From Wikipedia, Pauli has said in his Nobel lecture that "It is clear that this zero-point energy has no physical reality". This feels natural - I've always been slightly puzzled by the existence of the zero-point energy $\frac{\omega}{2}$ for the quantum harmonic oscillator. Yet the math seemed 'obvious' to me, until today.

Let $H$, $a$ and $a^\dagger$ be the Hamiltonian, annihilation and creation operators, respectively, defined by:

$$H = \frac{1}{2}(p^2 + \omega^2 x^2)$$ $$a = \frac{x}{\sqrt{2\omega}} + i\sqrt{\frac{\omega}{2}}p.$$

Then one can find the following relations:

$$H = \omega(a^\dagger a + \frac{1}{2})$$ $$H|n\rangle = (n + \frac{1}{2})\omega|n\rangle$$

Where $|n\rangle$ is the $n$'th eigenstate of $H$. This gives a zero point energy (ZPE) of $\omega / 2$. So far so good. Yet today I was reading David Tong's lecture notes on QFT. In the second chapter, he makes the remark that classically:

$$H_{\text{classical}} = \frac{1}{2}(p^2 + \omega^2 x^2) = \frac{1}{2}(\omega x - ip)(\omega x + ip).$$

Let $H'$ be the quantum analogue of the right hand side expression. Then:

$$H' = \frac{1}{2}(\omega x - ip)(\omega x + ip) = \omega a^\dagger a$$ $$H' = H - \frac{1}{2}\omega.$$

Thus yielding a different Hamiltonian operator, due to the fact that $x$ and $p$ do not commute. In particular, the ZPE is zero for $H'$:

$$H'|0\rangle = (H - \frac{1}{2}\omega)|0\rangle = 0.$$

I understand the Hamiltonian operator is ultimately defined as the generator of time evolution. There seems to be no difference between $H$ and $H'$ in that regard. Why is there a preference for $H$ over $H'$ in QM textbooks? And is this notion of strictly non-zero ZPE, which is often linked to the uncertainty principle, incorrect? What are the consequences for our interpretation of QM phenomena if the ZPE is indeed arbitrary?

EDIT: Thanks for all your responses. It's clear to me now that I was confusing zero-point energy and zero-point motion.