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  1. How can I test whether a wave function is normalizable?
  2. If you apply an operator to a wave function, sometimes the result will not be normalizable. But how can I find these wave functions that do not correspond to normalizable eigenstates of this operator?
  3. The Hamilton operator for the harmonic oscillator, I am told, has both a discrete and a continuous spectrum. The discrete spectrum is the eigenvalues, but how do I find the continuous spectrum? It's relevant because I am also told these are the non normalizable values of the spectrum.
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Don't believe the answer of A.O.Tell. Its point 2 is false! See my comment there. - The Hamiltonian of the harmonic oscillator has no continuous spectrum, but that of the free particle has - its unnormalizable eigenfunctions can be taken to be the plane waves. – Arnold Neumaier Oct 10 '12 at 15:35
@ArnoldNeumaier, I never said that the HO has a continuous spectrum. Where did you read that from? – A.O.Tell Oct 10 '12 at 15:52
You only said you were told it has:''The Hamilton operator for the harmonic oscillator, I am told, has both a discrete and a continuous spectrum''. - But the teller was ill-informed. – Arnold Neumaier Oct 10 '12 at 15:57
Arnold, I never wrote that. This is what the OP wrote. Will you please stop? – A.O.Tell Oct 10 '12 at 16:03
Oh, sorry, I though this comment came from the OP. But your comment was spurious, as I didn't claim you had said it. I only stated that your point 2 is false and shouldn't be believed! Your point 3 is ok. – Arnold Neumaier Oct 10 '12 at 16:16
up vote 2 down vote accepted
  1. You test a wave function for normalizability by integrating its square magnitude. If you get a finite result then it is normalizable. To spare you complicated integrations you can also take a simpler wave function that you know is normalizable and compare it using the usual arguments.

  2. An operator is not only defined by the mathematical operation it performs, but also by which space it acts on. The Hilbert space of square integrable functions is where quantum operators act on. So by definition they take a square integrable function and give you a square integrable function. It can happen though that such an operator has eigenfunction that are not in the set of square integrable functions. For some of these cases consistency requires that we extend the hilbert space (or its dual) we work with. The eigenfunctions of the position and momentum operators fall into this category. See the Gelfand construction or rigged Hilbert space for details.

  3. The hamiltonian of the harmonic oscillator as an operator on the quantum hilbert space has only a discrete spectrum. If you define it to be an operator on a more general space that admits functions that are not square integrable then the spectrum may in fact be continuous. This is the perfect example for why an operator must always be stated with the space it acts on.

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Does the state space of the harmonic oscillator admit functions that are not square integrable, and if so, how I can find them? – user714 Oct 10 '12 at 12:22
The state space of the harmonic oscillator is the hilbert space of square integrable functions, so by definition it does not contain any functions that are not square integrable. Maybe I misunderstand your question? – A.O.Tell Oct 10 '12 at 12:25
It could happen, as you said, that a quantum operator has an eigenfunction that is not square integrable. Could this happen for the Hamilton operator on the state space of the harmonic oscillator? Could I, without too much trouble, find an example of such an eigenfunction or the family of functions? – user714 Oct 10 '12 at 12:38
Like I said, the eigenfunction of an operator depend on the space it is defined on, which is the quantum hilbert space in this case. So a quantum operator always only comes with eigenfunctions in this space by definition. Rarely it makes sense to extend the space to get functions which are not physical solutions, but mathematically useful. Like the plane waves which are eigenfunctions of the momentum operator. But again, they are not physical solution, only mathematical tools. It's similar to introducing imaginary numbers to simplify a calculation that gives a real result in the end. – A.O.Tell Oct 10 '12 at 12:44
OK. Thank you very much! – user714 Oct 10 '12 at 14:21

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