# Calculation of 3-point function given a generating funcional $Z[J]$

With:

$$\ln Z[J]= \int dt \frac{J^2(t)}{2} f(t) + C \int dt \frac{J^3(t)}{3!}$$

I am asked to calculate the 3-point funcion.

Attempted solution:

The 3-point funcion is given by $$\frac{ \delta^3 }{\delta J(t_1)\delta J(t_2) \delta J(t_3)}Z[J]$$, which means that I need to evaluate

$$\frac{ \delta^3 }{\delta J(t_1)\delta J(t_2) \delta J(t_3)} exp \Bigg[\int dt \frac{J^2(t)}{2} f(t) + C \int dt \frac{J^3(t)}{3!}\Bigg]$$

When taking these functionals derivatives what I get is

$$\bigg(J(t_1)f(t_1) + C \frac{1}{2}J(t_1)^2\bigg)\bigg(J(t_2)f(t_2) + C \frac{1}{2}J(t_2)^2\bigg)\bigg(J(t_3)f(t_3) + C \frac{1}{2}J(t_3)^2\bigg)Z[J]$$

And when putting $$J=0$$ this gives zero for the 3-point function, in fact, it will give zero for all n-point functions if my reasoning is correct. But I don't think this exercise was supposed to be so trivial, that is why I am writing this question.

I would appreciate if someone can point out if I am making a mistake or I am incorrect with my reasoning.

• Your functional derivatives are wrong. Check again. – AccidentalFourierTransform Aug 10 at 21:02
• For example, $\frac{\delta}{\delta J(t_1)} \int dt \frac{1}{2}J(t)^2 f(t) = \int dt J(t) \frac{\delta J(t)}{\delta J(t_1)} f(t) = \int dt J(t)\delta (t-t_1) f(t) = J(t_1)f(t_1)$, where the functional derivative of $f(t)$ is zero because it doesn't have any $J$ dependence. Where is my mistake in this example? – Slayer147 Aug 10 at 21:56
• So far so good. Higher derivatives, though, ... – AccidentalFourierTransform Aug 10 at 22:18
• Would there be no integration to cancel the delta functions comming from $\frac{\delta J(t_1)}{\delta J(t_2)}$? – Slayer147 Aug 10 at 22:29
• As a hint, you should end up with only one term that doesn't vanish, and it will be proportional to something like $\delta(t_1 - t_2)\delta(t_2 -t_3)$. Other hint, it looks like you are failing to apply the second and third derivatives to the factor of $J(t_3) f(t_3) + \frac{1}{2} CJ(t_3)^2$ that comes down after the first derivative. – Luke Pritchett Aug 11 at 1:26