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:
- At room temperature, the deuterons cannot condense into a BEC.
- 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:
where N is the number of particles, Ψ is the many-particle wavefunction (in the position basis), 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 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 , 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:
where and is any complex function. This is a highly-restrictive assumption, i.e. a small family of candidates. In reality, 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:
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.)