In the classical (typically large occupation number) limit of QFT, the dynamics of, say, a scalar field can be approximated by its classical field equation. Let's consider the Klein-Gordon equation: $$ 0 = \ddot\phi - \nabla^2\phi + m^2\phi. $$ One can then relate the field bilinear $\phi^2$ to the occupation number: in the nonrelativistic limit, this is just $N = \int{\rm d}V\frac12m\phi^2$. If I were to extrapolate this equation to the single particle limit, I would find, roughly, $$ \phi^2\sim m^2, $$ where I've cut off the volume at a Compton wavelength.
Let us now consider quantum field theory, where my wavefunction has the following form $$ \phi = \int\frac{{\rm d}^3k}{(2\pi)^3\sqrt{2 E_k}}\tilde\phi(k) a_k^\dagger e^{-i k_\mu x^\mu} + {\rm h.c.} $$ where $a_k$ satisfy the commutation relations $$ [a_k,a_p^\dagger] = (2\pi)^3\delta^3(k - p) $$ I would naively expect that I could compute the RMS field excursion of the quantum field as $$ \langle\phi^\dagger(x)\phi(x)\rangle = \int\frac{{\rm d}^3 k}{(2\pi)^3}\frac{|\tilde \phi(k)|^2}{2 E_k}\,. $$ If I assume the mode functions are localized in momentum space as a Gaussian $$ \tilde \phi = \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}\frac{\vec k^2}{m^2}\right), $$ then I obtain $$ \langle\phi^\dagger(x)\phi(x)\rangle \sim m^2\,. $$ I think this makes sense, but I made an ad hoc assumption that the $\phi$ mode function had a particular form.
In general, do I expect single particle states to have field-excursions on the order of their mass? Is there a way to see this without specifying a particular form for the mode function?
