= Solution
For a normalized wavefunction, $\langle A\rangle_\psi$ is the statistical mean of measurements of the <quantum observable> $A$ over identically prepared systems, not necessarily the outcome of one measurement. Assume the state is in the required operator domains and the second moments are finite. Hermiticity gives
$$
0\le\|(A+i\lambda B)\psi\|^2
=\langle A^2\rangle+\lambda^2\langle B^2\rangle+i\lambda\langle[A,B]\rangle.
$$
The commutator expectation is purely imaginary, so $i\langle[A,B]\rangle$ is real. A real quadratic nonnegative for every $\lambda$ has nonpositive discriminant; if its quadratic coefficient vanishes, its linear coefficient must also vanish. Consequently
$$
\boxed{\langle A^2\rangle\langle B^2\rangle\ge\frac14|\langle[A,B]\rangle|^2.}
$$
Apply the same inequality to $A-\langle A\rangle I$ and $B-\langle B\rangle I$. Their commutator is unchanged and their second moments are the variances, yielding the <quantum uncertainty> relation
$$
\boxed{\Delta A\,\Delta B\ge\frac12|\langle[A,B]\rangle|.}
$$
For the <harmonic oscillator>, $p=-i\hbar\partial_x$ gives $[x,p]\psi=i\hbar\psi$ by the product rule. The uncentered inequality above therefore implies $\langle x^2\rangle\langle p^2\rangle\ge\hbar^2/4$. The <arithmetic-geometric mean inequality> now gives
$$
\langle H\rangle=\frac{\langle p^2\rangle}{2m}+\frac{m\omega^2\langle x^2\rangle}{2}
\ge\omega\sqrt{\langle p^2\rangle\langle x^2\rangle}
\ge\boxed{\frac{\hbar\omega}{2}}.
$$
No assumption that the position or momentum means vanish is needed.
Back to article page
