# About the calculation of one-particle-irreducible two-point diagram

This is derived from the answer and comments of this Phys.SE question concerning the calculation of two-point one-particle-irreducible diagram.

On the one hand, according to the discussion on P.236 of An introduction to quantum field theory by Peskin and Shroeder, the truncated two-point one-particle-irreducible diagram, $M^2(p^2)$, possesses an imaginary part for an unstable particle. Mathematically, this corresponds to the case when two internal lines go on-shell simultaneously, which can be visualized by a cut through the two propagators.

On the other hand, however, according to Eq.(11.90) in section 11.6 on P.381, the inverse propagator is related to the second functional derivative of the effective action. The matrix formed by the second order derivative is manifestly symmetric. In the classical case, the second derivative of the potential is directly related to the mass of the particle. When one considers the corrections of quantum fluctuations, the argument is similar, according to the textbook. In particular, if one takes $p=0$, the inverse propagator $p^2-m_0^2-M^2(p^2)$ ($=p^2-m^2$ when properly renormalized) gives $m^2$. Therefore, in the presence of quantum fluctuations, the physics is essentially the same. Mathematically, one can associate the second functional derivative of effective potential to the dressed mass of the particle by simply taking $p=0$.

Now, Adam's answer to the above-mentioned question seems to prove that the effective potential is always real via path integral calculations with the Wick rotation. To be specific, by changing the contour of the integral $t$ from $(-\infty\rightarrow+\infty)$ to $(+i\infty\rightarrow -i\infty)$, one shows that the exponent of the functional integral of Eq.(9.42) is positive definite in the sense that $\mathcal{L}_E(\phi)$ possesses the form of an energy which is real and bounded from below. Above positive definite matrix implies that its eigenvalue and subsequently the dressed mass is always positive, due to linear algebra. But now I have a contradiction, since the existence of resonance implies a pole in the complex plane, and therefore, if one substitutes $p=0$ into the truncated two-point one-particle-irreducible diagram, one gets its mass which is not real.

I understand that the Wick rotation is essentially a trick to evaluate a path integral by choosing a different contour for the time coordinate. So, before and after the Wick rotation, $Z$, $\Gamma(\phi_{cl})$ (as well as all its functional derivatives and their corresponding forms in momentum space) are the same functionals. Why do they seemly have different pole structures on the complex plane of $p$? How these two approaches can be understood consistently?

Edits

1. I updated the above question to clarify my doubt according to Adam's answer below.
2. It is noted that the assertion that $\mathcal{L}_E(\phi)$ possesses the form of an energy, when transformed to momentum space, implies the requirement that the components of four momentum being real, $p^2$ as well as the bare mass term being positive. This is not violated by taking $p=0$ afterwards.
3. Part of my misunderstanding lies in the incorrect statement that substituting $p=0$ into $\Gamma^{(2)}$ gives the mass, which is not consistent with the renormalization condition. To be specific, Eq.(11.17) gives $\frac{d}{dp^2}(p^2-m_0^2-M^2(p^2))=0$ or $1-\frac{dM^2}{dp^2}=0$ at $p^2=m^2$. Therefore to the first order one has $p^2-m_0^2-M^2(p^2)=(p^2-m^2)\left(1-\frac{dM^2(p^2)}{dp^2}\right)$. By substituting $p=0$, the second term is not necessarily real. The whole expression being real implies both $m^2$ and $\frac{dM^2(p^2)}{dp^2}$ may be on the complex plane simultaneously.
• The sentence "In particular, if one takes $p=0$, the inverse propagator $p^2−m^2_0−M^2(p^2)$ ($=p^2−m^2$ when properly renormalized) gives $m^2$" is clearly wrong. Could you correct that, to help find where the problem is. – Adam Sep 24 '17 at 14:50

The OP's problem comes from a slight misunderstanding on how to find the particle life-time and mass, and when the self-energy is real or complex.

After Wick rotation, such that $p_W^2\geq 0$ for all $p_W$, one can indeed show that the effective potential Hessian matrix is definite positive. For arbitrary momenta $p_W$, one has the inverse propagator (suppressing all internal indices) $$\Gamma^{(2)}(p_W)=p_W^2+m_0^2+M^2(p_W)$$ where $m_0$ is the bare mass, and $M^2(p_W)$ is the self-energy.

(Quasi)-particles and resonances corresponds to poles (in $p$) of the propagator, or equivalently, to zeros of $\Gamma^{(2)}$. (Side remark : there might not be any resonance if the theory is strongly coupled. For example, at a critical point, $\Gamma^{(2)}(p_W)\propto p_W^{2-\eta}$ with $\eta>0$.)

Going back to real momenta, i.e $i p_{W,0}\to E+i\epsilon$ (the imaginary part insures that we will obtain the retarded propagator), and at zero spatial momentum, we have in general $$\Gamma^{(2)}(E)=-E^2+m_0^2+M_R^2(E)+i M_I^2(E),$$ where now the self-energy might have obtained an imaginary part $M_I^2$.

The zeros of $\Gamma^{(2)}(E)$ might happen at either real or complex energies, corresponding to stable and unstable particles.

To address directly the OP's question, there is no inconsistency between the fact that the self-energy can be complex, and that on the other hand, the effective potential hessian is positive definite. One discussion is done before Wick rotation, the second after.

EDIT : Note that the pole of the stable or unstable particle is not in $p=0$, nor in $p^2=0$, but (in the case of zero spatial momentum) in $p^2=-E^2$, thus not applying directly to the effective potential, unless one is truly interested in a zero momentum and energy state (i.e. Goldstone mode).

• Many thanks for your patient explanation! I edited my question to clarify my doubts. – gamebm Sep 24 '17 at 14:07
• I see! so for a resonance, when going on-shell $p^2=-E^2$ contains both finite real and imaginary parts. $\Gamma^{(2)}$ is real when $p^2$ is real, and meanwhile it goes to zero at $p^2=-E^2$. Right? – gamebm Sep 24 '17 at 18:42
• Yes (though one should be careful when $p^2<0$, as $M^2(p)$ might have singularities right on the real axis). You are indeed looking for poles in the complex $E$-plane, and since it is a self-consistent equation (since $M^2$ depends on $E$), $\Gamma^{(2)}$ might in turn become complex. – Adam Sep 24 '17 at 20:57