= Solution
For the <uncertainty product of oscillator number states>, the two means vanish, so the <standard deviations> are the square roots of the second moments:
$$
\Delta x=\sqrt{\frac\hbar{m\omega}(n+\tfrac12)},\qquad\Delta p=\sqrt{m\hbar\omega(n+\tfrac12)}.
$$
Consequently,
$$
\boxed{\Delta x\Delta p=\hbar(n+\tfrac12).}
$$
Only the ground state $n=0$ saturates the <Heisenberg uncertainty principle> in this sequence of <energy eigenstates>; the excited number states have a larger uncertainty product.
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