# Lorentz transformation of Gamma matrices $\gamma^{\mu}$

From my understanding, gamma matrices transforms under Lorentz transformation $\Lambda$ as \begin{equation} \gamma^{\mu} \rightarrow S[\Lambda]\gamma^{\mu}S[\Lambda]^{-1} = \Lambda^{\mu}_{\nu}\gamma^{\nu} \end{equation} Where $S[\Lambda]$ is the corresponding Lorentz transformation in bispinor representation. So my question is: When we change from on frame to another, are we allowed to write $\gamma'^{\mu} = \Lambda^{\mu}_{\nu} \gamma^{\nu}$ where $\gamma'^{\mu}$ is the transformed version of $\gamma^{\mu}$? If yes, then when we write down $\gamma^{\mu}$ explicitly (in some representation) as we do in any standard QFT textbook like \begin{equation} \gamma^{\mu} = \begin{pmatrix} 0 & \sigma^{\mu} \\ \bar{\sigma}^{\mu} & 0 \end{pmatrix} \end{equation} do we assume any specific frame of reference? If so, which frame? Because if I apply Lorentz transformation such as a boost along $x$-direction I will have \begin{equation} \gamma'^{0} = \cosh(\eta)\gamma^0 + \sinh(\eta)\gamma^1 = \begin{pmatrix} 0 & \cosh(\eta) + \sigma^1 \sinh(\eta) \\ \cosh(\eta) - \sigma^1\sinh(\eta) & 0 \end{pmatrix} \end{equation} and similarly for $\gamma^i$. I understand that at the end any choice of reference frame does not matter because that's the point of relativistic theory like QFT. The term like $\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi$ in the theory will remain invariance under Lorentz transformation. But just like how we have the momentum shell condition $p^{\mu}p_{\mu} = -m^2$ in all frames but $p^{\mu}$ itself will change from one frame to the other and in particle rest frame we have $p^{\mu} = (m,0,0,0)$ it seems to me that by writing down $\gamma^{\mu}$ explicitly as above we are picking a specific frame. Could someone please clarify this to me?

• Note that in the usual approach to (say) the Lorentz covariance of the Dirac equation, the gamma matrices are taken to be scalars (Lorentz invariant) and so do not change under transformation. It is the Dirac spinor which transforms. There is an alternative (formally equivalent) approach which treats the spinor as a scalar and transforms the gamma matrices as components of a 4-vector. I was getting horribly muddled between these approaches earlier, which led to my (now deleted) totally erroneous answer. – tok3rat0r Jul 1 '15 at 20:04
• I can attempt to formulate an answer, if you would like, but can't guarantee even after extensive reading that I've really got a handle on it! So it would be up for discussion, rather than providing a definitive answer. – tok3rat0r Jul 1 '15 at 20:08
• So there are more than one formalisms to this? I didn't know that before. I'm now thinking if the transformation of $\gamma^{\mu}$ should be $\gamma^{\mu} \rightarrow S[\Lambda]\gamma^{\nu}S[\Lambda]^{-1}\Lambda^{\mu}_{\nu} = \gamma^{\mu}$. That is we transform both Lorentz indices and (hidden) spinor indices simultaneously, but they cancelled at the end by the properties of Clifford algebra. So basically $\gamma^{\mu}$ doesn't transform at all. That's just some ideas, not sure if it's correct. Any answers are welcome don't need to be definitive. – user113988 Jul 2 '15 at 1:24
• Indeed! I gather the transformation is exactly as you have written it, as seen from eq. (14) in the article. The spinors by contrast transform as $\psi^{\mu}(x)\rightarrow S[\Lambda]^{\mu}_{\nu}\psi^{\nu}(\Lambda^{-1}x)$. Defining the adjoint spinor $\bar{\psi}(x)=\psi^{\dagger}(x)\gamma^0$, where $\psi^{\dagger}(x)$ is just the Hermitian conjugate, we can then check that products of $\psi$ and $\gamma$ (e.g. the current, $j^{\mu}=\bar{\psi}\gamma^{\mu}\psi$) transform appropriately (as a 4-vector, in this case). – tok3rat0r Jul 2 '15 at 7:28
• (You can see from that expression for the probability current density how the fact that the $\gamma^{\mu}$ are not 4-vectors, despite their 'contravariant vector' index, causes much heartache! This set of notes is perhaps the clearest explanation I've been able to find online) – tok3rat0r Jul 2 '15 at 7:44

I think the clearest way to think about this is to say that the gamma matrices don't transform. In other words, the fact that they carry a vector index doesn't mean that they form a four vector. This is analogous to how the Pauli matrices work in regular quantum mechanics, so let me talk a little bit about that.

