of one. Are you thinking that one experimenter performs one measurement and then the other experimenter performs the same or a different measurement? In any case, what happens when a measurement is performed depends on your choice of interpretation; quantum mechanics says only that:
1) The result of the measurement will be an eigenvalue of the operator corresponding to the property we measure.
2) The subsequent evolution of the system will be as if the measurement had changed the particle's state to the corresponding eigenfunction of that operator.
These are sufficient to answer what happens when you have two observers making two measurements.

3. Jan 26, 2016

### entropy1

Maybe that was a needless complication of the questions. I was anticipating on asking a follow-up question, being: if the experimenters (of the same experiment!) are individually entangled with the measurement apparatus and in superposition as a result of that, how come they always agree about the outcome of the experiment. (I now seems a pretty odd question to me... )

4. Jan 26, 2016

### entropy1

I will be using some strange layman language now. Suppose we consider the lightcone propagating through the measurement apparatus. Consider cone C1 after time t=1 of the measurement and cone C2 after time t=2. C1 is a subset of C2. Now if C2 has evolved as if the particle has taken on it's eigenstate, C1 has too. We can take a subset of C1 and it will have evolved as is the particle has that eigenstate too, et cetera. So every part of the apparatus inside the lightcone of the measurement will behave as if the particle in fact has taken on that eigenfunction. So is the measurement yielding a completely deterministic result hen? If I am talking rubbish don't hesitate to tell me so.

5. Jan 26, 2016

### Staff: Mentor

The entire concept is meaningless because there's no such thing as "individual" entanglement. If you are choosing an interpretation that takes you down this path at all, then there is just one quantum system consisting (in this case) of a particle, two sets of experimental apparatuses, and two experimenters. When we say that the particle and the instruments are entangled, we're just saying that the possible states for that system are "particle is spin-up, observer one sees his apparatus reading spin-up, observer two sees his apparatus reading spin-up" and "particle is spin-down, observer one sees his apparatus reading spin-down, observer two see his apparatus reading spin-down" so the results are always properly correlated. (In practice, decoherence will very rapidly kill any macroscopic superposition of these two states).

6. Jan 26, 2016

### Staff: Mentor

You have the relationship backwards - C2 is a subset of C1. You should be thinking of the measurement at t=1 as the preparation procedure for the measurement at t=2.

With that said, there are several ways that indeterminacy enters into the picture.
1) The result of the measurement at T1 may, depending on how the particle was prepared, yield any of the possible eigenvalues of whatever we're measuring and leaving the particle in any of the possible corresponding eigenstates. That's the preparation procedure for the second measurement.
2) If whatever we're measuring at time t=1 does not commute with the Hamiltonian, then the particle will not stay in the eigenstate it entered at t=1, so by time t=2 even a second measurement of the same property will not deterministically produce the same result. (Any interaction with random air molecules or light that happens to be bouncing around the lab falls under this case).
3) If the observable that we're measuring at t=2 does not commute with the observable that we measured at t=1, then the result of the second measurement is necessarily not deterministic; the eigenstates of one are superpositions of the eigenstates of the other.

7. Jan 26, 2016

### entropy1

So decoherence (in the apparatus/environment) is (equivalent to) a series measurements that do not commute?

8. Jan 26, 2016

### bhobba

No. And how you arrive at such has me beat. As I said replying to another post of yours you need to put some thought into what you say.

Thanks
Bill

Last edited: Jan 26, 2016
9. Jan 27, 2016

### entropy1

I came to it as result from reading this from Nugatory:
However, he put an 'if' in front of it. I don't understand under which circumstances that 'if' is effective.

In the future, I will try to put some more thought in my questions.

Last edited: Jan 27, 2016
10. Jan 27, 2016

### bhobba

I simply cant follow the logic of that conclusion. Your reasoning totally escaprs me - if you can post more detail that would really help.

Thanks
Bill

11. Jan 27, 2016

### bhobba

Thanks - much appreciated. It wasn't meant as a put down - its just I was unable to follow your reasoning in this and another thread. Giving the chain of that reasoning not only helps others following it, but organises your thoughts a well.

Thanks
Bill

Last edited: Jan 27, 2016
12. Jan 27, 2016

### bhobba

Actually I am not sure I agree with that. I am thinking of wave-packet spreading as the result of a position measurement. It starts out in an eigenstate of position but later isn't. In principle one could devise an observation to detect that state so give a dead cert as the outcome.

However knowing Nugatory I suspect I am misinterpreting what he is saying and may feel a total fool when it's pointed out

Thanks
Bill

13. Jan 27, 2016

### Staff: Mentor

That's an example of #2: The position operator does not commute with the Hamiltonian, so if I were to measure the position at t=0 and get a result corresponding to a particular eigenstate of the position operator the particle will not stay in that eigenstate. Instead the state will spread out, evolving over time into a superposition of position eigenstates, so that when I make another position measurement at t=2 the outcome will not be certain.

(Because of the continuous spectrum there are some serious mathematical difficulties with this particular example - let's ignore them for now).

14. Jan 27, 2016

### Staff: Mentor

It's your thought experiment and you're the one who is asking about making two measurements. It makes a difference whether the two measurements are of commuting observables or not, and whether the first observable we're measuring commutes with the Hamiltonian or not.

No, and they are completely different things.

15. Jan 27, 2016

### entropy1

Ok, there seem to be a lot of misunderstandings going on here. This thread may be closed.

If anything, I now know I don't understand anything at all anymore. Maybe I'll just leave it.

Last edited: Jan 27, 2016