Given that $A\left|E_{n}\right\rangle=\alpha\left|E_{n-1}\right\rangle$ and $E_{n}=\left(n+\frac{1}{2}\right) \hbar \omega$, where the annihilation operator of the harmonic oscillator is
$$
A \equiv \frac{m \omega x+\mathrm{i} p}{\sqrt{2 m \hbar \omega}}
$$
show that $\alpha=\sqrt{n}$. Hint: consider $\left.|A| E_{n}\right\rangle\left.\right|^{2}$.