Adjusting the Zero Point Energy

Actually, the student is ignoring the finer details of why there's a non-zero energy in the first place.

If you apply the uncertainty principle for a particle in its ground state, say at the origin and with zero average momentum, it must be the case that

$\Delta x \Delta p \geq \frac{\hbar}{2}$ . Since the average x and p are zero, we have $\Delta x ^2 = <x^2>$ and $\Delta p^2 = <p^2>$ . Thus, the uncertainty principle implies that the first two terms of the Hamiltonian cannot be made simultaneously zero.

Thus, the kinetic and elastic terms are inversely proportional to each other. We all know how to minimize this type of function, and it's to make the terms equal to each other.

$\displaystyle \left<\frac{p^2}{2m}\right> = \left<\frac{m\omega^2}{2}x^2\right> = A \rightarrow \left<\frac{p^2}{2m}\right>\left<\frac{m\omega^2}{2}x^2\right>=\frac{1}{4}\omega^2\Delta x^2\Delta p^2\geq\frac{\hbar^2\omega^2}{16}$

Thus, $A = \frac{\hbar}{4}$ so the Hamiltonian is ALWAYS $\frac{\hbar}{2}$ higher than any classical energy level added to the Hamiltonian.

Another problem is as the system becomes large and classical, there is no mechanism for justifying the proposed offset, and we're stuck with the same problem were started with and the classical oscillator will have an offset with quantum parameters: $\hbar$ .

Accept the ZPE and realize that it's a natural consequence of the Heisenberg Uncertainty Principle.