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People say operator of an observable helps in measuring for an observable. We also know that measuring leads to collapse of wave function. But operator on wave function gives a number times same wave function (which of course is not a collapsed wave function!). All intuitions I made about operator, wave function, measures, collapse are all seeming to be inconsistent. If operator doesn't collapse a wave function then what it is for. Is it just for calculating expectation value of observable. What in physical sense it is?

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  • $\begingroup$ Welcome on Physics SE :) Your post contains a few misunderstandings about the way the measurement process works in the collapsing wave function picture - I'd think it useful for you to read that up again ;) $\endgroup$ – Sanya Oct 9 '16 at 16:21
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There are a few different points/distinctions that need to be made here.
1. The expectancy value when measuring an observable to which the operator $A$ can be assigned is $\langle A \rangle = \langle \Phi | A | \Phi \rangle$ for a system described by a wavefunction $|\Phi\rangle$.
2. If the value $a$ is measured, the collapse of the wavefunction means that we project $ |\Phi\rangle$ onto the eigenspace of the eigenvalue $a$ of the operator, i.e., the wavefunction is changed.
3. $A | \Phi \rangle = a | \Phi \rangle$ for $a \in \mathbb{C}$ holds only if $| \Phi \rangle$ is an eigenfunction of the operator $A$. In general, $A | \Phi \rangle$ does not need to be proportional to $| \Phi \rangle$ but can be a different wavefunction.
4. An operator $A$ determines the possible values of its measurement variable by its eigenvalues and determines the possible wavefunctions after it has been measured by its eigenstates/eigenspaces.
5. Applying an operator to a wavefunction does not describe the measurement process.

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  • $\begingroup$ Thank u so much. Its all clear now in those points. Just wanted to make sure that does it mean if operator  acts on a general wave function then resultant would be a wave function which would be a linear combination of several eigenfunctions of  with associated eigenvalues? Also is all this business of operator just a mathematical trick with no physically observable analogue? $\endgroup$ – Kumar Pranshu Oct 9 '16 at 17:33
  • $\begingroup$ If I am not completely mistaken, the spectral theorem tells us that the operator is basically a sum/integral over the projections on the eigenspaces multiplied by their eigenvalues, so I think the answer to your first question is yes. To your second question - physically observable is the spectrum of the operator as the possible measurement outcomes, so I'd say it is more than a mathematical trick. $\endgroup$ – Sanya Oct 9 '16 at 20:29
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You are right in pointing out that operating with an operator on a wavefunction gives a number times the wavefunction (assuming that it is indeed an eigenfunction of the said operator). Measuring the observable collapses the wavefunction. How a wavefunction collapses is still an open question in Quantum Physics. The job of the operator is to find out the possible eigenvalues of the wavefunction. This paper might help you with the progress that has been done in solving that open problem.

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  • $\begingroup$ Just a side note here. Acting with an operator on a wavefunction does not constitute a measurement right? Also tell me if the question says 'what is the value of the measurement of an operator say A on a wavefunction $\psi$, then i find its expectation value; then does finding the expectation value mean measurement which in turn means collapse of the wavefunction? Or is measurement just a vague physical process which messes with the instruments and find the expectation value is a separate mathematical calculation on paper that has nothing to do with measurement?? $\endgroup$ – Prasad Mani Oct 9 '16 at 16:02
  • $\begingroup$ That s a doubt I had as well. What I believe by 'measurement' they precisely mean finding out some definite quantitive parameters which potentially define state of a system and hence constitute 'psi'. Various systems with different 'psi' might turn out to be having same expectation value. I'm not sure though. $\endgroup$ – Kumar Pranshu Oct 9 '16 at 17:22

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