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In a free scalar field theory, Wick's theorem guarantees that $\langle \hat\phi(x)\rangle = 0$ and $\langle \hat\phi(x)^2\rangle = \infty$. Given that $\hat \phi(x)$ creates a particle at $x$, these have the relatively straightforward interpretations $$ \langle 0|\text{particle at x}\rangle=0 $$ and $$ \langle \text{particle at x}|\text{particle at x}\rangle \equiv \langle x|x\rangle = \infty $$ where the latter parallels the delta function normalization of position eigenkets in single-particle quantum mechanics.

My main question is—what implications do these calculations have when we treat $\hat \phi(x)$ as an observable? The first result is relatively unproblematic: the vacuum expectation of a free scalar field is zero. The second, however, seems to imply that the variance of the field is infinite. How should we interpret this? Since the calculation works the same way for a vector field, it seems to imply that the EM field has infinite variance in the vacuum, which (at least initially) seems kinda fishy.

Now, my hypothesis is that the above infinities should go away when you consider a more realistic measurement scenario, like measuring the average value of the field in some small region. Where $f(x)$ is some Gaussian peaked at the point of interest, the operator corresponding to this measurement should be something like $$ \hat\varphi(x)=\int d^4x'\,f(x-x')\, \hat \phi(x') $$ which creates a particle in a Gaussian distribution centered around the point. This will still have $\langle \hat \varphi\rangle = 0$, but instead of the variance diverging, we have $$ \langle 0 | \hat \varphi(x)^2 |0\rangle = \langle \text{particle in Gaussian distribution}|\text{particle in Gaussian distribution}\rangle = \text{finite} $$ since Gaussian distributions are normalizable. So even if "vacuum fluctuations" at a point are infinite, they wash out to a small, finite size at any measurable scale, as we'd expect. Is this intuition/explanation roughly correct?

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    $\begingroup$ Linked. $\endgroup$ Jun 27, 2020 at 22:48
  • $\begingroup$ Thanks @cosmas zachos—I learned QFT from Schwartz's textbook, and it looks like I followed him too closely on the one-particle state issue. I haven't done the math here, but my intuition is that the 1/x (as opposed to delta function) behavior could be tricker to deal with in constructing a "realistic" operator. $\endgroup$
    – laaksonenp
    Jun 27, 2020 at 23:05
  • $\begingroup$ Also linked. $\endgroup$ Jun 28, 2020 at 0:06

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Now, my hypothesis is that the above infinities should go away when you consider a more realistic measurement scenario, like measuring the average value of the field in some small region.

When it come to treating quantum field as "the average value of the field in some small region", two guys named H Epstein and V Glaser beat you to the punch.

In 1973, they published a paper titled "The role of locality in perturbation theory" (see here). In the paper, quantum fields are regarded as "operator-valued tempered distributions", whereby the annoying infinities in QFT can be bypassed.

Lately, intimate connections between Hopf algebra and Epstein/Glaser's approach to QFT have been discovered (see here). And it has since become a fertile research arena as an alternative approach to renormalization. If you are really interested, you can consult the introductory book "Finite quantum electrodynamics: the causal approach" by G Scharf (see here).

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Where $f(x)$ is some Gaussian peaked at the point of interest, the operator corresponding to this measurement should be something like $$ \hat\varphi(x)=\int d^4x\,f(x)\, \hat \phi(x) $$ which creates a particle in a Gaussian distribution centered around the point. This will still have $\langle \hat \varphi\rangle = 0$, but instead of the variance diverging, we have $$ \langle 0 | \hat \varphi(x)^2 |0\rangle = \langle \text{particle in Gaussian distribution}|\text{particle in Gaussian distribution}\rangle = \text{finite} $$ since Gaussian distributions are normalizable. So even if "vacuum fluctuations" at a point are infinite, they wash out to a small, finite size at any measurable scale, as we'd expect. Is this intuition/explanation roughly correct?

I think you mean that f(x) is a function peaked at some other value (e.g., y). So, we should write $f_y(x)$ to make that clear. For example, maybe: $$ f_y(x) = Ae^{a(x-y)^2} $$ or some such thing.

In this case: $$ \langle \varphi(x)\varphi(x)\rangle = \int d^4u \int d^4v f_x(u)f_x(v)\langle 0|\phi(u)\phi(v)|0\rangle $$ $$ = \int d^4u f_x(u)f_x(u) $$

In the above, I have assumed you can use: $$ \langle 0|\phi(u)\phi(v)|0\rangle = \langle u|v\rangle = \delta^4(u-v) $$

