Time ordering in QFT

1. Dec 18, 2013

aaaa202

I have asked this question once, but noone seemed to notice it, so I'll try again. In my book the time ordering operator is used to rewrite an operator product:

U(β,τ)A(τ)U(τ,τ')B(τ')U(τ',0) = T_τ(U(β,0)A(τ)B(τ'))

To refresh your memories the time ordering operator T_τ orders operators according to time such that:
T_τ(A(τ)B(τ')) = A(τ)B(τ') for τ>τ' and B(τ')A(τ) for τ'>τ
And the operator U(t,t') is a unitary operator that propagates a state from t' to t and has the property that:
U(t,t')=U(t,t'')U(t'',t')

I am still unsure how the rewriting is done though. One key ingredient is to use the property of the unitary above to write:
This way we have:
U(0,β)=U(0,τ')U(τ',τ)U(τ,β)
And i think the idea is then to insert in the expression and use time-order but I am not sure how to.

2. Dec 18, 2013

Avodyne

There is an implicit assumption here that $\beta>\tau>\tau'>0$.

$$T[U(\beta,0)A(\tau)B(\tau')]$$
Then substitute in
$$U(\beta,0)=U(\beta,\tau)U(\tau,\tau')U(\tau',0)$$
to get
$$T[U(\beta,\tau)U(\tau,\tau')U(\tau',0)A(\tau)B(\tau')]$$
Now rearrange the operators so that time labels decrease as you go left to right:
$$T[U(\beta,\tau)A(\tau)U(\tau,\tau')B(\tau')U(\tau',0)]$$
The labels are now in time-order, so the time-ordering symbol can be dropped:
$$U(\beta,\tau)A(\tau)U(\tau,\tau')B(\tau')U(\tau',0)$$
QED.

3. Dec 19, 2013

aaaa202

hmm okay I thought it was something like that, but I am still unsure though. Which time do you assing to the operator U(t1,t2)? It propagates a state from t1 to t2, so it is not really a function of one time - or am I missing something?

4. Dec 19, 2013

Qubix

Last edited by a moderator: May 6, 2017
5. Dec 19, 2013

Avodyne

You can think of it as a product of many operators at closely spaced times, and break it up as needed; this is what I did above.

6. Dec 19, 2013

aaaa202

So I should basically assign to U(t1,t2) a value of time between t1 and t2?