Why is it the finite piece of the self-energy often neglected to define the physical mass?

Question

Using the bare perturbation theory, for a $\lambda\phi^4-$theory in $d$-dimensions, the regularized self-energy turns out to be $$\Sigma=-\frac{\lambda_0 m_0^2}{16\pi^2\epsilon}+\text{finite}\tag{1}$$ where $\epsilon=4-d$. This contribution modifies the pole of the propagator from $$m_0^2\to m^2= m_0^2+\Sigma=m_0^2\Big(1-\frac{\lambda_0}{16\pi^2\epsilon}\Big)+\text{finite}\tag{2}$$ where $m^2$ is the physical mass. Why is it that the finite term often neglected and $m^2$ is just defined as $$m^2=m_0^2\Big(1-\frac{\lambda_0}{16\pi^2\epsilon}\Big)?$$

Answer 1

Why is it that the finite term often neglected

The finite term is NOT neglected. Rather, it could be absorbed by a counter term. If your text book just says "neglecting finite term", you should throw the book away immediately and ask for a refund.

To cancel or not to cancel the finite part by the counter term makes the all the difference between modified minimal subtraction ($\bar{MS}$) and minimal subtraction ($MS$) schemes. It's just a human convention that has no physics impact.

When it comes to a counter term, below is the rule of thumb,

• It has to be local, which means no additional momentum dependence other than that prescribed by the original Lagrangian term.
• It has to take care of all the divergences, which means the divergent portion is fixed and the finite/non-divergent portion could be set to any value you prefer.

Answer 2

This is the philosophy behind renormalization: one has a Lagrangian $\mathcal{L}(g_i)$ where $g_i$, $i=1,\ldots,N$ are some couplings (let me think of the mass as just another coupling). And then a machinery that takes as an input $\mathcal{L}(g_i)$ and outputs some observables $\Gamma^{(n)}(g_i;p_j)$, which will be typically correlation functions or scattering amplitudes ($p_j$ being the external momenta). This machinery consists in the computation of Feynman diagrams.

The experiments are able to fix the $\Gamma$'s for some configuration of the external momenta, for example one could say $$ \Gamma^{(n)}(g_i;p_j)|_{p_j\to \mu} = \tilde{g}_n\,,\quad n=1,\ldots,N \,.$$ Where $\tilde{g}_n$ is just a number given by experiments. So what we want to do is to tweak the $g_i$'s so that we get the desired result. As we know, after we regularize the theory, the answer is something of the form $$ \tilde{g}_n = f_n^{(0)}(g_i) + \frac{f_n^{(1)}(g_i)}{\varepsilon} + \frac{f_n^{(2)}(g_i)}{\varepsilon^2} + \cdots\,, $$ where the $f_n^{(l)}$ are some finite functions of the couplings. For any $\varepsilon>0$ this fixes the $N$ couplings $g_i$ in terms of the $N$ observables $\tilde{g}_n$ and we could just leave it at that. The $g_i$'s are functions $g_i(\varepsilon,\tilde{g}_n)$ and they can be fed to the Lagrangian, which in turns produces other observables and determines them in terms of the previous $N$ experiments as follows $$ \Gamma^{(m)}(g_i(\varepsilon,\tilde{g}_n);p_j)|_{p_j\to \mu} = \tilde{g}_m(\tilde{g}_n)\,,\quad m\neq1,\ldots,N\,. $$ It is crucial that the $\varepsilon$ dependence disappears in the end, but this is ensured by the theorem of power counting in renormalization theory.

I said that we could leave it at that because this is all we need from a theory, we want it to predict infinitely many experiments starting from $N$ of them as an input. However, for all practical purposes, it's better to compute once and for all the divergent part in $g_i(\varepsilon,\tilde{g}_n)$ and express it as $$ g_i(\varepsilon,\tilde{g}_n) = h_i^{(0)}(\tilde{g}_n) + \frac{h_i^{(1)}(g_i)}{\varepsilon} + \frac{h_i^{(2)}(g_i)}{\varepsilon^2} + \cdots\,. $$ We know how the divergent part works, so let's just focus on the finite piece. $h_i^{(0)}(\tilde{g}_n)$ is what we call the renormalized coupling. But this is totally arbitrary, we could also call $h_i^{(0)}(\tilde{g}_n) - 2\pi$ the renormalized coupling, as long as we specify that the divergent part is $2\pi + \mathrm{poles}$.

I'm sorry if I decided not to address your specific questions but rather give you a more general answer.