BEC 2: Coulomb barrier in a BEC

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In the previous post, I introduced Kim’s Bose-Einstein Condensate (BEC) theory of cold fusion. I said that the two biggest problems with the theory are:

  1. At room temperature, the deuterons cannot condense into a BEC.
  2. Even if the deuterons condensed into a BEC, they would not undergo nuclear fusion, for the same reason as usual: Because the Coulomb barrier prevents them from getting close enough.

In this post I will just talk about #2. So for the time being, please assume for the sake of argument that the deuterons really do condense into a BEC. The question is: Will that make the Coulomb barrier problem go away?

Coulomb barrier as a two-particle correlation

What is the effect of the Coulomb barrier? It’s simple. When two deuterons get very close, they repel very strongly. Because of this, it is extraordinarily unlikely to find two deuterons very close to each other.

A more technical way to say this is the Coulomb barrier introduces a two-particle correlation. “Correlation” here has the same meaning as correlations in statistics: Let’s say I simultaneously measure the position of all the particles (deuterons). A correlation means that if I tell you that one of the particles is at point A, it alters the conditional probability that a second particle is at a nearby point B. For example, if there is a deuteron at (0,0,0), then it is extraordinarily unlikely that there is a second deuteron at (0, 0, 0.01nm), because of Coulomb repulsion.

I’m going to pause here and repeat this. The only way to calculate the effects of Coulomb repulsion is to talk about two-particle correlations. Remember, Coulomb repulsion is by far the most important factor affecting fusion rates. So if you want to calculate fusion rates in the BEC, you need to do a calculation that accounts for the fact that the BEC has two-particle correlations. If you neglect that fact, it amounts to forgetting that Coulomb repulsion exists in the first place!

How can a BEC have two-particle correlations?

As an undergraduate, I learned (incorrectly, as we’ll see) that the wavefunction of a BEC is the following:

\Psi(\mathbf{r}_1,\mathbf{r}_2,\ldots,\mathbf{r}_N) = \psi(\mathbf{r}_1) \psi(\mathbf{r}_2) \cdots \psi(\mathbf{r}_N)

where N is the number of particles, Ψ is the many-particle wavefunction (in the position basis), \mathbf{r}_7 is the location of the 7th deuteron, and ψ is the single-particle wavefunction into which the deuterons have condensed. If you haven’t seen many-body wavefunctions before, keep in mind that it is a complex function, and that |\Psi(\mathbf{r}_1,\ldots,\mathbf{r}_N)|^2 is the probability density of finding the N deuterons at those N locations. (I am leaving out the spin degree of freedom for simplicity, it doesn’t affect this discussion.) This wavefunction is called the “Gross-Pitaevskii” (GP) wavefunction.

What are the two-particle correlations in the GP wavefunction? None. No correlations whatsoever. What does that mean? The GP wavefunction is not the actual wavefunction of a real-world BEC. It is just an approximation. (My undergraduate textbook lied to me!) A real-world BEC will always have correlations, as long as the pairs of particles have short-range attraction or repulsion.

This fact has long been known to BEC experts. You can read a standard review, say this one, and you will find equation (8.23) that describes how the real two-particle correlation is different than GP predicts, and can be calculated using a somewhat-more-accurate approximation to the BEC wavefunction, due to Bogoliubov, that involves quantities like |\mathbf{r}_i-\mathbf{r}_j|, and cannot be written in the GP form. Anyway, in the end, it’s just what you expect from common sense: If particles repel each other, then you are very unlikely to find two particles very near each other, even if they are part of the same BEC.

How does Kim conclude that Coulomb repulsion does not prevent nuclear fusion?

I started at this paper (Naturwissenschaften 2009) which refers back to this paper (Fusion Technology 2000) which I think is the core of the argument. It uses the variational method. So let’s talk about that next.

Pause for background: Variational method in quantum mechanics

The variational method is a way to approximate the ground state wavefunction of a quantum-mechanical system (i.e. the lowest-energy state). The way it works is: You take a family of candidate wavefunctions, and out of those candidates, and you find the one with the lowest energy (more precisely, lowest expectation value of energy). This one is the closest approximation to the actual ground state, out of the candidates. Here is a silly example showing the limitations of the variational method:

Example variational-method problem: Using the variational method, find an approximation to the most comfortable sleeping position possible.

