Finding conjugate momentum of field in Dirac Lagrangian

Question

Dirac Lagrangian density is is $L= \bar{\psi}(i\gamma_{\mu}\partial^{\mu}-m)\psi$.

Here $L$ is a number. $\psi$ is a $4\times 1$ matrix. Now to get momentum conjugate to $\psi$, we define $\pi=\frac{\partial{L}}{\partial{\dot \psi}}$. This kind of procedure we follow before promoting $\psi$ and $\pi$ to operators. I.e we first identity the field and find out its momentum conjugate and then we promote them as operators.

But in Dirac field theory case to start with the field $\psi$ is a column matrix. I don't understand how we use this matrix $\psi$ to differentiate $L$, to get $\pi$?

After finding out $\pi$, $\psi$, we promote each component of them to operators (since they are $1\times4$ and $4\times 1$ matrices respectively). I understand this part. But I don't understand the above asked paragraph part.

Answer 1

You simply derivate every component. You can write (sum over repeated indices implied, where $\alpha,\beta$ run from $1$ to $4$) $$ L = \psi^*_{\alpha} \gamma_0 (i (\gamma_{\mu})_{\alpha \beta} \partial^{\mu} -m) \psi_{\beta}. $$ Now the components $\psi_{\alpha}$ are just complex numbers (or operators in the quantum theory) and so is $$ \pi_{\alpha} \equiv \frac{\partial L}{\partial(\partial_0\psi_{\alpha})} = i\psi^*_{\alpha}. $$