Recent content by agingstudent

No, it's not. I got the right answer. Thanks for all your help!

This can't be right. The integral is between -a and a, and this would set <p^2> to 0. But <p> is 0, and that would make the variance 0, which would violate the uncertainty principle.

So the second derivative is just taking the derivative twice, right? Which means the answer is 2?

Is it just [tex](-2x)^2[\tex]

Thank you. I already found p, but my question is particularly about the part of the integrand that contains the partial derivative squared. \frac{d^{2}}{dx^{2}}(a^{2}-x^{2}) dx I don't know what to do with that.

Homework Statement I am trying to find the variance of p for a wave function \Psi(x,0)=A(a^2-x^2) I'm confused about how to set up the integral. it should be something like -i^2h^2\int_{-a}^a A(a^2-x^2) (\frac{\partial\Psi}{\partial x})^2 dx I'm confused about the partial...

Homework Statement The book says to consider the gaussian distribution p(x) = Ae-m(x-a)2 where A, m, and a are all positive, real constants. I have no idea how to evaluate this! The book says to look up the relevant integrals. I see the integral of e-x2is pi1/2 but I don't know...