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Doubts taking the second functional derivative of the Klein Gordon action

+ 3 like - 0 dislike
134 views

I have very little background with functional derivatives and I would like to clarify some issues.

I am trying to compute the second functional derivative of the Klein Gordon action expressed in real components

$S[\phi_1,\phi_2,\partial_{\mu}\phi_1,\partial_{\nu}\phi_2]=\int{}d^4x\,\mathcal{L}=\int{}d^4x(-\partial_{\mu}\phi_1\partial^{\mu}\phi_1-\partial_{\mu}\phi_2\partial^{\mu}\phi_2+$

$-m^2(\phi_1^2+\phi_2^2)-\lambda(\phi_1^2+\phi_2^2)^2)$

I know how to compute the first functional derivative 

$$\frac{\delta{}S}{\delta\phi_1(x)}=\frac{\partial\mathcal{L}}{\partial\phi_1}-\partial_{\mu}\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_1)}$$

which gives

$$2\left(\partial^2\phi_1(x)-m^2\phi_1(x)-4\lambda\phi_1(x)[\phi_1^2(x)+\phi_2^2(x)]\right)$$

from now on I have doubts. I wanna compute

$$\frac{\delta^2S}{\delta\phi_1(y)\delta\phi_1(x)}$$

 If there were no derivatives this would be trivial but I don't know how to deal with $\partial^2$. I have thought in introducing a delta to write this as an integral and use the formula I have used to compute the first functional derivative, but I am concerned about taking derivatives of something having a Dirac delta.

So, my question is, how am I supposed to compute this functional derivative?

asked Jun 30, 2015 in Theoretical Physics by Dmitry hand me the Kalashnikov (720 points) [ revision history ]

1 Answer

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If $S(\phi)$ is given, one gets the first and second functional derivatives by expanding $S(\phi+sf)$ as a power series in the real parameter $s$ up to second order.

The first order term can always be rearranged in the form $s\int dx f(x) S_1(x)$, which gives $\frac{\delta S}{\delta \phi(x)}=S_1(x)$. (Note: In your expression for the first functional derivative the big closing parenthesis should be after the $m^2\phi_1$ term.)

The second order term can always be rearranged in the form $\frac12 s^2\int dx dy f(x) S_2(x,y) f(y)$ with symmetric $S_2(x,y)$, which gives $\frac{\delta^2 S}{\delta \phi(x)\phi(y)}=S_2(x,y)$. Note that this is a differential operator, and if there are multiple fields, a matrix of differential operators. Alternatively, you could also substitute $\phi$ by $\phi+sg(x)$ in your formula for the first functional derivative, expand to first order to get $sS_2(x,y)g(x,y)$, and proceed as before. In both cases, the second functional derivative you are looking for is found to be $2\delta(x-y)\Big(\partial^2-m^2+4\lambda(3\phi_1^2(x)+\phi_2(x)^2)\Big)$. 

answered Jun 30, 2015 by Arnold Neumaier (12,355 points) [ revision history ]

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