- #1
sith
- 14
- 0
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:
T\left\{\psi(x)\bar{\psi}(x')\right\} = \theta(t - t')\psi(x)\bar{\psi}(x') - \theta(t' - t)\bar{\psi}(x')\psi(x)
The spinor indices are always left out. So should it be A:
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)
or B:
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)?
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.
T\left\{\psi(x)\bar{\psi}(x')\right\} = \theta(t - t')\psi(x)\bar{\psi}(x') - \theta(t' - t)\bar{\psi}(x')\psi(x)
The spinor indices are always left out. So should it be A:
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)
or B:
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)?
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.