Inner product on boson one-particle space

[schieghoven]

Hi,

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

 

[QuantumBend]

I say put all derivative operatorings with gauge covariant derivative operatorings, then get answer.

 

[schieghoven]

Could you be a little more direct in your answer? Suppose H is the space of solutions to the K-G equation, and \psi, \phi \in H. What is the inner product \langle \psi, \phi \rangle? Is it a constant of motion and Lorentz invariant? Is it positive definite?

 

[QuantumBend]

Positive definite ONLY if positive frequency solutions of the K-G. (time look like vector field).

 

[Fredrik]

This one might help: http://home.uchicago.edu/~seifert/geroch.notes/

 

[Fredrik]

I might as well ask it here instead of starting a new thread: Is there a book that explains these things? The sort of "things" I'm talking about is e.g. how to define the Hilbert space of one-particle states from the solutions of the field equation, and how to define operators on that space. Geroch does it pretty well, but there are things in his notes that I just don't get, so it might help to read something written by someone else.

 

[schieghoven]

Cheers for the help, in fact, I had read parts of Geroch's notes previously and forgot his take on the problem. I'm a little dissatisfied: restriction to the positive-frequency solutions is equivalent to specifying the equation of motion
<br /> i\frac{\partial}{\partial t} \psi = \sqrt{k^2 + m^2} \; \psi<br />
in Fourier space, and this isn't equivalent to a differential equation on spacetime. I'd prefer having a differential equation as conceptually more appealing, and it might suggest some way of making it generally covariant. The latter would be necessary if we consider quantizing the scalar field on a curved spacetime.

Or to phrase the same objection in a different way, does the solution space of a generally covariant K-G equation (replace partial derivatives with covariant derivatives) also admit the same separation into positive and negative-frequency solutions?

Any comments?

 

[Fredrik]

The covariant derivative and the partial derivative operator have exactly the same effect on a scalar.

\nabla_{\partial_\mu}\phi=\partial_\mu\phi

So there's no difference.

 

[schieghoven]

Yes, for the first covariant derivative, but no, not for the second derivative. The laplacian of a scalar on a Riemannian manifold is

<br /> \square \psi = \psi_{;i}{}^i =<br /> \frac{1}{\sqrt{\det g}}<br /> \left( g^{ij} \psi_{,i} \sqrt{\det g} \right)_{,j}<br />

 

[George Jones]

Or to phrase the same objection in a different way, does the solution space of a generally covariant K-G equation (replace partial derivatives with covariant derivatives) also admit the same separation into positive and negative-frequency solutions?

If spacetime is stationary, this separation can be made; see Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics by Wald.

 

[schieghoven]

Stationary spacetime is a heavy restriction! But I'll check it out. I can't get the book you mention but I found some papers by the same author with similar titles on SPIRES and the arxiv. Thanks.

 

[Fredrik]

Yes, for the first covariant derivative, but no, not for the second derivative. The laplacian of a scalar on a Riemannian manifold is

<br /> \square \psi = \psi_{;i}{}^i =<br /> \frac{1}{\sqrt{\det g}}<br /> \left( g^{ij} \psi_{,i} \sqrt{\det g} \right)_{,j}<br />

D'oh...the posts I write immediately before I go to bed seem to be much dumber than then ones I write at other times of the day.