# Infinite square well – minimum energy

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.3.

We’ve seen that the energy of a system must always be greater than the minimum of the potential function. As a specific example of this we can look at the Schrödinger equation for the square well, between ${x=0}$ and ${x=a}$:

$\displaystyle \frac{d^{2}\psi}{dx^{2}}=-\frac{2m}{\hbar^{2}}E\psi \ \ \ \ \ (1)$

If ${E=0}$, ${\psi"=0}$. Integrating gives ${\psi=Ax+B.}$ Attempting to satisfy the boundary conditions, we get ${\psi(0)=0}$ giving ${B=0}$. Then the condition ${\psi(a)=0}$ gives ${A=0}$, thus ${\psi(x)=0}$ and cannot be normalized.

If ${E<0}$, we solve the equation

$\displaystyle \psi"=-\frac{2mE}{\hbar^{2}}\psi\equiv k^{2}\psi \ \ \ \ \ (2)$

with ${k=\sqrt{-2mE/\hbar^{2}}}$. Since ${E<0}$, ${k}$ is real. The general solution is ${\psi(x)=Ae^{kx}+Be^{-kx}}$. Applying boundary conditions, we get ${\psi(0)=0=A+B}$, so ${A=-B}$. At ${x=a}$, we have

 $\displaystyle \psi(a)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Ae^{ka}+Be^{-ka}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle AC+\frac{B}{C}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A\left(C-\frac{1}{C}\right) \ \ \ \ \ (6)$

where ${C\equiv e^{ka}>0}$. Since ${C}$ is strictly positive (being an exponential) the only way we can get ${C-1/C=0}$ is for ${C=1}$, implying ${ka=0}$. However, neither ${k}$ nor ${a}$ is zero here, so ${C\neq1}$, so ${A=0=B}$ and ${\psi(x)=0}$ again. Thus if ${E\leq0}$, the wave function cannot be normalized and satisfy the boundary conditions.