Coherent state basis of (relativistic) particle Fock space

Question

For a neutral scalar bosonic particle of mass $m$, I consider a Fock space with an orthonormal basis of momenta eigenstates \begin{equation}\label{Fock-p-states} \left|p_1p_2\cdots p_n\right\rangle=\frac{1}{n!}\sum_\sigma\left|p_{\sigma(1)}\right\rangle\otimes\left|p_{\sigma(2)}\right\rangle\otimes \cdots\otimes\left|p_{\sigma(n)}\right\rangle\,, \end{equation} with a definite number of particles running from 1 to $\infty$, together with the vacuum state $|0\rangle$; the sum is over all permutations $\sigma$ of the particles. The normalization must be of the form $\langle p'|p\rangle=2E\delta(\vec{p}-\vec{p}')$, where $E=\sqrt{m^2+\vec{p}^2}$, for $\langle p'|p\rangle$ to be Lorentz invariant.

I define the creation operator $\hat{a}^\dagger(p)$ as \begin{equation*} \hat{a}^\dagger(p)|p_1p_2\cdots p_n\rangle=\sqrt{n+1}|p_1p_2\cdots p_np\rangle\,, \end{equation*} and a coherent state $|a\rangle$ as an eigenstate of the annihilation operator $\hat{a}(p)$ with eigenvalue $a(p)$ for every possible $p$: \begin{equation*} \hat{a}(p)|a\rangle=a(p)|a\rangle\:\:\:\forall p\,. \end{equation*} The set $\{|a\rangle\}$ of all coherent states is thus constructed through the functional ($\langle 0|a\rangle\equiv a_0$ is fixed by normalization) \begin{equation*} a(p)\mapsto|a\rangle=a_0|0\rangle +\frac{a_0}{\sqrt{n!}}\sum_{n=1}^\infty\int\!\!\bar{d}^3\!p_1\cdots\bar{d}^3\!p_n a(p_1)\cdots a(p_n)|p_1\cdots p_n\rangle\,,\hspace{5pt} \int\!\!\bar{d}^3\!p\,|a(p)|^2<\infty\,, \end{equation*} with a coherent state for each element of the set $A$ of all modulus-square-integrable complex functions $a(p)$; and \begin{equation}\label{square-integrable} \langle b|a\rangle=b^*_0a_0\exp\!\int\!\!\bar{d}^3\!p\,b^*(p)a(p)\,. \end{equation} Integrations are performed with the Lorentz invariant momentum element \begin{equation*} \bar{d}^3\!p_i\equiv\frac{d^3\!p_i}{2\sqrt{\vec p_i^2+m^2}} =\frac{d^3\!p_i}{2E_i}\,. \end{equation*}

Now, the question is if $\{|a\rangle\}$ is a basis, an overcomplete basis of the Fock space; more precisely, if there is a suitable definition of the measure $\mathcal{D}a$ of a modulus-square-integrable complex function $a(p)$ giving a functional integral \begin{equation}\label{identity-alpha} \int_A\!\!\mathcal{D}a|a\rangle\langle a|=\hat{1}\,, \end{equation} with $|a\rangle$ normalized to $1$, i.e., $|a_0|^2\exp\int\!\!\bar{d}^3\!p\,|a(p)|^2=1$. In the momenta basis, this is translated into \begin{equation}\label{identity-alpha-p} \delta_{\!mn}\sum_\sigma\prod_i2E_i\delta(\vec{p}_i-\vec{p}'_{\sigma(i)}) =\int\!\!\mathcal{D}a\,e^{-\int\!\bar{d}^3\!p\,|a(p)|^2}a(p_1)\cdots a(p_n)a^*(p'_1)\cdots a^*(p'_m)\,; \end{equation} \begin{equation}\label{identity-alpha-0} 1=\int\!\!\mathcal{D}a\,e^{-\int\!\bar{d}^3\!p\,|a(p)|^2}\,,\:\:m=n=0\,. \end{equation}

Answer 1

The answer is yes; the desired measure is a Gaussian measure. The construction works for any Fock space, with coherent states labelled by the 1-particle wave functions. For a rigorous, measure-free exposition in terms of reproducing kernel Hilbert spaces see, e.g., my paper

A. Neumaier and A. Ghaani Farashahi, Introduction to coherent quantization, arXiv:1804.01400.

This can be converted into a measure-theoretic construction using the Bochner-Minlos theorem.

Answer 2

A different approach that avoids some of the issues is to consider quadrature bases instead of a Fock basis. The eigenstates of the quadrature operators, given by $$ \hat{q}(\mathbf{p}) |q\rangle = |q\rangle q(\mathbf{p}) , $$ form complete orthogonal bases. In other words, $$ \langle q|q'\rangle = \delta[q-q'] , $$ where $\delta[q-q']$ is a Dirac delta functional. Moreover $$ \int |q\rangle\langle q| \mathcal{D}[q] = \hat{1} , $$ where the functional integral runs over the space of square integrable functions.

For the derivations see: PhysRevA 98/043841 and PhysRevA 101/019903.