In the Feynman Lectures (III-8-6 and 9-1) there is a nice discussion of the ammonia molecule as a prototypical example of the two-state system in quantum mechanics. Feynman begins by saying he's assuming there is no vibrational excitation but that the projection of the angular momentum (notated $K$?) along the symmetry axis is nonzero.

However, he never seems to make any explicit use of this assumption of $K\ne0$. He only seems to briefly use it in a figure, where he uses it to define a physical observable that distinguishes one inversion from the other: in state |1>, the angular momentum is parallel to the electric dipole moment, in |2> antiparallel. The stationary states, which he notates |I> and |II>, are superpositions of 1 and 2.

I believe that the true ground state of the system has total spin-parity $0^+$ (including all electron and nuclear spins, as well as the collective rotational degrees of freedom of the whole molecule). What happens in this case, and why does Feynman take such pains to exclude it? I assume there is still a splitting between the $0^+$ ground state and a $0^-$ first excited state...?

  • $\begingroup$ I could be wrong, but here is my rough guess. I think the state of zero angular momentum is one that is isotropic in the lab frame, so that the shape and orientation of the molecule are captured only in the correlations of the nuclei's positions with each other, not their correlation with the lab-fixed axes. So it may be that the multiplicity of this state is simply 1, not 2. In a state with nonzero angular momentum, we can have inversion or we can have a change of orientation w.r.t. the angular momentum, but both have an energy barrier...? $\endgroup$ – user4552 Nov 30 '17 at 1:36
  • $\begingroup$ I forgot that I had come across this description of the system, which is pretty detailed and useful: courses.washington.edu/phys432/NH3/ammonia_inversion.pdf I need to comb through it to look for this particular issue. $\endgroup$ – user4552 Nov 30 '17 at 17:04
  • $\begingroup$ I think the UW link does address this on p. 4: "[Re] the ground state, J = 0, K = 0, the splitting of this state which would result in an inversion line is not permitted. The splitting is ruled out for all rotational states with K = 0 due to symmetry considerations related to the fact that the H nuclei (protons) are fermions, and as such the total wavefunction describing the molecule must change sign when two H nuclei are interchanged [3]. This requirement is verified experimentally since no inversion line for any state with K = 0 has been observed [4]." $\endgroup$ – user4552 Nov 30 '17 at 18:26

I eventually tracked down the following discussion, which is visible in the Google Books peephole view of Townes and Schawlow, Microwave Spectroscopy, 1955, pp. 69-71. Here is a 3-page pdf of the relevant pages.

Other types of symmetry operations in addition to inversion about the center of mass may also be considered. For a symmetric top with a threefold axis of symmetry such as NH3 or BF3, a rotation of 120 degrees about the symmetry axis should leave the molecule essentially unchanged, and reasoning similar to that applied above to inversion about the origin shows that this rotation must either leave the wave function unchanged or change only its sign if the state is not degenerate.

[...] Any [...] wave function for NH3 changes sign when two H nuclei are exchanged.

Consider just rotation of 120 degrees around the symmetry axis of NH3. This is equivalent to exchanging two pairs of H nuclei, say first numbers 1 and 2, then 2 and 3. Since there are two exchanges, the wave function must be unchanged by a rotation of 120 degrees if H obeys either Fermi-Dirac or Bose-Einstein statistics. The only one of Euler's angles which changes with such a rotation is chi, which enters the wave function as $e^{iK\chi}$ or $e^{-iK\chi}$. Hence after a 120 degree [...] rotation, $\Psi' = \Psi e^{\pm(2\pi/3)Ki}$. If K is a multiple of 3, then the exponential ... equals 1, and $\Psi'=\Psi$, so that the wave function is symmetric. If K is not a multiple of 3, then $\Psi$ is neither symmetric nor antisymmetric. This indicates that the state is degenerate, which is true since the same energy is obtained for $+K$ as for $-K$. In order to make wave functions of the[...]

[Here there is a figure enumerating the 8 possible spin states of the 3 protons.]

When $K=0$, the [wavefunctions of a certain form] become zero when the $-$ sign is used, and hence no such wave function exists. This is the reason why half the levels are nonexistent when $K=0$ as indicated in Fig. 3-9. In the lowest inversion state, when $K=0$ and $J=0$, a $-$ sign [...] would be called for, but this makes the wave function zero. In the upper inversion state, however, when $K=0$ and $J=0$, a $+$ sign is called for and such a wave function is not zero. When $K=0$ and $J$ is odd, however, the ground inversion level requires the $+$ sign and hence is the state in which molecules can exist.

