4

Regarding measurements of an observable in a quantum system. My understanding, from the postulates of quantum mechanics, is that when we measure an observable quantity, the state of the system collapses to an eigenfunction of the linear Hermitian operator which corresponds to the observable: $$\hat{A}|\psi \rangle = y|y \rangle$$ where $y$ is the eigenvalue and $|y \rangle$ is the eigenstate. Then if we project onto the basis of the observable we get the dirac delta function. Let's consider the position operator for example, then:

$$\langle x| \hat{A}|\psi \rangle = y \langle x| y \rangle = y \delta(x-y).$$

From what I understand, in real world measurements the state of the system is not exactly a dirac delta function but rather some wave packet. What is the nature of this wave packet and what determines the shape and corresponding function thereof? Why can't the function be a dirac delta function in real world measurements?

Thanks.

  • Related to, and possibly a duplicate of http://physics.stackexchange.com/questions/218947/how-does-one-compute-the-state-of-a-quantum-system-following-imperfect-measureme – DanielSank Jul 29 '16 at 17:46
  • 1
    @DanielSank His question is way too advanced. I'm looking for a much more basic discussion, I'm just starting to learn QM. –  Jul 29 '16 at 17:51
  • I really think these questions are almost identical. Please note that my question is not more advanced than yours, it's more more specific (particularly in terms of notation) because I know a bit more about the subject so was able to ask a more focused question. I guess you're more asking why we don't get delta functions, while I was asking for an explanation of the final wave function, whatever it may be. – DanielSank Jul 29 '16 at 17:57
  • I think the already-existing answer by Ian is good. However, I would encourage someone who knows about such things to provide an explanation or at least a reference on how to actually calculate the evolution of a quantum state as it is being imperfectly measured. – DanielSank Jul 29 '16 at 18:02
  • @DanielSank :) Oh didn't see it was your question. So you know the answer to my question then? –  Jul 29 '16 at 18:02
  • @DanielSank You are a quantum physicist? We should be best friends, but it would be a bit of a one-sided friendship unfortunately at this stage... –  Jul 29 '16 at 18:06
  • 1
    The collapse postulate is quite controversial. Read Cini quantum measurement without collapse –  Jul 29 '16 at 19:03

1 Answers1

4

A Dirac delta function has a vanishing width. To "collapse" the wavefunction to a delta function, one's measuring apparatus would need to have infinite precision, i.e. zero uncertainty. Since no measuring apparatus is perfect, no measurement can force the wavefunction to have zero uncertainty, i.e. zero width. Therefore, measurement will collapse the wavefunction to a width related in some way to the uncertainty of the measuring apparatus.

The shape of the wavefunction following measurement depends on the nature of the measurement act. This would be impossible to model without knowledge about the measuring apparatus.

