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

[koolmodee]

\intdt' G(t-t') f(t') = 1/i \delta/\deltaf(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]

\intdt' G(t-t') f(t') = 1/i \delta/\deltaf(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)}=<br /> {1\over i}\,{i\over 2}\int dt\,dt&#039;\left[\left({\delta f(t)\over\delta f(t_2)}\right)G(t-t&#039;)f(t&#039;)+f(t)G(t-t&#039;)\left({\delta f(t&#039;)\over\delta f(t_2)}\right)\right]

{}\qquad\qquad={1\over 2}\int dt\,dt&#039;\Bigl[\delta(t-t_2)G(t-t&#039;)f(t&#039;)+f(t)G(t-t&#039;)\delta(t&#039;-t_2)\Bigr]

{}={1\over 2}\int dt&#039;\,G(t_2-t&#039;)f(t&#039;)+{1\over 2}\int dt\,f(t)G(t-t_2)

{}=\int dt&#039;\,G(t_2-t&#039;)f(t&#039;)

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&#039;, so that it is then the same as the first term.

 

[koolmodee]

thank you!