Dirac bilinears transform in the Lorentz indices as,

  • $\bar{\psi}\psi$ scalar
  • $\bar{\psi}\gamma^\mu\psi$ vector
  • $\bar{\psi}\sigma^{\mu\nu}\psi$ 2nd rank (antisymmetric) tensor
  • $\bar{\psi}\gamma^{\mu}\gamma^5\psi$ axial vector
  • $\bar{\psi}\gamma^5\psi$ pseudoscalar

The scalar as the simplest example,

$\bar{\psi}\psi$ is invariant under Lorentz transformations and is hence a scalar (source)

When referred to as a scalar does that mean I can arbitrarily boost and treat the quantity as a scalar (i.e. freely move the quantity in within terms) or does that just mean that the quantity transforms without any change in rank but cannot be moved about as a number.

Example with a scalar: Is the following permitted?

$$ (u\bar{u})\not pu = \not pu(u\bar{u}) $$ for an arbitrary spinor $u$ and $\not p = \gamma^\mu p_\mu$ as usual.

  • 3
    $\begingroup$ You're mixing up a bunch of things here. When you write down a Lagrangian, all five of these quantities can be treated as numbers, because they are trivial in spinor space. This has nothing to do with their Lorentz transformation properties. $\endgroup$
    – knzhou
    Apr 13, 2016 at 23:15
  • $\begingroup$ Basically you can move it around between structures in Dirac space because the Dirac structure closes, so it's just a number in Dirac space. But this would be true even if it were not Lorentz invariant. $\endgroup$
    – Orca
    Apr 13, 2016 at 23:20
  • $\begingroup$ @knzhou When you write down a Lagrangian, all five of these quantities can be treated as numbers, because they are trivial in spinor space is this unique to bilinears? $\endgroup$ Apr 13, 2016 at 23:25
  • $\begingroup$ It just works like regular matrix multiplication. The bilinears all look like (1x4 thing) (possibly other things) (4x1 thing), so the end result is 1x1 in spinor space. If you're not sure, just write out the dimensions of everything! $\endgroup$
    – knzhou
    Apr 13, 2016 at 23:28
  • $\begingroup$ Yeah I can always do block matrices but I'm trying to muddle through an exercise without using any explicit representations and it's revealing all my QFT fallacies in one 8 hour marathon... $\endgroup$ Apr 13, 2016 at 23:40

1 Answer 1


A scalar is, like other scalars, merely just a number. Think about their matrix representation: $$ \psi=(\phi_R\; \phi_L)^T$$ and $$\bar{\psi}=\psi^\dagger\gamma^0 =(\phi^*_L \; \phi_R^*).$$ It is clear that $\bar{\psi}\psi$ is a 1x1 matrix (scalar), and of course the operation is legitimate.

Those other forms are also 1x1 matrices. However under Lorentz transformation, an expression transforms like a scalar, or like a vector, etc. So it has nothing to do with the term "scalar" as a 1×1 matrix.

  • $\begingroup$ any examples for the others? $\endgroup$ Apr 14, 2016 at 0:34
  • $\begingroup$ The others are of course 1x1 matrices, but under a Lorentz transformation, they get those various names $\endgroup$
    – donnydm
    Apr 14, 2016 at 1:20

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