One possible solution (outline): In the variational method, I need a family of candidates, of which I will choose the best one. Here’s one: Sleeping while hanging upside down, in a room at temperature T (for any possible temperature T). I would do some mathematical analysis to decide that, out of these possibilities, T=20.2°C is the best. (I calculated the derivative of comfort with respect to T, and set it equal to 0, or whatever.) So now I have my answer. The most comfortable way to sleep (in this approximation) is by hanging upside down in a room whose temperature is 20.2°C.

Obviously, we see that the “best approximation” calculated by the variational method may be a really really bad approximation. I would rather sleep on a bed! The variational method is only as good as the candidates that you put into it.

Back to Kim’s argument

Again, we are looking at this paper. He is calculating the ground-state wavefunction of the deuteron BEC. (You may wonder why Kim is assuming that a room-temperature system is in the ground state. It’s a good question. But it is more related to the topic of whether a BEC forms in the first place, which I will discuss in a separate post.)

He uses the variational method. The family of candidates he considers is:

\Psi(\mathbf{r}_1,\mathbf{r}_2,\ldots,\mathbf{r}_N) = \tilde{\Psi}(\rho)

where \rho = [\sum_{i=1}^N \mathbf{r}_i^2]^{1/2} and \tilde{\Psi} is any complex function. This is a highly-restrictive assumption, i.e. a small family of candidates. In reality, \Psi could be any complex function of 3N variables (the x,y,z coordinates of the N deuterons), as long as it is permutation-symmetric (because they are indistinguishible bosons). But by writing that equation above, Kim is only considering a special subset of these functions, namely the ones that depend on the deuteron positions only through their combination ρ (related to the root-mean-square distance of the deuterons from the origin).

Here is a simple example with N=2. Every wavefunction in this family of candidates has the property:

\Psi((0,0,1\text{nm}), (0,0,1.01\text{nm})) = \Psi((0,0,1\text{nm}),(0,1.01\text{nm},0))

even though the real BEC wavefunction is not required to have that property.

If we think back to the discussion above, we expect the real ground-state BEC wavefunction to have a specific kind of strong two-particle correlation, because of Coulomb repulsion. We expect that it should be very unlikely to find a pair of deuterons very close to each other. But none of the candidate wavefunctions that Kim is looking at can possibly have this property. Just look at the equation above. On the left side, the deuterons are 0.01nm apart, and on the right side, they are 1.4nm apart. But the two configurations have the same probability density, in every candidate wavefunction that Kim is considering as a possibility in his variational-method calculation.

So Kim’s analysis involves forgetting that the Coulomb barrier exists in the first place, rather than showing that its effects are small or negligible.

For reasons described above I suspect that a correct analysis would show that the fusion rate is not substantially increased (if at all) by the condensation of deuterons into a BEC.

Bonus: Kim’s recent paper about correlations

Kim has posted a paper here which claims to consider correlations. I am not convinced. The whole paper is really bizarre. Maybe I am missing something. Anyway, here goes…

Remember, we expect there to be a real BEC ground state, which has strong two-particle correlations, so that if there is a particle at A then it is very unlikely to find a second particle 0.01nm away from A.

Instead, in Eq. (4), Kim writes down a huge number of different many-particle states, including a totally-uncorrelated state (4-1), an almost-uncorrelated state (4-2), a slightly-less-uncorrelated state (4-3), etc. But he is not saying that these are successively better approximations to the real BEC, or anything sensible like that. He says that these are all equally-good degenerate ground states. (After symmetrization anyway.) That’s what Eq (3) means. This claim is certainly incorrect. A correlated state would obviously have lower energy, because it avoids greatly reduces the otherwise-huge Coulomb-repulsion energy. That’s the whole point.

It gets weirder. He orthogonalizes all these states, and—for no reason that I can understand—he takes a specific superposition consisting of an equal mixture of all of them (Eq (8)), and calls that the “general solution”. (Does he know what “general” means?)

Finally, he says that the fusion rate is still high, because this “general solution” still has a piece of the totally-uncorrelated state (4-1) in it. (Not to mention the mostly-uncorrelated states (4-2), (4-3), etc.)

So, the paper tries to distract us by talking about correlations, but it’s still basically relying on the calculation that does not take correlations into account. Certainly, I see nothing in this paper that refutes what I wrote above.

