Energy of the harmonic oscillator

Question

We know the Hamiltonian of harmonic oscillator is $H={p^2\over{2m}} +{1\over 2} kx^2$. We then fix the value of the energy to the eigenvalues $E=(n+1/2) {h\omega\over{2π}}$ . But sometimes in the virial theorem, we set the value of the potential energy equal to ${1\over 2}k_BT$. Where did we get this energy from? And what will be value of the energy for 2D and 3D harmonic oscillator in $k_BT$ terms?

Answer 1

The quantum harmonic oscillator is "semiclassical", so its quantum behavior in phase space is remarkably similar to that of its classical counterpart/limit.

In the hamiltonian, $$ H=\frac{1}{2} \left (\frac{P^2}{m} +m\omega^2 X^2 \right)\equiv T+V, $$ the coordinates and the momenta are dually matched to each other, and the system, both classical and quantum, rotates in ellipses in phase space, normalized to circles if m and ω are suitably absorbed into the variables.

Fock's version of the quantum viral theorem examines $$ \frac{d}{dt} (X P)=\frac{i}{\hbar} [H,XP]= \frac {1}{\hbar} (P^2/m - m\omega^2 X^2). $$

Acting on stationary states, the Hamiltonian eigenvalue is the same on bras and kets in expectation values, so the expectation of the above commutator vanishes, $$ 0=\langle i[H,XP] \rangle = \langle P^2/m -m\omega^2 X^2 \rangle = 2\langle T\rangle- 2\langle V\rangle, $$ that is, $$ \langle T\rangle= \langle V \rangle , $$ so the potential energy is perfectly matched to the kinetic energy, and their respective averages are half of the energy: a complete bi-partition of the energy.

A cluster of non-interacting oscillators, so an oscillator in any dimension, will split into a xerox-copy collection of this identical relation, and so this relation obtains for arbitrarily large noninteracting systems.

It is up to you to apply it in the thermodynamic setting of your choice.