I am struggling to understand the nature of the components of the Dirac matrices.

If we view the four components of a Dirac spinor as $\psi^a$ with $a$ being a 'spinor' index, then if a gamma matrix acts on this to give another spinor, then it's indices would be ... ?? $\gamma^{\mu b}{}_{a}$ where $\mu$ selects the gamma matrix, and $a,b$ are spinor indices specifying the components of the 4x4 matrix ?

Since the current four-vector is $$J^\mu = \bar{\psi} \gamma^\mu \psi$$ that suggests the $\mu$ index is a vector index here. Writing all the indices gives $$J^\mu = \bar{\psi}_a \gamma^{\mu a}{}_b \psi^b.$$

However, $$\bar{\psi} = \psi^\dagger \gamma^0$$ which makes it seem like I shouldn't view that as a vector index, because the zeroth component is being used without the rest!?

I'm clearly confusing a lot of things. So how exactly should we view the components (and thus indices) of these objects?

$$\gamma^{\mu a}{}_b,\ \psi^a,\ \bar{\psi}_a \text{ ... or } \bar{\psi^a} \text{ ?}$$

  • 3
    $\begingroup$ Who cares if the zero component is used separately? The charge of a current is $Q = \int d^3x \, J^0$. Does that mean $J^\mu$ is not a four-vector? $\endgroup$
    – knzhou
    Commented Mar 4, 2018 at 11:39
  • $\begingroup$ @knzhou I don't see how that is relevant. If it is, please expand it into an answer. Notice that if $\gamma^\mu$ satisfy the algebra, then $\Lambda^\mu{}_\nu \gamma^\nu$ do as well. So if $\psi$ transforms such that $\psi^\dagger \gamma^0 \psi$ is a scalar, then so too does $\psi^\dagger \Lambda^0{}_\nu \gamma^\nu \psi$. Which then means $\psi^\dagger \gamma^1 \psi$ is a scalar. And I assume it would bother you if I claimed $\int d^3x\ J^\mu$ was a scalar for any $\mu$, as that would appear to only be true if $J^\mu$ could only be zero. $\endgroup$
    – Coconut
    Commented Mar 4, 2018 at 12:37
  • $\begingroup$ @Coconut $\psi^\dagger \gamma^1 \psi$ does not transform as a scalar. Remember $ \psi^\dagger \to \psi^\dagger \Lambda_{1/2}^\dagger$ and $ \psi \to \Lambda_{1/2}\psi$ and $\gamma^\mu$ does not transform (it's not a goddamn vector), so you see that $ \psi^\dagger \gamma^1 \psi \to \psi^\dagger\ (\Lambda_{1/2}^\dagger \gamma^1 \Lambda_{1/2})\ \psi$. It is not true that $\Lambda_{1/2}^\dagger \gamma^\mu \Lambda_{1/2} = \gamma^\mu$, unless $\mu = 0$. $\endgroup$ Commented Mar 4, 2018 at 13:22
  • $\begingroup$ @NanashiNoGombe There I was not doing a coordinate transformation. I was instead changing my choice of $\gamma^\mu$ that satisfy the algebra. So what you seem to be telling me is that $\bar{\psi} = \psi^\dagger \gamma^0$ is not true in all choices of the gamma matrices. That seems to be the missing piece. There is something special about that choice, and hence what makes the zero-th component special. $\endgroup$
    – Coconut
    Commented Mar 4, 2018 at 13:28
  • $\begingroup$ @Coconut Different choices of gamma matrices are not related by Lorentz transformations. ${\Lambda^\mu}_\nu \gamma^\nu$ does not land you into another choice of gamma matrices. It gives you $\Lambda_{1/2}^{-1} \gamma^\mu \Lambda_{1/2}$ which is not another gamma matrix. The definition $\bar{\psi} = \psi^\dagger \gamma^0$ does not mention any choice of convention for what $\gamma^0$ is. Why would you think this definition is convention-dependent? Get Peskin and Schroeder and study chapter 3 please. $\endgroup$ Commented Mar 4, 2018 at 13:31

2 Answers 2


Gamma matrices are defined by the Clifford algebra

$$ \{\gamma^\mu, \gamma^\nu\}= 2g^{\mu\nu}\mathbb I_n \,. $$

So, you see the index $\mu$ in $\gamma^\mu$ runs from $0$ upto $D-1$ where $D$ is the number of spacetime dimensions. It does not mean $\gamma^\mu$ is a vector. The $\mu$ index here only tells you how many gamma matrices are there. The dimensionality of the matrices themselves is $n= 2^{[D/2]}$ where $[\cdot]$ gives you the integer part of a number. For example, in $(1+2)-$dimensions, $D=3$ and hence the Dirac matrices are $2^{[1.5]}= 2$ dimensional, which you recognize are the Pauli matrices. The labels of the entries of the gamma matrices are known as spinor indices. So, in 3 dimensions, for example, the $a,b$ in $\gamma^\mu_{ab}$ would run from $1$ to $2$.

What is a $4$-vector? It is something that transforms like a vector under Lorentz transformations $\Lambda$. Namely, $X^\mu$ is a vector if it transforms like

$$ X^\mu\to {\Lambda^\mu}_\nu X^\nu \,. $$

That's the definition! Just having a $4$-dimensional column vector with Greek indices labelling its entries does not make it a Lorentz vector. It needs to transform the right way.

