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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?

Qmechanic
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Guy
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    Sadly it's not so simple, the two point function $\langle 0|\phi(x)\phi(x')|0\rangle$ diverges when $x=x'$. You can find an explicit formula for the propagator of a massive scalar field in many places, such as https://physics.stackexchange.com/a/308228/27732 – Andrew Mar 29 '24 at 21:07
  • Yeah that was my primary reason for asking. My reason for choosing the gaussian mode function was that it regulates the quadratic divergence at $x = x'$. It also has the position space interpretation of the scalar's Compton wavelength.

    If I just take the propagator for plane-wave states, then I can still regulate the divergence e.g. with dim reg -- since there is only one scale $m$, the result will also scale as $\langle\phi^2\rangle\sim m^2(1 + {\cal O}(log))$ but it's not clear to me what the renormalization scale means here.

     – Guy Mar 29 '24 at 21:40
  • What do you mean by a Gaussian mode function in this context? – Andrew Mar 29 '24 at 21:47
  • Oh I see, you've altered the mode expansion for $\phi(x)$ to include some additional function $\tilde{\phi}(k)$. So the interpretation of that would be that you've modified the original field operator $\phi(x)$, so the operator you are taking a two point function with isn't the original one. Normally that is done in position space by defining a smeared operator like $\phi_\sigma(x) = (2\pi\sigma^2)^{-1/2} \int dx' e^{-(x-x')^2/2\sigma^2} \phi(x')$ and then looking at a two point correlation function of $\phi_\sigma(x)$. This will regulate the expression but it will still be infinite... – Andrew Mar 29 '24 at 21:51
  • ... in the limit $\sigma\rightarrow 0$. – Andrew Mar 29 '24 at 21:51
  • Hm I see what you're saying - I suspect that going through the algebra should yield something qualitatively similar to the momentum space mode function I wrote down above, but I need to check.

    Unrelated to your point, an additional motivation I have for asking this question is that in ordinary quantum mechanics a particle in a quadratic well will have a gaussian wavefunction (in position space) with variance given by the curvature of the well. This is where my intuition that $\langle\phi^2\rangle\sim m^2$ for a single particle comes from, but the divergent nature of QFT is confusing.

     – Guy Mar 29 '24 at 21:57
  • Understood. One physical interpretation is that strict localization of a particle is impossible in QFT, because as you localize a particle to within its Compton wavelength (which is the real space version of the momentum space statement that you are considering particles with relativistic energies), it is energetically possible to create particle-antiparticle pairs, which means the uncertainty principle forces you to consider multi-particle states. – Andrew Mar 29 '24 at 22:19
  • A lot of paradoxes arise when you try to interpret field theory in terms of one particle states, which disappear when you realize you need to consider multi particle states. A classic example for fermions is Zitterbewegung. – Andrew Mar 29 '24 at 22:19
  • I agree with your point in general, however for the simple case of a non-interacting field theory (which is the case I'm interested in) there are no number-changing processes, so this subtlety of particle creation at short distances is not an issue. – Guy Mar 29 '24 at 22:43

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