# Renormalized mass

I am reading Schwarz QFT and I reached the mass renormalization part. So he introduces, after renormalization, a physical mass, defined as the pole of the renormalized propagator, and a renormalized mass which can be the physical mass, but can also take other values, depending on the subtraction schemes used. Are these masses, other than the physical one, observable in any way experimentally, or are they just ways to do the math easier (using minimal subtraction instead of on-shell scheme, for example)?

Also, in the case of charge renormalization, the explanation was that due to the vacuum polarization, the closer you are to the charge, the more you see of it, so the charge of the particle increases with the momentum you are testing the particle with. However, I am not sure I understand, from a physical point of view, why do you need to renormalize the mass. Is this physical mass (defined as the pole of the renormalized propagator) the same no matter how close you get, or it also changes with energy? And if it changes, what is shielding it at big distances?

• The 2 point correlator contains information about the propagation character (including its mass) of a field. The point is that it is different in free and interacting theories. In the presence of interactions, quantum effects (loops) can (and must) be included in the calculation of the full 2 point function. These quantum effects shift the mass. – Avantgarde Oct 18 '18 at 5:50
• I can accept this (although it is not clear to me how is the mass changed, compared to vacuum polarization, where the reason is pretty clear for charge). However, is this changed mass fixed? So for QED, the pole of the propagator is the actual mass of the particle no matter how close you are to the particle, or is it also a function of energy (as far as I can tell from Schwarz it is not a function of energy, which is different from the electric charge case)? – Alex Marshall Oct 18 '18 at 13:28

At least in a model with only one species, the mass is (inversely) related to the correlation length, so one way to build intuition about mass renormalization is to think about it in terms of how the interaction term affects the correlation length. This is meant to address the question "why do you need to renormalize the mass" in a relatively straightforward, mathematically clear way.

To be specific, consider the $$\phi^4$$ model. After replacing continuous $$D$$-dimensional space with a finite lattice to make everything mathematically unambiguous, the Hamiltonian may be written $$H = \frac{b^D}{2}\sum_\mathbf{x}\left(\dot\phi^2(\mathbf{x})+\sum_\mathbf{b}\left(\frac{\phi(\mathbf{x}+\mathbf{b})-\phi(\mathbf{x})}{b}\right)^2+\mu\phi^2(\mathbf{x})+\frac{\lambda}{12}\phi^4(\mathbf{x})\right)$$ where $$\mathbf{x}$$ is a lattice site, $$\mathbf{b}$$ is a lattice basis vector, and $$b$$ is the lattice spacing. The sum over $$\mathbf{x}$$ is a lattice version of an integral over all space, and the term with the sum over $$\textbf{b}$$ is a lattice version of the gradient term $$(\nabla\phi(\mathbf{x}))^2$$. The overhead dot on $$\dot\phi$$ denotes the time-derivative of $$\phi$$ (in the Heisenberg picture), as usual.

Now, here's the intuition. First suppose that $$\lambda=0$$. In this case, we know that the physical mass $$m$$ is related to the coefficient $$\mu$$ by $$\mu=m^2$$. The kinetic term, the only term that connects different lattice sites, is responsible for the fact that correlation function $$\langle 0|\phi(\mathbf{x})\phi(\mathbf{y})|0\rangle - \langle 0|\phi(\mathbf{x})|0\rangle\,\langle 0|\phi(\mathbf{y})|0\rangle$$ is non-zero for $$\mathbf{x}\neq\mathbf{y}$$. After re-scaling the field and the time-parameter to put the $$\lambda=0$$ Hamiltonian in the form $$H = \frac{b^D}{2}\sum_\mathbf{x}\left(\dot\phi^2(\mathbf{x})+\sum_\mathbf{b}\left(\frac{\phi(\mathbf{x}+\mathbf{b})-\phi(\mathbf{x})}{mb}\right)^2+\phi^2(\mathbf{x})\right),$$ we see that the correlation length is necessarily determined by the combination $$mb$$, so the correlation length must be $$1/m$$ in units of the lattice spacing $$b$$.

Now suppose that $$\lambda>0$$ and $$\mu=0$$. Although the 2-point correlation function cannot be computed in closed from any more, the same scaling argument indicates that the correlation length is determined by the combination $$b\sqrt{\lambda}$$. The relationship between correlation and physical mass (which is related to identifying the physical mass with the pole in the propagator) then tells us that the physical mass must be non-zero in this case, even though $$\mu=0$$. In other words, when $$\mu=0$$, the physical mass is induced entirely by the coupling term.

To see what happens when $$\mu$$ and $$\lambda$$ are both non-zero, choose any value $$\lambda\geq 0$$ and consider how $$\mu$$ must be tuned to make the correlation length infinite (which corresponds to zero physical mass). If $$\lambda=0$$, then we know that the choice $$\mu=0$$ makes the correlation length infinite. We just saw that if $$\lambda>0$$, then the correlation length is finite if $$\mu=0$$, so to make the correlation length infinite again we must choose $$\mu<0$$ to offset the effect of the interaction term. Letting $$\mu_c(\lambda)$$ denote this special (negative) value of $$\mu$$ that makes the correlation length infinite, this says that if $$\lambda>0$$, then choosing $$\mu>\mu_c(\lambda)$$ will give a non-zero physical mass (that is, a finite correlation length), even if $$\mu$$ is still negative!