Okay, so what is a spinor? A spinor is something that transforms like a spinor. Namely, $\psi$ is a spinor if it transforms, under a Lorentz transformation parametrized by $\omega_{\mu\nu}$, like

$$ \psi \to \Lambda_{1/2} \psi\, \qquad (\Rightarrow \overline\psi \to \overline\psi\ \Lambda_{1/2}^{-1}\ ) \,, $$

where $\Lambda_{1/2} = \exp{(-\frac i2 \omega_{\mu\nu} S^{\mu\nu})}$ and $S^{\mu\nu} = \frac i4 [\gamma^\mu, \gamma^\nu]$ generates an $n-$dimensional representation of the Lorentz algebra.

Let's make a remark on why we use something like $\overline \psi = \psi^\dagger \gamma^0$. Well, because we want to construct bilinear Lorentz scalars like $\psi^\dagger \psi$, but $\psi^\dagger \psi$ is not a Lorentz scalar precisely because the matrix $\Lambda_{1/2}$ is not unitary. Under a Loretz transformation,

$$ \psi^\dagger \to \psi^\dagger \Lambda_{1/2}^\dagger \ne \psi^\dagger \Lambda_{1/2}^{-1}\,.$$

However, we notice an interesting property of the gamma matrix $\gamma^0$.

$$ \boxed{ \Lambda_{1/2}^\dagger \gamma^0 = \gamma^0 \Lambda_{1/2}^{-1} }$$

This immediately tells us that defining something like $\overline \psi \equiv \psi^\dagger \gamma^0$ will do the job.

$$ \overline \psi \to (\psi^\dagger \Lambda_{1/2}^\dagger)\gamma^0 = \psi^\dagger \gamma^0 \Lambda_{1/2}^{-1} = \overline\psi \Lambda_{1/2}^{-1} $$

Because of this special property of $\gamma^0$, now we have that $\overline\psi\psi\to \overline\psi\psi$.

You can check that the gamma matrices also satisfy the relation

$$ \Lambda_{1/2}^{-1} \gamma^\mu_{ab} \Lambda_{1/2} = {\Lambda^\mu}_\nu \gamma^\nu_{ab}\,. $$

Understand that this is not a transformation of the gamma matrices under a Lorentz transformation. Gamma matrices are fixed constant matrices that form the basis of an algebra. They do not transform. The above is just a property of the gamma matrices due to them being generators of a particular representation of the Lorentz algebra.

However, this relation allows you to take the $\mu$ index in $\gamma^\mu$ "seriously". Because, due to this you can immediately see that under a Lorentz transformation, the current $J^\mu := \overline\psi \gamma^\mu \psi= \overline\psi^a \gamma^\mu_{ab} \psi^b$ indeed transforms like a vector.

$$ J^\mu \to {\Lambda^\mu}_\nu J^\nu \,.$$

  • $\begingroup$ Is that boxed property of $\gamma^0$ representation independent? I think something not fully general snuck in there, so it would be nice to see how you noticed the interesting property. $\endgroup$
    – PPenguin
    Commented Mar 7, 2018 at 22:47
  • $\begingroup$ @PPenguin You only need to assume that $\gamma^0$ is Hermitian and the other $\gamma-$matrices are anti-Hermitian, which means that $\gamma^\mu = \gamma^0 (\gamma^\mu)^\dagger \gamma^0$. Then it follows from the unboxed formula I wrote later $\Big(\Lambda_{1/2}^{-1} \gamma^\mu \Lambda_{1/2} = {\Lambda^\mu}_\nu \gamma^\nu\Big)$ up to a sign. The latter formula is representation-independent because it needs to hold in order for the Dirac equation to be Lorentz invariant. Check Schweber for a proof. $\endgroup$ Commented Mar 8, 2018 at 10:11
  • $\begingroup$ Hi, thanks for your excellent explanation, so could you explain more about where do we get the conclusion that the dimension of gamma matrix is $2^{[D/2]}$, is this a mathematica conclusion? $\endgroup$
    – Daren
    Commented Aug 24, 2022 at 11:33

Where did you get this formula $\bar{\psi}_a=\psi_a\gamma^0$? It is wrong, and one of the reasons is complex conjugation is missing.

So components of $\gamma$- matrices have index $\mu$ for different $\gamma$- matrices and spinor indices $a,b$. I usually write spinor indices as lower indices.

So the formula should be replaced by something like $\bar{\psi}_a=\psi_b^* \gamma^0_{ba}$.

  • 2
    $\begingroup$ Thank you for pointing out the missing complex conjugate, but it doesn't change the observation that the zero-th component is being used separate from the others. So how can the $\mu$ index be viewed as a vector index in some cases, but then a single component be used without the others in this case? $\endgroup$
    – Coconut
    Commented Mar 4, 2018 at 11:02
  • 1
    $\begingroup$ @Coconut : I appreciate that it looks strange, but it does not contradict anything. Why do we choose such a definition for the Dirac adjoint? Because it provides correct behavior under Lorentz transformations (en.wikipedia.org/wiki/Dirac_adjoint). $\endgroup$
    – akhmeteli
    Commented Mar 4, 2018 at 11:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.