(Thanks Mark Mitchison for help teaching me about BECs here.)

14 thoughts on “BEC 2: Coulomb barrier in a BEC”

  1. HG

    An interesting blog about LENR!

    You cannot understand the foundations of BEC if you don’t understand zero point energy. Unfortunately there is no consistent scientific explanation of zero point energy to be find. Therefore every theoretical answer about BEC at much higher temperatures is only some kind of “opinion”.

    Nevertheless, suppose there is BEC at “room temperature”. What about the Coloumb force?
    The energy of the Coloumb force is still existent, but it is no longer a strong barrier between the nuclei.

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    1. steve Post author

      There is no mystery about zero point energy (aka vacuum energy, dark energy, cosmological constant), it is a perfectly natural and non-mysterious part of quantum field theory and general relativity. And it’s not at all relevant to BECs.

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    2. Kevin M O'Malley

      KP Sinha and Muelenberg postulate that a tunneling effect occurs.

      According to Krit Prasad Sinha and Andrew Meulenberg, hydrogen isotopes form a sublattice in the bulk, on the surfaces, and at defects within the lattice of palladium (and other metals). Collective one-dimensional (phononic) motion of these atoms in the lattice(s) is the basis for tightly bound ground-state electron pairs providing super-strong screening of the nuclear Coulomb barrier. During the point of nearest approach in repeated collisions of hydrogen atoms (in pairs: one with electron(s) and one without), the bound electron(s) go deep into the combined nuclear-Coulomb-potential well (without radiating photons) and attain near-MeV energies. The energetic electron(s) remain tightly bound during nuclear penetration of the residual nuclear Coulomb barrier and, on entering the nuclear potential well with both nuclei, temporarily increase(s) the effective nuclear-binding energy. These electrons (before being ejected) reduce the nucleon energies toward or below fragmentation levels, mediate energy transfer from the excited nucleus to the lattice, possibly create neutrons through the pep reaction, and, if paired, allow the “neutralized” nucleus to penetrate neighboring nuclei and result in transmutation. Thus, within known physics, this model can qualitatively explain all experimental results observed in low-energy nuclear reactions.

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  2. Matt

    Hi Steve,

    Just found your blog last week and am going through it from the beginning – so far I’m loving it! Thanks so much for putting time into making the theory accessible to other physicists – it makes the barrier to entry a lot lower.

    I have a question about your comment on “Bonus: Kim’s recent paper about correlations”, specifically at the end of paragraph 3 “A correlated state would obviously have lower energy, because it avoids the huge Coulomb-repulsion energy. That’s the whole point”.

    My initial understanding from reading your post is that correlations are happening because of Coulomb-repulsion – I was therefore expecting you to say something like “An UNcorrelated state would obviously have lower energy”.

    Unless I am missing something…my quantum is a bit rusty

    Thanks

    Matt

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    1. steve Post author

      Dear Matt, thanks for the kind words!

      I think I wrote it correctly. I’m not 100% sure where you’re coming from but I have a guess, let me try… Coulomb repulsion means that the energy of a state is higher if two deuterons are often near each other. Since the temperature is less than infinity, real systems tend to disproportionately be in lower-energy states (Boltzmann factor), which means that in real systems, we are less likely than chance to find two deuterons close to each other, i.e. real systems have correlations. The comparison in that sentence is, on the one hand, the real system with its correlations, versus on the other hand, a hypothetical scenario in which the Coulomb repulsion still exists, but I have magically placed the deuterons into an unrealistic configuration where they are positioned as if Coulomb repulsion did not exist, i.e. I put them into a configuration where they often get very close to each other despite the very high repulsion. The latter has higher energy.

      (Or sorry if I’m being stupid.)

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  3. Matt

    Thanks for getting back to me so fast.

    I agree with everything you wrote, in particular the point that with the coulomb repulsion the energy is higher in the real world. But isn’t this real word situation with higher energy the one which must have correlations because of what you say at the beginning of the article:

    “Remember, Coulomb repulsion is by far the most important factor affecting fusion rates. So if you want to calculate fusion rates in the BEC, you need to do a calculation that accounts for the fact that the BEC has two-particle correlations. If you neglect that fact, it amounts to forgetting that Coulomb repulsion exists in the first place!”

