In the study of Quantum Field Theory and Group Theory for the spinor representation of $SO$ groups, we know the following correspondence:

$\chi C\psi$ scalar

$\chi C\gamma^\mu\psi$ vector

$\chi C\sigma^{\mu\nu}\psi$ rank 2 tensor

$\chi C \gamma^5\psi$ pseudoscalar


So is there any map that can link these vectors and tensors nicely with those that we are more familiar with? Because they really don't look like the tensors and vectors that we constantly deal with in other physics area. And Can you give the corresponding transformation rules for the above objects?

  • $\begingroup$ Try counting components: $\chi C\psi$ Looks like an inner product and produces a scalar. $\chi C\gamma^\mu\psi$ Has one free index, $\mu$, and as a result has the right number of components for a vector. $\chi C\sigma^{\mu\nu}\psi$ Two indices, $\mu$ and $\nu$, so it must be rank-2. $\chi C \gamma^5\psi$ Looks like a scalar, but has that $\gamma^5$. I'd say they do look like the vectors, scalars, and tensors that we see in other places. $\endgroup$ – Display Name Dec 10 '18 at 20:45
  • $\begingroup$ related topic here $\endgroup$ – MadMax Dec 10 '18 at 22:03
  • 4
    $\begingroup$ @DisplayName Not all objects with tensor indices behave like tensorial objects. $\endgroup$ – Bob Knighton Dec 11 '18 at 0:30

A Dirac spinor is an object which transforms in the $(\frac{1}{2},0)\oplus(0,\frac{1}{2})$ representation of the Lorentz group (if you are not familiar with this notation, have a look at Chapter 6 of Weinberg's QFT I book).

The product of two Dirac spinor decomposes as: $$[(\frac{1}{2},0)\oplus(0,\frac{1}{2})]\otimes [(\frac{1}{2},0)\oplus(0,\frac{1}{2})]= [2\times (0,0)]\oplus [2\times (\frac{1}{2},\frac{1}{2})]\oplus [(1,0)\oplus (0,1)].$$

The first two objects in square brackets are two scalars (or a scalar and a pseudoscalar, if you are considering the full Lorentz group).

The second two objects are two vectors (or a vector and a pseudovector).

Finally, the third objects are a self-dual and an anti self-dual tensor (or an antisymmetric tensor, again under the full Lorentz group).

The $\Gamma$ matrices are nothing but the Clebsch-Gordan coefficient of the above tensor product decomposition, written in conventional bases for the integer spin representations (i.e. bases obtained by taking tensor products of unit orthogonal vectors in $\mathbb R ^{1,3}$).

So the answer to the question "why do the $\gamma$ matrices behave like vectors?" is: "because they are constructed to do exactly so". Notice that you could also work the other way round: you could start from an ordinary vector $V^{\mu}$ and construct a matrix $V^{\mu} \gamma _\mu$. Such a matrix would furnish a perfectly equivalent representation of the same mathematical object.


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