Why is $\vert \phi \vert ^2$ infinite in QFT?
Yes, it is relatively easy to see that if you are familiar with canonical quantization of the real Klein-Gordon field (KG). It follows very closely reference [1]. I hope, you can still benefit from it inspite the large time span between the question and the answer.
First of all, in QFT fields $\phi$ respectively objects constructed of fields like $\phi^2(x)$ should be considered as operators acting on states of the Fock-space, a Hilbert space consisting of multi-particle states (of course zero particle and 1-particle states are also included and are the most often used ones). Therefore making a statement on $|\phi(x)|^2$, one should consider an expectation value of this object in a Fock space state, i. e. one should consider
$$\lim_{x\rightarrow y} \langle 0| \phi(x)\phi(y)|0\rangle $$
where for simplicity the most basic state of the Fock-space was chosen, the vacuum state (zero-particle state). Furthermore, as we already anticipate the infinite result it is computed by a limiting process.
With some basic ingredients of the KG-theory we will compute it. First we develop the KG field in plane waves and consider that $\phi$ has to be an hermitian operator in order to get real expectations values ($E_k =\sqrt{\mathbf{k}^2 +m^2}$):
$$\phi(x)= \int \frac{d^3k }{\sqrt{(2\pi)^3 2E_k }} [a(k) e^{-ikx} + a^\dagger(k) e^{ikx}] $$
We have upgraded the development coefficients $a$ and $a^\ast$ to operators $a(k)$ and $a^\dagger(k)$ as operators that act on the Fock space that fulfill the following the commutations relations:
$$[a(k),a(k')]=0\quad\quad [a(k),a^\dagger(k')]=\delta^{(3)}(\mathbf{k}-\mathbf{k}')\quad\quad [a^\dagger(k),a^\dagger(k')]=0$$
Very important property of the so-called ladder operators:
$$a(k)|0\rangle = 0 \quad \text{and its conjugate correspondence}\quad \langle 0|a^\dagger(k) = 0$$.
The rest is a rather straight forward calculation:
$$\lim_{x\rightarrow y} \langle 0| \phi(x)\phi(y)|0\rangle \sim \langle 0| aa'+ aa'^\dagger + a^\dagger a' + a^\dagger a'^\dagger |0\rangle = \langle 0| a a'^\dagger|0\rangle $$
If now the full integrals are included we get:
$$\lim_{x\rightarrow y} \langle 0| \phi(x)\phi(y)|0\rangle = \lim_{x\rightarrow y} \int \frac{d^3k d^3k'}{(2\pi)^3 \sqrt{2E_k \cdot 2E_k'}} e^{-ikx}e^{ik'y} \langle 0| a a'^\dagger|0\rangle $$
Now we get the commutations relations in:
$$a a'^\dagger = \delta^{(3)}(\mathbf{k}-\mathbf{k}') + a'^\dagger a$$
Operating $a'^\dagger a$ on the $|0\rangle$ gives zero, so we are left with the delta-function. Knowing that $\langle 0|0\rangle = 1$ (all Fock states are normed) we finally get:
$$\lim_{x\rightarrow y} \langle 0| \phi(x)\phi(y)|0\rangle = \lim_{x\rightarrow y} \int \frac{d^3k d^3k'}{(2\pi)^3 \sqrt{2E_k \cdot 2E_k'}} e^{-ikx}e^{ik'y} \delta^{(3)}(\mathbf{k}-\mathbf{k}') = \lim_{x\rightarrow y} \int \frac{d^3k}{(2\pi)^3 2E_k} e^{-ik(x-y)} = \int \frac{d^3k}{(2\pi)^3 2E_k} =\infty$$
Q.E.D.
Reference:
- J.D. Bjorken, S.D.Drell, Relativistic Quantum Fields, Mc Graw-Hill Book Company (1965)
It does not really say that. All the fields (not just scalars) are finite at any given point in space. The correct statement (in the context of the mentioned paper) is that for a scalar field, the correction to the squared mass term involves the average over all space of another field $\langle|A(x)|^2\rangle$ where <> indicates averaging over all space. Now the average over all space of any quantity might be divergent (infinity) and if this is the case then we have to make sure we understand what that infinite correction in the mass means. This is explained partly in the paper as well as in any QFT book that has a chapter on renormalization.