# Motivating the Unintuitive Properties of Spinors

In the usual treatment of (Dirac) spinors, one usually starts with "factoring" the energy-momentum relation, deducing the properties of the $$\gamma$$ matrices by requiring the cross terms to cancel, and going on to show that quantities like $$\bar{\psi}\gamma^\mu\psi$$ transform like four-vectors.

I'm wondering, essentially, if the same argument can be run "in reverse", and what assumptions are necessary to re-derive all of the desired properties. What I'm imagining goes something like this:

If we want a theory described by a Lagrangian, that Lagrangian needs to be composed of Lorentz-invariant terms. Since our dynamical terms will presumably include factors of $$\partial_\mu$$, we'll need something that looks like a four-vector so we can pair it $$\partial_\mu$$ and form a scalar. Motivated by that requirement, let's suppose we can cook up something of the form $$\bar{\psi}\gamma^\mu\psi \equiv \big\lbrace \bar{\psi}\gamma^0\psi,\;\bar{\psi}\gamma^1\psi,\;\bar{\psi}\gamma^2\psi,\;\bar{\psi}\gamma^3\psi \big\rbrace$$ that behaves like a four-vector under Lorentz transformations.

Using only that assumption, how much can we deduce about the behavior of $$\psi$$, $$\bar{\psi}$$, and the $$\gamma^\mu$$s? In particular, I'm interested in whether we can show the following:

1. The $$\gamma^\mu$$s satisfy the Clifford relations (and are therefore (at least) $$4\times4$$ matrices)
2. $$\psi$$ and $$\bar{\psi}$$ are related to each other by $$\gamma^0$$
3. $$\psi$$'s and $$\bar{\psi}$$'s discrete components (in the appropriate bases) correspond to eigenstates of a family of spin-like operators
4. $$\psi$$ and $$\bar{\psi}$$ transform under Lorentz transformations in the expected way (in particular, they require the expected rotation of $$4\pi$$ to return their original state).

I find the idea of deriving Property 4 in this way especially appealing, since it seems to suggest (albeit through a bogus "proof by notation") that the infamously unintuitive "$$4\pi$$" property is basically a consequence of the fact that our four-vector object, which behaves as we'd intuitively expect, contains 2 copies of $$\psi$$; that, in a sense, rotating the entire object naturally requires rotating $$\psi$$ twice, hence $$4\pi$$'s-worth of rotation to return to the original state.

Does that argument actually hold water? Or is it "freshman's dream" wishful thinking? If just "being a four-vector" isn't enough to derive all the desired properties, what other minimal assumptions do we need (e.g. unitarity-of-something for probability conservation, further transformation properties of tensor-like $$\bar{\psi}\sigma^{\mu\nu}\psi$$ quantities, etc.)?

• Factoring the energy-momentum relation may have been how spinors and $\gamma$s were developed historically (and still in many texts), but the modern perspective is more similar to what you described: observables are constructed from fields, but fields themselves don't need to be observables. A field $\psi$ can transform according to a covering group of the Lorentz group, instead of the Lorentz group itself. The covering group is naturally described using a (matrix) representation of the Clifford algebra. Then we can use $\psi$ to construct bilinear observables like $\overline\psi\gamma\psi$. – Dan Yand Dec 14 '18 at 3:44
• Ok, I think that makes sense... I guess what I'm really wondering is, can we construct a bilinear covariant out of anything besides spinors, or are they unique in that regard? – TheMac Dec 16 '18 at 0:50
• If derivatives are not allowed, then using spinors like you described is the only way I know how to construct bilinears that behave like a vector, but I don't have a proof or even a compelling argument that that's the only way (which is why I haven't posted a real answer). Hopefully somebody else will chime in with a more satisfying insight. – Dan Yand Dec 16 '18 at 1:26