Self-energies in position space

[RedX]

According to Wikipedia ( http://en.wikipedia.org/wiki/Propagator_(Quantum_Theory)#Feynman_propagator ) the propagator for the KG-field is:

$$\ \Delta_F(x,y) \ = \lim_{\epsilon \to 0} \frac{1}{(2 \pi)^4} \int d^4p \, \frac{e^{-ip(x-y)}}{p^2 - m^2 + i\epsilon} \ = \left \{ \begin{matrix} -\frac{1}{4 \pi} \delta(s) + \frac{m}{8 \pi \sqrt{s}} H_1^{(1)}(m \sqrt{s}) & \textrm{ if }\, s \geq 0 \\ -\frac{i m}{ 4 \pi^2 \sqrt{-s}} K_1(m \sqrt{-s}) & \textrm{if }\, s < 0. \end{matrix} \right$$

where $$s:= (x^0 - y^0)^2 - (\vec{x} - \vec{y})^2.$$

Since these are complicated functions, take the mass equal to zero, and using the properties of these functions for small arguments ( http://en.wikipedia.org/wiki/Hankel_function#Asymptotic_forms ) you get:

$$\ \Delta_F(x-y) = \left \{ \begin{matrix} -\frac{1}{4 \pi} \delta(s) - \frac{i}{4 \pi^2 s}} & \textrm{ if }\, s \geq 0 \\ -\frac{i }{ 4 \pi^2 (-s)} & \textrm{if }\, s < 0. \end{matrix} \right$$

This expression seems perfectly finite except at s=0. In a phi⁴ theory, there is only one vertex for the next order correction to the propagator, so call the coordinate of this vertex z, and then you'll get:

$$\Delta(x-y)=\Delta_F(x-y)+C*\int d^4z \Delta_F(x-z) \Delta_F(z-z) \Delta_F(z-y)=\Delta_F(x-y)+C*\Delta_F(0)\int d^4z \Delta_F(x-z) \Delta_F(z-y)$$

for a constant C proportional to the coupling (this expression can be gotten from the Feynman diagram, with a line going from x to z, z to z, and z to y). My question is what's going on with this integral:

$$\int d^4z \Delta_F(x-z) \Delta_F(z-y)$$

Qualitatively when the value of z goes to x in the integral, it can do so from two different directions - from a space-like direction, or a time-like direction. If the latter, then a delta function will be picked up, in addition to an infinity term 1/s (i.e., 1/(x-z)²).

But the infinity term is harmless, because the integral is of the order d⁴z which is greater than the denominator which is of order z².

So the only trouble is $$\Delta_F(0)$$, and that's taken care of by a counter-term $$-\frac{C\Delta_F(0)}{2} \phi^2$$ in the Lagrangian (hopefully that minus sign is correct and there are no terms of i: does anyone have a good way of memorizing all those little factors of i and -1?).

Now take a phi³ theory, forgetting for the moment that for such a theory we can't find the vacuum state (but we take it in 6 dimensions for it to be renormalizeable). For the next order correction in the propagator, there are two vertices z and q. The propagator is:

$$\Delta(x-y)=\Delta_F(x-y)+C*\int d^6z d^6q \Delta_F(x-z) \Delta_F(z-q) \Delta_F(z-q) \Delta_F(y-q)= \Delta_F(x-y)+C*\Delta_F(0) \int d^6z d^6q \Delta_F(x-z) \Delta_F(z-q) \Delta_F(y-q)$$

Again, the divergent $$\Delta_F(0)$$ term, but my question is now about the integral next to it:

$$\int d^6z d^6q \Delta_F(x-z) \Delta_F(z-q) \Delta_F(y-q)$$

This looks convergent, even if both z goes to x and q goes to y simultaneously in the integral.

So far, in all these cases, the infinities come from a $$\Delta_F(0)$$ term, and not the integral term next to it. Is it true to say that infinities in quantum field theory come only from $$\Delta_F(0)$$ terms, which correspond to self-interactions? That is, integrals of products of delta functions converge (which to me is not so obvious - with something like $$\frac{d^4x}{x^2}$$ I'm worried about infinity more than 0), but it is an overall $$\Delta_F(0)$$ that causes all the trouble?

Also, all this analysis is based on m=0. Would the propagator be radically different if m was not set equal to zero?

 

[eljose79]

Thank you for bringing up this interesting topic. I would like to address some of your questions and provide some insights on the propagator in quantum field theory.

First of all, it is important to note that the propagator is a mathematical object that helps us understand the behavior of a quantum field in space and time. It is not a physical quantity that can be directly measured. Therefore, any infinities or divergences that arise in the calculations of the propagator do not have a physical meaning and need to be removed through a process called renormalization.

In the case of the KG-field, the propagator is given by the expression you have mentioned from Wikipedia. As you correctly pointed out, for a massless field (m=0), the propagator is finite everywhere except at s=0. This singularity at s=0 is known as the massless pole and it is related to the self-interaction of the field. This means that at this point, the field is interacting with itself, leading to an infinite contribution in the calculations.

For a massive field (m≠0), the propagator is different and it includes a term involving the mass. This changes the behavior of the propagator and it is no longer finite at s=0. In fact, it now has a finite value at this point, which is related to the mass of the field.

Now, coming to your question about infinities in quantum field theory, it is not entirely true that they only come from self-interactions. In general, infinities can arise from various sources such as divergent integrals, infinite sums, or infinite terms in the Lagrangian. However, the self-interaction term (or \Delta_F(0) term) is a common source of infinities in quantum field theory calculations.

In summary, the propagator in quantum field theory is a powerful tool that helps us understand the behavior of a field in space and time. It is important to note that any infinities or divergences that arise in the calculations are not physical and need to be renormalized. The massless pole in the propagator is related to the self-interaction of the field, but infinities can arise from other sources as well. I hope this helps clarify some of your doubts. Keep exploring and questioning, that's what science is all about!