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

• Mayhem

#### 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.

Praise
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.

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

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