# Why does parahydrogen have a lower energy than orthohydrogen?

I found this on the wikipedia page on spin isomers of hydrogen:

Parahydrogen is in a lower energy state than is orthohydrogen.

It seem almost like it is obvious but I am having trouble reasoning it out.

Does it have to do with some magnetic dipole coupling? Is the difference in energy explained by the exchange energy?

• Contrast respective rotational spectra dictated by symmetry. May 8, 2017 at 18:56

I dont think we should consider the energy of the electrons. For both para- and ortho- hydrogens, the two electrons are sitting in the same ground state molecular orbital of the hydrogen molecule, so I dont think there is a difference.

The energy of a hydrogen molecule arises from its translational, rotational and vibration kinetic energy. At room temperatures, the vibrational kinetic energies are "frozen out", while the translational kinetic energy of ortho- and para- hydrogen are more or less equal. So we'll consider only the rotational kinetic energy.

According to this:

The lowest parahydrogen level is lower than the lowest orthohydrogen level by $\Delta E / k_b \simeq 175 K$

This suggests that there are actually many energy levels of parahydrogen and orthohydrogen. When that wikipedia statement was comparing energy levels, it was probably referring to the lowest energy levels.

But why is the lowest parahydrogen energy level lower than the lowest orthohydrogen energy level?... Because the parahydrogen can reach a lower energy rotational state than orthohydrogen!

To understand this, first, frame the problem as a quantum rigid rotor. Next, realize that the wavefunction solution must be antisymmetric under particle exchange because the wavefunction describes two identical fermions (the hydrogen nuclei).

Now, wash away the idea that we are dealing with two nuclei and focus on the wavefunction.

The wavefunction consists of two parts: the spin part and the spatial part. If the spatial part is antisymmetric, the spin part must be symmetric, in which case we call the molecule a orthohydrogen. If the spatial part is symmetric, the spin part must be antisymmetric, in which case we call the molecule a parahydrogen.

The rotational hamiltonian is $$\hat{H} = \frac{\hat{\mathbf{J}}^2}{2I}$$

The wavefunction solutions are the spherical harmonics $Y_{J,M}$, and the energy eigenvalues are $$E_J = \frac{\hbar^2J(J+1)}{2I}$$ with a degeneracy of $2J+1$

(I used to think that spherical harmonics are amplitudes of an electron around a nucleus. But now that they are used to represent two rotating particles surprised me.)

The lowest energy spherical harmonic is the $J=0$ state, which is symmetric, followed by the $J = 1,2,3...$ states. So the lowest energy level will be a parahydrogen, followed by a orthohydrogen, then a parahydrogen....

PS: I am answering my own question, so there is a high chance I got something wrong

• Yes, cf WP. May 8, 2017 at 18:55
• Yeah, that is correct. May 21, 2017 at 21:14
• A surprisingly nice reference, with the clarity of a source that's trying to figure things out for the first time, is "Orthohydrogen, parahydrogen, and heavy hydrogen," by A Farkas, 1935 (I think).
– rob
May 21, 2017 at 21:27

The average separation of the electrons (integrated over time) in parahydrogen state is more than the average separation of the electrons in orthohydrogen state. And whenever electrons are away from each other, by nature, they would feel less Coulomb force and hence a lower energy state. This handwaving explanation can be justified in quantum mechanics more rigorously.

This can also be understood in terms of the Pauli exclusion principle, according to the practice physics GRE answer key.

• Err, care to add any details?
– Chris
Jan 24, 2018 at 6:47
• I would, if I understood them. I would also add this as a comment on the original question, rather than an answer, if I had the necessary reputation. Perhaps someone more knowledgeable can chime in with an edit.
– DJG
Jan 24, 2018 at 6:54