# Variance of an interacting quantum field in its vacuum state

A non-interacting quantum field $$\hat{\phi}(x)$$ can be decomposed into $$a_{\textbf{k}}$$ and $$a_{\textbf{k}}^\dagger$$. This enables us to calculate the variance of a free field. For example, the variance of the free real scalar field, in the vacuum $$|0\rangle$$ of the theory, is computed to be (without a momentum cut-off) $${\rm Var}(\phi)_0=\langle0|\phi^2|0\rangle-\big(\langle0|\phi|0\rangle\big)^2\\=\int\frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{\textbf{k}^2+m^2}}\rightarrow \infty.$$ Now, consider an interacting quantum field theory described by the Hamiltonian $$H$$ and $$|\Omega\rangle$$ is the vacuum state of the interaction theory i.e. $$H|\Omega\rangle=0|\Omega\rangle\hspace{0.5cm}\big(\text{also,}~P^\mu|\Omega\rangle=0|\Omega\rangle\big).$$ Decomposing the field into creation and annihilation operators is no longer possible. So how does one compute the variance $${\rm Var}(\phi)_\Omega=\langle\Omega|\phi^2|\Omega\rangle-\big(\langle\Omega|\phi|\Omega\rangle\big)^2$$ in $$\lambda\phi^4$$ theory?

• If you work with Wightman QFT, a part of the definition of the QFT is the 2-point function $W(x, y) = \left< 0 \right| \phi(x) \phi(y) \left| 0 \right>$. That function always has a singularity at $x = y$ -- your question is ill-posed, the variance is always divergent. You can choose to regularize the theory by re-defining the behavior of the 2-point function at short distance, i.e. make it such that $W(x, x) = \mu$. Then the variance is equal to $\mu$ by definition. Which demonstrates that variance is unphysical (physical quantities have well-defined limits for $\mu \rightarrow \infty$). – Prof. Legolasov Apr 11 '19 at 15:30
• Neither your question nor your "computation" for the free field at the beginning makes sense to me. A quantum field is not an operator, it is an operator-valued distribution. $\phi$, without feeding anything to it, is not an operator and it makes as much sense to ask about its variance as it does to ask for "the value" of a function. Are you trying to compute the variance of $\phi(x)$? Why? If so, why is there no $x$ in your first equation? Note also that $\langle \phi(x)\phi(x)\rangle$ is "the probability for a particle present at $x$ to be detected at $x$". – ACuriousMind Apr 11 '19 at 21:00
• Yes. I'm trying to compute the variance. It should be obvious that I have merely suppressed the argument ''x". It's clearly written in the opening sentence. @ACuriousMind – SRS Apr 11 '19 at 22:30

As already mentioned by other people, $$\phi^2(x)$$ is not really well defined as it is because there is an UV divergence in taking $$x\to y$$ on $$\phi(x)\phi(y)$$. But in QFT we can give a meaning to $$\phi^2(x)$$ as a composite operator. The UV divergence we encounter can be subtracted order by order in perturbation theory and obtain a finite answer in the end.

Let me assume for simplicity that the expectation value of $$\phi(x)$$ is zero on the vacuum. Then we need to compute $$\langle \phi^2(x)\rangle$$. I am assuming you are familiar with the path integral formalism. Let us define a source $$L(x)$$ to which the composite operator $$\phi^2(x)$$ is coupled. We then have a partition function $$Z[L] = \int \mathcal{D}\phi\, \exp\left(-S[\phi] + \int \mathrm{d}^dx\,L_B(x)(\phi_B)^2(x)\right)\,,$$ where the subscript "$$B$$" stands for "bare". The field renormalizes with the usual wave function renormalization $$\phi_B(x) = \sqrt{Z_{\phi}}\phi(x)$$ and $$L$$ renormalizes as $$L_B(x) = Z_L L(x)$$. We have by definition $$(\phi^2)_B(x) = Z_L^{-1} (\phi^2)(x)\,,$$ where I use the parentheses to distinguish between the square of the operator $$\phi$$ and the operator $$\phi^2$$. Correlators with the operator $$(\phi^2)$$ can be computed as $$\langle (\phi^2)(x_1)\cdots (\phi^2)(x_n)\rangle = \frac{1}{Z}\frac{\delta^n}{\delta L(x_1)\cdots \delta L(x_n)} Z[L]\,.$$ If we want to consider higher point insertions of $$(\phi^2)$$, by power counting we would need to add also a term $$a\int L^2(x)$$ and renormalize the coupling $$a$$, but for this case we do not care.

