Group averaging - Inner product

1. Nov 12, 2007

kakarukeys

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
$$\int_Gdg \langle\phi|\hat{U}^{-1}(g)|\phi\rangle \geq 0$$
$$\int_Gdg \langle\phi|\hat{U}^{-1}(g)|\phi\rangle = 0 \Rightarrow \int_Gdg \langle\phi|\hat{U}^{-1}(g) = 0$$

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

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

$$\phi, \phi_1, \phi_2\in\Phi\subset H$$ are vectors in a dense subspace of the Hilbert space H.
$$\hat{U}(g)$$ is a unitary representation of G (densely defined) on the Hilbert space H.
$$dg$$ 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