For some quantum mechanical observable $Q$, prove that its uncertainty $(\Delta Q)$ satisfies

$$(\Delta Q)^{2}=\left\langle Q^{2}\right\rangle-\langle Q\rangle^{2}$$

A quantum mechanical harmonic oscillator has Hamiltonian

$$H=\frac{p^{2}}{2 m}+\frac{m \omega^{2} x^{2}}{2}$$

where $m>0$. Show that (in a stationary state of energy $E$ )

$$E \geqslant \frac{(\Delta p)^{2}}{2 m}+\frac{m \omega^{2}(\Delta x)^{2}}{2}$$

Write down the Heisenberg uncertainty relation. Then, use it to show that

$$E \geqslant \frac{1}{2} \hbar \omega$$

for our stationary state.