# How to know how the self-energy changes the mass?

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

$$$$G(k)=\frac{1}{k^2-m^2-\Sigma(k)}$$$$ where $$\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

$$$$\tilde{k}^2-m^2-\Sigma(\tilde{k})=0$$$$

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.

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).