# Homework Help: Particle in the box

1. Mar 7, 2014

### LagrangeEuler

1. The problem statement, all variables and given/known data
Particle is put in into the infinity potential well. How the probability of finding particle between different maxima of probability density is changing by increasing of $n$?

2. Relevant equations
$\psi_n(x)=\sqrt{\frac{2}{a}}\sin \frac{n\pi x}{a}$
$E_n=\frac{n^2\pi^2\hbar^2}{2ma^2}$

3. The attempt at a solution
$\frac{d}{dx}|\psi_n(x)|^2=\frac{1}{a}\frac{2n\pi}{a}\sin\frac{2n\pi x}{a}=0$
so
$\frac{2n\pi x}{a}=k\pi$
From that $x=\frac{ka}{2n}$.
To bi maximum
$\cos \frac{2n\pi x}{a}<0$.
Right?
What is easiest way to solve this?

2. Mar 8, 2014

### Simon Bridge

Write out the probability calculation for n=1,2,3 ... and spot the pattern.

3. Mar 9, 2014

### LagrangeEuler

For $n=1$ I get that $\frac{d}{dx}\psi_1^2(x)=0$ for $x=\frac{ka}{2}$. How I now that $k=1?$. If I take second derivative I get $\frac{d}{dx}\psi_1^2(x)=\frac{4\pi^2}{a^3}\cos \frac{2\pi x}{a}$. If $k=0$, $x=0$ and $cos(0)=1>0$, so $\psi_1(x=0)$ is minimum. For $k=1$, $x=\frac{a}{2}$ and $cos(\pi)=-1<0$ so for $x=\frac{a}{2}$ function has maximum. For $k=2$, $x=a$ function again has minimum.

For $n=2$ I get that $\frac{d}{dx}\psi_2^2(x)=0$ lead to $x=\frac{k}{4}a$. So I have five possibilities $k=0,1,2,3,4$. Two maxima are for $k=1$ and $k=3$.

For $n=3$ seven possibilities $k=0,1,2,3,4,5,6$. Three maxima are for $k=1,k=3,k=5$.

But I still don't have real generalisation. I have only behaviour. I see that the all maxima are for odd number $k$.

Is there any other easier way to find this?

4. Mar 9, 2014

### Simon Bridge

You are overthinking it:
- just sketch the probability densities for the first 3-4 energy levels, what sort of function are they?

- mark in two adjacent maxima for each and shade the area between them that you need to calculate.
Make a note of what the x values are (hint: try substituting $\theta=kx$ - what are the values of $\theta$ going to be for the first two maxima?)