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

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 $$m^2=m_0^2\Big(1-\frac{\lambda_0}{16\pi^2\epsilon}\Big)?$$

• I recognize that this doesn't answer your question, but life is so much better if you use renormalized perturbation theory. In that case the infinities cancel in such a more natural way, and one can also think about the finite pieces of the counter terms much more easily, too; see, e.g., Sterman p 285-287. – WAH May 28 '17 at 18:46
• Is it not for the same reason you neglect the higher order terms in the naive Kallen-Lehmann expansion -- because they're all branch cuts not poles? – gautampk May 28 '17 at 19:12

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.

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.