# Srednicki 7.14 ND 7.16

1. Nov 19, 2008

### koolmodee

$$\int$$dt' G(t-t') f(t') = 1/i $$\delta$$/$$\delta$$f(t)

where G(t-t') = i/ 2w exp (iw (t-t'))

I thought the RHS of the first equation is f(t). Can someone explain?

thank you

2. Nov 19, 2008

### Avodyne

This doesn't make sense, and I'm not sure where you got it. Something like it that is correct and that is used to get the 2nd line of (7.16) is

$${1\over i}{\delta\over\delta f(t_1)}\left[{i\over2}\int dt\,dt'\,f(t)G(t-t')f(t')\right] = \int dt'\,G(t_1 -t')f(t').$$

3. Nov 20, 2008

### koolmodee

Well, I thought what I write was implied in the equations in the Srednicki book.

But then I don't see how we get from the first line two the second in 7.16. with your equation.

Is the term in the brackets equal to one? And you mean t_2 instead of t_1, right?

4. Nov 20, 2008

### Avodyne

What you wrote does not make sense. The functional derivative on your right-hand is not acting on anything.

Let

$$Z(f)=\langle 0|0\rangle_f$$

From 7.11,

$$Z(f)=\exp K(f)$$

where

$$K(f)={i\over 2}\int dt\,dt'\,f(t)G(t-t')f(t')$$

By the chain rule,

$${\delta\over\delta f(t_2)}Z(f)={dZ\over dK}\;{\delta\over\delta f(t_2)}K(f)$$

and since $$Z=\exp K$$, $$dZ/dK = \exp K = Z$$. Now we use

$${1\over i}\,{\delta K(f)\over\delta f(t_2)}= {1\over i}\,{i\over 2}\int dt\,dt'\left[\left({\delta f(t)\over\delta f(t_2)}\right)G(t-t')f(t')+f(t)G(t-t')\left({\delta f(t')\over\delta f(t_2)}\right)\right]$$

$${}\qquad\qquad={1\over 2}\int dt\,dt'\Bigl[\delta(t-t_2)G(t-t')f(t')+f(t)G(t-t')\delta(t'-t_2)\Bigr]$$

$${}={1\over 2}\int dt'\,G(t_2-t')f(t')+{1\over 2}\int dt\,f(t)G(t-t_2)$$

$${}=\int dt'\,G(t_2-t')f(t')$$

where, to get the last line, we use $$G(t-t_2)=G(t_2-t)$$, and change the dummy integration variable in the 2nd term from $$t$$ to $$t'$$, so that it is then the same as the first term.

5. Nov 21, 2008

thank you!!!