Does a Particle in an Infinite Potential Well Obey the Uncertainty Principle?

[G01]

Homework Statement

I need to show that a particle in an infinite potential well in the nth energy level, obeys the uncertainty principle and also show which state comes closest to the limit of the uncertainty principle.

This means i have to calculate <x>, <x^2>, <p> and <p^2>

Homework Equations

The wavefunction for Energy n: $$\psi(x) = \sqrt{\frac{2}{a}}\sin(n\pi x/a})$$

uncertainty principle: $$\sigma_x\sigma_p >= \hbar/2$$

The Attempt at a Solution

I have calculated <x> and <x^2> and <p> and <p^2> :

$$<x> = \frac{2}{a}\int_0^a x\sin^2(\frac{n \pi x}{a})dx = a/2 +a/(4n^2\pi^2)$$$$<x^2> = \frac{2}{a}\int_0^a x^2\sin^2(\frac{n \pi x}{a})dx = a^2/3 - a^3/(4n^2\pi^2)$$

Im not so sure about this one. I did integration by parts twice to obtain it.

Heres what gets me:

<p> comes out to be 0

<p^2> come out to be $$\frac{h^2n^2\pi^2}{a^2}$$, which seems to be correct.

I can find the sigmas easily, the expectation values, while I know how to find them, I keep screwing up, so could someone please tell me if <p> is supposed to be zero? Also, if anything about the rest of the problem looks incorrect, please tell me. Thank you for your help. I'm trying to teach myself this material and this forum has been indispensable. I really appreciate the help.

 

[Warr]

The <p> and <p^2> are right. The <x> and <x^2> seem to have a problem.

I got $$<x>=\frac{a}{2}$$ and $$<x^2>=\frac{a^2}{3}-\frac{a^2}{2{\pi}^2n^2}$$

 

[G01]

Thank you for your help. I fixed my mistake, which had nothing to do with physics just calculation and distributing! Thanks a lot! I'm sure I'll be back with more problems though