In all that follows I'll be dealing with everything massless.

The free, massless propagator ($\mathcal{L} = \int d^{4}x \left(\partial \phi(x) \right)^{2} $) is supposedly given by $G_{0}(x,y) = c (x-y)^{-2}$, where I believe $c = \frac{1}{4\pi^{2}}$.

I'm trying to calculate the propagator in $\phi^{4}$-theory, specifically the contribution due to this diagram: enter image description here

Using the Feynman rules in position space, I believe that I should be getting a contribution of the form: $$ C(x_{1},x_{2}) = -i\lambda \int d^{4}u\ G_{0}(x_{1}, u) G_{0}(u,u) G_{0}(u,x_{2}) $$

However, here is my question: why do I get $G_{0}(u,u) = c (u-u)^{-2} = \mathrm{undefined}$? There's no way I can see to evaluate this integral.

How do I deal with this? Maybe I've got the wrong order of variables? I'm new to these kinds of calculations.


To define a $\phi ^4$ massless theory, you can't simply start from a lagrangian $$\mathscr L = \frac{1}{2}(\partial\phi)^2-\frac{g_0}{4!} \phi^4,$$ separate a kinetic term $\mathscr L _0= \frac{1}{2}(\partial\phi)^2$ and do perturbation theory around $g _0=0$. The reason is that, even in lowest order of perturbation theory, the $\phi^4$ term will give rise to divergent diagrams like the one you have just calculated, which are formally infinite. To treat such diagrams, you need:

  1. To introduce a regulator, like an UV cutoff $\Lambda$ in the propagators.
  2. To impose some (in this case, three) finiteness conditions on the Green's functions.
  3. To express all quantities in terms of the (finite) parameters introduced in step 2.

The procedure will work only if you start from a renormalizable lagrangian like: $$\mathscr L = \frac{1}{2}(\partial\phi)^2-\frac{m_0^2}{2}\phi^2-\frac{g_0}{4!} \phi^4.$$

The purpose of Step 1 is to make sense of divergent quantities like $G_0(x-x)$. A simple way that works for the $\phi^4$ theory is Pauli-Villars regularization: $$\tilde G_0(p)=\frac{1}{p^2-m_0^2+i\varepsilon}\to \tilde G_0(p)_\Lambda = \frac{1}{p^2-m_0^2+\frac{(p^2)^2}{\Lambda^2}+\frac{(p^2)^3}{\Lambda^4}+i\varepsilon}.$$

In step 2 we can at one time make the theory massless and eliminate the divergent diagram that you have calculated. We simply require that$^1$: $$\tilde G(p^2=0)^{-1}=0.$$ This ensures that the physical mass of the particles described by the theory is zero, and if you calculate the propagator to first order in $g_0$, you'll also find that the constant value of the divergent diagram is exactly canceled by the bare mass $m_0^2$.

$^1$ Actually, to first order in $g_0$, this condition suffices to renormalize the theory.

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    $\begingroup$ If you've never seen anything related to renormalization, this probably isn't going to make much sense, sorry. In that case, the gist of it is that your calculation is formally correct in the sense that it gives an undefined result. This needs to be treated with regularization and renormalization techniques. $\endgroup$ – pppqqq Jan 31 '17 at 21:05
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    $\begingroup$ I wasn't aware that you cannot have massless $\phi^4$ from the Lagrangian, perhaps that is why I have not seen that in class. Since the mass renormalization is multiplicative, the tadpole diagram must be zero in order for the theory to be massless. Thanks for this insight! :-) $\endgroup$ – Martin Ueding Jan 31 '17 at 21:12
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    $\begingroup$ @Martin Ueding I'm glad to help ! I'm also learning these things right now and my teacher focused on the "hard cut-off" scheme I outlined above for the $\phi ^4$ theory. In this scheme, the pole condition gives, to first order in $g_0$: $$m_0^2 +\frac{g_0}{2} \Delta (0;\Lambda)=0,$$ where $\Delta (0;\Lambda)$ is the regulated propagator at $x=0$. Forcing $m_0^2=0$ spoils the renormalizability of the theory even to first order, because it makes impossible to cancel a divergent term like $\Delta (0;\Lambda)$ in the propagator. $\endgroup$ – pppqqq Jan 31 '17 at 21:20

I have just tried to compute this diagram in momentum space and I think it gives a zero contribution in the massless case. Basically this is because within the loop you will get an integral like $$ \int \frac{\mathrm d^4 k}{(2\pi)^4} \frac{\mathrm i}{k^2 - m^2 + \mathrm i \epsilon} \,,$$ where $m = 0$. One professor said that this integral will give zero because for $m = 0$, there is no mass/energy scale that this loop integral could probe. I remember that the explanation did not satisfy me at the time, I unfortunately do not have a better one right now.

Then also with loops, it is not uncommon for those to give infinities. This will lead to regularization and eventually to renormalization which feels like Pandora's box. Your integral seems to diverge since there are eight powers of $u$ in the numerator (from the integration measure) but only six powers in the denominator (from the Green's functions). It might still be infinite. Then you have to go to momentum space and use dimensional regularization.

I can give you momentum space derivation of this tadpole diagram with dimensional regularization that I did a couple years ago as a homework assignment in the QFT class I took. You can download the original and reviewed PDF to read it in full. This is the excerpt:

enter image description here

enter image description here

The final result is proportional to $m$. Setting $m = 0$ will make this while thing vanish, therefore the result seems to be zero in the massless theory.

  • $\begingroup$ So in some sense an internal vertex that loops back to itself, is viewed as two internal vertices? (since we've got to integrate twice, as you say) $\endgroup$ – Greg.Paul Jan 31 '17 at 20:13
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    $\begingroup$ Wait, no. What I wrote does not make sense any more, to me. Thinking in momentum space, there is only one free loop momentum. And there is only one integral. Where did you get that propagator from, anyway? Shouldn't it be $\langle 0 | \phi(x) \phi(y) | 0 \rangle = \int \frac{\mathrm d^3 p}{(2\pi)^3} \frac1{2E_{\vec p}} \exp(- \mathrm i p \cdot (x - y))$? cf. (290). $\endgroup$ – Martin Ueding Jan 31 '17 at 20:29
  • $\begingroup$ I see, okay - I was thinking there should only be a single integral as well. Here's one link discussing this propagator physics.stackexchange.com/q/101886, there's also a bit on the propagator I've used in Birrell and Davies' QFT in Curved Spacetime. I think that the propagator you've written reduces to this form in the massless case somehow. $\endgroup$ – Greg.Paul Jan 31 '17 at 20:35
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    $\begingroup$ Looking through my own QFT notes I found that I have computed this diagram. It seems to vanish in the massless case, at least when one performs the computation with dimensional regularization. In the version of the propagator that you have, I still do not see how one should extract a sensible result. $\endgroup$ – Martin Ueding Jan 31 '17 at 21:00

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