# Measurement collapse as a perturbation?

1. Oct 26, 2009

### Gerenuk

Is it possible to describe the measurement process in QM simply as a perturbation to the system (either a small one or a very large one) so that effectively the eigenstates of the Hamiltonian evolve into eigenstates of an additional measuring operator?

So at t=0 the Hamiltonian H turns into H+M and the quantum state becomes one of the eigenstates of M (because M is much stronger?). The probability for the different eigenstates of M depends on the previous state.

2. Oct 26, 2009

### Bob_for_short

You imply that a measurement in QM is one-time event whereas it is an ensemble of measurements. Only an ensemble gives you the state picture (an interference pattern, for example), and it does not collapse anything.

Last edited: Oct 26, 2009
3. Oct 26, 2009

### Gerenuk

Not sure what you mean.

In the double slit experiment, when each electron is sent individually, every electron has to go through both slits at once to be able to ever give an interference pattern, when more electrons follow. However, each electron produces a single spot on the screen. One can stop after any number of events.

So I am asking if it works to assume that the screen is a very large perturbation to the electron wave function, such that for each single electron it will localize more and more at one particular point on the screen.

4. Oct 26, 2009

### Bob_for_short

If the number of events is small, such an experiment is good for nothing - it says nothing about the electron wave fucntion.
An electron-screen interaction is a strong perturbation for both - an individual electron and the screen material. But it is not a collapse of the electron wave function describing the whole pattern. Collecting points is how one gets infromation about the wave function, it is "building" the wave function piece by piece rather than its collapse.

5. Oct 26, 2009

### Gerenuk

Hmm, you might be refering to the "ensemble interpretation" that I've just read.

I have a single electron confined in a very small box infront of the screen. The electron will be released, go through a double-slit, hit the screen and then photons will continue from that spot to the eye.

Now if I ask you what you can say about my experiment, then you will take a wavefunction, do the calculation, include a wavefuntion collapse at the screen and then model a light ray going to the eye. Right?

You are not going to calculate a light interference pattern from all spots that can be hit by the electron. So you are including a collapse.

In any case, the statements you made actually don't refer to the question.
I'm not confident about the result of that perturbation problem. Will my perturbation calculation give the result I propose or not?

6. Oct 26, 2009

### Bob_for_short

No. Your experiment is like taking a photo with one-pixel camera.

Which wave function will collapse? This one or that one? How to distinguish? You see, one point is meaningless. You cannot say: "I made a double-slit experiment" with presenting one point on a screen.

Think of probability. It describes ensembles of measurements. One event is not a collapse of the probability. It's an element of the ensemble.

7. Oct 26, 2009

### Gerenuk

I asked a question in the previous post, that refers to you statement. I do a one electron experiment. Tell me the physics that you will use to describe what's going on. Of course you cannot say "Sorry, if you don't fire a few more electrons I cannot give you any description of what has been going on up to now." I doesn't matter if one can call that a "full double-slit experiment" at that point. Call it "single electron goes through two slits" if you wish.

In any case... I repeat... this discussion about interpretation of concepts has nothing to do with my initially posted concern and I find it bad practice to "question the question" when there is some way of answering it.

I hope some knows a good answer.

8. Oct 26, 2009

### Bob_for_short

Well, you are right. You slap the wave function on the hat and it collapses. Do you like this answer more?

9. Oct 26, 2009

### Gerenuk

Just read my first post again if you want an answer to that.

10. Oct 27, 2009

### zonde

In case if it's so these perturbation are quite definite.
For example if you place two polarizers in path of light ray and polarization angle is the same for both polarizers it seems like second polarizer has no effect on wave function.

And to clarify question can you tell what do you consider as measurement process? From my example above is interaction with polarizers a measurement process or only interaction with detector or both?

11. Oct 27, 2009

### Gerenuk

Polarizers are not making wavefunctions converge so much. Only some processes like screens or the eye interact so strongly that "one spot soaks up all the probability". You understand my thought?

I haven't studied the philosophy of the interpretations yet.

In my personal view I want to examine if the interaction with polarizers can be both.
Everything is interaction anyway. And those things that cause the wavefunction to (roughly) localize to a certain subset of states could be called measurement or observation. With that idea your brain would end with with large certainly thinking that a particle is at one spot, but a small chance that it is in another

12. Oct 27, 2009

### Bob_for_short

Saying "one spot soaks up all the probability" is wrong. It would be the case if you obtained the same spot in multiple measurements. Otherwise one point is an element of unknown yet ensemble described with ψ.

Do you understand that a wave exists in a volume and needs more than one point to picture it?

13. Oct 27, 2009

### Gerenuk

Sorry, Bob. You are not understanding any of the ideas here. Or I don't have any idea what you are talking about. And I do not want to write down an explicit mathematical example to explain.

