Understanding Srednicki's 7.14-7.16 Equations: G(t-t') and the RHS of f(t)

[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

 

[Avodyne]

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

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

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

 

[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?

 

[Avodyne]

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

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

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

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.

 

[koolmodee]

thank you!