How can the mass of an unstable composite particle become complex?

To show where the resonances in cross sections come from, one usually considers the exact propagator in the interacting theory, which for a scalar is $$iG(p^2)=\frac{i}{p^2-m_R^2+\Sigma(p^2)+i\epsilon}\tag{24.47}$$ where $$i\Sigma$$ is the sum of all 1PI self-energy graphs and $$m_R$$ is the renormalized mass. For an unstable particle $$\Sigma$$ becomes complex and one defines the pole mass by $$m_P^2=m_R^2-Re\Sigma$$ which then leads to the Breit-Wigner distribution in a cross section with width $$2m_P \Gamma$$, if $$\Gamma$$ is the decay rate of the particle.

However, I read chapter 24.3 "Polology" in Schwartz, where he shows that correlation functions always have poles when on-shell intermediate particles can be produced. If there is a one-particle state $$\lvert\Psi\rangle$$ of mass $$m_\Psi$$ in the theory then the correlation function $$G_n(p_1,...,p_n)$$ has a term proportional to $$\frac{1}{p^2-m_\Psi^2+i\epsilon}\tag{24.105}$$ for $$p\approx m_\Psi$$, if $$p$$ is the sum of all incoming 4-momenta. The state $$\lvert\Psi\rangle$$ doesn't have to be an elementary particle, it could also be a composite particle of the theory, but we assumed that $$m_\Psi$$ is the actual mass of this state, connected to the energy by $$m_\Psi^2=E_\Psi^2-\vec{p_{\Psi}}^2$$ so it is definitely real. Also we have already included all interactions in the calculation, no perturbation theory was used. Now this looks to me like the cross-section will have an actual divergence at $$p^2=m_\Psi^2$$ instead of a resonant peak of width $$\Gamma$$. How could the mass become complex to yield a Breit Wigner distribution?

In the beginning of section 24.3 of Schwarz, the assumption is that you have some one-particle state $$|\Psi\rangle$$ with mass $$m_\Psi$$ in your theory. But, unstable particles aren't actually one-particle states, they are not eigenstates of the Hamiltonian (since they decay).

You can check that as you take the width $$\Gamma \to 0$$, the complex pole of the unstable particle approaches the real line, the lifetime of the particle goes to infinity, and the Breit-Wigner distribution approaches looking like a pole as you would expect from a real particle. So, depending on the energies involved in an experiment, you can treat certain unstable particles as stable.

A resonance (i.e. an unstable particle) does not correspond to an asymptotic state in the strict sense. It will show up in the correlation functions of the asymptotic fields (associated with the asymptotic states of the theory) as a pole in the complex plane shifted away from the real axis. On the other hand, asymptotic states (even if they are a stable bound states) show up as poles on the real axis.

Example: Take pure QCD with the electroweak interaction turned off. In this "ideal" world, the pions are stable (corresponding to asymptotic states), but the $$\rho$$ is a resonance, as it decays into $$\pi \pi$$. Note that the (interpolating) asymptotic fields are not necessarily identical with the dynamical fields of the theory, QCD being an extreme example.

1. In this answer, we assume that OP already does understand that:

• the one-particle state $$|\Psi\rangle$$ in section 24.3 of Ref. 1 corresponds to an asymptotic state in the Hilbert space, not an unstable (possible composite) particle.

• the optical theorem relates the total decay rate $$\Gamma$$ to the imaginary part of the self-energy $$\widetilde{\Sigma}$$, cf. eqs. (24.13)+(24.48).

• the connected propagator (24.47) is related to the self-energy, cf. e.g this Phys.SE post.

2. Instead, OP question seems to be how a connected amputated Feynman diagram in momentum space that contributes to the self-energy $$\widetilde{\Sigma}$$ could become non-real? Well that's a good question. Let's list the imaginary items in the Feynman diagram:

• Each vertex has a factor $$\frac{i}{\hbar}$$.

• Each internal propagator has a factor $$\frac{\hbar}{i}$$.

• Wick rotation of each loop integral yields a factor $$i$$.
1. If these were the only sources of imaginary units $$i$$, then $$\Sigma$$ would be real: $$\frac{i}{\hbar}\widetilde{\Sigma} ~=~\left(\frac{i}{\hbar}\right)^V\left(\pm\frac{\hbar}{i}\right)^Ii^L\cdots ~=~\frac{i}{\hbar}\hbar^L\cdots,$$ where we used that $$L=I-(V-1)$$, cf. e.g. my Phys.SE answer here.

2. However, we should also include the Feynman $$i\epsilon$$ prescription from the Feynman (F) propagator \begin{align}\frac{i}{\hbar}\widetilde{\Pi}_F(k) ~\equiv~&\frac{-1}{k^2-m^2+i\epsilon}\cr ~=~&{\rm PV}\frac{-1}{k^2-m^2}+i\pi\delta(k^2-m^2)\cr ~=~&\frac{i}{\hbar}\widetilde{\Pi}_A(k)+2i\pi\theta(k^0)\delta(k^2-m^2),\end{align}\tag{24.27} or equivalently, from the advanced (A) propagator \begin{align}\frac{i}{\hbar}\widetilde{\Pi}_A(k) ~=~&\frac{-1}{k^2-m^2-i\pi{\rm sgn}(k^0)\epsilon}\cr ~=~&{\rm PV}\frac{-1}{k^2-m^2}-i\pi{\rm sgn}(k^0)\pi\delta(k^2-m^2),\end{align}\tag{24.28} where we used the Sokhotski-Plemelj theorem. Note the imaginary part with the Dirac delta distribution enforces a mass-shell condition. Tree-amplitudes will still be real, but for loop-amplitudes this leads to a possible non-real amplitude.

In subsection 24.1.2 this is tied to the Cutkosky cutting rules.

3. In subsection 24.1.1 there is given an explicit example, where an infinitesimal imaginary shift can induce to a finite imaginary shift by analytic continuation. The prototype example is $$\ln(A-i\epsilon)~=~\ln |A| - i\pi\theta(-A),\qquad A~\in~\mathbb{R}, \tag{24.19}$$ which then is matched with the decay of an unstable particle.

References:

1. M.D. Schwartz, QFT & the standard model, 2014; chapter 24. Uses the $$(+,-,-,-)$$ signature convention. NB: In early editions, the advanced (A) propagator is conflated with the retarded (R) propagator, cf. the errata.

2. M. Veltman, Diagrammatica, 1994; chapter 8. Uses the $$(-,+,+,+)$$ signature convention.

• Additional references: 3. G. Sterman, Intro to QFT, 1993; section 9.6. Uses the $(+,−,−,−)$ signature convention. Commented Oct 27, 2023 at 11:58