Why does the fermionic field have this time ordered contraction?

[Avogadro Number]

Hello,

I am struggling to see why for a fermionic field $\psi$, one has the time ordered contraction $<0|T(\psi(x)\psi(y))|0>$. Could someone offer an outline/hints to see this please? Thanks!

 

[Dewgale]

Essentially, it's to preserve causality. You have a time-ordering operator so that $\mathcal{O}(t_2)$ is applied after $\mathcal{O}(t_1)$. Remember that $\psi$ is a time-dependent operator in this case, and unless otherwise stated, $x^0$ needn't equal $y^0$.

 

[Avogadro Number]

Yes, thanks for this. Is there a way to see this mathematically based on the commutation rules?

 

[king vitamin]

Can you expand on your question? One can define time-ordering for either fermionic or bosonic fields, or one can consider other notions of ordering (anti-time ordering, retarded, advanced, and more complicated combinations if you're using a formalism with multiple time contours). Are you maybe asking how time-ordering shows up in specific places?