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&#039;exp(-|k&#039;|/k_0)exp(i(k-k&#039;)x)
&lt;x^2&gt;=\int_{-\infty}^{\infty}dx|\psi(x)|^2x^2
\int_{-\infty}^{\infty}dxx^2exp(i(k-k&#039;)x)=-\frac{d^2}{dk^2}\int_{-\infty}^{\infty}dxexp(i(k-k&#039;)x)=-2\pi\delta&#039;&#039;(k-k&#039;)

The Attempt at a Solution

&lt;x^2&gt;=\int_{-\infty}^{\infty}dk&#039;exp(-|k&#039;|/k_0)\int_{-\infty}^{\infty}dkexp(-|k|/k_0)\int_{-\infty}^{\infty}dxx^2exp(i(k-k&#039;)x)
&lt;x^2&gt;=\int_{-\infty}^{\infty}dk&#039;exp(-|k&#039;|/k_0)\int_{-\infty}^{\infty}dkexp(-|k|/k_0)(-2\pi\delta&#039;&#039;(k-k&#039;))
&lt;x^2&gt;=-2\pi\int_{-\infty}^{\infty}dk&#039;exp(-|k&#039;|/k_0) \frac{1}{k_0^2}exp(-|k&#039;|/k_0)
&lt;x^2&gt;=-\frac{2\pi}{k_0^2}\int_{-\infty}^{\infty}dk&#039;exp(-2|k&#039;|/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?

\frac{d}{dk}|k|=sg(k)[\tex]<br /> where sg(k) is 1 when k&gt; and -1 when k&lt;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