Suppose we have a Green's function of the typical form

\begin{equation} G(k)=\frac{1}{k^2-m^2-\Sigma(k)} \end{equation} where $\Sigma(k)$ is the self energy of that particle. How exactly can we figure out how the self-energy modifies the effective mass of the particle? Some books always refer to the mass as the pole in the Green's function so the mass would be equal to $m_{ef}=\tilde{k}^2$ where $\tilde{k}$ is just the pole

\begin{equation} \tilde{k}^2-m^2-\Sigma(\tilde{k})=0 \end{equation}

However, other books say that the change in the mass is only the part of the self-energy that is independent of $k$. And then also there's the whole problem of $\Sigma$ having real and imaginary parts.


1 Answer 1


In $\varphi^3$ theory the exact momentum-space propagator can be written as
$\mathbf{\tilde \Delta}(k^2) = \frac{1}{k^2 + m^2 -i \epsilon - \Pi(k^2)} \tag{1}$
where $\Pi(k^2)$ is the self-energy.

Yet, the Lehmann-Kaellén form of the exact momentum-space propagator shows
$\mathbf{\tilde \Delta}(k^2) = \frac{1}{k^2 + m^2 -i \epsilon} + \int^\infty_{4 m^2} ds \rho(s) \frac{1}{k^2 + s -i \epsilon} \tag{2}$
where $\rho(s)$ is the spectral density.
This form has an isolated pole at $k^2 = -m^2$ with residue one, just as the propagator in free-field theory.

Equations (1) and (2) are consistent if and only if
$\Pi(-m^2) = 0$
$\Pi^{'}(-m^2) = 0$
where the prime denotes a derivative with respect to $k^2$.

That means that the pole in equation (1) is exactly $k^2 = -m^2$.
Moreover, in order to fix the parameters of the Lagrangian specifying the interacting quantum field theory, the parameter $m$ is fixed by requiring it to be equal to the actual mass of the particle (equivalently, the energy of the first excited state relative to the ground state).


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