# Dirac Lagrangian and the meaning of $|\bar{\psi} \psi|^2$

Consider the following Dirac Lagrangian

\begin{align} L = \bar{\psi} (i \gamma^{\mu}{\partial_{\mu}} - m) \psi \end{align}

Suppose I extend $$\psi = \psi_L + \psi_R$$, with the projectors $$P_L, P_R$$ so that $$\psi_L = P_L \psi, \psi_R = P_R \psi, P_L + P_R=1, P_L^2=P_L, P_R^2=P_R$$, which is the projections $$P_{L,R} = \frac{1 \pm \gamma^{5}}{2}$$.

With this decomposition the Dirac Lagrangian becomes \begin{align} L= \bar{\psi}_{L} i \gamma^{\mu} \partial_{\mu} \psi_{L} + \bar{\psi}_{R} i \gamma^{\mu} \partial_{\mu} \psi_{R} - m (\bar{\psi}_{L} \psi_{R} + \bar{\psi}_{R} \psi_{L}) \end{align}

I want to know the physical meaning of $$|\bar{\psi} \psi|^2$$, under some conditions, i.e. when the case of $$\psi_L=0$$, or $$\psi_{R}=0$$ or both them are not zero.

Can we interpret this as the usual QM, i.e., the probability as in the Schrodinger equation?

How one can interpret the wave function or probability in Dirac Lagrangian?

If you consider the relativistic QM, the answer is yes, simply because in this case $$\psi$$ are not quantum fields but relativistic wave functions and $$\psi^{\dagger}\psi$$ satisfies a conservation law of probability current.

However, you consider the lagrangean, so I guess you are interested in quantum field theory. In this case, the answer is no. Quantum fields are not related to wave functions. In other wards, quantum fields is not a quantized wave function. The main difference is wave functions are state, and quantum fields are operator that acts to state.

The “operator” $$|\bar\psi\psi|^2$$ is one example of the interaction term ( see NJL model), where we note that $$|\bar\psi\psi|^2$$ is an operator not a wave function.

Also, since probabilities and wave functions are defined as a set with the Hilbert space in which the quantum field operator acts, looking only at the Lagrangian it does not include probabilities and wave functions in it.

This is as meaningless as looking at only the momentum operator $$\hat{P}$$ or the Hamiltonian operator $$\hat{H}$$ in QM and wondering what the wave function of this system is, because Lagrangeans, Hamiltonians and quantum fields are all operators, not states.

Thus, what we need to do is first define the appropriate state space; in the case of QFT, for example, the vacuum state is defined by $$\hat{H}|0\rangle=0$$, so we can use this to define a probability amplitude of transition amplitude like $$|\langle 0|\bar\psi\psi|0\rangle|^2.$$ One physical interpretation of the amplitude above is a transition amplitude from the vacuum $$|0\rangle$$ to the multi-particle state $$\bar\psi\psi|0\rangle.$$

If you want to know more about QFT, I recommend to read some famous textbook like Srednicki.