- #1

schieghoven

- 85

- 1

I'm interested in constructive QFT and I'd like to pose a question about construction of the one-particle Hilbert space of states for bosons.

For fermions satisfying the Dirac equation, the inner product is

[tex]

\langle \psi, \phi \rangle = \int d^3 x \; \psi^\dagger(x) \phi(x).

[/tex]

The inner product is a constant of motion (and Lorentz invariant) because the integrand is the 0th component of a conserved current. The inner product is positive definite, and defines the Hilbert space L^2.

Is there a similar inner product for bosons satisfying the Klein-Gordan equation?

I previously thought that boson state spaces required an indefinite inner product (i.e., existence of negative-norm states), which is a problem because the triangle inequality is broken. This basically ruins any chance of applying Hilbert space theory.

Recently I found a paper [1] which refers to an answer, but doesn't give it explicitly. To quote from the introduction:

The real normalizable solutions of the Klein-Gordan

equation: $\square \psi = m^2 \psi, m>0$; form a real

Hilbert space $K$. $K$ admits a non-singular skew 2-form

$B(\cdot,\cdot)$ which is uniquely determined, apart from

a scalar factor, by the condition that it be invariant under

the canonical action of the proper inhomogeneous Lorentz group

on $K$. There is an orthogonal transformation $\Lambda$ on $K$,

commuting with this action, which when interpreted as

multiplication by $i$ allows $K$ to be made into a complex Hilbert

space $H$ with $B(\cdot,\cdot)$ as the imaginary part of the

inner product. To quantize the Klein-Gordan Field, one needs only

$K$ and $B(\cdot, \cdot)$, or equivalently, $H$.

Can somebody help me out here?

Thanks very much,

Dave

[1] Shale, Trans. Am. Math. Soc. 103 (1962) 149, Linear Symmetries of Free Boson Fields.