Computing Heisenberg Uncertainty Value

[i.nagi]

Homework Statement

Consider a particle in a one dimensional box of length L, whose potential energy is V(x)=0 for 0<x<L, and infinite otherwise.Given the wave function at ground state ψ=sqrt(2/L)sin (pi*x/L) Compute ΔxΔp where

Homework Equations

Δx=sqrt(<x^2>-<x>^2) and Δp=sqrt(<p^2>-<p>^2)

The Attempt at a Solution

I have set the expected value for as <x^2> equal to the integral from 0 to L ψ1(x)(x^2)ψ1(x)dx, as done by my prof. I then evaluated this to 2/L*the integral from 0 to L x^2sin^2(pi*x/L)dx. However I am stuck here, and my prof's solution goes straight to L^2(1/3-1/2pi^2). I am confused as to how he evaluated this integral and the role of (x^2) in the equation, if steps could be shown that would help immensely, also when calculating <x>, is x included in the integral, and how might it be evaluated if included with sin^2(pi*x/L). Thanks in advance.

 

[Dick]

This is just integration not QM. But you first step is to use a trig identity sin(u)^2=(1-cos(2u))/2 to get rid of the power on the sin. Now multiply it out and start using integration by parts to deal with the x^2*cos part. You must have seen these things somewhere before.