 Solution

The standard deviations defining the position and momentum uncertainties satisfy $(\Delta x{)}^2=⟨x^2⟩-⟨x{⟩}^2$ and $(\Delta p{)}^2=⟨p^2⟩-⟨p{⟩}^2$. Part (a) makes both means zero. Using the second moments from part (b),Multiplication givesThe Heisenberg uncertainty relation requires only $\Delta x\Delta p\ge \hslash /2$. Among the oscillator energy eigenstates, equality occurs solely for the ground state $n=0$; the excited states have larger uncertainty products.