Zero-point energy of diatomic hydrogen (particle in a box)

[Mayhem]

Homework Statement

Can the zero-point energy of diatomic hydrogen be calculated as a sum of the zero-point energies of all particles in the system?

Relevant Equations

$E_1 = h^2/8mL^2$

If we take $H_2$ as a "particle" in a box, can the zero-point energy of the overall molecule be calculated as the sum of the zero-point energies of all particles in $H_2$?

That is $$E_ {1,H_2}=\frac{2h^2}{8m_{\mathrm{H^+}}L^2} + \frac{2h^2}{8m_{\mathrm{e^-}}L^2}= \frac{h^2}{4L^2}(1/m_{\mathrm{H^+}}+1/m_{\mathrm{e^-}})$$

My reasoning being that in our "ideal" box, the system is isolated, and thus energy must be conserved.

 

[anuttarasammyak]

One hydrogen molecule, two protons and two electrons system, has its ground state for chemical bond.
Say hydrogen molecules are in a box, since they are Bose particles, all remain in the ground state for molecule motion.
The ground state energy of hydrogen molecules in a box seems to be decomposed like that.

 

[Bystander]

You may want to consider the "ortho-para" equilibrium for "ground state" calculations.

 

[Mayhem]

You may want to consider the "ortho-para" equilibrium for "ground state" calculations.

Explain?