Behaviour of Dirac Bilinears

Question

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.

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.