# Does the hydrogen anion have bound excited states?

I'm having some trouble puzzling out the literature regarding the existence of bound excited states in the hydrogen anion, H$$^-$$.

Wikipedia claims that no such states exist, and that the subject is uncontroversial, stating that

H$$^−$$ is unusual because, in its free form, it has no bound excited states, as was finally proven in 1977 (Hill 1977)

and citing

1. R.N. Hill, "Proof that the H$$^−$$ Ion Has Only One Bound State". Phys. Rev. Lett. 38, 643 (1977)

There is a similar further paper by Hill,

1. R.N. Hill, "Proof that the H$$^−$$ ion has only one bound state. Details and extension to finite nuclear mass", J. Math. Phys. 18, 2316 (1977)

which extends the work to account for the finite mass of the proton.

On the other hand, upon chasing the highest-cited papers that reference the original ground-state-energy calculation by Bethe [Z. Phys. 57, 815 (1929)], I ran into the review

1. T. Andersen, "Atomic negative ions: structure, dynamics and collisions", Phys. Rep. 394, 157 (2004)

which tells a rather different story. In §4.1, on the hydrogen anion, Andersen states the following:

The H$$^−$$ ion has two bound states: the ground $$\rm 1s^2 \ {}^1S$$ state and the doubly excited $$\rm 2p^2 \ {}^3P$$ state. The latter has not been observed till date. It was predicted computationally nearly 40 years ago and its energy computed repeatedly, most recently and very precise by Bylicki and Bednarz [273]. There is no doubt about its existence, but the experimental non-appearance is linked to the lack of aninitial state from which it canbe reached [273].

Following the kicked can down to Bylicki and Bednarz,

1. M. Bylicki & E. Bednarz, "Nonrelativistic energy of the hydrogen negative ion in the $$\rm 2p^2 \ {}^3P^e$$ bound state". Phys. Rev. A 67 022503 (2003)

there are further self-assured statements that the state does exist,

The H$$^-$$ ion has only two bound states: the ground $$\rm 1s^2 \ {}^1S$$ state and the doubly excited $$\rm 2p^2 \ {}^3P$$. The former has been investigated both theoretically and experimentally. For references see recent papers of Drake, Cassar, and Nistor [1] and Sims and Hagstrom [2] where the ground-state energy has been computed with an extremely high precision. The other bound state of H$$^-$$, $$\rm 2p^2 \ {}^3P$$, has not been observed till date. It was predicted computationally [3] nearly 40 years ago. Its energy was computed repeatedly [4–8] and there is no doubt about its existence. The problem of its experimental nonappearance is due to the lack of an initial state from which it could be reached.

as well as links to a large set of references working on increasing the precision of the theoretical calculation of the energy of this presumed excited state of the system ─ some of them prior to Hill's 1977 work, but also several of them years or decades later from that publication, so they ought to be aware of the theorems in that paper that show that their calculations are impossible. And yet, that reference cluster seems to contain scant or no references to Hill's papers.

So, onwards to my question: what's going on here?

• Is there an actual conflict in the literature? Or are the two strands of work actually compatible with each other for some reason that I can't see yet?
• Say, do the rigorous theorems of Hill require some additional conditions that can actually be relaxed, and that's what's happening in the numerical calculations?
• Or do the calculations actually end up describing eigenfunctions that are so pathological that they shouldn't be counted as bound states?

There's something funny going on here, but I can't believe that the people writing here were unaware of the other side, so I imagine that there's some aspect of the discussion which is considered as 'obvious' and not mentioned too explicitly, and I'd like to better understand what that is.

