For a massive fermions, like the electron, chirality is not conserved in time. It is not a good quantum number although it is Lorentz invariant. The Dirac Hamiltonian (in particular, the mass term in it) is then supposed to change the chirality.

Is it possible for an electron to start out in the exclusively left-chiral state at $t=0$ i.e., $e_L=\frac{1}{2}(1-\gamma_5)e$ and get the chirality flipped to $e_R=\frac{1}{2}(1+\gamma_5)e$ at a later time?

But this is problematic. If the electron is at any moment either exclusively left-chiral or exclusively right-chiral, it cannot have a mass at that moment because Dirac mass requires both chiralities to be present at the same time (except neutrinos). Isn't it? I think at any instant of time an electron is a admixture of both left-chiral and right-chiral components.

  1. Then, what exactly is the mass term doing to the chirality? Is it in some sense "changing the proportion" in which left-chiral and right-chiral components mix/add up to make up the electron?

  2. Is there a way to mathematically see what is happening to the chiral projections of a massive fermion field with time? I was thinking along the following line. The fermion field evolves in time as $\psi(\textbf{x},t)=e^{-iH_Dt}\phi(\textbf{x},0)e^{iH_Dt}$ and $\psi(\textbf{x},0)=\psi_L(\textbf{x},0)+\psi_R(\textbf{x},0)$. The next thing involves exponentiating the Dirac Hamiltonian $H_D$ which looks like a formidable task.

EDIT: This confusion arose because often in Feynman diagrams people use a "cross symbol" to show that mass term flipping $e_L\rightarrow e_R$ or $e_R\rightarrow e_L$. I don't understand what do people mean by this.

I know that the mass term $m\bar{\psi}\psi=m(\bar{\psi_L}\psi_R+\bar{\psi_R}\psi_L)$ can be thought of as an interaction where the first term takes $e_R\rightarrow e_L$ and second term $e_L\rightarrow e_R$. But this is incomplete understanding. I want to understand what happens to the electron field as whole because at any instant of time it consists of both chiralities.

  • $\begingroup$ Write the equation of motion for a massive spinor in terms of the two chiral Weyl spinors. You get coupled equations telling you that classically, the two chiralities are not independent. Hence, the Hamiltonian is not diagonal in the Weyl basis. I'm not sure what exactly you want to know about this. $\endgroup$ – ACuriousMind Dec 12 '16 at 15:35
  • $\begingroup$ @ACuriousMind- Does an electron have a definite chirality at any instant of time? The answer is no. It is always a mixture of left and right-chiral components. Agree? If yes, my question is, what does it mean to say that mass flips chirality? There was no definite chirality to start with. $\endgroup$ – SRS Dec 12 '16 at 15:39
  • $\begingroup$ @ACuriousMind-Moreover, what do you mean by chirality is conserved? You might say, $[H_D,\gamma_5]\neq 0$. Well. But which quantity would you measure for the electron to show that its chirality is indeed changing? $\endgroup$ – SRS Dec 12 '16 at 15:44

Chirality is not well-defined for massive fields. A famous consequence of this fact are pion masses, which can be linked to Chiral Symmetry Breaking.

In the Lagrangian, you can define left- and right-handed Weyl fermions independently. A mass term will mix these, giving a massive Dirac fermion. Weyl fermions fulfill either $$ P_{L} \psi_L = \psi_L, \quad \text{or} \quad P_R \psi_R = \psi_R$$ but a Dirac fermion is not an eigenstate of the projection operators $$ P_{L,R} \psi_D \neq \alpha \psi_D. $$

There is a computational trick called a "mass insertion", which can be confusing in this regard:

A Dirac fermion can be considered as a coupled system of two Weyl fermions, where the mass is the coupling parameter. If a fermion's mass is small compared to the energy of a given process, one can approximate the Dirac fermion by its two (massless) Weyl components. The advantage is that for massless fields, loop integrals usually take much simpler forms.

Corrections to the massless case can then be included by adding a Feynman rule for the mass term in the Lagrangean, which is a biliear coupling between the left- and right-handed Weyl fermions. If you were to resum all possible mass insertions, the result is the same as if you had started with the massive Dirac fermion from the start. Since the underlying assumption of the approximation is that the mass is small compared to other energy scales in the theory, the corrections are usually small, though.

Sometimes, the diagram including a mass insertion is computed in order to show that the error induced by neglecting the mass is small indeed.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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