# Does ammonia emit microwaves?

Ammonia has an electric dipole moment which means that there is a charge located on one side of the molecule.

The molecule can also exist in an inverted form with the charge located on the other side of the molecule. According to this article on nitrogen inversion the energy barrier between the two states allows for tunnelling.

If the ammonia molecule has some energy (eg thermal room temperature) then it will oscillate between the two states at a frequency of 10GHz.

Does this charged particle moving through space thus interact with the EM field and emit 10GHz microwaves?

I am aware that ammonia can also be described as being in a superposition between the two states. Presumably this arrises from the Heisenberg uncertainty principle. However, if this is the case then surely the ammonia dipole location is also in a superposition and can be averaged over the two points.

Which of these descriptions is more correct? And does the observation of the ammonia, and projection into eigenstates lead to a classical-like oscillation of charge?

• Yes, it does emit noisy microwaves. I believe the are used in radio astronomy to some extent – KF Gauss Feb 6 at 5:35

## 1 Answer

I believe the latter description, in which each ammonia molecule is in a superposition of the nitrogen-up and nitrogen-down orientations, is more correct for ammonia found in nature. The two lowest-energy eigenstates of the Hamiltonian for an isolated ammonia molecule are even and odd superpositions of $$\mathrm{H_3N}$$ and $$\mathrm{NH_3}$$, i.e. (reference: 1):

$$|\psi_+\rangle = (|\mathrm{H_3N}\rangle + |\mathrm{NH_3}\rangle)/\sqrt{2}$$

$$|\psi_-\rangle = (|\mathrm{H_3N}\rangle - |\mathrm{NH_3}\rangle)/\sqrt{2}$$

The energy splitting between these two eigenstates is of order $$10^{-4}\,{\mathrm{eV}}$$ (references: 2, 3), which is much smaller than the energy per degree of freedom of a typical particle at room temperature $$(k_B \times 273\,\mathrm{K}/2 \approx 1 \times 10^{-2}\,\mathrm{eV})$$, so a population of ammonia molecules at room temperature will be approximately evenly distributed between these two energy eigenstates. These ammonia molecules can transition between the $$|\psi_+\rangle$$ and $$|\psi_-\rangle$$ states by absorbing and emitting microwave radiation at a few tens of GHz.

In the absence of an applied electric field to break the symmetry of the Hamiltonian, no particle has a true permanent dipole moment in the lab frame: the eigenstate of the system always exhibits reflection (parity) symmetry. However, as you suggest, if you project ammonia into either the $$|\mathrm{H_3N}\rangle$$ or the $$|\mathrm{NH_3}\rangle$$ state (and then remove the polarizing field), the molecule will invert back and forth between the two geometric configurations with a frequency given by the energy splitting (few 10s of GHz). Both classically and quantum mechanically, this oscillation results in microwave emission. In the quantum picture, the radiation occurs in the form of microwave photon emission when the excited state decays back to the ground state.

As a side note, this rapid oscillation occurs for ammonia because of the low and narrow energy barrier between its two inversion states. For larger and more complicated molecules, the tunneling time can be long enough that one orientation of the species is effectively stable (reference: 4). For example, right- and left-handed DNA molecules are related by a reflection symmetry analogous to ammonia's, and the absolute ground state of DNA would be a superposition of the two chiralities. However, biological processes assemble DNA in the right-handed configuration, and it would take many lifetimes of the universe for it to tunnel to the left-handed configuration. Therefore, unlike ammonia, DNA is geometrically stable.

This question might also be helpful: Superposition of quantum states.