# Hydrino (Deep Dirac) Levels

In high school chemistry and 1st-semester quantum mechanics we all learn that “1s” is the ground state of a hydrogen atom. Wake up sheeple! A few brave souls have argued that hydrogen has a lower-energy state than 1s. According to some, it’s a whopping 500keV lower! These alleged lower-energy states are called “deep Dirac levels” or “hydrino” states.

My assessment? The ground state is 1s. The textbooks are correct. Boring but true.

This post, like the previous two posts, has nothing to teach us about cold fusion, except that this is a dead-end to avoid. But it’s still an interesting story about what can go wrong when solving differential equations.

The two papers I’ll be discussing

I will not be discussing in this post anything by Randell L. Mills, who started a company and raised \$75 million on the premise that hydrinos exist and that he knows how to harness them. To learn more about his theories, see my previous post.

The first paper I want to discuss is the 1993 paper “Electron Transitions on Deep Dirac Levels” by Maly and Vávra. Here is Part I, with the main idea and theory, and here is Part II (1995) with consequences. Vávra has continued to push this theory as recently as 2013 — see this arxiv paper.

The second paper is this 2005 arxiv paper by Jan Naudts, This was taken seriously enough for at least two serious physicists to write refutations in serious mainstream journals. (Links below.) Naudts referred to the sub-1s states as “hydrino states”, following Mills.

Experimental situation

Before I get into these papers let’s do a reality check.

If sub-1s levels really existed, it is implausible—bordering on ludicrous—that they would not be everywhere, really obvious, and well known since the dawn of chemistry. There are so many hydrogen atoms, and they’ve been around for so long, and the 500keV energy release is so irreversible, that the world would be awash in hydrinos unless the 1s-to-hydrino decay lifetime were insanely long. Don’t forget, we’re really good at isolating trace chemicals! I think we can safely say that if a hydrino state existed, it better take at least a billion billion billion billion years (=$10^{43}$ seconds) on average for a 1s hydrogen to decay to it, in order for it to have escaped discovery in the 19th century, let alone the 20th . No atomic transition could possibly be that slow. It’s an electromagnetic interaction, there isn’t really any symmetry “protecting” it, and even if there were, it would be broken when the atoms form compounds, collide with stuff, etc.

So we know from the get-go that these papers have to be wrong. But that’s not enough for me. I want to find the error! Let’s dive in.

Main idea

In quantum mechanics, when we want to understand the “spectrum” (every possible energy level) of an atom or other system, we have to solve a differential equation called the “time-independent Schrodinger equation” (TISE).

Fig. 1: I made this animation for wikipedia many years ago. This is a 1D harmonic oscillator in classical mechanics (A-B) and quantum mechanics (C-H). When you “solve the time-independent Schrodinger equation”, you solve a differential equation in order to discover all the standing wave states (e.g. C,D,E,F, but not G or H). The oscillation frequency of each standing wave is proportional to an energy level of the system.

If you look in introductory quantum mechanics textbook about how to calculate the spectrum of a hydrogen atom, you’ll find that it uses a non-relativistic approximation (i.e., the approximation that the electron is moving much much slower than the speed of light). In more advanced textbooks, you’ll find various more accurate relativistic calculations, which (it turns out) gives almost the same results, but in which some of the energy levels are slightly shifted (cf. fine structure, Lamb shift.)

Neither of these two papers (Naudts and Maly&Vávra) dispute that the standard nonrelativistic calculation is done correctly. Instead they say that there are solutions to the relativistic equation that have no analog in the nonrelativistic version, and that the textbooks erroneously neglect or discard them.

To understand the key issues let’s start with a much simpler math problem.

Simpler math problem: Solving $\psi'' = \psi$ in 1D

Here ψ is a complex-valued function of 1 real variable x, and ψ″ is its second derivative. I want to search more specifically for a solution satisfying the following requirements:

• $\psi'' = \psi$
• $\psi$ is even, i.e. $\psi(x) = \psi(-x)$
• $\psi(x)$ decays to 0 as $|x| \rightarrow \infty$
• …and I’m not counting the stupid solution $\psi(x) = 0$.

(If you’re a physicist, you’ll immediately know that no such ψ exists, but be quiet, don’t ruin the suspense for everyone else!)

