- #1
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Consider the famous harmonic oscillator. Now imagine I use my freedom in choosing the origin for the potential and choose, instead of the usual thing, [itex] V(x)=\frac{1}{2}m \omega^2 x^2-\frac{1}{2} \hbar \omega [/itex], so the TISE becomes:
[itex]
-\frac{\hbar^2}{2m}\frac{d^2 \psi}{dx^2}+\frac 1 2 m\omega^2x^2 \psi=(E+\frac{1}{2} \hbar \omega) \psi
[/itex]
So its obvious, that the energy levels become [itex] E_n= n\hbar \omega [/itex]. So now the ground state has zero energy!
But this can't be right. Where is zero-point energy? If it is a physical concept, then it shouldn't depend on our choices. So there should be a way that it shows up here too. But I fail to see that way. Any ideas?
Thanks
[itex]
-\frac{\hbar^2}{2m}\frac{d^2 \psi}{dx^2}+\frac 1 2 m\omega^2x^2 \psi=(E+\frac{1}{2} \hbar \omega) \psi
[/itex]
So its obvious, that the energy levels become [itex] E_n= n\hbar \omega [/itex]. So now the ground state has zero energy!
But this can't be right. Where is zero-point energy? If it is a physical concept, then it shouldn't depend on our choices. So there should be a way that it shows up here too. But I fail to see that way. Any ideas?
Thanks