Why is the product of Dirac spinors a 4x4 matrix?

[tommy01]

Hi togehter.

I encountered the following problem:

The timeordering for fermionic fields (here Dirac field) is defined to be (Peskin; Maggiore, ...):

$$T \Psi(x)\bar{\Psi}(y)= \Psi(x)\bar{\Psi}(y) \ldots x^0>y^0$$
$$= -\bar{\Psi}(y)\Psi(x) \ldots y^0>x^0$$

where $$\Psi(x)$$ is a Dirac spinor and $$\bar{\Psi}(y) = \Psi(y)^\dagger \gamma^0$$ it's Dirac adjoint so that

$$S(x-y) = \langle 0|T{ \Psi(x)\bar{\Psi}(y)}|0 \rangle$$

is the Feynman propagator which is a 4x4 matrix.
But there is my problem: while it is clear that $$\Psi(x)\bar{\Psi}(y)}$$ is a 4x4 matrix, $$\bar{\Psi}(y)\Psi(x)$$ is a scalar.

I would be glad for an explanation.
Thanks.
Tommy

 

[Avodyne]

The indices are not contracted, and the propagator is a 4x4 matrix:

$$T \Psi_\alpha(x)\bar{\Psi}_\beta(y)= \Psi_\alpha(x)\bar{\Psi}_\beta(y) \hbox{\ if\ } x^0>y^0 \hbox{\ and\ } -\bar{\Psi}_\beta(y)\Psi_\alpha(x) \hbox{\ if\ } y^0>x^0$$

$$S_{\alpha\beta}(x-y) = \langle 0|T{ \Psi_\alpha(x)\bar{\Psi}_\beta(y)}|0 \rangle$$

 

[dextercioby]

In other words, the equations must be read component-wise, or you can think of a tensor product of the 2 spinors. That's the only way you can make sense of a product in which the barred spinor appears to the right of an un-barred one.

 

[tommy01]

Thanks a lot ...