I can guess a promising function off the top of my head: $\psi(x) = e^{-|x|}$. This has a problem: It’s not differentiable at $x=0$ (more on which shortly). OK, well maybe this can just be the solution for, say, $|x|>1$. Then here we are so far:

Fig. 2

Things are looking promising: We have a viable solution for almost the entire infinite domain except for one little interval, and it’s satisfying all the requirements that I listed. So far so good! But here’s the problem: What do we put in the middle? You might think: “Oh, we’ll find something-or-other to glue in here!” But it turns out that there is no flexibility at all. The differential equation, when combined with the boundary conditions that we already established at $x=1$, is 100% constraining. We have to walk along the same function ($e^{-|x|}$) all the way to $x=0$:

Fig. 3

What’s so bad about this? A budding mathematician would say: It’s not differentiable at x=0! But that’s not the problem. Physicists would not hesitate to differentiate or twice-differentiate this function. In fact here it is:

$\psi''(x) = e^{-|x|} - 2 \delta(x)$

where δ is the Dirac delta function. So actually $\psi''$ is negative-infinity at $x=0$!  No problem!

I’ll just give another random example: In talking about dipole fields here I was breezily discussing the derivatives of undifferentiable, discontinuous, even infinite functions.  Again, this is no big deal for physicists.

So the fact that this function is not differentiable at the origin (in the traditional sense) is not the problem. What is the problem? It doesn’t satisfy the differential equation at x=0! After all, we want $\psi''=\psi$, but at the origin, $\psi''$ is negative-infinity while $\psi$ is 1.

Conclusion: Fig. 2 was actually a dead end!

I went through this simple example to draw out two lessons:

• Lesson 1: If you solve a differential equation everywhere except some region around the origin, you cannot necessarily extend it to the origin to get a true solution.
• Lesson 2: If you try anyway, by extending your partial solution out the origin, the most important question to ask is “Does this function actually satisfy the differential equation (even at the origin)?” I would ask this question first, before I start asking whether the solution is finite at the origin, differentiable, normalizable, or anything else.

OK now let’s move on to the two papers, one by one.

Maly&Vávra’s paper

In Maly&Vávra’s paper, they are discussing electron wavefunctions in hydrogen whose amplitudes go to infinity at the nucleus. In fact, they go to infinity so fast that they are not normalizable (i.e., square-integrable). Well, Maly&Vávra point out (correctly) that the textbooks are using an oversimplified equation, where the nucleus is treated as an infinitesimal point, rather than being spread out over a femtometer or two. They argue that in the latter situation, the wavefunction would not go to infinity after all. The issues are: (1) They don’t prove it. I believe that would-be wavefunctions can go to infinity even when the potential does not. (2) More importantly, as we learned above, even if the wavefunction does stay finite at the origin, that does not prove that it satisfies the differential equation at the origin. Again, think of Fig. 3.

The first thing to note is that Maly&Vávra do not even try to write out a complete solution valid all the way to the origin, let alone confirm that it really satisfies the differential equation at the origin. Here is one of the relevant excerpts:

This potential is finite inside the nucleus (at r=0, its value is $-3Ze^2/2R$) and the solution of the relativistic Schrodinger equation inside the nucleus is completely different from the solution F(ρ) mentioned earlier. This solution has to be matched on the surface of the nucleus with the external solution Eqs. (8) through (12) (to have the same values on the nuclear surface). After matching both solutions, the wave function of the electron can be fully normalized in the entire region 0 < r < ∞.

For our discussion, however, it is only necessary that levels with both plus or minus signs in the parameters are valid solutions. This should be verified in further experimental work.

You have to read the paper to understand all the terminology but it’s actually kinda funny. The first paragraph describes more-or-less describe a perfectly correct way to proceed: Solve the wave equation both inside and outside the nucleus, and then search for combinations that are compatible, in the sense that they can be glued together at the mutual boundary into a complete solution. …But then the second paragraph says that they’re not going to do that! It’s too much work. Instead they are going to just find the solutions outside the nucleus, and assume that they can find something inside the nucleus to which it can be glued! As we learned from the simple 1D example above, that is a dubious assumption!

So, Maly&Vávra do not actually demonstrate that these alleged states are complete solutions to their differential equations (Eq. (2) or (14) in their paper), including inside the nucleus and including at the origin. What they do is analogous to getting up to Fig. 2 above, and then declaring victory.

OK so you might say, “Even if they didn’t demonstrate it, that doesn’t mean it’s not true anyway!” Well, dear reader, you are insufferably optimistic! I for one feel quite confident that these states cannot be extended into complete solutions (including at the origin). (This comes from thinking about the differential equation, not from my above-stated belief that hydrinos cannot exist for experimental reasons.) It’s too much for this blog post to go though that argument, but anyway you will soon see a few different reasons to believe me.

