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?


  • $\begingroup$ Related to, and possibly a duplicate of physics.stackexchange.com/questions/218947/… $\endgroup$
    – DanielSank
    Jul 29, 2016 at 17:46
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    $\begingroup$ @DanielSank His question is way too advanced. I'm looking for a much more basic discussion, I'm just starting to learn QM. $\endgroup$
    – user100411
    Jul 29, 2016 at 17:51
  • $\begingroup$ 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. $\endgroup$
    – DanielSank
    Jul 29, 2016 at 17:57
  • $\begingroup$ 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. $\endgroup$
    – DanielSank
    Jul 29, 2016 at 18:02
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    $\begingroup$ The collapse postulate is quite controversial. Read Cini quantum measurement without collapse $\endgroup$
    – user29978
    Jul 29, 2016 at 19:03

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – user100411
    Jul 29, 2016 at 18:04
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    $\begingroup$ 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. $\endgroup$
    – Ian
    Jul 29, 2016 at 18:09
  • $\begingroup$ 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). $\endgroup$
    – Ian
    Jul 29, 2016 at 18:11
  • $\begingroup$ @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. $\endgroup$
    – CuriousOne
    Jul 29, 2016 at 18:41
  • $\begingroup$ @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. $\endgroup$
    – Ian
    Jul 29, 2016 at 18:56

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