Why Does My Calculation of <x^2> Yield a Negative Result?

[facenian]

Homework Statement

Evaluate <x^2> for the wave function $\psi(x)=\int_{-\infty}^{\infty}dk exp(-|k|/k_0)exp(ikx)$
My calculation yields a negative answer and I can't find my error

Homework Equations

$$|\psi(x)|^2=\int_{-\infty}^{\infty}dkexp(-|k|/k_0)\int_{-\infty}^{\infty}dk'exp(-|k'|/k_0)exp(i(k-k')x)$$
$$<x^2>=\int_{-\infty}^{\infty}dx|\psi(x)|^2x^2$$
$$\int_{-\infty}^{\infty}dxx^2exp(i(k-k')x)=-\frac{d^2}{dk^2}\int_{-\infty}^{\infty}dxexp(i(k-k')x)=-2\pi\delta''(k-k')$$

The Attempt at a Solution

$$<x^2>=\int_{-\infty}^{\infty}dk'exp(-|k'|/k_0)\int_{-\infty}^{\infty}dkexp(-|k|/k_0)\int_{-\infty}^{\infty}dxx^2exp(i(k-k')x)$$
$$<x^2>=\int_{-\infty}^{\infty}dk'exp(-|k'|/k_0)\int_{-\infty}^{\infty}dkexp(-|k|/k_0)(-2\pi\delta''(k-k'))$$
$$<x^2>=-2\pi\int_{-\infty}^{\infty}dk'exp(-|k'|/k_0) \frac{1}{k_0^2}exp(-|k'|/k_0)$$
$$<x^2>=-\frac{2\pi}{k_0^2}\int_{-\infty}^{\infty}dk'exp(-2|k'|/k_0)=-\2\pi/k_0$$

 

[arkajad]

You have |k|. How are you differentiating it when you move '' from delta?

 

[facenian]

You have |k|. How are you differentiating it when you move '' from delta?

[tex]\frac{d}{dk}|k|=sg(k)[\tex]
where sg(k) is 1 when k> and -1 when k<0 and in the second order differenciation sg(k) behaves as a constant. Of course it is not defferentiable at k=0.

 

[arkajad]

Shouldn't second differentiation produce delta(k)?

 

[facenian]

I think the equation is
$$\int_{-infty}^{infty} f(x)\delta^{(n)}(x)=(-1)^nf^n(0)$$

 

[arkajad]

And sg'[k]=2delta[k].

 

[facenian]

I didn't know that. Where can I find a proof?

 

[arkajad]

Check http://en.wikipedia.org/wiki/Heaviside_step_function"

The formal proof can be found in any text on distributions, for instance: http://www.mae.ufl.edu/~uhk/HEAVISIDE.pdf"

sg[x]=H[x]-H[-x]

therefore sg'[x]=2 delta(x).

 

[facenian]

ok, thank you very much arkajad