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I found this footnote in the appendix (on path integral page 333) of j.polchinsky’s string theory book.can you explain this?

I found this footnote in the appendix (on path integral page 333) of J. Polchinski’s string theory book. can you explain this?

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  • $\begingroup$ Ultimately this might be better suited to mathematics, you might find a more convincing argument from a mathematical point of view instead. $\endgroup$
    – Triatticus
    Oct 7, 2020 at 0:10
  • $\begingroup$ Related and also. $\endgroup$ Oct 7, 2020 at 17:15

1 Answer 1

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Recall the definition, $$ \delta F[q;\phi]= \int \!\! dt~\frac{\delta F}{\delta q(t)} \phi(t) = \lim_{\varepsilon\to 0}\frac{F[q+\varepsilon \phi]-F[q]}{\varepsilon} = \left [ \frac{d}{d\varepsilon}F[q+\varepsilon \phi]\right ]_{\varepsilon=0}, $$ where $\phi (t)$ specifies the direction of the functional derivative in $\delta q(t)$.

Now take a special $F[q]\equiv \int \!\! d\tau ~~\delta(\tau-t') q(\tau)=q(t')$, so $$ \frac{\delta F}{\delta q(t)}= \delta(t-t'). $$

It's easiest if you think of t as the continuum limit of a discrete integer index i, so a vector $q_i \to q(t)$. Note that the vector calculus gradient, $\partial q_i/\partial q_j = \delta_{ij} $ tends to the above expression.

Then, for example, the gradient of a scalar function goes to a vector, $$ \frac{\partial (q_jq_j)}{\partial q_i} =2q_i ~~~\to ~~~\frac{\delta \int\!\!dt' ~q(t')q(t') }{\delta q(t)}= 2q(t). $$

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