# Inner product on boson one-particle space

• schieghoven
In summary, the inner product for bosons satisfying the Klein-Gordan equation is a constant of motion and Lorentz invariant.
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
$$\langle \psi, \phi \rangle = \int d^3 x \; \psi^\dagger(x) \phi(x).$$
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.

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

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?

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

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

Last edited by a moderator:
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.

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
$$i\frac{\partial}{\partial t} \psi = \sqrt{k^2 + m^2} \; \psi$$
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?

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.

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

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

schieghoven said:
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.

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.

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

$$\square \psi = \psi_{;i}{}^i = \frac{1}{\sqrt{\det g}} \left( g^{ij} \psi_{,i} \sqrt{\det g} \right)_{,j}$$
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.

## 1. What is an inner product on boson one-particle space?

An inner product on boson one-particle space is a mathematical operation that takes two vectors in the space and produces a scalar value. It is used to measure the similarity or "closeness" between two vectors and is an essential tool in quantum mechanics.

## 2. How is an inner product defined for boson one-particle space?

An inner product on boson one-particle space is defined as the sum of the products of the corresponding elements in the two vectors. It also follows the properties of linearity, conjugate symmetry, and positive definiteness.

## 3. What is the physical significance of an inner product on boson one-particle space?

The inner product on boson one-particle space has physical significance in quantum mechanics as it allows us to calculate the probability of a particle being in a particular state. It also helps us determine the expectation value of observables and the time evolution of a system.

## 4. How is the inner product related to the creation and annihilation operators in boson one-particle space?

In boson one-particle space, the inner product is related to the creation and annihilation operators through the commutation relations. These relations define the behavior of bosonic particles and are an essential aspect of quantum mechanics.

## 5. Can the inner product be extended to multi-particle boson states?

Yes, the inner product can be extended to multi-particle boson states by using the tensor product of individual inner products. This allows us to calculate the similarity between two multi-particle states and is crucial in studying systems with multiple bosonic particles.

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