• Interesting (ties in well with Arpad's question on doubly-negative hydrogen). There is a Drake sole-author article as well, journals.aps.org/prl/pdf/10.1103/PhysRevLett.24.126 – Jon Custer Oct 22 '19 at 12:57
• One lead to follow up on, which didn't fit into the question very neatly: Hill, with coworkers, does refer (in Phys. Rev. A 41, 1247 (1990)) to both strands in the same paper, and he describes the $(2p)^2$ $^3P^e$ state as "of unnatural parity imbedded in the natural parity continuum". This makes no sense to me - as far as I can tell, states of that parity are reported by NIST's ASD for He, Li$^+$ and Be$^{2+}$. How is this "unnatural parity"? – Emilio Pisanty Oct 22 '19 at 13:02
• sciencedirect.com/science/article/pii/S0370157305000943 (EAG Armour et al., Physics Reports 413 1-90 (2005) says: "This triplet Pe state is often described as a bound state.. However, it should more properly be called a metastable state as it does not satisfy the criterion for stability ... namely that its energy must be below the energy ofthe lowest continuum threshold of the system... There is some confusion over this point in the review article on the stability of three-body atomic and molecular ions by Armour and Byers Brown [89]. – Jon Custer Oct 22 '19 at 13:04
• @JonCuster Yeah, that's one of the papers in the cluster that follows from chasing Bylicki & Bednarz. Apparently it is now safe to criticize Drake's work, btw ;-). – Emilio Pisanty Oct 22 '19 at 13:05
• On your second comment ─ that's an interesting point. At least some of the literature does call the $^3P^e$ state 'metastable' ─ cf. e.g. Jáuregui & Bunge, J. Chem. Phys. 71, 4611 (1979). – Emilio Pisanty Oct 22 '19 at 13:08

A reasonable answer is given in

1. A.R.P. Rau, "The negative ion of hydrogen". J. Astroph. Astron. 17, 113 (1996)

where Rau explains as follows:

Of particular interest among the $$Ν = 2$$ states is the lowest one of $$^3 P^e$$ symmetry, described in independent-electron terms as $$2p^2$$. This is bound below the $$\mathrm H(N = 2)$$ threshold with about $$9.6 \:\rm meV$$. The only one-electron continuum at this energy being $$\mathrm H(N= 1) + \text{electron}$$ which cannot form a state with quantum numbers $$^3 P^e$$, this state is forbidden to autoionize. It can only decay into this continuum by also simultaneously radiating a photon along with the electron, these two particles sharing the excess energy of $$\approx 10.2\:\rm eV$$ (Drake 1973).

In other words, the energy of the $$2p^2\ {}^3P^e$$ state, $$E=-0.125\,355\,451\,24 \:\rm a{.}u.$$ as calculated by Bylicki and Bednarz, is strictly below $$E= -\frac18\:\rm a{.}u.$$, which is the minimal energy needed to get to the $$2s$$ or $$2p$$ states of neutral hydrogen coupled with a free electron, so that continuum is not energetically available.

Instead, the only available continuum is the $$N=1$$ continuum, i.e. a neutral hydrogen in the ground $$1s$$ state coupled with a free electron. Since this is energetically available (with the continuum starting at $$E= -\frac12\:\rm a{.}u.$$), the $$2p^2\ {}^3P^e$$ state could in principle be autoionizing, i.e., it could in principle spontaneously fly off into states in that continuum. However, for this to happen directly, the relevant continuum state needs to share the same quantum numbers ($$^3 P^e$$, i.e., a triplet $$S=1$$ state, total angular momentum of $$L=1$$, and even parity under spatial inversion) and this is no longer possible.

• The triplet state with $$S=1$$ is not a problem by itself, beyond its parity under electron exchange (even), which forces the orbital sector to be antisymmetric under exchange.
• Since the bound electron is in the $$1s$$ state with $$\ell_1=0$$ and we want a global $$P$$ state with $$L=1$$, the continuum electron needs to be in a $$p$$ wave with $$\ell_2=1$$. (Thankfully, the Clebsch-Gordan spiel is simple here - only a single combination is available.)
• Thus, at this stage, the wavefunction is required to have the form $$\Psi(\mathbf r_1,\mathbf r_2) = \frac{ \varphi_{1s}(r_1)\,\chi(r_2)Y_{1m}(\hat r_2) - \chi(r_1)Y_{1m}(\hat r_1)\,\varphi_{1s}(r_2) }{\sqrt{2}} \tag 1$$ with $$\chi(r)$$ the desired continuum wavepacket.
• And, unfortunately, by this stage everything is lost, since the wavefunction in $$(1)$$ has odd parity under spatial inversion, and we're forced to a term of the form $$^3P^o$$.