I assume that "ground inversion level" is spectroscopists' jargon for the positive parity state, and "upper inversion level" means the negative parity state. So the basic idea seems to be that because of the Fermi statistics of the three protons, we only get states with $J,K^\pi=0,0^-$ or, when $J$ is odd, $J,K^\pi=J,0^+$. I haven't worked out the logic in detail myself, but this is what I'm getting from Townes. Thus transitions from a state to its inversion will never occur when $K=0$. So the actual explanation seems to be quite a bit more specific and intricate than any of us, including me, had been imagining. It really depends on the fact that it's a three-fold symmetric rotor with identical fermionic nuclei on the three rotating atoms, and there is a specific rule that determines which parity state exists.

If I'm understanding correctly, then the ground state is $J,K^\pi=0,0^-$, with the orientation part of the wavefunction being totally constant, the negative parity being due to the spin states of the protons. As Sean Lake has observed, the system as a whole is fermionic if you consider the nuclei, so I assume the integer values of $J$ and $K$ in Townes refer to only the degrees of freedom apart from the nuclear spins. It seems that the nuclear spin enters into the discussion crucially due to statistics, but not dynamically in terms of angular momentum.


I believe Feynman is treating the ammonia molecule as a rigid rotor with intrinsic symmetry. In this case, wavefunctions for the system are linear combinations of full Wigner $D$-functions: $$ \Psi_{LM}(\Omega)=\sum_K c_K \chi_K D^L_{KM}(\Omega) $$ with coefficients $c_K$ to be determined by the intrinsic symmetry of the molecule and $\chi_K$ an intrinsic wave function.

In the case of $NH_3$, the intrinsic symmetry group is the group $D_\infty$ of rotations in the plane of the three hydrogen molecules, plus the reflection through this plane. For $K\ne 0$, the intrinsic wave function $\chi_K$ (in the body-fixed frame) satisfies \begin{align} R_y(\pi)\chi_K=\chi_{-K}\, ,\qquad &R_y(\pi)\chi_{-K}=\chi_{K}&\quad (\hbox{inversion}=R_y(\pi)) \tag{1}\\ R_z(\phi)\chi_{K}=e^{-iK\phi}\chi_{K}\, ,\qquad & R_z(\phi)\chi_{-K}=e^{iK\phi}\chi_{-K}\, ,&\quad (\hbox{rotations})=R_z(\phi)\, . \end{align} The combinations of intrinsic and $D$-functions that satisfy the symmetry properties of the molecule is $$ \Psi_{KLM}(\Omega)= \sqrt{\frac{2L+1}{16\pi^2}}\left(\chi_K D^L_{KM}(\Omega) +(-1)^{L-K}\chi_{-K} D^L_{-K,M}(\Omega)\right) \tag{2} $$ where the phase $(-1)^{L-K}$ comes up as a result of the reflection of $D^L_{KM}(\Omega)$.

When $K=0$, we have instead \begin{align} R_y(\pi)\chi_{0,\pm}&=\pm\chi_{0,\pm}\, , \tag{3}\\ R_z(\phi)\chi_{0,\pm}&= \chi_{0,\pm} \end{align} with the major change that the intrinsic wave function is an eigenstate of $R_y(\pi)$. In this case we have $$ \Psi_{\pm,LM}(\Omega)= \sqrt{\frac{2L+1}{16\pi^2}}(1\pm (-1)^L)\chi_{0,\pm}D^L_{0M}(\Omega)\, . \tag{4} $$

Tunnelling is possible in both types of systems are possible. The difference is that $K=0$ states are singly degenerate, whereas for $K\ne 0$ configurations states are doubly degenerate, and the rotational states in the two bands have different parity.

Tunnelling can occur between symmetric and antisymmetric $(\chi_{0,\pm})$ states in the $K=0$ configuration: the symmetric state has a slightly lower energy than the antisymmetric one (curvature argument). As far as I know, you are right in assuming the ground state is $0^+$ and tunnelling would then occur between $0^+$ and $0^-$. It can also occur in $K\ne 0$ cases; for these however it is clear from the structure of $D^L_{KM}$ that $L\ge K$, so the lowest energy state of the band is not an $L=0$ state.


  1. Aage Bohr and Ben R. Mottelson, Nuclear Structure (Vol.II: Nuclear Deformations), (World Scientific 1998), section 4.2 and more specifically 4.2f. Please be aware that Bohr and Mottelson do not use the standard definition of $D$-functions: the phases in my write up are those of Varshalovich et al. (Bohr and Mottelson is the best discussion I know.)

