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Help with expected value of non-hermitian operators

  1. Jan 2, 2007 #1
    I know that with an Hermitian operator the expectation value can be found by calculating the (relative) probabilities of each eigenvalue: square modulus of the projection of the state-vector along the corresponding eigenvector.
    The normalization of these values give the absolute probabilities.
    Alternatively it's possible to calculate directly the expectation value by the compact formula <A>=<psi|A|psi>.

    I got stumbled considering what adjustment to take in case of an operator that is not Hermitian.
    In this case the eigenbasis is not orthonormal. But I feel that there should be some way to calculate the expectation value of the operator nonetheless.

    Am I wrong? Or there's no meaningful way to define it in the non hermitian case?

    If the procedure is more or less the same with some adaptation to make then:

    Which projection to take? the components in the eigenbasis or the orthogonal projections of psi along the eigenvectors?
    Is there an analog of the formula <A>=<psi|A|psi> that is correct in the non.orthonormal basis?

    Any help appreciated
    Thanks
     
  2. jcsd
  3. Jan 2, 2007 #2
    Well, the problem with the eigenbasis is even worse than you think it is. Let's use the harmonic oscillator as an example to illustrate a few points:

    The creation operator has no eigenkets! To see this, let's take a look at what a normal eigenket would look like in the harmonic oscillator basis:
    [tex]
    | \lambda \rangle = \sum_n c_n |n \rangle
    [/tex]
    So, you act the raising operator on it, but where do you get the lowest nonzero term? Gotta create it from somewhere, but there's nothing there!

    That said, the expectation value of an operator [tex]\mathcal{O}[/tex] for a state [tex]|\Psi\rangle[/tex] is pretty much defined as
    [tex]
    \langle \mathcal{O} \rangle = \langle \Psi | \left ( \mathcal{O} | \Psi \rangle \right )
    [/tex]
    For hermitian operators, the fact that the operator acts on the ket is irrelevent, but for non-hermitian operators, there is a possibly big difference, so you just have to be careful about what you do with non-hermitian operators moreso than hermitian operators.
     
  4. Jan 2, 2007 #3
  5. Jan 4, 2007 #4
    lemma28:” Help with expected value of non-hermitian operators”

    Quantum River:” there seems to be a lot of physics in the non-hermitian operators”

    Notice that entire QFT is formulated in terms of non-hermitian operators (second quantization).
     
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