I'm seeing many conflicting statements about Dirac's gamma matrices. Some say it's a four-vector, some say it isn't, some say it's an invariant four vector. I know the Dirac matrices satisfy the identity

$$\Lambda^{-1}_{1/2}\gamma^\mu \Lambda_{1/2}=\Lambda^{\mu}_{~\nu}\gamma^\nu$$

where $\Lambda_{1/2}$ denotes the spin-1/2 representation of the Lorentz transformation, and $\Lambda^{\mu}_{~\nu}$ is the usual vector representation. I've seen in the literature people say not to confuse this identity with a transformation law, but that is exactly what it looks like to me. If we multiply both sides by spin-1/2 Lorentz transformations and the inverse transformation, we get


The gamma matrices carry a spacetime index $\mu$ and 2 suppressed spin indices $a,b$. To me this identity seems like it's saying is that when you ask how an object that carries both Lorentz and spin indices transforms, you have to appropriately transform all of the indices in their corresponding representation of the Lorentz group. That is, you're missing part of how $\gamma^\mu$ transforms if you only hit it with the vector or spin-1/2 rep of the Lorentz transformation, you need to hit it with both. For example, if I constructed some object by forming the product of some vector and some other object with two spin indices $$M^\mu_{ab}\equiv V^\mu\psi_{ab}$$ (maybe one of the latin indices should be upper) and then ask how $M^\mu_{ab}$ transforms, I say it would transform with two spin-1/2 Lorentz transformations and 1 vector transformation. But that's exactly how it looks like the $\gamma^\mu$ transform. Obviously the gamma matrices satisfy $$\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}\times 1$$ which supports the statement that it's an invariant four-vector with spin indices since the metric in flat space is also Lorentz invariant. What am I missing?

Edit: If we pick on the Dirac equation (in some primed frame) $$(i\gamma'^\mu\partial'_\mu-m)\psi'=0$$, and we substitute in the transformation rules $$\partial'_\mu=\partial_\nu(\Lambda^{-1})^\nu_{~\mu}, \quad\psi'=\Lambda_{1/2}\psi$$ and the "transformation rule" defining how I currently believe the gamma matrices transform in equation 2 which follows from the isomorphism, we get $$(i\gamma^\mu\partial_\mu-m)\psi=0$$. $\gamma^\mu$ transforming (albeit trivially) is absolutely essential to cancel the $(\Lambda^{-1})^\nu_{~\mu}$ coming from the derivative and the $\Lambda_{1/2}$ from the spinor. I don't see how immediately taking $\gamma'^\mu= \gamma^\mu$ because the gamma matrices "don't transform" would leave the Dirac equation Lorentz invariant. I imagine this is a matter of semantics. Are people just equating "Doesn't transform" = "Transforms Trivially", or are they suggesting inserting the isomorphism relation doesn't count as transforming $\gamma^\mu$?

  • 1
    $\begingroup$ What do you mean by $\gamma'$ in your fourth equation? $\endgroup$
    – jacob1729
    Sep 29, 2021 at 13:46
  • 9
    $\begingroup$ Does this answer your question? What does $\Lambda^{-1}_{\frac{1}{2}}\gamma^\mu\Lambda_{\frac{1}{2}}=\Lambda^\mu_{\phantom{\mu}\nu}\gamma^\nu$ mean? $\endgroup$
    – jacob1729
    Sep 29, 2021 at 13:48
  • 3
    $\begingroup$ I've always thought of the dirac matrices as a map between the lorentz space and the spin space, so, calling it a "vector"seems wrong. Maybe something like a "spin-representation-valued-vector" or something, but not a vector. But I guess the argument is basically semantics. $\endgroup$ Sep 29, 2021 at 16:49
  • $\begingroup$ @jacob1729 The prime means the gamma matrix as viewed from the frame moving with velocity $v$ specified in $\Lambda^{\mu}{~\nu}$. The link you shared does help me better phrase my confusion so I'll edit my question a bit. $\endgroup$
    – Adots005
    Sep 29, 2021 at 18:56
  • 2
    $\begingroup$ Dirac matrices are Clebsch-Gordan coefficients for converting between spinor space and real space. They don't transform at all, as long as you use consistently defined coordinates for all the spaces at play, just like ordinary Clebsch-Gordan coefficients don't. $\endgroup$
    – knzhou
    Sep 29, 2021 at 19:19

1 Answer 1


This is just an example of an important property of the GL(N) Lie Group tensor operators. It is said that the tensor operator $\gamma^{\mu}$ transforms like a 4-vector under conjugation.

If you want to know the matrix elements of a tensor operator (in this case $\gamma^{\mu}$) in some representation (in this case the spin=1/2 rep) transformed to another frame (in this case by a Lorentz transformation $\Lambda$) then you can do it one of two ways. First, you could transform the kets using a matrix for their representation (in this case the 2 x 2 matrix $\Lambda_{\frac{1}{2}}$) or secondly, you could leave the kets alone and just transform the tensor operator as a tensor by its indices (in this case $\gamma^{\nu}$ by the 4 x 4 matrix $\Lambda^{\mu}_{\nu}$). Both methods give the same answer for the $\gamma'^{\mu}$ matrix elements of the tensor operator $\gamma^{\mu}$ in the new frame. $$ \left< a \right|\gamma'^{\mu} \left| b \right> =\left< a \right| \Lambda^{-1}_{KetRep} \gamma^{\mu} \Lambda_{KetRep}\left| b \right> = \Lambda^{\mu}_{\nu}\left< a \right| \gamma^{\nu} \left| b \right> $$ Notice that when you wish to transform a tensor operator to another frame it is usually easier to transform the tensor operator with the rep you already know (in this case the 4x4 matrix $\Lambda^{\mu}_{\nu}$) instead of the possibly 1000 x 1000 matrix $\Lambda_{KetRep}$ that you must figure out how to compute.


Your Answer

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

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