Ian
  • 1,336
  • Okay thanks. Would it be a wave packet of some sort which resembles a dirac delta function? What is the nature of this wave packet? –  Jul 29 '16 at 18:04
  • 1
    I think it depends on the nature of the measurement. If the measurement aims to pinpoint the position of the particle, I would expect the "collapsed" wave function to look something like a very slightly broadened delta function. If the measurement simply tests whether the particle is in Appleton or Cherryville, I would expect the collapsed wave function to be a very broad function covering one of those towns that barely resembles a delta function. – Ian Jul 29 '16 at 18:09
  • And just for fun, consider positions A, B, C, and D. If the measurement detects the particle in one of two states (A or B) or (C or D) then I would expect the collapsed wavefunction to be a linear combination of two slightly broadened delta functions centered at either (A and B) or (C and D). – Ian Jul 29 '16 at 18:11
  • @JohnDoe: You would have to add classical statistical errors to the measurement both in classical physics and in quantum mechanics. In quantum mechanics this is usually done with a mathematical object called the "density matrix", which describes both the mixed states of the quantum mechanical system as well as any statistical uncertainty that we may have on top of that. There is no direct analog in classical mechanics because of the lack of linearity of classical systems, but in QM the density matrix is, as far as I know, the most complete description of our knowledge about a system's state. – CuriousOne Jul 29 '16 at 18:41
  • @CuriousOne Good point, but for the sake of a simplified discussion would you agree that we can consider the issue of measurement without considering classical statistical errors? I think the OP's question is still an interesting one even if we only consider pure states. – Ian Jul 29 '16 at 18:56
  • @Ian: The OP is asking for non-ideal measurements, and a real measurement device will always put the system into a (slightly) mixed state. I think he can handle the truth about that. We could withhold it, but he would inevitably be back in ten minutes (or a couple weeks) with a follow up question about that... :-) – CuriousOne Jul 29 '16 at 19:21
  • @CuriousOne Classical measurement error is not the key to OP's question, I think. I think the essence of the question comes from e.g. the fact that you can't possibly measure the momentum of a particle to infinite precision because, for example, you don't have an infinitely long laboratory. The measurement operators we have access too don't allow delta functions to come out of the measurement, but it's not because of noise. Well ok, the practical limits may come from noise, but I mean there's another limit there too, I think. – DanielSank Jul 30 '16 at 07:27
  • @DanielSank: Just because it's not in the OP's question doesn't mean that one is not required to think about it if one wants to understand what is really going on. The measurement operators allow distributions just fine, you just have to massage the functional analysis a little bit. What doesn't allow delta-functions is the fact that a measurement device can only expend a finite amount of mass-energy on a measurement. It's relativity at work, not some fine point of mathematics. – CuriousOne Jul 30 '16 at 08:03
  • @CuriousOne I guarantee you that you would find out you can't make delta functions via measurement even in a non-relativistic universe. Relativity has nothing to do with this question. – DanielSank Jul 30 '16 at 09:02
  • @DanielSank: There is no non-relativistic universe outside of the approximations of quantum mechanics who are stuck at the level of Schroedinger in 1926. Is that approximation self-consistent? No. It was never meant to be and the physicists of the time like Dirac moved beyond it at nearly the speed of light. I could give a speech about this, but why bother, Nima Arkani-Hamed has done it plenty and it's all on YouTube. I would invite you to listen to his ideas about what happens with quantum measurements in the really high energy limit. :-) – CuriousOne Jul 30 '16 at 09:09
  • @CuriousOne I'm not familiar with the relativity argument. Could you expand on this in an answer? – Ian Jul 30 '16 at 11:53
  • @Ian: See e.g. https://www.youtube.com/watch?v=87lz3U4CkPU around 12 minutes into the lecture. – CuriousOne Jul 30 '16 at 12:28
  • @CuriousOne Is the fact that we can't get a delta function just a limit of the measuring device? As I understand the delta function is not normalizable, but taking the real world uncertainty of the measurement, the resultant state after measurement (which is something possibly similar but not a dirac delta function) becomes normalizable? So if there was a perfect measuring device we could get a state which couldn't exist because it is not normalizable, is this right? :/ –  Jul 31 '16 at 19:18
  • @John Doe The problem here is that you are considering a perfect measuring device rather than the limit of a perfect measuring device. In the limit as the device becomes perfect, the normalizable wave function approaches a delta function. But it stays normalizable as you take the limit. If you insist on considering an unphysical perfect measuring device rather than its limit, than sure we can have nonnormalizable wave functions since we are considering an unphysical scenario anyway. – Ian Jul 31 '16 at 19:33
  • @Ian Yeah I understand the thinking should be in terms of a limit because such a device could not exist. It is just interesting that the theory relies on the lack of precision of measurement to be a complete description of the physical reality. Do you find it interesting? –  Jul 31 '16 at 19:40
  • I do find that interesting and I don't have a good response. Maybe @CuriousOne has something to say about that. – Ian Jul 31 '16 at 19:49
  • @Ian: At the end of the day you are over-thinking this issue. If you want to compress matter into a really tiny volume on Earth, you need an accelerator. The ultimate machine we have, right now, is LHC. That's it. You can get 14TeV in the center of mass system and you can calculate a length scale that is associated with that in quantum field theory. Everything else is merely a bunch of mathematical artifacts. If you want to know how the delta function really works, get a book about functional analysis and it will tell you everything you want to know, but nothing about physics. – CuriousOne Jul 31 '16 at 22:16
  • @CuriousOne This is a great argument and it explains why we can't have a perfect measuring device. But John Doe was asking us to consider a perfect measuring device. This is a question/answer about the theory, not about the real world. So my response considers the mathematical limit as the measuring device approaches perfection, i.e. as we can squeeze the particle into a vanishing length scale with a machine infinitely more powerful than the LHC. Whether or not QM applies to the real world in this limit, I was addressing a question about QM as a self-contained theory. – Ian Aug 01 '16 at 01:57
  • @JohnDoe Following CuriousOne's comment, when you say "relies on...to be a complete description of physical reality" the logic isn't quite correct since no one claims that QM is physically realistic even in the limit of vanishing uncertainty, let alone for uncertainty actually equal to zero. There is an upper bound on the energies where QM applies. Therefore, everywhere where QM applies, we have imprecise measurement anyway. Nobody has to enforce an axiom of imprecise measurement for QM to be physically realistic, since within the bounds of QM nature already gives us imprecise measurement. – Ian Aug 01 '16 at 02:06
  • @Ian: Then it's a question about mathematics and the answer can be found in books about functional analysis, which is an approx. 100 year old field by now. It's kind of interesting and I sat trough three mathematics major courses about it when I was in university, but as a mathematical discipline it is a "physics free zone". – CuriousOne Aug 01 '16 at 21:09