Parity violating Dirac particle We normally write down the Dirac Lagrangian as 
\begin{equation} 
{\cal L} _D = \bar{\psi} ( i \partial _\mu \gamma ^\mu - m ) \psi 
\end{equation} 
but are the Lagrangian's, 
\begin{equation} 
\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
\end{equation} 
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?
 A: 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:
\begin{equation} 
\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 
\end{equation} 
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:
\begin{equation} 
{\cal L} _{mass} = i \left| m \right|  \left( \psi _L ^\dagger \psi _R ^c - \psi _R ^{ c \, \dagger } \psi _L \right) 
\end{equation} 
We can consider a field redefinition, $ \psi _R ^c \rightarrow - i \psi _R ^c  $,
\begin{equation} 
{\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 
\end{equation} 
Thus the parity-violating mass term is actually equivalent to the canonical one after a redefinition of the fields.
