The gamma matrices are defined by their anticommutation relations, which are symmetrical in permutations of $\gamma_1, \gamma_2, \gamma_3$. Given this symmetry, why is the change conjugation operator $\gamma_2$, rather than some symmetrical expression in $\gamma_1, \gamma_2, \gamma_3$?

# [Physics] Charge conjugation operator and gamma matrices

charge-conjugationdirac-equationdirac-matrices

#### Related Solutions

The most general Lorentz transformation that is connected to the identity is given by the conjugation by $\exp(-A)$ where $$ A = \frac 12 \omega_{\mu\nu} \gamma^\mu \gamma^\nu $$ and $\omega_{\mu\nu}$ is an antisymmetric tensor containing $D(D-1)/2$ parameters. The group of all such transformations is isomorphic to $Spin(D-1,1)$. If $\omega$ only contains one component $0\mu$, then it is a boost, and the nonzero numerical value of $\omega$ is the rapidity - the "hyperbolic angle" $\eta$ such that $v/c=\tanh\eta$.

If only one doubly spatial component of $\omega$ is nonzero, then this component $\omega_{\mu\nu} = -\omega_{\nu\mu}$ is obviously the angle itself. Note that the spatial-spatial terms in $A$ are anti-Hermitean, producing unitary transformations; the mixed temporal-spatial terms in $A$ are Hermitean and they don't product unitary transformations on the 4-component space of spinors (but they become unitary if they're promoted to transformations of the full Hilbert space of quantum field theory).

In 4 dimensions, a general antisymmetric matrix $4\times 4$ contains 6 independent parameters and has eigenvalues $\pm i a, \pm i b$, so in 3+1 dimensions, one can always represent a general Lorentz transformation as a rotation around an axis in the 4-dimensional space followed by a boost in the complementary transverse 2-plane. This is the counterpart of the statement that any $SU(2)$ rotations in 3 dimensions is a rotation around a particular axis by an angle.

If you allowed $A$ to contain something else than $\gamma^{\mu\nu}$ matrices which generate the Lorentz group, you could get other groups. Only for a properly chosen subset of allowed values of $\omega$, you would get a closed group from the resulting exponentials (under multiplication). In particular, if you allowed $A$ to be an arbitrary complex combination of any products of gamma matrices, well, then you would allow $A$ to be any complex $4\times 4$ matrix, and its exponentials would produce the full group $GL(4,C)$ - surprising, Carl? ;-) It's not a terribly useful groups in physics because actions are usually not invariant under this "full group", are they?

Also, there are not too many groups in between $Spin(3,1)$ and $GL(4,C)$ - I guess that there's no proper group of $GL(4,C)$ that has a proper $Spin(3,1)$ subgroup. Obviously, there are many subgroups of $Spin(3,1)$ - such as $Spin(3)$, $Spin(1,1)\times Spin(2)$, and others.

## Best Answer

The charge conjugation operator $C$ cannot be expressed as a representation-invariant polynomial in $\gamma^0, \gamma^1, \gamma^2, \gamma^3$. Proof: Under a spinor basis change $U$, the gamma matrices transform as $\gamma^\mu \rightarrow U \gamma^\mu U^{-1}$, so any polynomial $P$ will transform likewise. But the charge conjugation operator transforms as $C \rightarrow U^* C U^{*-1}$, so cannot be expressed by any $P$.

In the Dirac representation, $C$ happens to be given by $\gamma^2$. This is a coincidence due to our choice of basis - in another basis it will not be true. As shown above, no polynomial expression can hold for every basis, no matter whether the expression is symmetrical in $\gamma^{1,2,3}$ or not.