    So in my mind i have:

    Coulomb force (ie higher energy) => correlations

    Which if i take the contrapositive gives:

    No correlations (ie uncorrelated) => no coulomb force (ie lower energy)

    So when you say that the correlated state will obviously have lower energy it feels like it contradicts this logic that I have in my mind….Which of course might be wrong.

    Does that make more sense?

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    1. steve Post author

      “No Coulomb force” is not relevant. Deuterons are positive and they repel each other, and no one (neither me nor Kim) is interested in the hypothetical scenario in which deuterons do not repel each other. So forget the possibility of “no Coulomb force”.

      In reality, since deuterons repel each other, their positions become (anti)correlated. In Kim’s faulty physics analysis, deuterons repel each other but their positions nevertheless stay uncorrelated. 😀

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      1. Matt

        Hi Steve,

        I’m sorry, this whole thing has got a bit confused now, i was trying to clarify my thoughts but seem to have made it worse lol. I do of course understand that the coulomb force does exists and that no one should be interested in simply ignoring it.

        My contrapositive statement was just to try and illustrate that an uncorrelated state would seem to have lower energy, in contradiction to what you wrote here:

        “A correlated state would obviously have lower energy, because it avoids the huge Coulomb-repulsion energy. That’s the whole point.”

        I guess another way of stating my query is: How does a correlated state obviously avoid coulomb repulsion when earlier you described correlations as being a necessary part of having coulomb repulsion? (See your quote below)

        “Remember, Coulomb repulsion is by far the most important factor affecting fusion rates. So if you want to calculate fusion rates in the BEC, you need to do a calculation that accounts for the fact that the BEC has two-particle correlations. If you neglect that fact, it amounts to forgetting that Coulomb repulsion exists in the first place!”

        Sometimes it can be hard to translate what’s in our heads into someone elses so i appreciate your time in discussing this with me.

        Matt

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      2. steve Post author

        Hmm, I’m not sure I understand where you’re coming from, but I’ll try again. Coulomb repulsion is a term in the Hamiltonian proportional to 1/|r – r’| where r and r’ are two of the particle locations. So the closer that pairs of particles get to each other, the higher the energy from Coulomb repulsion. I think you are mixing up three questions:
        (1) What are the possible configurations (many-particle wavefunctions) of deuterons?
        (2) What is the energy of each possible configuration?
        (3) Which configuration(s) are actually likely to occur?
        For question (1), the answer is “Any wavefunction is possible (as long as it satisfies boson statistics), regardless of the existence of Coulomb repulsion or anything else.” The uncorrelated states don’t stop existing in the presence of Coulomb repulsion! They still exist, they just have higher energy. That brings us to the next topic: The relation between questions (2) and (3) is given by the Boltzmann factor. At infinite temperature, all configurations are equally likely. At a realistic finite temperature, the very high-energy states (including the uncorrelated states) are vanishingly unlikely to occur. Kim agrees with the ground rules here, i.e. he really does try to find the lowest-energy state, and then he says that this lowest-energy state is the most likely to occur. The only problem is, when he tries to find the lowest-energy state, he doesn’t do a good job.

        When I wrote that neglecting the possibility of 2-particle correlations “amounts to forgetting that Coulomb repulsion exists in the first place”, I meant more specifically that: When Kim tries to calculate the likely configuration of deuterons (i.e. the answer to question (3)) but neglects the possibility of 2-particle correlations, he will probably get a similar “answer to question 3”, and hence a similar fusion rate, as if Kim had forgotten that Coulomb repulsion exists in the first place. Kim would not get the same answer to (2), just (3), if he had hypothetically forgotten to include Coulomb repulsion. Again, I don’t think Kim made any mistakes in the process of answering question (2) (well, I don’t remember whether he did or not), but rather his mistake was not even trying to answer question (2) for any correlated state. An analogy is that if Kim were trying to find the lowest elevation point on earth, I’m not complaining that he has faulty geographical surveying equipment, I’m complaining that he did all of his measurements in the Himalayas and only succeeded in finding the lowest point of the Himalayas, and seems unaware that it might be worth checking other places, e.g. the Dead Sea.

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      3. Matt

        Hi Steve,

        Thanks for expanding on a lot of your points, it helps with general understanding and getting on the same page so to say.

