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The question is rather simple. We all hear all the time that an observable of a quantum mechanical system (for some observable that is given by a some self-adjoint operator, let's call it $H$) is given by the eigenvalues of this operator $H$. One also often hears (reads) that the set of all the possible values you can measure (in your experiment for example) is given by the spectrum of an operator $H$. I have a conundrum here. which of the two is it? When we measure something in quantum mechanics are we measuring the eigenvalues or the spectrum?

It is important to note that these two things are $\textbf{NOT}$, in general, the same. For finite-dimensional operators sure, they happen to correspond, but for non-finite operators, they are categorically different. One guarantees that $\exists \lambda$ such that for some $\psi_\lambda$ we have $H\psi_\lambda=\lambda \psi_\lambda$ and the other that the matrix $H-\lambda I$ is non-invertible.

Am I missing some assumption here? I know there are examples of operators with a nonvanishing spectrum that have no eigenvalues (I think multiplication by a sine function is one example if I am not mistaken).

What is the definition (formally) of the observables of a QM system? and also why is it defined in that way and not the other (I suspect there is some theorem/result/definition that only works if we use one and not the other).

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  • $\begingroup$ I think your question is definitely not simple :). The theory of measurement in QM is nowhere from being understood as far as I understand. Regarding the question itself: I think that in reality you don't need to consider the difference between the spectrum and eigenvalues. This is because for example the energy-scale of your measurement makes you able to totally neglect a lot of possible states of your system. For example: when was the last time someone cared about the 100th excited state of a H atom? $\endgroup$
    – zltn.guba
    Commented Dec 29, 2021 at 13:51
  • $\begingroup$ I agree, but doesn't it make all the difference if (conjuring up a pathological example) the difference is in that there is an element of the spectrum that is in between two eigenvalues, one would certainly care. On a more concrete note, this is quite important when you consider whether a system is gapped, eigenvalues might be discrete and the spectrum can be compact. I believe there are concrete examples. I think the Fibonacci hamiltonian might be one, I'm not entirely sure tho (I know there is something strange that goes on with the spectrum vs the eigenvalues in that example). $\endgroup$ Commented Dec 29, 2021 at 14:06
  • $\begingroup$ I see your point and I agree. One more thing I would like to point out is that in a real measurement you cannot measure continuous values. This makes me think that perhaps in reality it does not matter how you define the possible set of possible values of your measurement. After all, physics is an experimental science. (I know that this probably does not help you but maybe it will be a good starting point for someone else's answer.) $\endgroup$
    – zltn.guba
    Commented Dec 29, 2021 at 14:14
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    $\begingroup$ An observable is not an eigenvalue: it is a self-adjoint operator. The possible outcomes of measuring this observable are the eigenvalues. $\endgroup$ Commented Dec 29, 2021 at 14:36
  • $\begingroup$ @ZeroTheHero, does self adjoint and hermitian mean the same thing? $\endgroup$
    – Kashmiri
    Commented Aug 11, 2022 at 17:39

1 Answer 1

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  1. An observable is a self-adjoint operator (but not all self-adjoint operators are observables, see this question).

  2. To every self-adjoint operator, there is an associated projection-valued spectral measure. Possible outcomes of measurements of the observable are subsets of its spectrum, the corresponding resulting state after measurement is the original state projected through the integral of the spectral measure over the subset.

  3. For operators with a discrete spectrum consisting solely of eigenvalues, this reduces to the usual Born rule when we measure single eigenvalues. For operators with less nice spectra - for instance the position operator with its purely continuous, eigenvalue-less spectrum - most physics treatments still pretend there are "eigenvectors" $\lvert x\rangle$ and just treat it like the discrete case. Even though this is rigorously deeply questionable (a mathematical formulation of this would involve rigged Hilbert spaces, see this question), it works surprisingly well in practice.

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