What is Time Ordered Product T(AB) at Equal Times?

[zombiegirl]

So the time ordered product T(AB) just rearranges A and B so that the one with the earliest time co-ordinate goes to the right.

Does anybody know what the time ordered product of two fields at equal time is?

Because they're often written using a heaviside step function, it's difficult to tell - I can't seem to find a straight answer as to what the heaviside function does at zero (0, 1/2 or 1 seem to be the possibilities).

What I'm actually trying to find is the Feynman progagator, which is the time ordered product enclosed between to ground states: <0|T(AB)|0>. It would make my calculations really nice if it was zero, but I'd have thought it more likely that it's just the <0|AB|0> as though there was no T function there...

Any help appreciated!

Thanks.

 

[RedX]

I'm inclined to say use the value of 1/2 for the heaviside step function. The reason is that in your calculation <0|AB|0> should equal <0|BA|0> for equal times. So since 1/2+1/2=1, so you can use either <0|AB|0> or <0|BA|0>.

 

[Avodyne]

If A and B commute at equal times, then the ordering doesn't matter. If the ordering does matter, then one usually takes the step function to be 1/2 at zero.