$\newcommand{\ud}[1]{{#1^{\dagger}}} \newcommand{\bra}[1]{\left\langle #1\right|} \newcommand{\ket}[1]{\left| #1\right\rangle} \newcommand{\braket}[2]{\left\langle #1 | #2\right\rangle} \newcommand{\braketo}[3]{\left\langle #1 \right| #2 \left| #3\right\rangle} \newcommand{\ketbra}[2]{\left| #1\right\rangle\left\langle #2\right|} \newcommand{\angle}[1]{\left\langle #1\right\rangle}$
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Potential Wells

Infinite Box Well

The infinite box well, $0 \leq x\leq L$, has the following potential function: $$V(x)=\begin{cases} \infty,&x < 0\\ 0,&0 < x < L\\ \infty,& x > L \end{cases}$$ Solving the energy eigenvalue equation, we get the following equations:
  • Outside well: $\phi_E(x)=0$
  • Inside well: $\frac{d^2}{dx^2}\phi_E(x)=-\frac{2mE}{\hbar^2}\phi_E (x)=-k^2\phi_E(x),\quad k^2=\frac{2mE}{\hbar^2}\geq 0\text{ since }E\geq 0$
The general solution to the wavefunction is: $$\phi_E(x)=A\sin kx + B\cos kx$$ We fix the coefficients, $A$ and $B$, by imposing the following boundary conditions:
  1. Wavefunction has to be continuous
  2. Wavefunction must go to $0$ at the walls
    • At $x=0$, $B=0\implies \phi_E(x)=A\sin kx$
    • At $x=L$, $\sin kL=0\implies kL=n\pi\implies k_n=\frac{n\pi}{L}$ $$E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}$$
To get $A$, we normalize the wavefunction $1=\int_{-\infty}^\infty\phi_n^*(x)\phi_n(x)dx$ to get $A=\sqrt{\frac{2}{L}}$ Thus, for a particle in an infinite box well, $$\phi_n(x)=\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L}$$

Finite Square Well

The potential energy function for a finite square well, $-a \leq x \leq a$, is: $$V(x)=\begin{cases} V_0,&x< -a\\ 0,& -a\leq x\leq a\\ V_0,&x>a \end{cases}$$ Once again solving the energy eigenvalue equation, we obtain:
  • Inside well: $k=\frac{\sqrt{2mE}}{\hbar}$ $$\frac{d^2\phi_E(x)}{dx^2}=-k^2\phi_E(x)$$
  • Outside well: $q=\frac{\sqrt{2m(V_0-E)}}{\hbar}$ $$\frac{d^2\phi_E(x)}{dx^2}=q^2\phi_E(x)$$
Thus, we get the general solution: $$\phi_E(x)=\begin{cases} Ae^{qx}+B^{-qx},&x< -a\\ C\sin kx +D\cos kx,& -a\leq x\leq a\\ Fe^{qx}+Ge^{-qx},& x > a \end{cases}$$ We eliminate some of these coefficients by imposing the following conditions:
  • Wavefunction must be normalizable (must go to $0$ at the endpoints). So, $B=F=0$.
  • Wavefunction is continuous.
  • Wavefunction is differentiable (or $\frac{d\phi_E(x)}{dx}$ is continuous).
We split the problem into the odd and even cases of the wavefunction: $$\phi_{E,\text{ even}}(x)=\begin{cases} Ae^{qx},&x < -a\\ D\cos kx,&-a\leq x\leq a\\ Ae^{-qx},&x > a \end{cases},\quad\phi_{E,\text{ odd}}(x)=\begin{cases} Ae^{qx},&x < -a\\ C\sin kx,&-a\leq x\leq a\\ -Ae^{-qx},&x > a \end{cases},$$
Ex 1: A particle in an infinte square well has an initial state of $\psi(x)=\sqrt{\frac{30}{a^5}}x(1-x)$. Calculate $\angle{H}$.
Use the SE recipe.
  1. We find the eigenvalues and eigenstates. $E_n=\frac{n^2\pi^2\hbar^2}{2mL^2},\quad\phi_n(x)=\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L}$
  2. Write $\psi(x)$ as a superposition of eigenstates $\phi_n(x)$. $$\begin{align*}c_n&=\int_0^L\phi_n^*\psi(x)dx\\&=\frac{4\sqrt{15}(1-\cos n\pi)}{n^3\pi^3}\\&=\begin{cases}0,&n\text{ even}\\\frac{8\sqrt{15}}{n^3\pi^3},&n\text{ odd}\end{cases}\end{align*}$$ $$\psi(x)=\sum_n c_n\phi_n(x)$$
  3. Add time dependence. $$\psi(x,t)=\sum_{n\text{ odd}}\frac{8\sqrt{15}}{n^3\pi^3}\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L}e^{-\frac{in^2\pi^2\hbar t}{2mL^2}}$$
  4. Calculate $\angle{H}$. $$\begin{align*} \angle{H}&=\sum_{n\text{ odd}}|c_n|^2E_n\\ &=\sum_{n\text{ odd}}\bigg(\frac{8\sqrt{15}}{n^3\pi^3}\bigg)^2\frac{n^2\pi^2\hbar^2}{2mL^2}\\ &=\frac{64(15)}{\pi^6}\frac{\pi^2\hbar^2}{2mL^2}\sum_{n\text{ odd}}\frac{1}{n^4}\\ &=\frac{64(15)}{\pi^6}\frac{\pi^2\hbar^2}{2mL^2}\bigg(\frac{\pi^4}{96}\bigg)\\ \angle{H}&=\frac{5\hbar^2}{mL^2} \end{align*}$$