The Feynman rules are simple, just add to the rules for $$S[\phi]$$ a new vertex with an $$L$$ leg and two $$\phi$$ legs. The function we need $$\langle (\phi^2)(x)\rangle$$ is the sum of all Feynman diagrams with one external $$L$$ leg. At one loop in dim-reg this is

$$(\textbf{Fig. 1})= \int \frac{\mathrm{d}^d p}{(2\pi)^d} \frac{1}{p^2+m^2} =\frac{m^4 \mu^{-2\varepsilon}}{2(4\pi)^3 \varepsilon}\left(\frac{4\pi \mu^2}{m^2}\right)^\varepsilon + (\mathrm{finite})\,.$$

You can then absorb that pole in $$\varepsilon$$ in the definition of $$Z_L$$ to obtain a finite answer in the $$\mathrm{MS}$$ scheme. Note that if the field is massless, this integral identically vanishes in dim-reg.

Fig. 1

## $$$$ Damiano Anselmi, Renormalization. 14B1

• It is not clear to me that the "composite $\phi(x)^2$" is meaningfully related to what we usually mean by the variance of a random variable: When you define the $\phi^2$ here, you're basically defining a new multiplication rule for the field operators (and one that is not associative - e.g. to learn how $\phi(y)^2 \phi(x)^2$ relates to $\phi(x)^4$ as $y\to x$ one needs to perform an operator product expansion), and it is not clear that the standard definition of the variance $\langle (x - \langle x\rangle)^2\rangle$ works for such a non-standard product. Could you comment on that? – ACuriousMind Apr 14 '19 at 7:43
• My definition for $\phi^2$ is physicallly the same as doing point splitting: $\lim_{x\to y}\phi(x)\phi(y) - \mathrm{divergence}$, but with a different scheme of regularization. Also, in perturbation theory normally operators have zero expectation value on the vacuum. If there is some spontaneous symmetry breaking tipically one redefines $\phi(x) = \langle\phi\rangle + \chi(x)$. So I think that the formula for the variance should work. Lastly, with this formalism you can write composite operators only of fundamental fields, so it's not really a multiplication rule. There is no $(\phi^2)^2(x)$ – MannyC Apr 14 '19 at 14:45
1. You don't need to compute the variance of a scalar field to see that it will always diverge: The expression $$\langle \phi(x)\phi(x)\rangle$$ is essentially the limit of the propagator for $$y\to x$$ and $$\langle \phi(x)\rangle$$ is constant because it's a Lorentz invariant. The propagator must diverge for $$y\to x$$ since otherwise it would predict a non-unit probability for a particle to propagate from event $$x$$ to event $$x$$, which would be non-sensical.

2. In the interacting theory, any attempted "computation" would have to proceed by computing the Dyson-resummed propagator to the desired accuracy and then taking the limit $$y\to x$$. Which, as argued above, will always diverge, so it is pointless to attempt.

3. Bonus fact: The ill-definedness/divergence of $$\langle \phi(x)^2\rangle$$ is a reflection of the fact that a quantum field is an operator-valued distribution and you can't square a distribution in a mathematically rigorous way.

• Ah... right. Do you have any idea about why the vacuum fluctuations of the electric field turns out to be finite in quantum optics but not in QFT? It seems that it is because that they find the vacuum fluctuation for a single mode? Here is the link of a 5 min video by Alain Aspect. youtube.com/watch?v=jXxW82L6os8 I am not sure why is the vacuum fluctuation of a single mode is experimentally meaningful @ACuriousMind – SRS Apr 14 '19 at 14:25
• @SRS Please do not use comments for follow-up questions. If you have a new question, ask a new question. Note that the electric field is a) not a scalar and b) not a "quantum field" (in the sense that the exp. value of its square would be related to a propagator) since the dynamical variable of the EM Lagrangian is the potential, not the field strength, so it has nothing to do with the variance of a scalar quantum field your question here asks about. – ACuriousMind Apr 14 '19 at 14:34