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Apr 10, 2011 at 11:41 comment added Marek @Luboš: ah, right; I didn't read the question carefully enough, I am afraid :(
Apr 10, 2011 at 10:17 comment added Luboš Motl The question whether "it's verifiable" was really the original question of the OP. If it were not verifiable even in principle - and in non-relativistic non-gravitating QM with a fixed potential, it's not - then the OP would be right that we can't really claim that there is a physical zero-point energy because it depends on the way how we write it.
Apr 10, 2011 at 10:15 comment added Luboš Motl Dear @Marek, right, physics contains many important non-measurable concepts. But one must distinguish whether a quantity is unmeasurable just "directly" - but it has physical consquences - from the case when it's unmeasurable in principle. In the latter case, it's literally unphysical. In non-relativistic non-gravitating quantum mechanics with a fixed potential etc., the additive energy shift is unmeasurable even in principle because it may be incorporated into a redefinition of $V$. This is not the case in SR; GR; or when we may change $V$ or $H$ and compare the energies.
Apr 10, 2011 at 9:02 comment added Marek @Luboš: well, measuring these values is certainly a problem. Nevertheless, QM tells us that $E > V_{\rm min}$ for any bound state localized around $V_{\rm min}$, right? It's no problem that it's not verifiable. Theory can (and must) surely produce lots of results we can never measure.
Apr 10, 2011 at 8:40 comment added Luboš Motl Note that if you calculate the energy as $T(p)+V(x)$ out of measured values of $x$ and $p$, the uncertainty principle makes the error of the energy exceed $\hbar\omega/2$ or so, anyway. In this sense, $V_{\rm min}$ cannot be measured separately from $E_0$.
Apr 10, 2011 at 8:39 comment added Luboš Motl Right, @Marek, $E_0-V_{\rm min}$ is independent of conventions. However, one may always imagine that $V(x)$ was different by $\hbar\omega/2$ than we thought and we will produce the same energy levels. Of course, then we must ask whether $E_0$ and $V_{\rm min}$ may be measured independently. It depends what tools we have to measure them. You have to assume that we can - $V_{\rm min}$ may be measured by localizing the electron, except that then it has a huge kinetic energy.
Apr 10, 2011 at 8:24 history answered Marek CC BY-SA 3.0