By the way, choosing $$\mu<\mu_c(\lambda)$$ gives spontaneous symmetry breaking (of the discrete $$\phi\rightarrow -\phi$$ symmetry).

This whole picture is confirmed by numerical calculations, some of which can be found in Luscher and Weisz (1987), "Scaling laws and triviality bounds in the lattice $$\phi^4$$ theory (I). One-component model in the symmetric phase", Nuclear Physics B 290: 25-60, and some in Hasenbusch (1999), "A Monte Carlo study of leading order scaling corrections of $$\phi^4$$ theory on a three dimensional lattice", https://arxiv.org/abs/hep-lat/9902026.

You are confused because you think that the counterpart of renormalized charge is renormalized mass. It's wrong (at least, it's not accurate)! Actually, the counterpart of renormalized charge $$e$$ is self-energy $$\Sigma$$. While the renormalized charge $$e(p^2)$$ is momentum dependent , so is self-energy $$\Sigma(\not{p})$$.

Some clarifications:

• More precisely, we should be talking about momentum dependence of renormalized coupling/1PI vertex in place of renormalized charge. We are using OP's terminology here. Note that one could instead define renormalized charge at a fixed momentum only.
• The renormalized charge is usually phrased as renormalization scale $$\mu$$ dependent $$e(\mu)$$ in the renormalization group setting. But I personally think that the angle of "momentum dependent" $$e(p^2)$$is more elucidating.

Let's look at the fermion propagator $$G = \frac{i}{\not{p}-m_0 - \Sigma(\not{p})+i\epsilon}$$ where self energy can be generally expressed as $$\Sigma(\not{p}) = a(p^2) + b(p^2)\not{p}.$$ To simplify our discussion, let's assume that (which means there is no wave function renormalization) $$b(p^2) = 0.$$ If we further expand self energy as $$\Sigma(p^2) = a(p^2) = m_0' + c_1p^2 + c_2p^4 + ...$$ we will find out that $$m_0'$$ is divergent, while $$c_1$$ and $$c_2$$ are finite. The whole (mathematically shady) mass renormalization business is hinging on the assumption that $$m_r = m_0 + m_0'$$ is finite (or equivalently, $$m_0 = m_r - m_0'$$, regarding $$m_0'$$ as mass counter term), so that the fermion propagator $$G = \frac{i}{\not{p}-m_0 - \Sigma(p^2)+i\epsilon}$$ $$= \frac{i}{\not{p}- (m_r + c_1p^2 + c_2p^4 + ...) + i\epsilon}$$ is finite and well defined.

The physical mass $$m_p$$ is simply determined by the pole of the fermion propagator $$m_p^2 = (m_0 + \Sigma(m_p^2))^2 = (m_r + c_1m_p^2 + c_2m_p^4 + ...)^2.$$

Note that $$m_p$$ (or $$m_r$$) can only be determined by experiment, whereas $$m_0$$ and $$m_0'$$ are divergent.

On the other hand, $$c_1$$ and $$c_2$$ can be calculated (is that cool!), so that we know how self-energy $$\Sigma(p^2)$$ (mutatis mutandis, renormalized charge $$e(p^2)$$/coupling/vertex as well) runs with momentum/energy $$p^2$$.

And, by definition, $$m_p$$ does not run.

A side note: you can entertain the notion that the counterpart of physical mass $$m_p$$ (corresponding to the pole of renormalized fermion propagator) is the Landau pole energy scale $$\Lambda_{Landau}$$ (corresponding to the Landau pole of renormalized charge/coupling), with the understanding that the latter is a spurious one, indicating the failure (or triviality) of perturbative QFT at high energy scale.

• The renomalised charge is not momentum-dependent. In $\mathrm{OS}$ it is a constant ($\approx0.3$), and in $\mathrm{MS}$ it depends on the mass $\mu$. More generally, it may depend on the cutoff scale or any other mass scale introduced by the renormalisation scheme; but it is not, in any case, momentum-dependent. – AccidentalFourierTransform Oct 18 '18 at 15:19
• Thank you @AccidentalFourierTransform, see updates. – MadMax Oct 18 '18 at 15:41
• To be clear, what you call $e(p^2)$ is what is usually denoted by $\Gamma^\mu(p_1,p_2)$, right? The 1PI vertex function associated to $\langle\psi(p_1)A^\mu(p_2)\bar\psi(p_3)\rangle\delta(p_1+p_2+p_3)$? – AccidentalFourierTransform Oct 18 '18 at 15:47
• Right. please see update. – MadMax Oct 18 '18 at 15:49