  2. L. Landau and E. Lifshitz, Quantum Mechanics, (Pergamon Press), Chapters XI, XII and XIII

  • $\begingroup$ This is nice, thanks for writing it up. There were a couple of things I was unclear about on this answer, and maybe clarifying them would improve the answer. When you say "singly degenerate," do you mean simply that the state is not really degenerate? In Townes' treatment, he has a long and complicated treatment of the statistics of the three hydrogen nuclei. I don't see any discussion of that i your answer. Is there a spot where that enters into your argument, but it's not explicitly worked out? $\endgroup$ – user4552 Dec 7 '17 at 23:37
  • $\begingroup$ @BenCrowell I’m away from my notes right now and the write up took some care. I will look it up (it’s a good question) and will clarify by middle of next week once I’m back. $\endgroup$ – ZeroTheHero Dec 8 '17 at 2:39
  • $\begingroup$ That would be cool, thanks. I have a note on my calendar to check back next Friday. $\endgroup$ – user4552 Dec 8 '17 at 4:28

He wasn't looking for the ground state. He was showing that

Every one of the possible states of the molecule, whatever energy it has, is “split” into two levels. We say every one of the states because, you remember, we picked out one particular state of rotation, and internal energy, and so on. For each possible condition of that kind there is a doublet of energy levels because of the flip-flop of the molecule.

As for why he chose to direct the angular momentum along the axis of symmetry, it's probably a question of not wanting to confuse the picture. If the atom is rotating about some other axis, then the nitrogen atom is going above and below the plane of the hydrogen atoms due to the rotation, alone. Whilst if the angular momentum is along the axis of symmetry, any transition between above and below has to be purely from tunneling.

Once he's established that tunneling splits those states, it's more plausible that it splits all states, though he doesn't explicitly show it. The insight needed to show that any state will have that split comes from the fact that any rotational state can be represented as a superposition of rotational eigenstates along a single axis, though this neglects the fact that for $|J_z|$ large, the ammonia molecule should flatten, narrowing the split between the states.

  • $\begingroup$ Hm, thanks, but this doesn't convince me. He's specifically talking about avoiding the special case of zero angular momentum, but your reasoning doesn't address why zero angular momentum would be special. $\endgroup$ – user4552 Nov 30 '17 at 1:32
  • $\begingroup$ @BenCrowell Can ammonia even be in a $j=0$ state (i.e. is it a boson)? I think N14 an an even number of deuteriums is, or N15 and an odd number of deuteriums. Regardless, I didn't find any mention of $K$ not being zero, only that it has some particular value. What edition are you reading? I'm using: Feynman, Richard P.; Leighton, Robert B.; Sands, Matthew. The Feynman Lectures on Physics, Vol. III: The New Millennium Edition: Quantum Mechanics: Volume 3 (Feynman Lectures on Physics (Paperback)) (Kindle Locations 7694-7696). Basic Books. Kindle Edition. $\endgroup$ – Sean E. Lake Nov 30 '17 at 2:38
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    $\begingroup$ Section 8-6, p. 8-11 has this: "We will consider only that the molecule is spinning around its axis of symmetry (as shown in the figure), that it has zero translational momentum, and that it is vibrating as little as possible." He is explicitly ruling out the case of $K=0$. The figure confirms this by showing the direction of rotation as counterclockwise. There are 10 electrons, so if we assume the coupling to the nuclear spins is negligible, then it's a boson. It's true that if you add in the nuclear spins, it's a fermion. $\endgroup$ – user4552 Nov 30 '17 at 17:03

I think the answer to this question is rather simple. As stated by Feynman himself an ammonia molecule has infinitely many states; electronic, vibrational, translational, rotational, etc. But since he wants to study only a two state system he assumes that

  1. molecule does not move (no translation)
  2. molecule does not vibrate (no vibration)
  3. molecule is in the electronic ground state (this I add myself, he does not say this but I think it is necessary to add)

But he has to assume that the molecule is rotating, that is, angular momentum is not zero. Because for an ammonia molecule which does not rotate around the axis that he shows in the drawing, there is no difference between N up or N down states. In other words, an ammonia molecule in its electronic ground state, not moving, not vibrating, not rotating (angular momentum zero) would be a single state system.

For analogy let's for a moment assume that the electron spin is really a result of electron spinning around its axis. Then one can say that any electronic energy level splits into two; one for clockwise (CW) spinning electron and one for counter clockwise (CCW) spinning electron. Now observe that moving N up or down for a rotating ammonia molecule is equivalent to changing the direction of rotation. Therefore just like CW and CCW spinning electrons create a two state system, N up and N down geometries create a two state system for a rotating ammonia molecule. Electron spin manifests itself in certain cases such as spin-orbit coupling or Stern-Gerlach type of experiments. I guess, there must be manifestations of N up and N down states in certain experiments. I would not be surprised if, for example, one absorbs light with certain polarization while other does not.

  • $\begingroup$ In other words, an ammonia molecule in its electronic ground state, not moving, not vibrating, not rotating (angular momentum zero) would be a single state system. This may be correct, but what is the justification for this statement? $\endgroup$ – user4552 Nov 30 '17 at 16:57
  • $\begingroup$ Just read your answer below (or above) and I do not understand it. I will edit my answer to include your comment. $\endgroup$ – physicopath Nov 30 '17 at 19:49

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