# Partial derivatives in the derivation of a Dirac Spinor

As per JF132's answer to Conservation of the axial current using Dirac equations of motion,

"since the gamma matrices $$\gamma^\mu$$ are $$4\times 4$$ matrices, and the conjugate Dirac spinors $$\bar{\psi}$$ are a $$1\times 4$$ row vectors, the following are not equivalent because of the rules of matrix multiplication,"

\begin{align} i\partial_\mu\bar{\psi}\gamma^\mu+m\bar{\psi} &=0 \\ i\gamma^\mu\partial_\mu\bar{\psi}+m\bar{\psi} &=0 \\ (i\gamma^\mu \partial_\mu + m)\bar{\psi} &= 0 \end{align}

My confusion is regarding how the partial derivative works here. It seems to me that $$\partial_\mu \bar{\psi}$$ is a scalar, and so I'm free to multiply $$\gamma^\mu$$ on the right instead of on the left if I want.

What is the flaw in my understanding?

Working out all the indices, the partial derivative has one Lorentz index $$\mu$$ and a Dirac index (spinor index) $$a$$ so that $$\partial_\mu \bar{\psi} \equiv \partial_\mu\bar{\psi}_a$$ So this quantity is not a Lorentz scalar, neither a Dirac scalar. The gamma matrices have also two Dirac indices $$\gamma^\mu \equiv \left(\gamma^\mu\right)^a_{\;b}$$ so you need to be careful on how you move things.
• Taking these spinor indices into account, $$i\partial_\mu\bar{\psi}_a(\gamma^\mu)^a{}_b+m\bar{\psi}_a =0$$ it still is not clear to be how in that equation, you aren't allow to switch the order of the partial psi and gamma matrix. Can you expound on how clarifying spinor indices helps you here? Commented Mar 19, 2020 at 13:15
• Strictly speaking, if $\psi$ is a column vector, $\bar\psi$ is a column vector. Since the product. $\bar{\psi}\gamma^\mu$ depends only on a low index, the product has to be a row vector. The only way of taking a product between a row vector and a matrix to get a row vector, is to multiply with the vector on the left. Commented Mar 19, 2020 at 13:38
• Still struggling a bit with $\partial_\mu \psi^a$. If $\psi^a$ with a Dirac index, is a column vector, what is the object $\partial_\mu \psi^a$ with its one curved index and one Dirac\spinor index? I think I'm realizing why the vielbein/tetrad formalism is needed to couple fermions with gravity hahah Commented Mar 26, 2020 at 14:10