3 ms·

All observable quantities are eigenvalues of some operator, which are real numbers but discrete. How can they contain infinite amount of information?

nyssos 3y ago
There are operators with continuous spectra. The previous commenter was accidentally half-right, in that the usual intro QM picture where everything lives in L2 really isn't fully rigorous, but this is fairly easy to resolve. The correct setting is a rigged Hilbert space: given an algebra of operators A on a Hilbert space H, let S be the maximal subspace of H such that |sa| is finite for any s in S, a in A. These are your states. Operators in A don't necessarily have eigenvectors in H, but they do have eigenvectors in the space S* of all continuous linear functionals on S. So <x|, for instance, is just the map `psi -> delta_x(psi)`.
- eigenket 3y ago
  I take minor issue with the phrase "correct" here. Thats one way you can do things but its also works completely fine to not do that. Another way of setting these things up has your states be honest elements of L2, and says observables are just POVMs (i.e. maps from a space of measurable sets to positive operators which obey some natural restrictions like additivity). Then given a measurable subset A of the spectrum of some operator the Born-rule probability is just given by an inner product like < phi | P phi> where P is the projector you get if you integrate the spectral measure of the operator over A. This has the advantage of not having any funky "rigged" states suddenly appearing in your calculations and is also exactly how we deal with non-projective measurements in finite dimensional quantum mechanics. See here, for example https://en.wikipedia.org/wiki/POVM https://en.wikipedia.org/wiki/POVM
  - nyssos 3y ago
    Sure, I should have been clearer: a rigged Hilbert space is the right setting for bras and kets. You can also get rid of them entirely. In my experience QM classes unfortunately tend to split the difference by slinging around suggestive nonsense like \int_{x} |x><x|.
eigenket 3y ago
Consider the Hilbert L^2([0,1]) associated with a physical particle that has position somewhere between 0 and 1, the corresponding multiplication operator X which takes a wavefunction f and maps it to Xf where (Xf)(x) = x f(x). Then X is a bounded self-adjoint operator. It doesn't have any eigenvalues or eigenvectors but it's spectrum is exactly the set of numbers [0,1] as you'd expect (prefect measurements of position return real numbers in [0,1]). The spectral theorem, rather than decomposing X in terms of a sum of eigenvectors & eigenvalues instead decomposes it as an integral over the spectrum with respect to the (spectral) projection-valued measure. Now it is fair to question whether this "observable" is really observable, but it certainly works out mathematically consistently in the normal way we do things in quantum mechanics.
deleted 3y ago
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nyssos 3y ago

There are operators with continuous spectra. The previous commenter was accidentally half-right, in that the usual intro QM picture where everything lives in L2 really isn't fully rigorous, but this is fairly easy to resolve. The correct setting is a rigged Hilbert space: given an algebra of operators A on a Hilbert space H, let S be the maximal subspace of H such that |sa| is finite for any s in S, a in A. These are your states. Operators in A don't necessarily have eigenvectors in H, but they do have eigenvectors in the space S* of all continuous linear functionals on S. So <x|, for instance, is just the map `psi -> delta_x(psi)`.

eigenket 3y ago
I take minor issue with the phrase "correct" here. Thats one way you can do things but its also works completely fine to not do that. Another way of setting these things up has your states be honest elements of L2, and says observables are just POVMs (i.e. maps from a space of measurable sets to positive operators which obey some natural restrictions like additivity). Then given a measurable subset A of the spectrum of some operator the Born-rule probability is just given by an inner product like < phi | P phi> where P is the projector you get if you integrate the spectral measure of the operator over A. This has the advantage of not having any funky "rigged" states suddenly appearing in your calculations and is also exactly how we deal with non-projective measurements in finite dimensional quantum mechanics. See here, for example https://en.wikipedia.org/wiki/POVM https://en.wikipedia.org/wiki/POVM
- nyssos 3y ago
  Sure, I should have been clearer: a rigged Hilbert space is the right setting for bras and kets. You can also get rid of them entirely. In my experience QM classes unfortunately tend to split the difference by slinging around suggestive nonsense like \int_{x} |x><x|.

eigenket 3y ago

Consider the Hilbert L^2([0,1]) associated with a physical particle that has position somewhere between 0 and 1, the corresponding multiplication operator X which takes a wavefunction f and maps it to Xf where (Xf)(x) = x f(x). Then X is a bounded self-adjoint operator. It doesn't have any eigenvalues or eigenvectors but it's spectrum is exactly the set of numbers [0,1] as you'd expect (prefect measurements of position return real numbers in [0,1]). The spectral theorem, rather than decomposing X in terms of a sum of eigenvectors & eigenvalues instead decomposes it as an integral over the spectrum with respect to the (spectral) projection-valued measure. Now it is fair to question whether this "observable" is really observable, but it certainly works out mathematically consistently in the normal way we do things in quantum mechanics.

deleted 3y ago

[deleted]