jkwiens.com: Answer: Infinite Potential Well

Answer: Infinite Potential Well

Saturday, 23 of February , 2008 at 10:43 am

Question: A quantum particle is inside a one-dimensional infinitely deep potential well.

Find the wavefunction that satisfies Schrödinger’s equation.

$i \hbar \frac{\partial}{\partial t} \Psi (\vec{r}, t)= \frac{-\hbar^2}{2m} \nabla^2 \Psi (\vec{r}, t) + V(\vec{r}) \Psi (\vec{r}, t)$

Answer:

First assume that $\Psi (\vec{r}, t)$ can be represented as $\Psi (\vec{r}, t) = \phi (\vec{r}) \Phi (t)$ . This means that Schrödinger’s equation can be manipulated as followed:

$i \hbar \frac{\partial}{\partial t} \Psi (\vec{r}, t)= \frac{-\hbar^2}{2m} \nabla^2 \Psi (\vec{r}, t) + V(\vec{r}) \Psi (\vec{r}, t)$
$i \hbar \phi (\vec{r}) \frac{\partial}{\partial t} \Phi (t)= \frac{-\hbar^2}{2m}\Phi (t) \nabla^2 \phi (\vec{r}) + V(\vec{r}) \phi (\vec{r}) \Phi (t)$
$\frac{i \hbar}{\Phi (t)} \frac{\partial}{\partial t}\Phi (t)= \frac{-\hbar^2}{2m \phi (\vec{r})} \nabla^2 \phi (\vec{r}) + V(\vec{r})$

Since the this equation holds for all $\vec{r}$ and , both sides of the equation must equal a constant.

$\frac{i \hbar}{\Phi (t)} \frac{\partial}{\partial t}\Phi (t)= \frac{-\hbar^2}{2m \phi (\vec{r})} \nabla^2 \phi (\vec{r}) + V(\vec{r}) = E$

Time Component
Next, we will solve for the time component of the above equation.

$\frac{i \hbar}{\Phi (t)} \frac{\partial}{\partial t}\Phi (t) = E$
$i \hbar \int \frac{d\Phi}{\Phi (t)} = E \int dt$
$i \hbar ln(\Phi (t)) = E t$
$\Phi (t) = e^{\frac{-iE}{\hbar}t}$

Time-Independent Component
The time-independent component of the Schrödinger equation would be:

$\frac{-\hbar^2}{2m} \nabla^2 \phi (\vec{r}) + \phi (\vec{r})V(\vec{r}) = E\phi (\vec{r})$

The solution to this problem can be easily solved for the case $x \le 0$ and $x \ge L$ . Since $V(x) = \infty$ , the only way to solve

$\frac{-\hbar^2}{2m} \frac{d^2}{dx^2} \phi (x) + \phi (x)V(x) = E\phi (x)$

is to have $\phi (x) = 0$ .

For the case when , we know that . Therefore, the time-independent equation will become:

$\frac{-\hbar^2}{2m} \frac{d^2}{dx^2} \phi (x) = E\phi (x)$

This equation should look rather familiar because it is the differential equation describing a simple harmonic oscillator. Therefore, the general solution to this differential equation would be:

$\phi (x) = A sin(\frac{\sqrt{2mE}}{\hbar}x) + B cos(\frac{\sqrt{2mE}}{\hbar}x)$

Therefore, the solution would be:

$\phi (x)=\begin{cases}0 & x \le 0~and~x \ge L \\A sin(\frac{\sqrt{2mE}}{\hbar}x) + B cos(\frac{\sqrt{2mE}}{\hbar}x)& 0<x<L\end{cases}$

If we add a requirement that $\phi (x)$ needs to be piecewise-smooth, we can add boundary conditions $\phi (0) = 0$ and $\phi (L) = 0$ to the problem:

$\phi (x) = A sin(\frac{\sqrt{2mE}}{\hbar}x) + B cos(\frac{\sqrt{2mE}}{\hbar}x)$

Therefore, when $\phi (0)$ , we can see that because

$\phi (x) = A sin(0) + B cos(0) = B = 0$

Now that we know , we can apply the second boundary condition $\phi (L)=0$ to the remaining portion of the equation.

$\phi (L) = A sin(\frac{\sqrt{2mE}}{\hbar}L) = 0$
$\therefore ~ \frac{\sqrt{2mE}}{\hbar}L = n \pi$ where $n \in \mathbb N$

This implies

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

Therefore, we can see that the general solution can be reduced to:

$\phi (x)=\begin{cases}0 & x \le 0~and~x \ge L \\A sin(\frac{n\pi}{L}x) & 0<x<L\end{cases}$

The last step is to normalize this wave-function

$1 = \int_{-\infty}^{\infty} |\phi (x)|^2 dx$
$= |A|^2 \int_{0}^{L} sin^2(\frac{n\pi}{L}x) dx$
$= |A|^2 \frac{L}{n\pi}(-\frac{1}{2}cos(\frac{n\pi}{L}x)sin(\frac{n\pi}{L}x) + \frac{1}{2}\frac{n\pi}{L}x)\arrowvert_{0}^{L}$
$= |A|^2 \frac{L}{n\pi}(\frac{1}{2}n\pi) = |A|^2 \frac{L}{2}$

Therefore

$A = \sqrt{\frac{2}{L}}$

Therefore, the final solution would be

$\Psi (x, t)=\begin{cases}0 & x \le 0~and~x \ge L \\ \sqrt{\frac{2}{L}} sin(\frac{n\pi}{L}x)e^{\frac{-iE}{\hbar}t} & 0<x<L\end{cases}$
where $E = \frac{n^2 \pi^2 \hbar^2}{2mL^2}$

Category: Answers, Quantum Mechanics

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