# Parity violating Dirac particle

We normally write down the Dirac Lagrangian as $${\cal L} _D = \bar{\psi} ( i \partial _\mu \gamma ^\mu - m ) \psi$$ but are the Lagrangian's, $$\bar{\psi} ( i \partial _\mu \gamma ^\mu \gamma ^5 - m ) \psi , \quad \bar{\psi} ( i \partial _\mu \gamma ^\mu - m \gamma ^5 ) \psi , \quad {\cal L} _D = \bar{\psi} ( i \partial _\mu \gamma ^\mu \gamma ^5 - m \gamma ^5 ) \psi$$ all equally as good but just don't agree with Nature? Furthermore, how would you check that the propagator is or isn't Parity invariant?

• I have a stupid guess. $\gamma^5$ is a pseudo scalar. Thus your first two Lagrangians are mixture of scalar and pseudo scalar. Since parity is experimentally conserved in electromagnetics, these two Lagrangians are rule out. – user26143 Feb 16 '14 at 16:09
• The last Lagrangian is a pseudo scalar. If we write the Hamiltonian, $$H= -i \bar{\psi} \partial_i \gamma^i \gamma^5 \psi + m \bar{\psi} \gamma^5 \psi$$, it is also a pseudo scalar. Consider the eigen equation, $H |n \rangle = E_n | n \rangle$. Applying parity transformation, $P H P^{-1} P|n \rangle = E_n P | n \rangle$, thus $- H P|n \rangle = E_n P | n \rangle$ thus $H P|n \rangle = (- E_n) P | n \rangle$, which implies the energies will no longer be positive definite. – user26143 Feb 16 '14 at 16:10
• @user26143: Thanks for your response and good point about the last Lagrangian! While the mixture of scalar and pseudo scalar would be intuitive, I'm not sure how to see that mathematically. – JeffDror Feb 16 '14 at 18:30
• Peskin & Schroeder has a few parity transformation for spinors on p. 65. The extra term above P&S I guess is the transformation on $\partial_{\mu} \psi$. Since the parity operator is implemented on the creation and annihilation operators, we may modify P&S Eq. (3.124) as $$P \partial_{\mu} \psi(x) P = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2E_\mathbf{p}}} \sum_s \left( \eta_a a_{\mathbf{-p}}^s u^s(p) (-ip^{\mu}) e^{-ipx} + \eta_b^* b^{s\dagger}_{\mathbf{p}} v^s (p) (ip^{\mu}) e^{ipx} \right)$$ – user26143 Feb 17 '14 at 7:32
• By changing variables as $\tilde{p}=(p^0,−\mathbf{p})$, we got one extra minus sign for $\mu=1,2,3$ as we may anticipate that $\partial_i$ is odd under partity. P.S. I made a mistake in my second comment, the Hamiltonian is not a pseudo scalar, but the $00$ component of energy-momentum tensor. Nevertheless it is still odd under parity transformation (by directly verify). – user26143 Feb 17 '14 at 9:42

All the alternatives to the Dirac Lagrangian are actually forbidden by the requirement of requiring the hamiltonian to be well behaved (bounded from below and unbounded from above) and hermiticity of the action. To see this most simply we write the Lagrangian in terms of the fundamental left and right handed fields, $\psi \equiv \left( \begin{array}{c} \psi _L \\ \psi _R ^c \end{array} \right)$. For the modified kinetic term, \begin{align} i\bar{\psi} \partial _\mu \gamma ^\mu \gamma ^5 \psi & = i \psi _L ^\dagger \partial _\mu \bar{\sigma} ^\mu \psi _L - i \psi _R ^{c, \dagger } \partial _\mu \bar{\sigma} ^\mu \psi _R ^{ c} \end{align} We see that the left handed kinetic term is well behaved but the right handed kinetic term has the wrong sign. Having a negative in front of the kinetic term results in an unphysical spectrum (see for example, this question).
Now we consider the proposed modification of the mass term. The complex conjugate of this term is: $$\left( m \bar{\psi} \gamma ^5 \psi \right) ^\dagger = m ^\ast \psi ^\dagger \gamma ^5 \gamma ^0 \psi = - m ^\ast \bar{\psi} \gamma ^5 \psi$$ Therefore, in order for this term to be hermitian we must have purely imaginary $m$ (one could imagine having real $m$ and just adding an hermitian conjugate but then the term vanishes identically). To see the meaning of this mass term we write it in terms of two component fields: $${\cal L} _{mass} = i \left| m \right| \left( \psi _L ^\dagger \psi _R ^c - \psi _R ^{ c \, \dagger } \psi _L \right)$$ We can consider a field redefinition, $\psi _R ^c \rightarrow - i \psi _R ^c$, $${\cal L} _{ mass} \rightarrow \left| m \right| \left( \psi _L ^\dagger \psi _R ^c + \psi _R ^{ c \, \dagger } \psi _L \right) = \left| m \right| \bar{\psi} \psi$$ Thus the parity-violating mass term is actually equivalent to the canonical one after a redefinition of the fields.