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Group averaging - Inner product

  1. Nov 12, 2007 #1
    1. The problem statement, all variables and given/known data
    Does anyone know how to show the inner product obtained from group averaging is positive definite?

    Show that
    [tex]\int_Gdg \langle\phi|\hat{U}^{-1}(g)|\phi\rangle \geq 0[/tex]
    [tex]\int_Gdg \langle\phi|\hat{U}^{-1}(g)|\phi\rangle = 0 \Rightarrow \int_Gdg \langle\phi|\hat{U}^{-1}(g) = 0[/tex]

    2. Relevant equations
    [tex]\langle\langle \eta\phi_1, \eta\phi_2\rangle\rangle_{phys} = \int_Gdg \langle\phi_2|\hat{U}^{-1}(g)|\phi_1\rangle[/tex]
    [tex]\eta[/tex] is the rigging map (not important here, just ignore the L.H.S.)

    [tex]\eta: \Phi \longrightarrow \Phi'[/tex]
    [tex]\eta\phi = \int_Gdg \langle\phi|\hat{U}^{-1}(g)[/tex]

    [tex]\phi, \phi_1, \phi_2\in\Phi\subset H[/tex] are vectors in a dense subspace of the Hilbert space H.
    [tex]\hat{U}(g)[/tex] is a unitary representation of G (densely defined) on the Hilbert space H.
    [tex]dg[/tex] is a bi-invariant measure on G

    3. The attempt at a solution
    No idea at all. Not sure what kind of maths is needed.
     
    Last edited: Nov 12, 2007
  2. jcsd
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