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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 $$ \hat\varphi(x)=\int d^4x'\,f(x')\, \hat \phi(x') $$ 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. $\endgroup$ – Cosmas Zachos Jun 27 '20 at 22:48
  • $\begingroup$ Your equation for "varphi" does not make sense. There is an "x" on the LHS that is not a dummy variable, but on the RHS the "x" is a dummy integration variable. Where is the "x" dependence on the LHS coming from? The LHS is a constant... I think you mean that the f(x) is peaked at some "y" and the LHS is a function of "y" (not x). $\endgroup$ – hft Jun 27 '20 at 23:04
  • $\begingroup$ 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 '20 at 23:05
  • $\begingroup$ Also linked. $\endgroup$ – Cosmas Zachos Jun 28 '20 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 be 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: $$ <\varphi(x)\varphi(x)> = \int d^4u \int d^4v f_x(u)f_x(v)<0|\phi(u)\phi(v)|0> $$ $$ = \int d^4u f_x(u)f_x(u) $$

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

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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 '20 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 '20 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 '20 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 '20 at 1:14
  • $\begingroup$ Tong explicitly endorses the KG propagator linked in that question in his (2.90). It agrees with most QFT texts. $\endgroup$ – Cosmas Zachos Jun 29 '20 at 19:29

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