Maly&Vávra argue with Y.E. Kim et al.

This paper provoked a lively exchange: Comment, Response to comment reply to response to comment. One of the three comment authors was Y.E. Kim, whom I’ve previously criticized but I think he’s pretty much on target here. The way I would summarize it is: Kim and coauthors take seriously the program of building solutions valid all the way to the origin, and find that the sub-1s states are not among them. In Maly&Vávra’s response, it seems to me that they still haven’t gotten the possibility that something can go wrong when extrapolating a wavefunction to the origin, and therefore they just don’t understand what Kim and coauthors are doing or why or how. So it’s a lot of talking past each other.

Naudts’s paper

Now we move on to the 2005 arxiv paper by Jan Naudts. The paper is similar in many ways to Maly&Vávra, and I haven’t really parsed the differences if any. Like Maly&Vávra, he considers two different relativistic variants of the Schrodinger equation: The Klein-Gordon equation and the Dirac equation. Here are some of the key points in the paper:

1. He says the Klein-Gordon equation has sub-1s states that diverge at the origin but are nevertheless normalizable;
2. He says the Dirac equation has similar states, they also diverge at the origin, but these are not normalizable.
3. He includes a funny discussion of whether 1 or 2 is more important: “Is it an axiom of physics that the spinor solutions of Dirac’s equation are square integrable? Or is Dirac’s equation just a convenient way to handle the Klein-Gordon equation … and is the basic requirement that the solution of the latter equation is square integrable?” This is not rhetorical: He does not know the answer!
4. Then he brings up, as Maly&Vávra do, that the nuclear charge is not truly a point source. “The solutions of the Klein-Gordon equation or of the Dirac equation with smeared-out Coulomb potential are expected not to diverge at the origin. Hence the problem of square integrability is not a physical problem…”

My criticism here is exactly the same as before: I don’t care whether the wavefunction is normalizable or finite or whatever, it’s not even worth talking about until you first convince me that the wavefunction actually satisfies the differential equation everywhere (including at the origin). As we learned in the simple 1D example above, you can have a wavefunction that satisfies the differential equation away from the origin, and is finite at the origin, but nevertheless fails to satisfy the differential equation at the origin.

Two papers against Naudts

I’m aware of two papers published directly in response to Naudts. The first is by Norman Dombey: Official linkfree arxiv version. There’s a funny quote in the paper:

[Hydrino states’] obvious failings are that
(1) they lack non-relativistic counterparts even for arbitrarily small coupling;
(2) the states persist even when the coupling is turned off;
(3) the strength of the binding increases as the coupling strength α decreases. The maximum binding occurs for α=0 when the potential has disappeared completely. [footnote: We could call these anomalous states homeopathic states because the smaller the coupling, the larger the effect.]

(“Coupling” here is jargon for “electrical attraction between electron and nucleus”.) LOL! Indeed, for all these reasons and more, my physicist intuition is screaming at the top of its lungs that these alleged sub-1s states are not real quantum states. But, he says, that’s not enough. We still want to know how we went astray mathematically. So he continues:

We now demonstrate that, if the point charge of the nucleus is replaced by a charge extending over an arbitrarily small but finite radius R, then the anomalous functions … cease to satisfy the appropriate wave equation…

I’ll leave out the math which I believe is sound. I think the message here is consistent with what I’m saying in this blog post. At the end of the day, the functions do not actually satisfy the differential equations. If the nucleus were a point charge, you could (presumably) prove this by manipulating delta functions, maybe with the help of some integral transforms. Nobody has done this as far as I know, but I’m quite confident it could be done. This paper takes the alternate (arguably better) approach of examining the case of a slightly-spread-out nuclear charge, and then the solutions are finite and you can just see explicitly that the sub-1s states are not among them.

The second paper is by Antonio S. de Castro: Official versionfree arxiv version. The paper is elaborating on an  appendix at the end of Dombey’s paper (which in turn was inspired by previous work from de Castro himself). Anyway, it is a basic fact of linear algebra that if two eigenvectors of a Hermitian operator have different eigenvalues, then they are orthogonal to each other. The Hamiltonian is a Hermitian operator and the wave equation is an eigenvalue problem. Yet, the sub-1s states are not orthogonal to the normal (1s, 2s, 2p, etc.) states. So if the normal states truly satisfy the differential equation, then the sub-1s states are impostors! Again, I’m pretty confident that this paper is sound physically and mathematically. I think of this as a super-indirect, but nevertheless compelling, proof that the sub-1s states do not satisfy the wave equation at the origin.

