Measurement immediately after wave function collapse

[Royston]

I did a search, but could not find anything on this specific problem, so apologies if this has already been answered.

 

I just received my marked QM assignment (got a distinction, thought I'd slip that in ). In one of the essay style questions, I deliberately added a bugbear to prompt tutor feedback. However, it was not particularly satisfactory, and no different to what I have already read.

 

Supposing I made a measurement of position on a system, that caused collapse of the wave function to a given eigenstate. If I make a measurement immediately after the collapse, (i.e there is no time evolution due to the Schrodinger equation) I should obtain the same value, say x₀.

 

However, the very act of measurement, perturbs the system. Regardless of whether the Hamiltonian commutes with the operator or not, I can't see how the act of measurement does not throw up a spread of values, due to perturbation.

 

I'm sure this is a classic question, and I was wondering if this fell into the 'interpretation of measurement' camp. I've spent the last hour trying to come up with a mathematical treatment to the problem (this may even come up later on in the course) but it seems more of a conceptual one, perhaps.

 

I would rather a mathematical answer over a conceptual one, because I find that a lot easier to see where the problem lies, rather than an interpretational approach.

 

EDIT: Dirac notation is fine BTW

Edited April 13, 2012 by Royston

[questionposter]

I don't think it's possible to measure a particle "Immediately" after collapse because that would require light or forces to instantaneously travel to your eyes then instantaneously have the electrical signal sent to your brain. In order to happen you would also have to have the time the information about the particle's position a photon would carry to have happened infinitely close to the moment of a particle's collapse, which if space and time is quantized anyway, that definitely can't happen.

 

If your talking about something like entanglement, then the measurement of one should instantaneously effect the other.

 

Otherwise though, even if you do somehow instantaneously measure it after collapse, I think that the uncertainty principal and a few other equations say that you can't base the next position of a particle that you measure off of another position either. But, I guess it could be wrong, maybe I'm going outside the parameters too much.

Edited April 14, 2012 by questionposter

[swansont]

However, the very act of measurement, perturbs the system. Regardless of whether the Hamiltonian commutes with the operator or not, I can't see how the act of measurement does not throw up a spread of values, due to perturbation.

 

Since you don't know what measurement is going to be done, this is not part of the theory. AFAIK it's added on, if necessary. Sometimes the perturbation is expected and you do not need to account for it because you are doing a destructive measurement, e.g. a photon hitting a target. You get the position information, but there is no photon left.

 

It's an issue if you want to do a subsequent measurement.

[Royston]

I appreciate the response questionposter, but (perhaps impossible to not sound offensive) I really need an answer from somebody who know's what they're talking about.

 

There are several *flaws in what you've stated, but addressing them, would be veering off topic. Again, no offense intended.

 

 

*It is more than likely there is a flaw with what I've stated, for instance the commutation of the operator with the Hamiltonian isn't really relevant here...albeit I thought I had deleted that part, when I posted.

 

Since you don't know what measurement is going to be done, this is not part of the theory. AFAIK it's added on, if necessary. Sometimes the perturbation is expected and you do not need to account for it because you are doing a destructive measurement, e.g. a photon hitting a target. You get the position information, but there is no photon left.

 

It's an issue if you want to do a subsequent measurement.

 

Thanks, swansont.

 

The last two points have made it a lot clearer. I got caught up with information in a system can only be obtained with making an interaction with the system...however if there is no time evolution, then the same information can be extracted due to a measurement made immediately before. Extracting information is still a measurement, albeit it seems slightly abstract in this instance.

Edited April 15, 2012 by Royston

[questionposter]

I appreciate the response questionposter, but (perhaps impossible to not sound offensive) I really need an answer from somebody who know's what they're talking about.

 

There are several *flaws in what you've stated, but addressing them, would be veering off topic. Again, no offense intended.

 

 

*It is more than likely there is a flaw with what I've stated, for instance the commutation of the operator with the Hamiltonian isn't really relevant here...albeit I thought I had deleted that part, when I posted.

 

 

 

Thanks, swansont.

 

The last two points have made it a lot clearer. I got caught up with information in a system can only be obtained with making an interaction with the system...however if there is no time evolution, then the same information can be extracted due to a measurement made immediately before. Extracting information is still a measurement, albeit it seems slightly abstract in this instance.

 

Wait, maybe I'm not understanding something, but is this actually suggesting I can instantaneously make two separate measurements of the same thing and count it as the same result because the particle wouldn't change or "evolve" over time between the measurements because they are instantaneously consecutive? Even if there is no time evolution, isn't there still uncertainty with measurements?

Wouldn't that imply that you can in fact predict future position because there would be only one possible result for one specific time if two different yet simultaneous measurements yield the same result?

 

Also, just a random question I thought of:

Does an infinite summation of possible energy levels = an infinitely localized probability wave?

Edited April 15, 2012 by questionposter