Chapter 7, Problem 59P

Show that the two lowest energy states of the simple harmonic oscillator, ${\psi }_0\left(x\right)$ and ${\psi }_1\left(x\right)$ from Equation 7.57, satisfy Equation 7.55.

${\psi }_n\left(x\right)=N_{n}e-{\beta }^2{x}^2/2H_n\left(\beta x\right),n=0,1,2,3,\mathrm{....}$

$\frac{h}{2m}\frac{d^2\psi \left(x\right)}{d{x}^2}+\frac{1}{2}m{\omega }^2{x}^2\psi \left(x\right)=E\psi \left(x\right).$