I'm trying to understand how Pauli Villars Regularization works. I know we add ghost particles, but I want to see more precisely. To do this, we'll work with $\phi^3$ theory. The Lagrangian is $$ {\cal L} = \frac{1}{2}(\partial_\mu\phi)^2 - \frac{1}{2} m^2 \phi^2 + \frac{\lambda}{3!}\phi^3 $$ We wish to calculate the correction to the scalar propagator. We know that this integral diverges. We then introduce a Pauli Villars ghost in the Lagrangian $$ {\cal L}' = - \frac{1}{2} (\partial_\mu \phi')^2 + \frac{1}{2}M^2 \phi'^2 + \frac{\lambda}{3!}\phi'^2 \phi $$ This Lagrangian has a negative norm. We can see this by calculating the contribution of the free part to the Hamiltonian $$ {\cal H}' = -\frac{1}{2} {\dot \phi'}^2 -\frac{1}{2} \pi'^2 - \frac{1}{2} M^2 \phi'^2 $$ Because it has a negative norm, such a particle is called a GHOST PARTICLE. Now, from what I understand from current text, this field has a propagator $$ D_F'(x-y) = \int \frac{d^4p}{(2\pi)^4} \frac{e^{ i p (x-y) } } { p^2 - M^2 + i \epsilon} $$ This is where I'm having so much trouble. How can we prove that this is the propagator? I'm trying to use the usual method to find the propagator, and I seem to be stuck. Any help?

  • $\begingroup$ I think your propagator is missing a $-i$ (from $e^{iS}$), no? $\endgroup$ Aug 8, 2012 at 2:02
  • $\begingroup$ That is a convention. $\endgroup$ Aug 8, 2012 at 2:05

1 Answer 1


The ghost propagator connected with your Lagrangian has the opposite sign to the standard one. This makes the sum of both propagators well-defined.

How can we prove that this is the [free] propagator?

The free propagator is the green function of the free equation of motion (with the suitable boundary conditions given by the $i\epsilon$ terms), so applying the Klein-Gordon operator to the free propagator you must get the Dirac delta modulo a $i$ factor which depends on the convention one is using.

Added: Regularizing an integral is to replace a sick-defined integral by a well defined one. This process entails the introduction of a dimension-full parameter (and in some cases like in dimensional regularization a dimension-less parameter as well). There are several ways to do this and one usually —but not always— chooses the most symmetric one according to the problem in question. However, none of the known methods to deal with ultraviolet divergences in relativistic QFT is physical, that is, none of them corresponds to a physical effect. Some of them improve the behaviour of the integrand for high (ultraviolet) momenta: imposing a sharp cut-off (step function in the integrand) or a smother one like a gaussian $e^{-p^2/M^2}$. Likewise one can replace the propagator $\frac{1}{k^2+m^2}$ with:

$$\frac{1}{k^2+m^2}\frac{M^2}{k^2+M^2}$$ that is equal to ($M>>m$) $$\frac{1}{k^2+m^2}-\frac{1}{k^2+M^2}$$ So that one can think of the new term as something equivalent to add a very massive scalar particle with the wrong sign in the kinetic term (and therefore something unphysical). Or maybe one prefers to think of it as adding a very massive scalar particle with the wrong statistics in order to get a minus sign in each closed loop... each interpretation may be more convenient depending on the particular diagram one is regulating, but the interpretations are unnecessary because they are not physical. The significant is to define an expression that was undefined. At least, until somebody finds a physical regularization. Some people think that quantum gravity (through a violation of Lorentz invariance, for example) may provide a physical regulator for the UV divergencies of QFT. We do not know yet if quantum gravity is able to give us that gift.

  • $\begingroup$ What do you mean by "It is a convention"? I don't think that's it. We add a new field to the Lagrangian so as to cancel out the divergence in the integral. Also, I don't agree with your statement that-The ghost propagator has opposite sign to the standard one. This is because if that were true, the net contribution to the one-loop diagram is $$ \int \frac{d^4k}{(2\pi)^4} \left[ \frac{i}{k^2 - m^2 + i \epsilon} \frac{ i } { (p-k)^2 - m^2 + i \epsilon} + \frac{ -i}{k^2 - m^2 + i \epsilon} \frac{ -i } { (p-k)^2 - m^2 + i \epsilon} \right] $$ That wouldn't regulate the divergence at all! $\endgroup$
    – Prahar
    Aug 8, 2012 at 2:30
  • $\begingroup$ I meant that for a general field one can define the propagator as a vacuum expectation value of time ordered fields or as $i$ times the same thing as long as one be consistent. In PV regularization, depending on the graph one wants to regularize, one can introduce fields with the wrong sign in the kinetic term or with wrong statistics. The purpose is to replace a sick defined integral by a well defined one, the rest is blah, blah. Regarding your expression, the net contribution of P-V field in the loop should go with a minus sign to cancel high energy modes. $\endgroup$ Aug 8, 2012 at 7:16
  • $\begingroup$ (cont.) With this end one chooses the PV field to be a fermi field. It has a wrong statistic but nothing happens because it decouple s and is not physical. But in this case one obviously needs a complex field and not the real one the OP wrote. I did not realize that in my answer. $\endgroup$ Aug 8, 2012 at 7:17
  • $\begingroup$ @Prahar And the graph you are considering can also be regularized changing $\phi$ by $\phi '$ in the interaction term of $\mathcal{L'}$. $\endgroup$ Aug 8, 2012 at 7:29
  • $\begingroup$ Ahh! I see now. Thanks a lot. Cleared a lot of things up. $\endgroup$
    – Prahar
    Aug 8, 2012 at 13:36

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