        Forgive me, but I am a bit confused by the confusion. In my last reply I restated my original question as:

        “I guess another way of stating my query is: How does a correlated state obviously avoid coulomb repulsion…”

        This was in response to your original statement which I was struggling to understand when I wrote my first reply:

        “A correlated state would obviously have lower energy, because it avoids the huge Coulomb-repulsion energy. That’s the whole point”.

        As I can see, my question (“How does a correlated state obviously avoid coulomb repulsion [energy]”) is well posed, so I’m struggling to understand what part you are confused with.

        This makes me think I’m really totally misunderstanding the meaning of some of your words in original statement:

        “A correlated state would obviously have lower energy, because it avoids the huge Coulomb-repulsion energy. That’s the whole point”.

        So, let me have another go – I think my issue is with the word “avoid” – i’m conflating it (perhaps incorrectly) with neglect, i.e avoid makes me think that you mean the correlated state is somehow neglecting coulomb repulsion. That would seem like a total contradiction with other things you wrote. But…. by avoid are you just referring to the real world situation where:
        1) A correlated state would necessarily include the effect of Coulolmb repulsion
        2) This means the positive charges would really physically avoid being close to each other because they are trying to reach a lower energy.
        In contrast an uncorrelated state just wouldn’t care about the Coulomb force pushing the charges apart so the charges would be perfectly happy to sit close to each other which would technically have a higher energy if you put all the positions into the formula for Coulomb potential energy.

        If at this point I’m still not getting it, don’t worry too much. I do still need to brush up on a few thing so perhaps i’ll do that and come back to the post later.

        Thanks again for your time.

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      4. steve Post author

        I agree 100% with what you said here, especially the part after “So let me have another go”. Sorry for a poor choice of words (I just edited it) and sorry if we were talking past each other before, thanks for your patience. 😀

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  4. Matt

    Whoop, it makes me happy that we got there in the end. I actually learnt a bunch of stuff along the way so it was all good for me ?. By the way, I think you write exceptionally well, so this was definitely a brain malfunction on my part https://media.giphy.com/media/X1QcEzYEfbxXq/giphy.gif

    Thanks again for taking the time.

    ps I couldn’t reply to your reply this time, so i had to create a new one

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  5. Abd ul-Rahman Lomax

    You have made what may be a classic mistake, relying on a “sense” about what something would be, without actually calculating it. Or not. We’ll see. The Coulomb barrier is a rate issue.

    As you know full well, tunneling “through” the barrier is possible, and fusion rate never becomes zero, but is a number that may be so small that the practical effect is impossibility. We have no experimental data on BEC fusion (unless we assume that the FP Heat Effect is caused by it, which is way premature except for discussion). I have not read further in your blog than this page, as of right now, but you are writing about Kim’s work, which is relatively general and vague, whereas Takahashi actually bit the bullet.

    Having found experimentally that 3-body fusion in condensed matter occurred at a rate 10^26 higher than simple plasma theory predicted, he started looking at multibody possibilities to explain LENR. He studied, in particular, the possibility of a configuration of 4 deuterons, arranged in a tetrahedron, with low relative momentum (necessary for condensate formation), and how long it would take to collapse (this is really just two deuterium molecules, and it includes the electrons).

    He found it to take a femtosecond. What this implies is that the low relative momentum need not last for long, and hence the common objection about temperature is not relevant. Could low enough relative momentum exist in a two-molecule system *confined, perhaps, in a tetrahedral trap,* for a femtosecond? If so. then BEC formation is possible. I have claimed that BECs form routinely at higher temperatures than thought, but simply do not last long enough to be detected. Just like ice forms at room temperature, if we define “ice” as meaning two water molecules stuck together for a little time.

    Now, then, the question becomes *how close* do the nuclei come? And for how long? The barrier does not magically disappear. The electrons collapse with the nuclei, so there may be some level of shielding, but Takahashi does the collapse and fusion rate calculations and came up with a femtosecond from collapse to fusion. That’s a calculation, and you can read his papers. If you have trouble finding them, I’ll provide them to you or links.

    He is also generally available for useful discussions. The world needs cogent criticism of cold fusion theory, and it could be published. Of course, this might actually bring us closer to the day that cold fusion research accelerates, so WE MIGHT ALL DIE!

    So you might consider the ethical issues. I suggest, in fact, that you go all the way with the consideration. Don’t stop with terrified paralysis. That is no place to live, or even to die. Do both with clarity and we will all be much happier.

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