# Quantum mechanics HW problem on infinite square well.

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1. Oct 6, 2016

### quellcrist

1.
$<x>= \int_{0}^{a}x\left | \psi \right |^{2}dx$
$\psi (x)=\sqrt{\frac{2}{a}}\sin\frac{n\pi x}{a}$
then $<x>= \frac{2}{a} \int_{0}^{a}x \sin\frac{n\pi x}{a}dx$

2. Relevant equations

1) $y=\frac{n\pi x}{a}$ then $dy=\frac{n\pi}{a}dx$
and
2)
$y=\frac{n\pi x}{a}$ then $dx=\frac{a}{n\pi}dy$

then
$\psi (x)=\sqrt{\frac{2}{a}} \sin(y)$
$<x>= \frac{2}{a}\int_{0}^{a=n\pi}y \sin^{2}ydy \times \frac{a}{n\pi} \times \frac{a}{n\pi}$

3. The attempt at a solution

I don't need help solving the general problem for the expectation value of x...I have the solution manual. The question I have is about how/why they chose to solve the integral this way by substituting y for (n*pi*x)/(a)? I understand how 1) works but I need help clarifying how 2) works.

I need a general walkthrough of why they are doing this integral this way.

Thank you

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2. Oct 6, 2016

### BvU

Hi quell,
It is pretty customary to work around integrands with $\sin ax$ to get integrands with $\sin y$ : it makes it easier to get this factor a outside the integral, especially when higher powers occur and/or partial integrations are involved.