Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Green, Schwarz & Witten Superstring theory, Vol. I, page 141, I don't understand how pulling the derivative inside the Time-ordered product can give an Equal-time Commutator:

$$\tag{3.2.44} \partial_- \langle T \big( T_{++}(\sigma, \tau) T_{++}(\sigma', \tau') \big) \rangle ~=~ \frac12 \delta(\tau - \tau') \langle \big[ T_{++}(\sigma, \tau), T_{++}(\sigma', \tau) \big] \rangle$$

Is there any (rigorous) proof for this?

share|cite|improve this question
up vote 2 down vote accepted

Write out the time ordering explicity with step functions:

$$ \mathrm{T}\left(A(t_1) B(t_2)\right) = \Theta(t_1 - t_2) A(t_1) B(t_2) + \Theta(t_2 - t_1) B(t_2) A(t_1).$$

Now just differentiate. When the derivative hits the step functions you get a delta function:

$$ \partial_{x} \Theta(x) = \delta(x). $$

Just use the chain rule and product rule like normal and it should fall out.

share|cite|improve this answer
    
Thank you, everything is clear to me now! – Trung Phan Jun 24 '13 at 13:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.