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In Griffiths, the average potential energy for the quantum harmonic oscillator is given as

$$<V>=\frac{1}{2}\hbar \omega(n+\frac{1}{2})$$

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? Thank you in advance.

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3 Answers 3

OK, let's unpack this a bit. The result in Griffiths is, if we write it out more carefully, $$\langle n | \hat{V} | n \rangle = \frac{\hbar \omega}{2}(n + 1/2)$$ where $\hat{V}$ is the potential energy operator $\frac{1}{2} m \omega^2 x^2$ and $|n\rangle$ is the eigenstate of the Hamiltonian, $\hat{H} |n\rangle = E_n |n \rangle$, with $E_n = \hbar \omega(n + 1/2)$.

"Is the potential energy of the quantum harmonic oscillator always one half the oscillator's total energy?"

No. Remember, a state only has a definite value of an operator if it is an eigenstate of that operator - the state $|n\rangle$ does not have a well-defined potential energy, since $\hat{V}$ and $\hat{H}$ do not commute. If we attempt to measure the potential energy $V$ of a quantum harmonic oscillator, we would get a value $V$ with a probability proportional to $|\langle V | n \rangle|^2$, and collapse the oscillator into the eigenstate of potential energy (also the eigenstate of position for a harmonic oscillator).

"Does this mean that the average potential energy for said oscillator is always equal to the average kinetic energy?"

Yes, it does. $\hat{H} = \hat{T} + \hat{V}$, which (with what we know from above) shows $\langle n | \hat{T} | n \rangle = \langle n | \hat{V} | n \rangle$.

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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)

$$ \tag{2} \langle T \rangle ~=~\langle V \rangle.$$

I 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.

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Correction to the last sentence: I QM should read In QM. –  Qmechanic Sep 29 '13 at 12:21

I believe that this is simply $ 1/2 E_n $ where $E_n$ is the total energy. In a harmonic oscillator, the energy is constantly switching between kinetic and potential energy (as in a spring-mass system) and therefore, the average will be 1/2 the total energy. Mind you this is just the average in time, so if you sat there and recorded the potential energy over a long period of time, you would get readings ranging from 0 to $E_n$ and they would average to $1/2 E_n$. I hope you can see now the difference between averaging and taking an instantaneous reading.

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