Infinite Square Well

#Physics
$$
U(x)=
\begin{cases}
0, & 0\leq x\leq L \\\infty, & x<0\text{ or } x>L
\end{cases}
$$

• $\displaystyle L$ is the length of the square potential

$\displaystyle \psi_{n}(x)=\sqrt{ \frac{2}{L} }\sin\left( \frac{n\pi x}{L} \right),~n=1,2,3,\ldots$

• $\displaystyle n$ is the energy level. It can't equal $\displaystyle 0$ because otherwise, normalization would not be satisfied, demonstrating Zero-Point Energy
• Desmos Demo
• As $\displaystyle n\rightarrow\infty$, the PDF approaches the classical case of a continuous uniform distribution
• Solutions for different $\displaystyle n$ are orthogonal with respect to the Square Integrable Vector Space

$\displaystyle E_{n}=\frac{n^{2}\pi ^{2}\hbar ^{2}}{2mL^{2}}$

• $\displaystyle E_{n}$ is the energy of the $\displaystyle n$th state of the function

Parameter Values

$\displaystyle {\left\langle{x}\right\rangle}=\frac{a}{2}$

$\displaystyle {\left\langle{x^{2}}\right\rangle}=a^{2}\left( \frac{1}{3}-\frac{1}{2(n\pi)^{2}} \right)$

$\displaystyle {\left\langle{p}\right\rangle}=0$

$\displaystyle {\left\langle{p^{2}}\right\rangle}=\left( \frac{n\pi \hbar}{a} \right)^{2}$

$\displaystyle \sigma ^{2}_{x}=\frac{a}{2}\sqrt{ \frac{1}{3}-\frac{2}{(n\pi)^{2}} }$

$\displaystyle \sigma_{p}=\frac{n\pi \hbar}{a}$