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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?

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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.

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Thank you, everything is clear to me now! – Trung Phan Jun 24 '13 at 13:29

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