Energy-momentum relation for dressed particles and how interactions change mass

Question

When a free real scalar field $\phi(x)$ described by the Lagrangian \begin{equation}\mathscr{L}=\frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\frac{1}{2}m^2\phi^2\end{equation} is quantized, a quanta with energy E and momentum $\textbf{p}$ can be shown to obey the dispersion relation $$E^2-\textbf{p}^2=m^2$$ which enables us to identify $m$ as the mass of the quanta.

However, it is not possible to exactly solve the interacting theories such as the $\phi^4$ theory described by \begin{equation}\mathscr{L}=\frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4.\end{equation}

But is there a way to understand that in presence of the interaction, the mass of the dressed quanta will be different from $m$? I know that the pole of the dressed propagator changes in an interacting theory and therefore, the mass also shifts from the value $m$. But is it possible to see that by looking at the dispersion relation of the dressed particles?

In Peskin and Schroeder (the paragraph below Eqn. 7.1), a different mass $m_\lambda$ is assumed for the dressed particles. But from the mathematics, I can't see why, in general, $m_\lambda$ cannot be equal to $m$.

Answer 1

But from the mathematics, I can't see why, in general, $m_\lambda$ cannot be equal to $m$.

The physical mass $m_\lambda$ may be equal to $m$, the bare mass. This is precisely what happens in some SUSY theories. The point is that we may or we may not have $m=m_0$, and this can only be decided a-posteriori, once we calculate the two-point function and look at the position of the pole. A-priori, we cannot know if the Lagrangian parameter $m$ has anything to do with something measurable; it is a free parameter that we must fit to experiments.