Calculating some functional derivative

Question

I am reading Mark Srednicki's quantum field theory, p.50~p.52 (Part I section 7).

In the section, he derives a the formula for the ground state to ground state transition amplitude of harmonic oscillator in the presence of an external force as

$$<0|0>|_f = exp \big[{i \over 2} \int_{-\infty}^{\infty} dt dt' f(t)G(t-t')f(t') \big], \tag{7.11}$$

where $G(t-t')$ is a Green's function for the oscillator equation of motions.

If furthur information is needed, I will upload continuosly.

My question is, why next functional derivative

$$ {1 \over i} {\delta \over \delta f(t_2)}<0|0>_f |_{f=0} = \big[ \int_{-\infty}^{\infty} dt' G(t_2-t')f(t') \big]<0|0>_f |_{f=0} $$

is true?

In his book p.46, he states that the functional derivative satisfies that $${\delta \over \delta f(t_1)}f(t_2) = \delta(t_1 - t_2).\tag{6.14}$$

A point that I can't understand is why the ${1 \over 2}$ in the formula for $<0|0>|_f$ is disappeared through the derivative.

My question is originated from following underlined statement in his book:

Why the underlined derivation is true? Can any one provide derivation in detail?

Anyone helps?

Answer 1

A point that I can't understand is why the $\frac{1}{2}$ in the formula for $\langle 0|0\rangle_f$ is disappeared through the derivative.

It happens because of symmetry. You have $G(t-t')=G(t'-t)$. Using the chain rule we just need to differentiate the exponent, but there the symmetry plays a role. Using the product rule we have

\begin{eqnarray}\dfrac{\delta}{\delta f(t_2) }\left[\frac{i}{2}\int_{-\infty}^\infty dtdt' f(t)G(t-t')f(t')\right]&=&\frac{i}{2}\int_{-\infty}^\infty dtdt' \delta(t-t_2)G(t-t')f(t')\\ &&+\frac{i}{2}\int_{-\infty}^\infty dtdt' f(t)G(t-t')\delta(t'-t_2).\end{eqnarray}

Now you can easily check the two terms are the same. Integrate over $t$ on the first using $\delta(t-t_2)$ and over $t'$ on the second using $\delta(t'-t_2)$. Then relabel $t'$ as $t$, combine the integrals and finally use the symmetry property $G(t-t_2)=G(t_2-t)$. You'll get what he says.