Conclusion

All lines of evidence point to the same conclusion: The alleged sub-1s states do not actually satisfy the wave equation. They might work away from the origin (Fig. 2), but if you extend them to the origin, they do not satisfy the equation there. It doesn’t matter whether or not they are finite and/or normalizable (Fig. 3): If they don’t satisfy the equation everywhere then there’s no point in discussing them. This conclusion holds whether the nuclear charge is treated as spread-out (as it is in real life) or simplified to a point charge.

So in conclusion, I am left with no reason to believe that sub-1s states exist, and many very convincing reasons to believe that they don’t.

## 11 thoughts on “Hydrino (Deep Dirac) Levels”

1. DM

I agree with everything you wrote. I do wonder though if lower than ground states could be conceived that are like the hydrino states but are evanescent somehow. I also tried to read the Mills book and wonder if he’s not saying that the differential equation that needs to be solved is actually different than the one you are treating. Care to comment?

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1. steve Post author

I don’t know what you mean by “evanescent somehow”. All bound states have an evanescent (exponentially-decaying) spatial tail. If you mean exponentially decaying in time (imaginary energy?), such states cannot exist because they would violate conservation of probability (though it can come up in certain approximations.)

Since Mills rejects all of quantum mechanics (see previous post), it’s unsurprising that Mills is NOT trying to solve the normal quantum-mechanical equations (in this context: Dirac or Klein-Gordon equation). The important questions are: (1) Is Mills actually solving a specific, well-defined, generally-applicable differential equation? [I see no evidence that Mills is even capable of solving a 3D differential equation, or that he has done so. I think he just writes down formulas that look right to him, and misleadingly refers to them as “solutions” / “derivations” / etc. when they are not.] (2) If so, is there any objective reason to believe that this equation is the correct equation to solve? Again, see my previous post.

FWIW, you might think that Naudts and Maly & Vávra papers offer support / endorsement of Mills. They don’t. They are contradicting Mills by predicting a very different sort of hydrino than Mills predicted, with dramatically different energy. To my knowledge, no trained physicist has ever endorsed Mills’s argument in favor of hydrinos.

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1. Robert Virkus

Your statements regarding Mills are inaccurate. Many competent scientists agree Mills has discovered those states he calls Hydrino’s . He knows how to solve equations and your casual disregard of his abilities disqualifies your argument as having any merit. Simply put, you’re wrong about Mills.

(Steve: All discussion of Randell Mills should happen at the previous post.)

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2. Glen Brisebois

Such a ground state could be considered to exist, but at that point the electron and proton fuse into a neutron and spit out a neutrino. So perhaps the undiscovered “hydrino” country is just a neutron. (No offense to neutrons. I love neutrons.)

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1. steve Post author

This is wrong: You need to add a huge amount of energy to a normal hydrogen atom to create a neutron+neutrino, whereas you would need to subtract energy from a regular hydrogen atom to create a hydrino.

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3. M2

Recently, in an experiment to LENR, scientists have found the annihilation peak of 511 keV. What is likely due to the formation of a compact hydrogen atom.

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1. steve Post author

M2 has linked a video where the folks at the Martin Fleischmann Memorial Project claim to see a gamma ray peak at 511keV (if I understand correctly … I didn’t watch the whole thing).

I don’t know if that measurement is reliable, but for the sake of argument, let’s say they really did get a gamma ray peak at 511keV. This peak is certainly NOT related to hydrinos, because hydrinos don’t and can’t exist. It could be electron-positron annihilation instead. How did the positrons get there? I don’t know. I can’t imagine any way, but that doesn’t mean that I’m confident that it’s impossible. Maybe the positrons could appear by some weird process I didn’t think of. It’s not the kind of thing you can rule out categorically. On the other hand, I *am* extremely confident that there is no such thing as hydrinos. Hydrinos are categorically ruled out by both the laws of quantum mechanics and overwhelming empirical evidence (cf. the “Experimental situation” subheading).

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1. M2

Steve, you’re right, I also think that the hydrino theory wrong. But I’m not talking about the hydrino states. On the contrary, I’m talking about quantum mechanics. You probably have not read my second article http://vixra.org/abs/1609.0086 . There is a section “Compact hydrogen atom and the annihilation peak”. Look at this interesting.

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4. steve Post author

I’m deleting comments about Randell Mills here, because they belong at the previous post. Again, the papers discussed in this post (Maly and Vávra and Naudts) are not endorsing Mills, but rather contradicting him; they propose wildly different hydrino energy levels, for entirely different reasons.

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