What is the probability that a particle in the first excited $\mathbf{(}\mathit{n}\mathbf{=}\mathbf{2}\mathbf{)}$ state of an infinite well would be found in the nuddle third of the well? How does this compare with the classical expectation? Why?

Probability to be found in the middle third of the well. That is, between $a/3$and $2a/3$as per the Figure $2$.

Consider the infinite potential well given below (Figure localid="1657602770659" $1$).

localid="1657602788132" $V\left(x\right)=\{0$, if localid="1657602793033" $0\pounds x\pounds a\ne$otherwise

Figure $1$: Infinite potential well and width are ‘a’.

Figure $2$: First excited state when $n=2$

The solutions are given for an infinite well as:

$\Psi n\left(x\right)=2a\sin \left(n\pi ax\right)$

For $n=2$

$\Psi 2\left(x\right)=2a\sin \left(2\pi ax\right)$

Then, the probability of finding the particle could be determined from the below formula:

$P=\int \left|2a\sin \right(2\pi ax\left)\right|2dx$

$P=\int \left|2a\sin \right(2\pi ax\left)\right|2dx$

Let $\mathbf{k\ae }2$

Then, apply the limits $a/3$and $2a/3$substitute:

$P=2a\int a/32a/3\sin 2kxdx$

Integrating the right-hand side, and simplifying:

$P=2a\left\{\int a/32a/3(1-\cos 2kx)2dx\right\}$

$P=2a\left\{\int a/32a/3dx2-\int a/32a/3\cos 2kx2dx\right\}$

We can take $1 / 2$ outside and divide the 2 outside curly brackets.

$P=1a\left\{\int a/32a/3dx-\int a/32a/3\cos 2kxdx\right\}$

Now integrate and apply upper and lower limits:

$\sin 8\pi 3=\sin \left(480\pi \right)=\sin (360\pi +120\Omega )=\sin \left(120Z\right)$

$P=1a\left\{\right[x]a/32a/3-[\sin 2kx2k]a/32a/3\}$

Now substitute and simplify:

$P=1a\left\{\right[2a3-a3]-[\sin 4\pi a\times 4\pi a]a/32a/3\}$

$P=\{13-[\sin 8\pi 3-\sin 4\pi 34\pi \left]\right\}$

Here, Substituting the values and simplifying we will get

$P=\{13-[\sin 8\pi 3-\sin 4\pi 34\pi \left]\right\}=[13-0.137]=0.196$

$P=0.196$