Either way... I hope zonde can clarify.

14. Oct 27, 2009

### Bob_for_short

Oh, I understand you quite well. You think: "Before the screen (measurement) the electron has a wave function spread out in a space and after measurement is collapses in one point. So it should be a strong attractive potential that "sucks" the entire wave function into this point", right?

15. Oct 27, 2009

### zonde

I think I understand.
Let's check. Wave function is used to describe ensemble but it seems that it is not unusual that people perceive it as adequate description of single event (particle). As I understand that is exactly how you treat wave function.
I am not sure that I agree with that. I would treat that a bit differently.
Wave function describes collection of different states that particle can take i.e. phase space. However interaction can happen only in certain states (eigenstates) of particle. And so interaction process alters each particle so that it shifts to one of (two) eigenstates and interaction can take place. Moreover interaction is different depending from eigenstate the particle has taken.

But here I am again back at question I asked about what you consider measurement process.
When light is interacting with polarizer and after that with detector there are clearly two pieces of equipment and two interaction processes.
Speaking about interpretations I do not have clear understanding if there are interpretations that treat those two things I mentioned separately. As far as I understand there isn't.

16. Oct 27, 2009

### Gerenuk

Almost - the causality is wrong here. I think there is a strong potential for sure, that will suck in most of the wavefunction as time progress to one of the points on the screen. Which point on the screen exactly will depend on the details of the initial state of the wavefunction. One would find this just by solving the Schrödinger equation. And that the question in the whole thread:

Given
$$\hat{H}\psi=-\mathrm{i}\psi_t$$
$$\psi(t<0)=\psi_i$$
$$\hat{H}=\begin{cases} \hat{H}_0 & \text{for }t<0\\ \hat{H}_0+\hat{M} & \text{for }t\geq 0 \end{cases}$$
Given that $\hat{M}\gg \hat{H}$, is it true that $\psi(\infty)$ will be almost an eigentstate of M written as $|\psi^{(M)}_n>$ with a probability $<\psi^{(M)}_n|\psi_i>$ (probably one can write that somehow with density matrices)?

I really do not know how much more precise I can be and how much I can repeat what the actual question is so that you refer to the actual question.
You made some philosophical objections which might be interesting later, but off-topic here.

Last edited: Oct 27, 2009
17. Oct 27, 2009

### Gerenuk

Maybe you mean that I imagine that there is only a delocalized wavefunction and never an actual particle at a single spot. Everything evolves with these wavefunctions and also your brain might be in a state of yes and no together.
Yes, that's my proposal or question.

I think I know what you mean. Everyone can prefer his own interpretation and I cannot argue with that
Here I only wanted to find out, if my other picture is at least mathematically correct. Only thereafter one can discuss which picture is more appealing.

In that picture there is no such thing as a distinguished measurement process.
If you wish you can call a process, that turns the wavefunction into an eigenstate of an operator of your desire, measurement process. That's subjective tough...

18. Oct 27, 2009

### zonde

In that case I have a question. Do you take that mathematical form of wavefunction of ensemble is equal with mathematical form of wavefunction of single particle?
Because physically they are clearly two very different things. One lives in Hilbert space other lives in real space.

19. Oct 27, 2009

### Gerenuk

I'm not absolutely confident with the in-depth mathematical details. I hope the following answers your question.
I want to make a mathematical model for a one electron process which will give probabilistic results. To get the answers for what will happen with many electrons, I would repeat my model. So I do not consider the many electron experiment as a whole.
About the Hilbert space thing... I'm not sure of the abstract meaning you mention, but I guess my answer would be I always stay in Hilbert space?!
Ensemble means in an abstract way that you are basically repeating the experiment with equal initial conditions? Well, in that case I would repeat the experiment with equal initial conditions. Maybe its is possible to inject this this mathematics into the electron picture. I haven't thought about that. I suppose density matrices would work as well.
But whatever that really means, I have a system to find answers to physics questions, right? As long as I can predict the final distribution it's all fine? And if the suggestion works, I'd find it much more intuitive than the other interpretation I've read - at least the ones I understand. I have no idea what quantum foam means

Last edited: Oct 27, 2009
20. Oct 27, 2009

### Bob_for_short

If you have a sudden change of the Hamiltonian, the state does not change, it remains the same just after the sudden perturbation (see the sudden perturbation theory).
What changes is the basis. You can expand the original wave function in a spectral sum of the new Hamiltonian and find "projections" - amplitudes of finding the particle in the corresponding new eigenstates. In particular, in a double-slit experiment you observe the particle position eigenstates. As soon as the original wave function is not a position eigenstate but a superposition of such, you observe different spot density (distribution) on the screen rather than a repeating spot.

Last edited: Oct 27, 2009