In Chap. 7.1 of Peskin & Schroeder, the electron mass difference between the physical mass, $m$, and the Lagrangian bare mass, $m_0$ was calculated to be $$m-m_0 = \delta m \rightarrow \frac{3\alpha}{4\pi} \hspace{1mm} m_0\,\log\biggl(\frac{\Lambda^2}{m_0^2}\biggl)\tag{7.29}$$ which diverges when $\Lambda \rightarrow \infty$.

P&S then go on to explain this logarithmic dependence on $\Lambda$ conceptually by saying on p. 221

$\hspace{1cm}$"Suppose that $m_0$ were set to $0$. Then the two helicity components of the electron field $\psi_L$ and $\psi_R$ would not be coupled by any term in the QED Hamiltonian. This would imply that perturbative corrections could never induce a coupling of $\psi_L$ and $\psi_R$, nor, in particular, an electron mass term."

Why is it that setting $m_0=0$ causes the failure of the coupling between the helicity components? And how does this then lead to perturbative method to be no longer valid?

  • $\begingroup$ Does $\lambda$ here have mass dimension of $m^2$ here? Just asking as I don't have a copy of this book and the logarithm looks like the argument is not dimensionless. $\endgroup$
    – Triatticus
    Aug 13, 2020 at 10:04
  • $\begingroup$ @Triatticus $\Lambda$ has dimension of mass m here. Sorry I missed out the square factor on the $\Lambda$- I have now edited this. $\endgroup$
    – Student 1
    Aug 13, 2020 at 10:08

2 Answers 2


This post follows very closely reference [1]. A Dirac-spinor $\psi(x)$ (for simplicity the dependence of the spinors on $x=(t,\mathbf{r})$ is suppressed in the following)

$$\psi = \left(\begin{array}{c} \psi_L \\ \psi_R \end{array}\right) $$

can be decomposed in its chiral components $\psi_L$ and $\psi_R$ by the means of the $\gamma^5$ matrix:

$$\gamma^5 = \left(\begin{array}{cc} 0 & \mathbb{1} \\ \mathbb{1} & 0\end{array}\right)$$

(where $\mathbb{1}$ is a 2x2 unit matrix) in the following way:

$$\psi = \psi_L + \psi_R = \frac{1}{2}(1-\gamma^5)\psi + \frac{1}{2}(1+\gamma^5)\psi$$

The Lagrangian of the electron field can be written like:

$${\cal L} = \overline{\psi}(i\gamma^\mu \partial_\mu - m_0)\psi= \overline{\psi}_L i\gamma^\mu\partial_\mu \psi_L + \overline{\psi}_R i\gamma^\mu\partial_\mu \psi_R - m_0(\overline{\psi}_L \psi_R + \overline{\psi}_R \psi_L)$$.

Therefore if $m_0=0$, there is no coupling anymore between the chiral components $\psi_L$ and $\psi_R$. And the mass renormalisation does not destroy this decoupling, if it is then the case, since the perturbative correction is proportional to the bare mass which is zero. The perturbation theory remains valid, but the mass correction will be zero too. So the mass renormalisation does not change a zero mass.

This result actually is very good, because a massless Dirac theory enjoys an additional symmetry, the chiral symmetry that keeps the massless Lagrangian invariant:

$$\psi' = e^{i\gamma^5\phi} \psi$$

The found result shows that the chiral symmetry of the massless Lagrangian would not be destroyed by the perturbation theory.

EDIT: The mass renormalisation for fermions is nice, because $\delta m \propto m_0$ and it is even nice if $m_0\neq 0$ because it makes the mass correction term $\delta m \sim log(\Lambda/m)$. The latter is nice because even if one chooses as cut-off 10 times the mass of the universe, $\delta m$ remains tiny compared to $m_0$.

[1]: A. Zee, Quantum Field Theory in a Nutshell, Princeton University Press (2002)

  • $\begingroup$ Thank you for your answer. What I am not quite seeing, however, is why the lack of coupling between the helicity components is relevant for $\delta m$ to vanish? $\endgroup$
    – Student 1
    Aug 13, 2020 at 11:12
  • $\begingroup$ $\delta m = \frac{3\alpha}{4\pi}m_0 \log(\Lambda^2/m^2) =0$ if $m_0=0$. And if $m_0=0$, then ${\cal L}=\overline{\psi}_L i\gamma^\mu\partial_\mu \psi_L + \overline{\psi}_R i\gamma^\mu\partial_\mu \psi_R$, so the coupling term has gone. $\endgroup$ Aug 13, 2020 at 12:24
  • 1
    $\begingroup$ What I already said: The decoupling of the helicity components is not destroyed by the perturbation theory, or in different words, the chiral symmetry is maintained. In modern physics it's all about symmetry. Moreover, something that is already out of scope of the post: This symmetry makes the mass renormalisation "nice". However, for a theory where such a symmetry does not exist, the mass renormalisation is "ugly". This theory is that of a scalar field, its incarnation is the Higgs boson, and its "ugliness" is the hierarchy problem. (this explanation requires another post). $\endgroup$ Aug 13, 2020 at 12:36

Experimentally we observe the mass $m_\text{e}$. Then we write some approximate equations to describe it. First, they are Newton equations in external fields, and they work fine except for small inacuraccy. Then we move to QM with the same $m_\text{e}$. Finally we advance QED equations with the same $m_\text{e}$, and, at first sight, it works too, but the perturbative corrections surprizingly modify the mass term. Obviously, such a modification was not foreseen, and some theorists invent the bare mass $m_0$. They imply that the infinite (or no) corrections to the experimental mass are good, but the original mass (or, better, the original equations) are wrong; thus this bla-bla about the value of $m_0$. Zero is not good, according to them, but strongly negative value is good (?!). As to me, I think the initial approximation and interaction term in QED contain wrong parts; thus divergences. All the mainstream bla-bla is supposed to get rid of these divergent addenda to $m_\text{e}$, as a matter of fact.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.