The upshot is that, if the universe under consideration consists only of the proton and the two electrons, this state is stable: it is square-integrable and an eigenstate of the hamiltonian. If you prepare the system in that state, it will sit there indefinitely. If you add arbitrary (small) perturbations, it will shift slightly, but it will remain there. (In other words, it is not an autoionizing state.)

However, the real world has more things in it than just two electrons and a proton, and, in particular, it contains the electromagnetic field. This opens up the possibility of radiative transitions:

• The first clear target is to the $$1s \, Ep$$ continuum described above, with term $$^3P^o$$, where the parity flip is absorbed by the emission of a photon.
• In addition to that, there is also a resonance with structure $$2s\,2p\ ^3P^o$$ (for which the symmetry considerations are identical to the continuum described above), which can also be the recipient state of a radiative transition. (However, since this state's symmetry, $$^3P^o$$, matches that of the $$1s \, Ep$$ continuum, it's an autoionizing state and will spontaneously come apart; hence its classification as a resonance.)

The availability of radiative decay means that this bound state is not truly stable, so it is better termed a metastable bound state. (This does get used in the literature ─ cf. R. Jáuregui & C.F. Bunge, J. Chem. Phys. 71, 4611 (1979).) The question of whether the bound state "exists", though, is ultimately fairly subjective, and depends on what you want those terms to mean and what your tolerance is to radiative transitions down from that state.

In that sense, the $$^3P^e$$ state is similar to, say, the $$2p$$ states of neutral hydrogen, which also radiatively decays down to other states. However, the $$^3P^e$$ state of H$$^-$$ seems to be fairly unique in atomic physics in that it's a stable bound state in the absence of radiative transitions, but their introduction allows it to decay to a continuum state.

So, what's the deal with Hill [1,2], then? Do the rigorous variational numerical calculations, from Drake through Bylicki and Berdnarz, mean that there is a problem with the theorem. I'm inclined to say that there is no problem, particularly because Hill himself reviews the $$^3P^e$$ work (in Phys. Rev. A 41, 1247 (1990)) without finding it at all problematic:

The H$$^-$$ ion also has a genuinely bound (square-integrable) doubly excited state, the $$(2p)^2\ ^3P^e$$ state, of unnatural parity imbedded in the natural parity continuum; thus this state is discrete within its symmetry subspace.

From what I can understand of Hill's theorem, its methods rely exclusively on keeping tabs on the global spectrum, which means that it understands bound states exclusively as point eigenvalues that are isolated from any continuum, and this kicks out the $$2p^2\ {}^3P^o$$ state as it's embedded in the $$1s\,Ep$$ continuum. As far as I can tell, Hill's methods can't really tell that there is a parity selection rule forbidding transitions to that continuum, so its conclusions are compatible with the existence of an bound state in a Hilbert-space sector that's cut off from that continuum.

(Furthermore, there is additional rigorous work on the $$^3P^e$$ sector [H. Grosse & L. Pittner, J. Math. Phys. 24, 1142 (1983)] which shows that there is indeed only one bound state within this "unnatural parity" subspace.)

Does that mean that rigorously speaking there is only one bound state, then, or that there are two? Well, as above, it depends on what matters to you when giving a precise definition of "bound state."

Now, finally: does this state, like, actually exist, in the actual real world of experiments?

• In laboratory experiments, there appears to be a solid consensus that this excited state of H$$^-$$ has never been observed. This doesn't mean that it's impossible to make it ─ it just means that it is very hard and that our capacity to make it (multiplied by the interest of the problem) has not yet hit the mark.
• There does seem to be at least some evidence of astrophysical observation. Rau [5] mentions in particular spectroscopic observations of Zeta Tauri [S.R. Heap & T.P. Stecher, Astrophys. J. 187, L27 (1974)], though given their age and the relative paucity of other similar observations (from what I can tell ─ I'm not an astrophysicist so I don't know what the standards are) I'm tempted to take that reported observation with a grain of salt.
• Seems to be a reasonable summary. Is it 'Pau' or 'Rau'? – Jon Custer Oct 22 '19 at 20:13
• Hmmm, you're right. I'll fix it as soon as reasonable. Good catch! – Emilio Pisanty Oct 22 '19 at 20:56