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Homework Help: Can an operator without complete eigenstates be measured?

  1. Mar 22, 2012 #1
    Say you have some operator A with an incomplete set of eigenstates, but the state of the system is such that it happens to be expressible as a sum (possibly infinite, or integral, whatever) of the eigenstates of A, and lets say the eigenvalues are real and whatever is necessary....we may assume that A is a dynamical variable. Can A be measured?

    The justification that an observable must have complete eigenstates is that if you make a measurement corresponding to the observable on a system, the system will jump into one of the observables eigenstates. But why must any state be expressible as a sum of an operators eigenstates, rather than just the current one?
  2. jcsd
  3. Mar 22, 2012 #2
    In order for an operator to represent an observable it must be Hermitian. Hermitian operators always have eigenfunctions that form a complete set, so in order for something to be an observable it must be Hermitian and that implies that its eigenfunctions are complete.
  4. Mar 22, 2012 #3
    I believe there are subtleties. I don't think the operators are necessarily hermitian on the whole Hilbert space. xp+px is hermitian, but I don't believe it has a complete set of eigenstates. So maybe the operator A isn't defined on the whole Hilbert space..
  5. Mar 22, 2012 #4
    I do not believe xp+px is considered an observable because you would not be able to measure an exact eigenstate due to the uncertainty principle.
  6. Mar 22, 2012 #5
    You wouldn't be measuring x and p separately and then putting them together to get xp+px...you would just be making one measurement, xp+px.
  7. Mar 22, 2012 #6
    Yes I understand this. I am saying that I do not believe that xp+px is measurable because it does not have a complete set of eigenstates. I used the uncertainty principle to offer an explanation as to why it may not be able to be measured.
  8. Mar 22, 2012 #7
    Right. But that is my question. If the system happens to be in an eigenstate of xp+px (I am not claiming they are complete, there just needs to be one), then could you make a measurement of xp+px?
  9. Mar 22, 2012 #8
    Sorry I confused myself, in Griffiths it is stated as an axiom that all observables with Hermitian operators have a complete set of eigenfunctions. If xp+px does not have a complete set, it is not observable.
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