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  • $\begingroup$ Thanks for pointing out the error in the OP. The post Cosmas Zachos linked to indicated that the vacuum expectation of phi(u) phi(v) isn't actually a delta function, so I wonder how much that affects what you wrote here (and the strategy I applied in the above). $\endgroup$
    – laaksonenp
    Jun 27, 2020 at 23:13
  • $\begingroup$ If the vev for phi(u)phi(v) isn't a delta function then the result changes. If <phi(u)phi(v)> is peaked at (u-v), and is narrow compared to f(x), and integrates to one then the result doesn't change much. But otherwise the result will change. $\endgroup$
    – hft
    Jun 27, 2020 at 23:25
  • $\begingroup$ Right, I'm mainly interested in if the integral ends up being convergent (which it would be if the inner product were a delta). Interestingly, both Tong and Schwarz state that the field operator creates a particle in a delta function distribution (Tong is very firm about this—see eq. 2.52 and 2.116), which makes me wonder if the post Cosmas Zachos linked to is in error. $\endgroup$
    – laaksonenp
    Jun 28, 2020 at 0:07
  • $\begingroup$ I don't think the linked-to post is in error, I think it just has a different normalization choice. The integral looks convergent to me, for the example f I've given, but of course it depends on what the function "f" is. $\endgroup$
    – hft
    Jun 28, 2020 at 1:14
  • $\begingroup$ Tong explicitly endorses the KG propagator linked in that question in his (2.90). It agrees with most QFT texts. $\endgroup$ Jun 29, 2020 at 19:29
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For a free field we can actually carry through these computations analytically. Suppose we have a standard free Klein-Gordon field of mass $m$ in $\mathbb{R}^{d+1}$ and define the Gaussian smearing function $$ f_\sigma(x) = \frac{1}{(\sqrt{2\pi}\sigma)^{d+1}}\exp(-\Vert {\bf x}\Vert_E^2/2\sigma^2)\delta(t) $$ where $\Vert {\bf x}\Vert_E^2 = \sum_{i=1}^d (x^i)^2$ is the Euclidean norm on $\mathbb{R}^{d}$. Now we smear our field with $f_\sigma$: $$ \phi(f_\sigma)=\int_{\mathbb{R}^{d}}\phi(x)f_\sigma(x)\,\mathrm{d}^{d+1}x. $$ Note that because of the $\delta(t)$ in $f_\sigma$ we're not smearing the field in time. Strictly spacelike smearing is enough in this free theory to cure divergences, but as far as I know timelike smearing may also be required in other theories. Anyway, we could also carry through the sequel with timelike smearing, but for simplicity let's content ourselves with this. Now, since $f_\sigma$ is real, $\phi(f_\sigma)$ is hermitian so it potentially a bona-fide observable. Suppose we measure this smeared field. How to compute the probability distribution of the measurements? Since diagonalizing the operator seems very difficult, here's an alternative approach: let's calculate all the moments of this probability distribution. Then $n^{\mathrm{th}}$ moment is given as $$ m_n = \langle \phi(f_\sigma)^n\rangle_0. $$ with $\langle \ldots\rangle_0$ being the vacuum expectation. We'll evaluate this with Wick's theorem, so first let's write out $\phi(f_\sigma)$ with ladder operators (I'll be following Tong's QFT conventions): \begin{align*} \phi(f_\sigma)&=\int\mathrm{d}^{d}{\bf x}\,\mathrm{d}t\,f_\sigma({\bf x},t)\int\frac{\mathrm{d}^d{\bf p}}{(2\pi)^d}\frac{1}{\sqrt{2\omega_{\bf p}}}(a_{\bf p}e^{-ip\cdot x} + h.c)\\ &=\int\frac{\mathrm{d}^d{\bf p}}{(2\pi)^d}\frac{1}{\sqrt{2\omega_{\bf p}}}\int\mathrm{d}^{d}{\bf x}\,\mathrm{d}t\,f_\sigma({\bf x},t)(a_{\bf p}e^{-ip\cdot x} + h.c)\\ &=\int\frac{\mathrm{d}^d{\bf p}}{(2\pi)^d}\frac{1}{\sqrt{2\omega_{\bf p}}}e^{-\sigma^2{\bf p}^2/2}(a_{\bf p}+a_{\bf p}^*). \end{align*} By Wick's theorem only even moments will be nonzero: $$ m_{2n}=(2n-1)!!\bigg(\iint\frac{\mathrm{d}^d{\bf p}}{(2\pi)^d}\frac{\mathrm{d}^d{\bf p'}}{(2\pi)^d}\frac{1}{\sqrt{4\omega_{\bf p}\omega_{\bf p'}}}e^{-\sigma^2({\bf p}^2+{\bf p'}^2)/2}\mathcal{C}(a_{\bf p}+a_{\bf p}^*,a_{\bf p'}+a_{\bf p'}^*)\bigg)^{\!n}. $$ Here $\mathcal{C}(a_{\bf p}+a_{\bf p}^*,a_{\bf p'}+a_{\bf p'}^*)$ is a Wick contraction which equals $(2\pi)^d\delta({\bf p}-{\bf p'})$, so $$ m_{2n} = (2n-1)!!\bigg(\int\frac{\mathrm{d}^d{\bf p}}{(2\pi)^d}\frac{1}{2\omega_{\bf p}}e^{-\sigma^2{\bf p}}\bigg)^{\!n}=(2n-1)!!\,I^n $$ where $I$ has the obvious meaning. It turns out that these are precisely the moments of a Gaussian distribution with mean $0$ and variance $I$. The Hamburger moment problem theory tells us that this is actually the unique probability distribution with these moments. Hence the probability density for the measurement of $\phi(f_\sigma)$ to give outcome $s$ is $$ p(s)=\frac{1}{\sqrt{2\pi I}}e^{-s^2/2I}. $$ Finally let's compute $I$ (let's fix $d=1$ to simplify the integrals, I just threw this into Wolfram Alpha): $$ I = \frac{1}{2\pi}\int_0^\infty\frac{1}{\sqrt{p^2+m^2}}e^{-\sigma^2p^2}\,\mathrm{d}p=\frac{1}{4\pi m}e^{-m^2\sigma^2/2}K_0(m^2\sigma^2/2), $$ where $K_0$ is a Bessel function. Observe that as we remove the smearing ($\sigma\rightarrow 0$), $I$ diverges. Conversely, if we observe the field at distances much larger than the scale set by the mass $m$, then the field is very well restricted to $0$.

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