# spinor vs vector indices of Dirac gamma matrices

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?

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

• 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? – knzhou Mar 4 '18 at 11:39
• @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. – Coconut Mar 4 '18 at 12:37
• @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 $\mu = 0$. – Nanashi No Gombe Mar 4 '18 at 13:22
• @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. – Coconut Mar 4 '18 at 13:28
• @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. – Nanashi No Gombe Mar 4 '18 at 13:31

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.

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

• 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. – PPenguin Mar 7 '18 at 22:47
• @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. – Nanashi No Gombe Mar 8 '18 at 10:11

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 $\bar{\psi}_a=\psi_b^* \gamma^0_{ba}$.

• 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? – Coconut Mar 4 '18 at 11:02
• @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). – akhmeteli Mar 4 '18 at 11:24