# Is $\displaystyle{\not} p$ a Lorentz Scalar?

I am a bit confused about something. $\gamma^\mu$ is a (Lorentz) vector (c.f. Pesking & Schroeder chapter 3), and so is $p^\mu$, therefore I’d expect their product $\displaystyle{\not}p \triangleq \gamma^\mu p_\mu$ to be a scalar.

The problem is that $\displaystyle{\not}p$ does not commute with the vector $\gamma^\nu$:

$$[\displaystyle{\not}p, \gamma^\nu] = -2p^\nu \neq 0$$

Shouldn’t a scalar always commute with a vector?

• $\gamma^\mu$ is not a vector. $\not\! p$ is neither a vector nor a scalar; it is a rank-$(1,1)$ spinor. Apr 8, 2018 at 13:40
• To add on to the previous comment, $\gamma^\mu_{\alpha\beta}$ is not a Lorentz vector. It transforms as a spinor$\otimes$anti-spinor$\otimes$vector. Apr 8, 2018 at 13:51
• @AccidentalFourierTransform this is not true, check Perskin & Schroeder chapter 3, $\gamma^\mu$ does transform like a vector under Lorentz transformation.
– Y2H
Apr 8, 2018 at 17:28
• @Y2H I don't know what P&S say or do not say, but $\gamma^\mu$ is not a vector. Apr 8, 2018 at 18:33
• @AccidentalFourierTransform it’s fine, I looked at diracula’s answer.
– Y2H
Apr 8, 2018 at 18:35

As pointed out already in the comments, the $\gamma^\mu$ do not transform as the components of a Lorentz four-vector. In fact it is (I think) most natural to say that they do not transform at all under a Lorentz transformation. In particular, $\gamma^\mu p_\mu$ is not a Lorentz scalar.

The usual reason for loosely saying that the $\gamma^\mu$ transform as the components of a Lorentz four-vector is the behaviour of so-called bilinears in Dirac spinors $\psi$ which are formed using the $\gamma^\mu$ matrices. For example, the bilinears $$\psi^\dagger\gamma^0\psi \,, \quad \psi^\dagger\gamma^0\gamma^\mu\psi \,, \quad \mathrm{and} \quad \psi^\dagger\gamma^0\gamma^\mu\partial_\mu\psi \,,$$ transform under a Lorentz transformation as a scalar, four-vector, and scalar respectively. Hence the presence of $\gamma^\mu$ changes the transformation properties of bilinears in a way reminiscent of a four-vector. For completeness, below I include a brief explanation of how under a Lorentz transformation the above bilinears transform as claimed, even though the $\gamma^\mu$ do not transform.

The last part of your question, about commutators, seems to confuse two notions. We must distinguish between something being a Lorentz scalar/four-vector etc, which refers to the behaviour under a Lorentz transformation, and something being a number/vector/matrix etc, which refers to whether we have collected several numbers into some list and agreed to multiply these objects according to the rules of for example matrix multiplication. For example, a particular gamma matrix, $\gamma^\mu$, is a Lorentz scalar, but it is also a matrix, so the commutator $[\gamma^\mu,\gamma^\nu] = \gamma^\mu\gamma^\nu - \gamma^\nu\gamma^\mu$ is an ordinary matrix commutator, and is not simply zero.

Additionally, even if $\gamma^\mu$ did transform as the components of a four-vector, the object $[\gamma^\mu,\gamma^\nu]$ would not be commutator between two four-vectors: there is no notion of a commutator of four-vectors, rather we would have chosen one component from each four-vector and computed the commutator between them.

Appendix

Here is an explanation of how under a Lorentz transformation Dirac spinor bilinears transform as claimed, even though the $\gamma^\mu$ do not transform. Let $\psi$ be a Dirac spinor, and let $v^\mu$ be the components of a four-vector. Under a Lorentz transformation, the two are transformed with different matrices $\Lambda_{\frac{1}{2}}$ and $\Lambda$, $$\psi \to \Lambda_{\frac{1}{2}} \psi \,, \quad v^\mu \to \Lambda^\mu_{~~\nu}v^\nu \,,$$ where I wrote the transformation for the four-vector in component form as this is what we will use. It is also easy to show that $$\psi\gamma^0 \to \psi \gamma^0 \Lambda_{\frac{1}{2}}^{-1}$$ The key to determining the transformation properties of spinor bilinears is the following identity, $$\Lambda_{\frac{1}{2}}^{-1}\gamma^\mu\Lambda_{\frac{1}{2}} = \Lambda^\mu_{~~\nu}\gamma^\nu \,,$$ which expresses how the $\gamma^\mu$ matrices combine with the Lorentz transformation of the Dirac spinors to give the transformation matrix for a four-vector. This is not a statement about a transformation of the $\gamma^\mu$ matrices under a Lorentz transformation, it is merely a matrix identity. Using this identity, and the properties of the $\gamma^\mu$ matrices, it is then straightforward to show the Lorentz transformation properties of the above Dirac spinor bilinears.

• Thank you very much for this thorough answer. I really appreciate it!!
– Y2H
Apr 8, 2018 at 17:30