# Can Dirac equation be reformulated in an equivalent tensor form?

1) Can Dirac equation (including bispinors) be represented by a tensor formalism?

2) If yes, what kind of tensors could be the components of the wave function in Dirac equation in such formulation? and what are their essential (notable) properties?

Also introducing any relevant new book (or paper) which concerns this question, is highly appreciated.

I'm not sure what is it exactly that you're asking, so I'll cover everything that comes to my mind.

First of all, let's be clear we're talking about the same thing. Tensors are objects that transform like tensors, i.e.

$$T^{i j k \dots} (x) \longrightarrow R^{i j k \dots}_{a b c \dots} T^{a b c \dots} (x)$$

$R$ is the appropriate transformation. If you're talking about Lorentz-tensors, then it is the Lorentz transformation $\mathbf{\Lambda}$ and, for example, a tensor with two Lorentz-indices transforms as

$$T^{\mu \nu} \longrightarrow \Lambda^{\mu}_{\, \, \alpha} \Lambda^{\nu}_{\, \, \beta} T^{\alpha \beta}$$

If this is what you mean, then the Dirac equation (using natural units $c=\hbar=1$) is simply

$$(i {\partial\!\!\!/} - m) \Psi = 0$$

Here, $\Psi (x)$ is a bispinor and ${\partial\!\!\!/} \equiv \gamma^\mu \partial_\mu$, where $\gamma^\mu$ stands for Dirac matrices.

Note that Lorentz 4-vectors (and indices) are elements of a vector space considered as the representation space of the $(\frac{1}{2}, \frac{1}{2})$ representation of the Lorentz group.

On the other hand, a bispinor (Dirac spinor) is an element of a vector space considered as the representation space of the $(\frac{1}{2}, 0) ⊕ (0, \frac{1}{2})$ representation of the Lorentz group.

This means that bispinors can also be written in a "tensor form", where we use latin letters to denote spinor indices:

$$\psi^{a} \longrightarrow S^{a}_{\, \, b} \psi^{b}$$

$S$ is the appropriate Lorentz transformation for spinors.

When working with Yang-Mills gauge theories, one also has the Lie algebra indices which transform with the appropriate transformation matrices.

Bottom line, if it transforms like a tensor, you can slap indices on it, if you want.

P.S. Check out this notation for Weyl ($(\frac{1}{2}, 0)$ or $(0, \frac{1}{2})$) spinors: http://en.m.wikipedia.org/wiki/Van_der_Waerden_notation

• This word, tensor form, seems confusing to me. Is not the word tensor deserved only to objects that behaves as a tensor product of vectors: $\Lambda^{\mu}_{\nu}\Lambda^{\lambda}_{\rho}...$? The Latin letters are spinorial indices, a simple manifestation that the representation is on a linear space, i.e. Lorentz transformations are linear transformations. – Nogueira Oct 30 '16 at 18:49
• That's exactly why I started with the appropriate definition of a tensor. Spinors do transform like tensors. What I meant by tensor form simply means explicit indices with a tensorial transformation rule. I'm sure I could have worded it better, but I'm not sure how. – user20250 Oct 30 '16 at 18:57
• I mean, we could go into spin structure and spin bundles, but I wanted to keep it simple. Hopefully, it's not too confusing. – user20250 Oct 30 '16 at 19:04
• As far as I know, from the chapter 5 of Weinberg QFT book, there is no way to represent what Dirac equation represents (spin $1/2$ massive particles) by tensor, i.e. tensor products of (n,n). – Nogueira Oct 30 '16 at 19:24
• Not in the usual sense, which was Weinberg's context too. That's absolutely right and I do acknowledge it :) – user20250 Oct 30 '16 at 20:21

My articles may qualify as "new papers that concern this question": http://akhmeteli.org/wp-content/uploads/2011/08/JMAPAQ528082303_1.pdf (Journal of Mathematical Physics, 52, 082303 (2011)) and https://arxiv.org/abs/1502.02351 . For an arbitrary constant (not depending on a point in spacetime) eigenvector $\xi$ of $\gamma^5$ I derive an equation for one component $\bar{\xi}\psi$ of the Dirac bispinor $\psi$ (so the component is a scalar function). This equation is equivalent to the Dirac equation (please see the caveats).