# Hilbert basis for one-particle Hilbert space of a scalar field

While thinking about whether the fock space of a free scalar field is separable, I distilled the initial question into the following problem.

Consider the one-particle hilbert space for a scalar field. If we go through the motions of solving the free scalar-field equation, we will find that the creation and annihilation operators are indexed by $$k$$, let us say for simplicity it is a real number describing the momentum of the particle being created. Knowing that, then the one-particle Hilbert space should be the completion of the vector space spanned by the vectors : $$\left|k\right> = a_k^\dagger\left|0\right>$$ Where $$a_k^\dagger$$ is the creation operator and $$\left|0\right>$$ is the vacuum, defined as the state or vector being annihilated by all $$a_k$$'s. I already have a problem with what should be the associated scalar product. I guess it should be inherited from the relations $$\left\propto\delta(k-k')$$, but I don't really know what to do with the delta function.

Now, I would like to find a countable Hilbert basis for this Hilbert space (the $$\left|k\right>$$ are clearly a Hilbert basis, but uncountable). Namely I would like to find a countable set of vectors $$\left|i\right>$$ such that their span is dense in the Hilbert space, w.r.t. to the above norm (which admittedly I am not sure how to define).

So to summarize : is the Hilbert space for a single particle free scalar field even well-defined ? If yes, is it separable (which implies iff that there is a countable Hilbert basis)?

• Isn't it just $L^2(\mathbb{R}^3)$, as in, the space of square integrable wavefunctions, just like in regular QM? Mar 23, 2022 at 18:26

The creation operators $$a^\dagger_k$$ are operator-valued distributions; therefore, $$a^\dagger_k |0\rangle$$ is not an element of the Fock space $$\bigoplus_{n \in \mathbb{N}} L^2(\mathbb{R}^d)^{\otimes n}$$. The solution to your problem is to smear $$a_k^\dagger$$ with elements $$f\in L^2(\mathbb{R}^d)$$: $$a^\dagger(f) := \int_{\mathbb{R}^d} \hat{f}(k) a^\dagger_k \ \mathrm{d} k,$$ where $$\hat{f}$$ is the Fourier transform of $$f$$. You can easily verify that $$a^\dagger(f) |0\rangle = f \in L^2(\mathbb{R}^d)$$ (i.e. $$a^\dagger(f)$$ creates one-particle states).
The Fock space $$\bigoplus_{n \in \mathbb{N}} L^2(\mathbb{R}^d)^{\otimes n}$$ is separable because the $$n$$-particle subspaces $$L^2(\mathbb{R}^d)^{\otimes n} \cong L^2(\mathbb{R}^{dn})$$ are separable, and the direct sum of countably many separable Hilbert spaces is a separable Hilbert space.