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.)?

  • $\begingroup$ 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 $\overline\psi\gamma\psi$. $\endgroup$ – Dan Yand Dec 14 '18 at 3:44
  • $\begingroup$ 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? $\endgroup$ – TheMac Dec 16 '18 at 0:50
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    $\begingroup$ 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. $\endgroup$ – Dan Yand Dec 16 '18 at 1:26

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