Measurement in QM

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  • #26
PeterDonis
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what happens if we don't see it there?
Then you have the two other peaks left. Since you started with a combination of three delta functions, it doesn't make sense to talk about normalization, since delta functions are not normalizable. So the best you can do is to say that the measurement eliminates one of the three delta functions, leaving the other two.
 
  • #27
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Then you have the two other peaks left. Since you started with a combination of three delta functions, it doesn't make sense to talk about normalization, since delta functions are not normalizable. So the best you can do is to say that the measurement eliminates one of the three delta functions, leaving the other two.
Thank you for your reply. This is true if we had 3 delta functions. But I was asking for a more general case. In my example I was talking about a particle in a box with 3 regions (so a sin function). If we don't see the particle at the top of one of the regions, we can't assume that the whole region goes to 0. So I was wondering how does the wavefunction changes after the measurement.
 
  • #28
PeterDonis
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If we don't see the particle at the top of one of the regions, we can't assume that the whole region goes to 0.
So what can you assume? How much is excluded by an actual position measurement? (Hint: the answer is not "just one point"; such a measurement is impossible.)
 
  • #29
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So what can you assume? How much is excluded by an actual position measurement? (Hint: the answer is not "just one point"; such a measurement is impossible.)
Well yeah, it is not just one point it is a region dx. So ##\psi## in that region becomes 0. So what happens to the rest?
 
  • #30
PeterDonis
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So what happens to the rest?
You just renormalize to a total measure that now excludes the region ##dx## that was ruled out by the measurement.
 
  • #31
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You just renormalize to a total measure that now excludes the region ##dx## that was ruled out by the measurement.
But the shape of the wavefunction will not be the same as before, minus the region dx. So I can't just multiply everything by a constant.
 
  • #32
Orodruin
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But the shape of the wavefunction will not be the same as before, minus the region dx. So I can't just multiply everything by a constant.
Why would you think so? Did you read post #7?
 
  • #33
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Why would you think so?
Because the wavefunctions still needs to be continuous. If we just remove the region dx then the right limit as ##x \to x_0-dx## will be the value before the measurement while the left limit will be 0 (as all the region between ##x_0-dx## and ##x_0+dx## is 0). So if we just remove that region and multiply everything else by a normalization constant, we will have 2 discontinuities at ##x_0 \pm dx##, right?
 
  • #34
Orodruin
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Because the wavefunctions still needs to be continuous
Says who? I suggest you start working out the case of a finite dimensional Hilbert space instead of looking at the case of a wavefunction in position representation. It is conceptually simpler.
 
  • #35
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You say "obviously the wavefunction changes" but what's so obvious about it? I'm not sure I follow your logic... how can this part be the obvious part, but then you don't understand how it changes??

Instead of using a cop out by saying obviously x occurs, I'd ask you to show us WHY you think it should change.
 
  • #36
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You say "obviously the wavefunction changes" but what's so obvious about it? I'm not sure I follow your logic... how can this part be the obvious part, but then you don't understand how it changes??

Instead of using a cop out by saying obviously x occurs, I'd ask you to show us WHY you think it should change.
I am not sure what you mean. Before the measurement the wavefunction had a non-zero value for ##x_0##. After the measurement, if the particle is not found at ##x_0##, the value of the function at that point becomes 0. So it is pretty obvious that the wavefunction is not the same as before, so it changes. Isn't it?
 
  • #37
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Says who? I suggest you start working out the case of a finite dimensional Hilbert space instead of looking at the case of a wavefunction in position representation. It is conceptually simpler.
Wait I am confused. You mean the wavefunction doesn't always need to be continuous? I mean I am quite new to QM, but this is one of the thing our professors kept repeating during the whole course, that the wavefunction needs to be continuous (even if the potential is not).
 
  • #38
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I am not sure what you mean. Before the measurement the wavefunction had a non-zero value for ##x_0##. After the measurement, if the particle is not found at ##x_0##, the value of the function at that point becomes 0. So it is pretty obvious that the wavefunction is not the same as before, so it changes. Isn't it?
More likely is that if you start with a Gaussain WF you'll end with another Gaussian with different center and smaller variance.
 
  • #39
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More likely is that if you start with a Gaussain WF you'll end with another Gaussian with different center and smaller variance.
Why is that? If you remove a central part of the gaussian, don't you get at least 2 other gaussians on the left and right of the initial one?
 
  • #40
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Why is that? If you remove a central part of the gaussian, don't you get at least 2 other gaussians on the left and right of the initial one?
Subtracting something from a WF does not make sense. Evolving to a different shape does not mean that points not measured cannot be measured in future.

It is easy to find a unitary transformation that 'morphs' a Gaussian.
 
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  • #41
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Subtracting something from a WF does not make sense. Evolving to a different shape does not mean that points not measured cannot be measured in future.
I know. The whole point of this post was to make me understand how can I describe mathematically what happens to the wavefunction, after we measure the position of a particle, without finding the particle at that position (how the rest of the wavefunction changes). When I said subtracting I just meant that it gets equal to 0, as we know for sure it is not there. So my question (to which I still didn't get an answer to have my problem clarified) is to what does the wavefunction evolve?
 
  • #42
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I know. The whole point of this post was to make me understand how can I describe mathematically what happens to the wavefunction, after we measure the position of a particle, without finding the particle at that position (how the rest of the wavefunction changes). When I said subtracting I just meant that it gets equal to 0, as we know for sure it is not there. So my question (to which I still didn't get an answer to have my problem clarified) is to what does the wavefunction evolve?
That is your false assumption. You know there is a very low probability of it being there because the far away locations are now in the tail of a new Gaussian.
 
  • #43
atyy
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As far as I understand, if you measure the position of a particle, the wavefunction of that particle changes into a delta function, and thus the particle gets localized. Now, let's say we have a particle in a box in a state with 3 main peaks (##\psi_3##). If we look at the main peak let's say (so the center of the box) and see the particle, the wave function collapsed at that point. But what happens if we don't see it there? Obviously, the wavefunction changes, as we know for sure that the probability of the particle being at that point is 0 now, but how is the wave function changed? Does it turn into a delta function at a random point, different from the one where we measured, or what?
An analogous situation is discussed in https://arxiv.org/abs/1406.5535, Quantum Measurements: a modern view for quantum optics experimentalists, Aephraim M. Steinberg (p5):

On the other hand, if you wait one half-life and observe no photons – what state should you conclude the atom is in? Since it hasn’t decayed, you might suspect that it is still in |+. This can clearly be seen to be incorrect by going to extremes: if you waited for 100 years and still observed no photon, you would presumably reason that any excited atom would have decayed by now, so the atom must have been in the ground state all along. Even though no photon was emitted, your description of the quantum state should change from |+〉to |g. Observing nothing is also an observation.

...

This example can also be seen as a simple case of a more general observation: not all measurements are projections. By observing no photon to be emitted for some length of time, we have acquired some information about the state, but a limited amount. This modifies the state vector, but it does not project it onto one of two alternative eigenstates; in fact, the final states which result from the two possible measurement outcomes are not even orthogonal to one another.
 

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