# Finding conjugate momentum of field in Dirac Lagrangian

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.

• Is the question about column vs. row vectors? – Qmechanic Nov 12 '20 at 13:44
• I am asking mainly for $\psi$ , how we use this to differentiate L .Since $\psi$ is a $4\times 1$ matrix. – P Rakesh Kumar Dora Nov 12 '20 at 14:43

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 $$\pi_{\alpha} \equiv \frac{\partial L}{\partial(\partial_0\psi_{\alpha})} = i\psi^*_{\alpha}.$$