What does $\Lambda^{-1}_{\frac{1}{2}}\gamma^\mu\Lambda_{\frac{1}{2}}=\Lambda^\mu_{\phantom{\mu}\nu}\gamma^\nu$ mean?

Question

\begin{equation} \Lambda^{-1}_{\frac{1}{2}}\gamma^\mu\Lambda_{\frac{1}{2}}=\Lambda^\mu_{\phantom{\mu}\nu}\gamma^\nu \end{equation}

In P&S, p. 42:

Equation (3.29) says that the $\gamma$ matrices are invariant under simultaneous rotations of their vector and spinor indices. In other words, we can “take the vector index $\mu$ on $\gamma^\mu$ seriously,” and dot $\gamma^\mu$ into $\partial_\mu$ to form a Lorentz-invariant differential operator.

(3.29) is the equation above. I know l.h.s. is about a spinor and r.h.s. is about a vector since $\Lambda_{\frac{1}{2}}$ is about spinor rotation (and boost) and $\Lambda$ is about vector, but don’t understand what simultaneous rotations are and what this equation mean.

1. What does the equation above mean?
2. What are simultaneous rotations?

Answer 1

Usually when we see an object with a Greek index like $\gamma^\mu$, we assume that the object contains the components of a vector and that the way to rotate it involves a sum over the index $\mu$. Since $\{\gamma^0, \gamma^1, \gamma^2, \gamma^3\}$ are matrices, we can transform them by multiplying them by matrices from the left and right. They are chosen such that matrix multiplication by a particular form of matrix from the left and its inverse from the right coincides with a transformation of the form $\Lambda^\mu_{\phantom\mu\nu} \gamma^\nu$, as if the $\gamma^\mu$ were components of a vector. (They aren't; they're more like guide posts telling the derivative operators in $\sum_\mu \gamma^\mu \partial_\mu$ which components of the field on the right actually point along the direction in which $x^\mu$ increases.)

Simultaneous rotation means transforming using the summation over Greek index and multiplication by matrices at the same time:

$\sum_\nu \Lambda^\mu_{\phantom\mu\nu}(\Lambda_{\tfrac12}\gamma^\nu \Lambda_{\tfrac12}^{-1}) = \gamma^\mu$

It's equivalent to rotating/boosting in one direction and then reversing it.

Answer 2

$γ$ is an injective map from $\mathbb R^{1,3}$ into the Clifford algebra of $\mathbb R^{1,3}$, that takes each vector to itself. Since it's a linear map, you can think of it as a rank-2 tensor. Suitably interpreted, it's the identity tensor, so transforming it equivalently on both sides leaves it unchanged. That is essentially what that equation says, albeit in a confusing way.

The Clifford algebra of $\mathbb R^{1,3}$ is a pretty mathematical object that was reinvented in a rather ugly form by Dirac. Abstractly, a Clifford algebra is a free noncommutative algebra of vectors from some normed vector space modulo S+S and V+V addition, SS and SV multiplication, and V² squared norm. From this you can derive all other properties, including the existence of matrix representations like Dirac's.

It can be understood as an algebra of reflections, in which a vector represents a reflection through a hyperplane normal to itself and products of vectors represent compositions of reflections. Any rotation in a plane can be written as a composition of two reflections. If you rotate one of the mirrors through 180°, it reflects in the same direction as before, but its normal points in the opposite direction, so the corresponding rotation (by 360°) picks up a factor of $-1$ in the Clifford algebra. This is the geometric reason behind the double covering.

From the interpretation as an algebra of reflections, if you accept it as correct, you can derive that reflections act on vectors by conjugation, and therefore so do products of reflections, including all rotations.

$Λ_\frac12$ is a representation of an arbitrary rotation in $\mathbb R^{1,3}$ as a product of two or four vectors. If you think of $γ^μ$ as a vierbein, then conjugation by $Λ_\frac12$ rotates each of its vectors independently. On the other side, ${Λ^μ}_νγ^ν$ treats $γ$ as an orthonormal basis for $\mathbb R^{1,3}$ (though it is really made of Clifford-algebraic vectors, not 4-vectors), and transforms it by mixing them. The result is the same either way.

Answer 3

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

Please see my answer to "Do the Dirac matrices form a proper four-vector?" which might have been better posted here.