Can the Energy of Ammonia Molecule Be Defined in a Non-Eigenstate State?

[Lebnm]

We can consider the ammonia molecule $NH_{3}$ a two level quantum system, because the $N$ atom can be either above or below the plane formed by the three $H$ atoms. We call these states $|+ \rangle$ and $|- \rangle$, respectively.
The hamiltonian in the basis $(|+ \rangle, |- \rangle)$ is
$$H =
\begin{pmatrix}
E & -A \\
-A & E
\end{pmatrix}
$$
so its eigenvalues are $E + A$ and $E - A$, and the related eigenkets are $$|\phi_{1}\rangle = \frac{1}{\sqrt{2}}( |+ \rangle + |- \rangle),$$ $$|\phi_{2}\rangle = \frac{1}{\sqrt{2}}( |+ \rangle - |- \rangle).$$I am thinking it's strange, because the both states above are linear combinations of the states $|\pm \rangle$, so, if the system is in $|\phi_{1} \rangle$ or $|\phi_{2} \rangle$, the $N$ atom will be both bellow and above the plane? Is this interpretation correct? And, supposeing I know that the system is in the state $| + \rangle$ (for example), can I say that the energy is not defined? (because this state is not a eigenstate of the hamiltonian).

 

[PeterDonis]

I highly recommend Feynman's discussion of the ammonia molecule in his lectures:

http://www.feynmanlectures.caltech.edu/III_09.html

I am thinking it's strange, because the both states above are linear combinations of the states $|\pm \rangle$, so, if the system is in $|\phi_{1} \rangle$ or $|\phi_{2} \rangle$, the $N$ atom will be both bellow and above the plane?

No. This is a good example of how the common description of states like this as "the atom being in two places at once" is misleading. A better description would be that the $N$ atom doesn't have a definite position relative to the $H$ atoms. But even that requires some further clarification, because what the $| \phi_1 \rangle$ and $| \phi_2 \rangle$ states are telling you is that there are two ways of the $N$ atom not having a definite position relative to the $H$ atoms, and those two ways have different energies. There isn't a classical analogue to this, which is why it seems counterintuitive.

supposeing I know that the system is in the state $| + \rangle$ (for example), can I say that the energy is not defined?

You can say the system doesn't have a definite energy, yes. But it's also important to note that if the system starts out in this state, it won't stay in the same state. Whereas if the system starts out in the $| \phi_1 \rangle$ or $| \phi_2 \rangle$ state, it will stay in that state.

 

[Lebnm]

I understood. Thank you!

 

[DrDu]

The idea that molecules have a well defined shape given by the position of its atomic nuclei is an approximation (the Born - Oppenheimer approximation) which can be justified by noting that the nuclei are much more heavy than the electrons. On a more fundamental level, the atomic nuclei are as delocalized as the electrons in an energy eigenstate. However, the energetic splitting between these eigenstate is in most cases so small, that it is unobservable. Hence the non-eigenstates, which correspond to a well defined molecular structure, will change only on the order of millions of years. The ammonia molecule is exceptional, as the hydrogen nuclei are very light, so that the transformation between the two isomers is rapid.