Doubt while computing the second functional derivative of the Euclidean action of a point particle

Question

I am reading the chapter on instantons on Coleman's Aspects of Symmetry 8http://www.amazon.com/Aspects-Symmetry-Selected-Erice-Lectures/dp/0521318270).

Consider the Eucidean action of a point particle in 1+1 dimensions, under a potential

$$S[x]=\int{}dt\,(\frac{1}{2}\dot{x}^2+V)$$

I wanna get the second functional derivative of the action

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

I have been able to compute the first one

$$\frac{\delta{}S}{\delta{}x(t)}=-\ddot{x}+V'(x)$$

but I am clueless about how to proceed. Any hint will be much apreciated.

Answer 1

If you take the functional derivative of first order, i.e $\frac{\delta S[x(t)]}{\delta x(t)}=- \ddot{x}(t)+V'(x)$.

Then express these with dirac deltas i.e:

$\int \ddot{x}(s)\delta(s-t')ds= \int x(s)\ddot{\delta}(s-t')ds$ where I have used to integration by parts and boundary conditions obviously vanish.

So if you take another functional derivative you get:

$$\int \frac{\delta x(s)}{\delta x(t)} \ddot{ \delta}(s-t') ds = \int \delta(t-s) \ddot{\delta}(s-t') ds = \ddot{\delta}(t-t')$$

And for the other part also:

$$V'(x(t')) = \int V'(x(s)) \delta(s-t')ds$$

So now we have:

$$\int \frac{\delta V'(x(s))}{\delta x(t)} \delta(s-t')ds = \int V''(x(s)) \delta(s-t)\delta(s-t') ds = V''(x(t))\delta(t-t')$$

So if we gather all the terms we get:

$$-\ddot{\delta}(t-t')+V''(x(t))\delta(t-t')$$

Which is a rigorous treatment of the maple calculation.

Comments

Very nice computation, congratulations and many thanks. 

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thanks

Answer 2

Using Maple I am obtaining the obvious result:

$$\left( D^{ \left( 2 \right) } \right)  \left( V \right)  \left( x \left( \tau \right)  \right) {\it Dirac} \left( \tau-t \right) -{\it Dirac} \left( 2,\tau-t \right)$$

which in a more standard notation reads

$$\frac{\delta{}^2S}{\delta{}x(\tau)\delta{}x(t)}= V''(x)\delta(\tau-t)-\ddot{\delta}(\tau-t)$$