How do I find the variance of p for a given wave function?

[agingstudent]

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 derivative squared. My technique was to set up the integral like this:

$$hA^3\int_{-a}^a (a^2-x^2) (-2x^2)^2 dx$$ but I'm pretty sure my answer is way wrong. I don't know how to deal with the partial derivative. please help.

Homework Equations

The Attempt at a Solution

 

[kuruman]

For the variance, you need to find <p> and <p²>, then σ²=<p²>-(<p>)². With

$$\psi(x)=A(a^{2}-x^{2})$$

You should use

$$<p^{2}>=A^{2}\int(a^{2}-x^{2})(-\hbar^{2})\frac{d^{2}}{dx^{2}}(a^{2}-x^{2}) dx$$

and

$$<p>=A^{2}\int(a^{2}-x^{2})(-i\hbar)\frac{d}{dx}(a^{2}-x^{2}) dx$$

Take the second or first derivatives under the integral sign, before you integrate.

 

[agingstudent]

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.

 

[agingstudent]

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

 

[gabbagabbahey]

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.

Forget about the dx on the end of that expression until you are ready to integrate.

Just take the second derivative of $a^2-x^2$ with respect to $x$...that's exactly what $\frac{d^{2}}{dx^{2}}(a^{2}-x^{2})$ means.

 

[agingstudent]

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

 

[gabbagabbahey]

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

Yes, $\frac{d^2}{dx^2}f(x)$ is just another way of writing $f''(x)$.

 

[kuruman]

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

- 2.

 

[agingstudent]

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.

 

[gabbagabbahey]

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.

No,

$$\langle p^2\rangle=2\hbar^2A^2\int_{-a}^{a} (a^2-x^2)dx\neq 0$$

 

[agingstudent]

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