In Griffiths, the average potential energy for the quantum harmonic oscillator is given as
$$\langle V\rangle~=~\frac{1}{2}\hbar \omega\left(n+\frac{1}{2}\right).$$
Is the potential energy of the quantum harmonic oscillator always one half the oscillator's total energy? Does this mean that the average potential energy for said oscillator is always equal to the average kinetic energy? Are there any cases when it is not exactly one half?
Best Answer
In Newtonian mechanics, one may derive a virial theorem, which says that the (time-averaged) kinetic and potential energies are related as
$$ \tag{1} 2\langle T \rangle ~=~p\langle V \rangle ,$$
if the potential $V(r) \propto r^p $ is a power law. Thus for the classical harmonic oscillator (HO)
$$ \tag{2} \langle T \rangle ~=~\langle V \rangle.$$
In QM the averaging procedure is of a different nature, but even so, it turns out that there are no quantum mechanical corrections to (2) in the HO case.