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Spectral theory in quantum mechanics

  1. Jan 11, 2009 #1
    Can someone please tell me what the integrand in the below equation mean?
    [tex] 1 = \langle \psi | \psi \rangle = \int_{-\infty}^{\infty} d \langle \psi |E_\lambda \psi \rangle [/tex]
    [tex]E_\lambda[/tex] is an increasing (and absolutely continuous) function of projection operators such that [tex]\int_{-\infty}^{\infty} dE_\lambda = I [/tex]

    ( I read the integrand as a differential (or measure) of a complex constant, which should have been zero!? So I am certainly wrong in interpreting it)
  2. jcsd
  3. Jan 11, 2009 #2
    I would say that [tex]E_\lambda[/tex] corresponds to an element of a continuous set of projection operators i.e. [tex]E_\lambda=|\lambda\rangle\langle\lambda|[/tex] with the completeness relation
    [tex]\int dE_\lambda\equiv \int d\lambda |\lambda\rangle\langle\lambda|=1[/tex].

    In this sense the equation you wrote simply means

    [tex]1=\langle\psi|\psi\rangle=\langle\psi|(\int dE_\lambda )|\psi\rangle =\int d\underbrace{\lambda\langle\psi|\lambda\rangle\langle \lambda|\psi\rangle}_{\langle \psi |E_\lambda\psi\rangle}[/tex]

    Hope this helps
    Last edited: Jan 11, 2009
  4. Jan 11, 2009 #3
    Thanks!.. I get the drift.

    As in, in the case of [tex]\lambda[/tex] belonging to the spectrum of position operators,
    [tex]\langle \psi | E_\lambda \psi \rangle = \int_{-\infty}^{\lambda} |\psi |^2 dx[/tex]
    which is absolutely continuous and its differential is well defined ([tex]= | \psi |^2 [/tex])

    But, is writing [tex]E_\lambda=|\lambda\rangle\langle\lambda| [/tex] valid even for [tex]\lambda \in [/tex] spectrum, which can be continuous?
    I was of the opinion that it is valid only when [tex]\lambda[/tex] is an eigenvalue (and hence discrete).

    And when you wrote [tex]E_\lambda=|\lambda\rangle\langle\lambda| [/tex], I assume you meant the following:
    [tex] E_\lambda=\sum_{\lambda_i \le \lambda} |\lambda_i\rangle\langle\lambda_i| [/tex]
  5. Jan 12, 2009 #4
    Actually I think the answer I gave was not really what you were looking for. I just took a shot in the dark. To answer what the integrand means I would have to know in what context it appears. And I have a feeling even then I probably am not the right person to answer.

    I think that, in a "rigged" Hilbert space, continuous eigenvalues are allowed and one can safely write [tex]E_\lambda=|\lambda\rangle\langle \lambda[/tex] and even [tex]1=\int d\lambda |\lambda\rangle\langle\lambda|[/tex] but I don't know how to prove this or where proof could be found.

    Sorry for not being able to help you. I hope somebody comes along who can.
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