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In his lecture notes on String theory, David Tong derives Ehrenfest theorem using the path integral:

$$S = \frac{1}{4\pi \alpha'}\int d^2\sigma\ \partial_\alpha X\ \partial^\alpha X\tag{4.19}$$

$$ 0 = \int DX \frac{\delta}{\delta X(\sigma)} e^{-S} = \int DX \ e^{-S} \left[\frac{1}{2\pi \alpha'} \partial^2 X(\sigma)\right].\tag{p.77}$$

How to do this calculation? When I did it myself, I tried something with partial integration, but got instead: $$\int DX \frac{\delta}{\delta X(\sigma)} e^{-S} = \int DX \ e^{-S} \left[\color{Red}{\int d^2\sigma}\ \frac{1}{2\pi \alpha'} \partial^2 X(\sigma)\right].$$

Also, how is the functional derivative defined here? (In the past I put some effort into understanding the functional derivative, but I found several definitions, which makes it a rather confusing concept.)

Qmechanic
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Jens Wagemaker
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1 Answers1

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Dozens of questions on this site remind you that $$ \frac{\delta}{\delta X(\tau)}~\int\!\! d^2\sigma ~ \partial_\alpha X(\sigma) \partial^\alpha X(\sigma) =-2\partial^2 X(\tau) .$$

Crudely, you may use $\frac{\delta X(\sigma)}{\delta X(\tau)}= \delta (\tau-\sigma)$.

Cosmas Zachos
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