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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 $\left<k'\right|\left|k\right>\propto\delta(k-k')$, 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)?

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

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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.

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  • $\begingroup$ Thank you for the comment on smearing the operators. Using that formulation then indeed makes it clear that the Hilbert space is separable. $\endgroup$
    – Frotaur
    Commented Mar 24, 2022 at 13:55

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