I need to ask something regarding the Dirac equation (for a charged particle in an electromagnetic field) with the slash notation, which i fail to understand. We have the Dirac equation with the slash notation:

($\gamma^\mu p_\mu -m$)$\psi$=0

In the presence of an electromagnetic field, we get:

($\gamma^\mu (p_\mu -eA_\mu) -m$)$\psi$=0

I know that we derivative the above expression by multiplying with $\beta$ matrix.

And,correct me if i am wrong $\gamma^\mu$ is a vector matrix, what i mean is that $\gamma^\mu$=($\gamma^0,\vec\gamma)$.

Now, in my script we change the Dirac equation by using $\overline \psi$.

I know that $\overline \psi$= $\psi^\dagger\gamma^0$. Therefore we initially conjugate transpose the equation and then multiply it with $\gamma^0 \gamma^0$:

($\gamma^\mu p_\mu -m$)$\psi$=0.

($\gamma^\mu (p_\mu -eA_\mu) -m$)$\psi$=0

($\gamma^\mu (i\partial_\mu - eA_\mu) -m$)$\psi$=0. I think here there must be a mistake because $p^\mu = i\hbar\partial^\mu$, where the index is up and not down,as shown in the equation. But this is not my main question, thought I'd happily accept an explanation or clarification about this thing.

Now we transpose conjugate:

$\psi^\dagger [(\gamma^\mu)^\dagger (-i\partial_\mu -eA_\mu) - m] $ = 0.

Now in order to transform this into an equation where $\overline \psi$ is present, we initially multiply with $\gamma^0 \gamma^0$ (as I stated above). Then we have:

$\psi^\dagger \gamma^0 \gamma^0 [(\gamma^\mu)^\dagger (-i\partial_\mu -eA_\mu) - m] $ = 0.

$\overline \psi \gamma^0 [(\gamma^\mu)^\dagger (-i\partial_\mu -eA_\mu) - m] $ = 0.

Now these is the part that i don't get at all. As i mentioned above $\gamma^\mu$ is a vector matrix of the form $\gamma^\mu$=($\gamma^0,\vec\gamma)$. Then in the skript is said that the following relation holds true:

$(\gamma^\mu)^\dagger = \gamma^0 \gamma^\mu \gamma^0$

How is this possible?

What we are doing here is this:

$(\gamma^\mu)^\dagger = \gamma^0 (\gamma^0,\vec\gamma) \gamma^0$

How does this even work? We have a matrix multiplying a vector, whose components are matricies? How? Like i don't know the rules on how to solve things like these. Nothing about the necessary algebra was taught to us.

Please tell me if i have a mistake in my equations, or reasoning!


  • 2
    $\begingroup$ In the very first line, it should be $p_\mu$ not $p^\mu$. $\endgroup$
    – J. Murray
    Mar 16, 2021 at 17:34
  • $\begingroup$ yes, you are right $\endgroup$
    – imbAF
    Mar 16, 2021 at 17:35

1 Answer 1


The expression $(\gamma^\mu)^\dagger = \gamma^0 \gamma^\mu \gamma^0$ is shorthand for $$(\gamma^0)^\dagger = \gamma^0 \gamma^0 \gamma^0$$ $$(\gamma^1)^\dagger = \gamma^0 \gamma^1 \gamma^0$$ $$(\gamma^2)^\dagger = \gamma^0 \gamma^2 \gamma^0$$ $$(\gamma^3)^\dagger = \gamma^0 \gamma^3 \gamma^0$$

It's sloppy - but standard - to write that $\gamma^\mu = (\gamma^0,\vec \gamma)$. $\gamma^\mu$ is the $\mu^{th}$ gamma matrix, where $\mu=0,1,2,$ or $3$. If anything, we should write something like $\boldsymbol \gamma = (\gamma^0,\vec \gamma)$ to denote that we are collecting the four gamma matrices into a matrix-valued 4-vector.

This is exactly the same as saying that $V^\mu$ is a vector. $\mathbf V=(V^0,V^1,V^2,V^3)$ is a vector, while $V^\mu$ is its $\mu^{th}$ component. It's important not to get mixed up and confuse vectors with their components.

Explicitly, $\psi^\dagger \gamma^\mu p_\mu$ means $\psi^\dagger(\gamma^0p_0 + \gamma^1 p_1 + \gamma^2 p_2 + \gamma^3 p_3)$. Inserting $\gamma^0\gamma^0 = \mathbb I$ yields $$\psi^\dagger\gamma^0\gamma^0(\gamma^0p_0 + \gamma^1 p_1 + \gamma^2 p_2 + \gamma^3 p_3)$$ $$=\overline{\psi}\gamma^0(\gamma^0p_0 + \gamma^1 p_1 + \gamma^2 p_2 + \gamma^3 p_3)$$

Multiplying by $\gamma^0$ from the right and noting the shorthand I mentioned above, this becomes

$$\overline{\psi}\bigg((\gamma^0)^\dagger p_0 + (\gamma^1)^\dagger p_1 + (\gamma^2)^\dagger p_2 + (\gamma^3)^\dagger p_3\bigg)$$

  • $\begingroup$ Ok, but in the dirac equation you have a summation sign between the $\alpha$ matricies and the gradient ones. And by multiplying with the beta vector you get the gamma vector with components. In my equation the $\gamma^\mu$ is the vector, no? $\endgroup$
    – imbAF
    Mar 16, 2021 at 17:45
  • $\begingroup$ The dirac equation has the summation sign, and when you multiply with beta, you must get all the gamma vectors,right? $\endgroup$
    – imbAF
    Mar 16, 2021 at 17:47
  • $\begingroup$ @imbAF $\gamma^\mu$ is the $\mu^{th}$ matrix, not a collection of matrices, as I said. It may help to write everything explicitly and in full rather than using index notation - you'll see what I mean. $\endgroup$
    – J. Murray
    Mar 16, 2021 at 17:51
  • $\begingroup$ So if i understand it right, the contracted indices imply the summation, whose Sign is not shown(Einstein convention ), and that the notation $ \gamma^\mu = (\gamma^0 ,\vec \gamma)$ is not an accurate notation? He is basically taking into consideration every multiplication between $\gamma^0$ and every component of the matrix-valued 4 vector ? $\endgroup$
    – imbAF
    Mar 16, 2021 at 17:55
  • $\begingroup$ @imbAF I've edited my answer to be more explicit. $\gamma^\mu=(\gamma^0,\vec \gamma)$ is an extremely common abuse of notation which persists because, once you are comfortable with the convention and know what you're doing, it is extremely convenient to use. $\endgroup$
    – J. Murray
    Mar 16, 2021 at 18:04

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