6
$\begingroup$

I've read here¹ that for a scalar field $\phi$, the square $\vert \phi \vert ^2$ is infinite (which gives an infinite contribution to mass), more precisely:

the square of the field – a quantity which diverges in QFT as necessary consequence of the commutation rules of the theory and unitarity.

It seems that this is related to the fact that $\phi$ is a distribution and the square would be the correlation function.

if we compute in quantum field theory a correlation function like $\langle A(x)A(y)\rangle$ and let $x\to y$, we find a divergent quantity.

Is there a simple way to see why is this true, with little knowledge of QFT?

  1. Inertial Mass and Vacuum Fluctuations in Quantum Field Theory, Giovanni Modanese, PACS: 03.20.+i; 03.50.-k; 03.65.-w; 03.70.+k 95.30 Sf
$\endgroup$
0

2 Answers 2

5
$\begingroup$

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:

  1. J.D. Bjorken, S.D.Drell, Relativistic Quantum Fields, Mc Graw-Hill Book Company (1965)
$\endgroup$
3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Thanks. And why does he mention the commutation relations? $\endgroup$
    – jinawee
    Commented Mar 11, 2014 at 20:49
  • $\begingroup$ @jinawee : This is not easy to explain without going into the dirty details. The big picture is that a QFT is a quantum theory so all the observables are turned into operators. The commutation relations between these operators are a fundamental ingredient of the theory and will of course affect many quantities in the theory. As a general comment, I would say you can only learn that much about renormalization from a single paper. I would recommend having a look in some QFT books for more details. $\endgroup$
    – Heterotic
    Commented Mar 12, 2014 at 8:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.