The harmonic oscillator and the rigid rotator are traditional examples in any quantum mechanics text. The former can represent the vibrations of a diatomic molecule while the latter can represent its rotation. By solving the time-independent Schroedinger equation for the two systems, one obtains:(adsbygoogle = window.adsbygoogle || []).push({});

E[itex]_{n}[/itex] = const. (n+[itex]\frac{1}{2}[/itex]) , where n=0,1,2,.... for the harmonic oscillator, and:

E[itex]_{J}[/itex]=const. J(J+1) , where J=0,1,2,.... for the rigid rotator.

One can see that in the former case there is zero point energy (at n=0), while in the latter there is not (J=0 [itex]\Rightarrow[/itex] E=0). In one text I came across the following explanation for the appearance of the zero point energy in the harmonic oscillator:

If E=0 , Kinetic energy =0 [itex]\Rightarrow[/itex] momentum=0 AND potential energy =0 [itex]\Rightarrow[/itex] x=0 . Hence, Both Δx=0 and Δp=0 violating the uncertainty principle.

I tried to follow this logic on the rigid rotator for which the potential energy is zero by construction. Hence, E=0 implies p=0 but the position has infinite uncertainty ( I guess it may be better to talk about angular momentum and angle here instead of p, x).

I tried to conclude from this that zero point energy arises from potential energy. For potential-free systems , it should not arise. Am I right in my conclusion? Any insight will be appreciated.

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# Zero Point Energy

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