- #1

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## Main Question or Discussion Point

I guess the answer to this question actually should be pretty obvious, but I still have problems getting it right though. I wonder about the definition of the time ordered product for a pair of Dirac spinors. In all the books I've read it simply says:

[tex]T\left\{\psi(x)\bar{\psi}(x')\right\} = \theta(t - t')\psi(x)\bar{\psi}(x') - \theta(t' - t)\bar{\psi}(x')\psi(x)[/tex]

The spinor indices are always left out. So should it be

[tex]T\left\{\psi_\alpha(x)\bar{\psi}_\beta(x')\right\} = \theta(t - t')\psi_\alpha(x)\bar{\psi}_\beta(x') - \theta(t' - t)\bar{\psi}_\beta(x')\psi_\alpha(x)[/tex]

or

[tex]T\left\{\psi_\alpha(x)\bar{\psi}_\beta(x')\right\} = \theta(t - t')\psi_\alpha(x)\bar{\psi}_\beta(x') - \theta(t' - t)\bar{\psi}_\alpha(x')\psi_\beta(x)[/tex]?

I personally think the

[tex]T\left\{\psi(x)\bar{\psi}(x')\right\} = \theta(t - t')\psi(x)\bar{\psi}(x') - \theta(t' - t)\bar{\psi}(x')\psi(x)[/tex]

The spinor indices are always left out. So should it be

**A**:[tex]T\left\{\psi_\alpha(x)\bar{\psi}_\beta(x')\right\} = \theta(t - t')\psi_\alpha(x)\bar{\psi}_\beta(x') - \theta(t' - t)\bar{\psi}_\beta(x')\psi_\alpha(x)[/tex]

or

**B**:[tex]T\left\{\psi_\alpha(x)\bar{\psi}_\beta(x')\right\} = \theta(t - t')\psi_\alpha(x)\bar{\psi}_\beta(x') - \theta(t' - t)\bar{\psi}_\alpha(x')\psi_\beta(x)[/tex]?

I personally think the

**A**definition feels more natural, but when I use it in my derivations I get strange results. On the other hand, the**B**definition gives more reasonable results. It could simply be that I've done some mistakes in the derivations, but before I dig into those I want to know if I've got the definition right in the first place.