Infinite square well potential

Question:
There is an infinite potential well of width $a$. Find the ground state energy.

Answer:
The potential energies for each domain are
\begin{eqnarray}
V(x) &=& \infty \quad & &(x \leqq 0, \ x \geqq a)  \\
V(x) &=& 0 \quad & &(0 < x <  a)
\end{eqnarray}
For $V(x) = \infty$, obviously there is no particle since it gives zero for the wave function. For the other domain, the hamiltonian is
\[
H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}
\]
Therefore the Schroedinger equation becomes
\begin{eqnarray}
-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} &=& E\psi   \\
\frac{d^2\psi}{dx^2} + \frac{2mE}{\hbar^2}\psi &=& 0  \\
\frac{d^2\psi}{dx^2} + k^2\psi &=& 0
\end{eqnarray}
where $k^2=\frac{2mE}{\hbar^2}$.
This differential equation has the general solution.
\begin{equation}
\psi(x) = A\sin kx + B\cos kx
\end{equation}
The wave function is a continuous function; and both edges must be zero due to the infinite potential.
\begin{equation}
\psi(0) = \psi(a) = 0
\end{equation}
This indicates that the wave function must be odd because $0<x<a$. Thus, from equation (6), only the sine function survives, so the wave function is $\psi(x)=A\sin kx$. From (7),
\begin{equation}
A\sin ka = 0
\end{equation}
This gives
\begin{equation}
k_n a = n\pi, \quad n=0,1,2,...
\end{equation}
Since $k=\frac{n\pi}{a}$, the wave function becomes
\begin{equation}
\psi(x) = A\sin \left(\frac{n\pi x}{a}\right)
\end{equation}
Let us now normalize the function.
\begin{eqnarray}
A^2\int^a_0 \sin^2\left(\frac{n\pi x}{a}\right) dx = 1
\end{eqnarray}
Put $\theta=\frac{n\pi x}{a}$, so $dx=\frac{a}{n\pi}d\theta$ and $0<\theta<n\pi$.
\begin{eqnarray}
& &\frac{A^2a}{n\pi}\int^{n\pi}_0 \sin^2\theta d\theta = 1  \\
& &\frac{A^2a}{n\pi}\left[\frac{\theta}{2}-\frac{\sin 2\theta}{4}\right]^{n\pi}_0 = 1   \\
& &\frac{A^2a}{n\pi}\frac{n\pi}{2} = 1   \\
& &A = \sqrt{\frac{2}{a}}
\end{eqnarray}
Hence, the wave function is completed as
\begin{equation}
\psi(x) = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)
\end{equation}
Now, we obtain the energy eigen values. From equations $k^2=\frac{2mE}{\hbar^2}$ and $k=\frac{n\pi}{a}$, we have
\begin{equation}
E_n = \frac{\hbar^2 k^2_n}{2m} = \frac{\hbar^2 n^2 \pi^2}{2ma^2}
\end{equation}
$n=0$ corresponds to $E=0$, so the ground state energy is when $n=1$.

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