Suppose you have a spin $1/2$ particle in some state $|\psi\rangle$. You can calculate the mean value of $\sigma_x$ by doing $\langle \psi | \sigma_x | \psi\rangle$. Now let's say you rotate your particle by an angle $\theta$ around the $z$-axis. (Warning: There is about a 50% chance my signs are incorrect.) You now describe your particle with a different ket, given by $|\psi'\rangle = \exp(-i \sigma_z \theta /2)$. Remember that we are leaving the coordinates fixed and rotating the system, as is usually done in quantum mechanics. Now the expectation value is given by

$$\langle \psi' | \sigma_x | \psi' \rangle = \langle \psi |\, e^{i\sigma_z \theta /2}\, \sigma_x\, e^{-i \sigma_z \theta / 2}\, | \psi\rangle$$

There is a neat theorem, not too hard to prove, that says that

$$e^{i\sigma_z \theta /2}\, \sigma_x\, e^{-i \sigma_z \theta / 2} = \cos \theta\, \sigma_x -\sin \theta\, \sigma_y$$

So it turns out that the expectation value for the rotated system is also given by $\langle \psi |\, \cos \theta\, \sigma_x -\sin \theta\, \sigma_y \, |\psi\rangle = \cos \theta\, \langle \sigma_x \rangle - \sin \theta\, \langle \sigma_y \rangle$. It's as if we left our particle alone and rotated the Pauli matrices. But note that if we apply the rotation to $|\psi\rangle$, then we don't touch the matrices. Also, I never said that I transformed the matrices. I just transformed the state, and then found out that I could leave it alone and rotate the matrices.

The situation for a Dirac spinor is similar. The analogous identity is that $S(\Lambda) \gamma^\mu S^{-1}(\Lambda) \Lambda^\nu_{\ \mu} = \gamma^\nu$. This is just something that is true; nobody said anything about transforming $\gamma^\mu$. There's no $\gamma^\mu \to \dots$ here.

Now let's take the Dirac equation, $(i \gamma^\mu \partial_\mu - m)\psi = 0$, and apply a Lorentz transformation. This time I will change coordinates instead of boosting the system, but there's no real difference. Let's say we have new coordinates given by $x'^\mu = \Lambda^\mu_{\ \nu} x^\nu$, and we want to see if the Dirac equation looks the same in those coordinates. The field $\psi'$ as seen in the $x'^\mu$ frame is given by $\psi' = S(\Lambda) \psi \iff \psi = S^{-1}(\Lambda) \psi'$, and the derivatives are related by $\partial_\mu = \Lambda^\nu_{\ \mu} \partial'_\nu$. Plugging in we get $(i\gamma^\mu \Lambda^\nu_{\ \mu} \partial'_\nu-m) S^{-1}(\Lambda)\psi' = 0$, which doesn't really look like our original equation. But let's multiply on the left by $S(\Lambda)$. $m$ is a scalar so $S$ goes right through it and cancels with $S^{-1}$. And in the first term we get $S(\Lambda)\gamma^\mu S^{-1}(\Lambda) \Lambda^\nu_{\ \mu}$, which according to our trusty identity is just $\gamma^\nu$. Our equation then simplifies to

$$(i\gamma^\mu \partial'_\mu - m)\psi'=0$$

This is the same equation, but written in the primed frame. Notice how the gamma matrices are the same as before; when you're in class and the teacher writes them on the board, you don't need to ask in what coordinate system they are valid. Everyone uses the same gamma matrices. They're not really a four-vector, but their "transformation law" guarantees that anything written as if they were a four vector is Lorentz invariant as long as the appropiate spinors are present.

• is it reasonable to summarize this as: the gamma matrices are such that $\bar\psi\psi$ is a scalar and $\bar\psi\gamma^\mu\psi$ is a four-vector, but the gamma matrices themselves are don't transform. – innisfree Sep 22 '15 at 0:30
• @innisfree: I don't know if that summarizes everything, but it is an important property. In my point of view, it is the spinors that transform, and the matrices turn that into a Lorentz transformation. – Javier Sep 22 '15 at 21:09

The gamma matrices do not change if one does not apply a change of representation (e.g., chiral -> standard) along with the Lorenz transformation. Recall that you can write Dirac's equation in any frame with gamma-matrices in the same (e.g., chiral) representation. If you change the representation of them by using an invertible matrix $\gamma^\mu \to S\gamma^\mu S^{-1}$ (which is not a Lorenz transformation, i.e. a scalar in this sense) the spinor also transforms by S: $\Psi \to S\Psi$. This is the source of confusion: is $S=I$ then your equation

$$\gamma^\mu\to S[Λ]\gamma^\mu S[Λ]^{-1} =Λ^\mu_\nu \gamma^\nu$$

$$S[Λ]\gamma^\mu S[Λ]^{-1} = Λ^\mu_\nu \gamma^\nu.$$

I hope it helped to exorcize the confusion.

• I forgot to mention the reason for this: the representation of lorenz group by $S[\Lambda]$ is not unique. – Valery Shchesnovich Oct 1 '15 at 13:55

The problem here is just bad notation.

$$γ$$ is a linear injection from $$\mathbb{R}^{3,1}$$ to the Clifford algebra of $$\mathbb{R}^{3,1}$$, which takes each vector to itself. If you're a fan of abstract index notation, it should have a vector index and a Clifford index.* If you aren't a fan of abstract index notation, it should have no indices. It certainly shouldn't have one index.

But particle physics has settled on using explicit indices for $$T(V)$$ and hidden indices for $$C\ell(V)$$, and now we have to live with that forever, along with all of the other unfortunate frozen accidents of mathematical notation.

This by itself doesn't tell you what the transformation rule for $$γ$$ should be; it just means it's uninteresting. It's merely a matter of convention and not a deep property of spinor geometry. One could adopt the convention that the Dirac matrices represent four fixed physical directions in spacetime, in which case $$γ$$ would transform and the spinors wouldn't. But the usual convention is that they represent whatever basis you're currently using in the vector notation, and then $$γ$$ transforms like an identity matrix, since that's more or less what it is.

* Instead of one Clifford index, it should probably have two spinor indices, but I didn't want to say that outside of a footnote for fear of distracting from my point.