The point of the ZPE is not that it is non-zero, because that could indeed be undone by redefining the potential energy as $$ V\rightarrow \overline{V}=V-E_0 \ . $$ The resulting potential energy has now a somewhat unintuitive dependence on $\hbar$, but the new ZPE is indeed $\overline{E}_0=0$.

However, the important thing is not whether $E_0$ or $\overline{E}_0$ is zero or not, but that the difference $E_0-V_\text{min}=\overline{E}_0-\overline{V}_\text{min}$ is greater than zero, where $V_\text{min}=V(\vec{r}_\text{min})$. This has the important meaning that a particle in an energy eigenstate cannot be strictly localized in the mimimum of the potential (or in its arbitrarily small neighborhood), and this is where the usual arguments with the uncertainty principle come in to play. There is obviously no such restriction classically, where a particle could be standing still in $\vec{r}_\text{min}$ with zero kinetic energy.

This may not seem such a big deal for the harmonic oscillator, but consider instead a hydrogen atom, where the minimum of the potential energy is either $-\infty$ or a very large negative number, depending on whether a point-like or a small finite nucleus is assumed. The zero line of the energy could be again shifted in any way, but the shift-invariant $E_0-V_\text{min}>0$ means that the electron cannot fall into the nucleus. (This just means that the electron cannot be strictly localized at the nucleus with a delta-function-like density, not that it does not have detection probability at that region; actually, $S$-states always have a non-zero detection probability around the nucleus.)

The point of the ZPE is not that it is non-zero, because that could indeed be undone by redefining the potential energy as $$ V\rightarrow \overline{V}=V-E_0 \ . $$ The resulting potential energy has now a somewhat unintuitive dependence on $\hbar$, but the new ZPE is indeed $\overline{E}_0=0$.

However, the important thing is not whether $E_0$ or $\overline{E}_0$ is zero or not, but that the difference $E_0-V_\text{min}=\overline{E}_0-\overline{V}_\text{min}$ is greater than zero, where $V_\text{min}=V(\vec{r}_\text{min})$. This has the important meaning that a particle in an energy eigenstate cannot be strictly localized in the mimimum of the potential (or in its arbitrarily small neighborhood), and this is where the usual arguments with the uncertainty principle come into play. There is obviously no such restriction in classical mechanics, where a particle could be standing still in $\vec{r}_\text{min}$ with zero kinetic energy.

This may not seem such a big deal for the harmonic oscillator, but consider instead a hydrogen atom, where the minimum of the potential energy is either $-\infty$ or a very large negative number, depending on whether a point-like or a small finite nucleus is assumed. The zero line of the energy could be again shifted in any way, but the shift-invariant $E_0-V_\text{min}>0$ means that the electron cannot fall into the nucleus. (This just means that the electron cannot be strictly localized at the nucleus with a delta-function-like density, not that it does not have detection probability at that region; actually, $S$-states always have a non-zero detection probability around the nucleus.)

The point of the ZPE is not that it is non-zero, because that could indeed be undone by redefining the potential energy as $$ V\rightarrow \overline{V}=V-E_0 \ . $$ The resulting potential energy has now a somewhat unintuitive dependence on $\hbar$, but the new ZPE is indeed $\overline{E}_0=0$.

However, the important thing is not whether $E_0$ or $\overline{E}_0$ is zero or not, but that the difference $E_0-V_\text{min}=\overline{E}_0-\overline{V}_\text{min}$ is greater than zero, where $V_\text{min}=V(\vec{r}_\text{min})$. This has the important meaning that a particle in an energy eigenstate cannot be strictly localized in the mimimum of the potential (or in its arbitrarily small neighborhood), and this is where the usual arguments with the uncertainty principle come in to play. There is obviously no such thing classically, where a particle could be standing still in $\vec{r}_\text{min}$ with zero kinetic energy.

This may not seem such a big deal for the harmonic oscillator, but consider instead a hydrogen atom, where the minimum of the potential energy is either $-\infty$ or a very large negative number, depending on whether a point-like or a small finite nucleus is assumed. The zero line of the energy could be again shifted in any way, but the shift-invariant $E_0-V_\text{min}>0$ means that the electron cannot fall into the nucleus. (This just means that the electron cannot be strictly localized at the nucleus with a delta-function-like density, not that it does not have detection probability at that region; actually, $S$-states always have a non-zero detection probability around the nucleus.)

The point of the ZPE is not that it is non-zero, because that could indeed be undone by redefining the potential energy as $$ V\rightarrow \overline{V}=V-E_0 \ . $$ The resulting potential energy has now a somewhat unintuitive dependence on $\hbar$, but the new ZPE is indeed $\overline{E}_0=0$.

However, the important thing is not whether $E_0$ or $\overline{E}_0$ is zero or not, but that the difference $E_0-V_\text{min}=\overline{E}_0-\overline{V}_\text{min}$ is greater than zero, where $V_\text{min}=V(\vec{r}_\text{min})$. This has the important meaning that a particle in an energy eigenstate cannot be strictly localized in the mimimum of the potential (or in its arbitrarily small neighborhood), and this is where the usual arguments with the uncertainty principle come in to play. There is obviously no such restriction classically, where a particle could be standing still in $\vec{r}_\text{min}$ with zero kinetic energy.

This may not seem such a big deal for the harmonic oscillator, but consider instead a hydrogen atom, where the minimum of the potential energy is either $-\infty$ or a very large negative number, depending on whether a point-like or a small finite nucleus is assumed. The zero line of the energy could be again shifted in any way, but the shift-invariant $E_0-V_\text{min}>0$ means that the electron cannot fall into the nucleus. (This just means that the electron cannot be strictly localized at the nucleus with a delta-function-like density, not that it does not have detection probability at that region; actually, $S$-states always have a non-zero detection probability around the nucleus.)