Position and momentum in the infinite well

The probability density $|\phi_n(x)|^2=\frac2L\sin^2(n\pi x/L)$ is symmetric about the midpoint of the well, since $\sin^2(n\pi(L-x)/L)=\sin^2(n\pi-n\pi x/L)=\sin^2(n\pi x/L)$. The expectation value of the position is therefore $\langle\hat X\rangle=L/2$.

For $\langle\hat X^2\rangle$, let $k=2n\pi/L$ and use $\sin^2u=(1-\cos2u)/2$:

$$\langle\hat X^2\rangle=\frac2L\int_0^Lx^2\sin^2\Bigl(\frac{n\pi x}{L}\Bigr)dx=\frac1L\int_0^Lx^2\,dx-\frac1L\int_0^Lx^2\cos(kx)\,dx .$$

The first integral is $L^3/3$. For the second, two integrations by parts give, using $\sin(kL)=0$ and $\cos(kL)=1$,

$$\begin{aligned} \int_0^Lx^2\cos(kx)\,dx&=\Bigl[\frac{x^2\sin kx}{k}\Bigr]_0^L-\frac2k\int_0^Lx\sin(kx)\,dx\\ &=-\frac2k\left(\Bigl[-\frac{x\cos kx}{k}\Bigr]_0^L+\int_0^L\frac{\cos kx}{k}dx\right)=\frac{2L}{k^2}=\frac{L^3}{2n^2\pi^2}. \end{aligned}$$

Thus

$$\langle\hat X^2\rangle=L^2\Bigl(\frac13-\frac{1}{2n^2\pi^2}\Bigr), \qquad \sigma_X^2=\langle\hat X^2\rangle-\frac{L^2}{4}=L^2\Bigl(\frac{1}{12}-\frac{1}{2n^2\pi^2}\Bigr).$$

The function $\phi_n$ is real. From the exercise on probability current, or by direct calculation, $\langle\hat P\rangle=-i\hbar\int\phi_n\phi_n'\,dx=-\frac{i\hbar}{2}\bigl[\phi_n^2\bigr]_0^L=0$, since $\phi_n$ vanishes at both boundaries. For $\langle\hat P^2\rangle$, the simplest approach is to use the Hamiltonian: inside the well, $\hat H=\hat P^2/(2m)$, so $\langle\hat P^2\rangle=2mE_n=n^2\pi^2\hbar^2/L^2$. Alternatively, integration by parts gives $\langle\hat P^2\rangle=\hbar^2\int_0^L|\phi_n'|^2dx=\hbar^2\frac2L\frac{n^2\pi^2}{L^2}\frac L2$, which yields the same result. Hence

$$\sigma_P=\frac{n\pi\hbar}{L}.$$

Physically, the particle is in a superposition of two plane waves with momenta $\pm n\pi\hbar/L$ and zero mean momentum.

The product is

$$\sigma_X\sigma_P=\hbar\,n\pi\sqrt{\frac{1}{12}-\frac{1}{2n^2\pi^2}}=\hbar\sqrt{\frac{n^2\pi^2}{12}-\frac12}.$$

It is an increasing function of $n$, with its minimum for the ground state $n=1$: $\sigma_X\sigma_P=\hbar\sqrt{\pi^2/12-1/2}\simeq0{,}568\,\hbar$. This product is greater than $\hbar/2$, in accordance with the Heisenberg relation, but does not attain it: only Gaussian wave packets saturate the inequality, and the function $\phi_1$ is not Gaussian.

As $n\to\infty$, the term in $1/n^2$ vanishes and $\sigma_X^2\to L^2/12$. A classical particle bouncing between the walls at constant speed spends the same amount of time in each interval of equal length: its position, observed at a random time, is uniformly distributed over $[0,L]$, with density $1/L$. Its mean is $L/2$ and its variance is

$$\frac1L\int_0^L\Bigl(x-\frac L2\Bigr)^2dx=\frac{L^2}{12}.$$

Thus the classical limit is recovered for large quantum numbers, in accordance with the correspondence principle. For finite $n$, the quantum density $\frac2L\sin^2(n\pi x/L)$ oscillates rapidly about the mean value $1/L$; when $n$ is large, these oscillations become indistinguishable on the scale of an apparatus with finite resolution.