0
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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.

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  • 2
    $\begingroup$ Your functional derivatives are wrong. Check again. $\endgroup$ – AccidentalFourierTransform Aug 10 at 21:02
  • $\begingroup$ 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? $\endgroup$ – Slayer147 Aug 10 at 21:56
  • $\begingroup$ So far so good. Higher derivatives, though, ... $\endgroup$ – AccidentalFourierTransform Aug 10 at 22:18
  • $\begingroup$ Would there be no integration to cancel the delta functions comming from $\frac{\delta J(t_1)}{\delta J(t_2)}$? $\endgroup$ – Slayer147 Aug 10 at 22:29
  • 1
    $\begingroup$ 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. $\endgroup$ – Luke Pritchett Aug 11 at 1:26

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