1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Can an operator without complete eigenstates be measured?
  1. Operator eigenstates (Replies: 6)

Loading...