# Spin-boson interaction term scaling

For those who haven’t been following along, cold fusion is a set of disputed experiments, in which there is (allegedly) indirect evidence that nuclear fusion is occurring, under circumstances where it seems impossible for nuclear fusion to occur. One problem: Everyone knows that when nuclear fusion releases energy, the energy usually winds up as the kinetic energy of fast-moving particles, but such particles are not seen in cold fusion experiments; the only energy that anyone sees is heat energy. This leads us to the lossy spin-boson model of cold fusion, which says that the fusion energy gets directly transferred into creating a billion phonons in a single phonon mode. It seems crazy, but the authors (MIT Professor Peter Hagelstein and collaborators) have produced intricate arguments in favor, which I have been gradually working my way through over many blog posts.

Anyway, here is another post on some details of the theory: Why cooperative effects over large volumes are not really helpful for making this reaction more likely to start.

The spin-boson model involves many nuclei and one phonon mode. The nuclei emit into the mode cooperatively, analogous to superradiance in optics. The larger the phonon mode is, the more nuclei cooperate. But the larger it gets, the interaction between the phonon and each individual nucleus goes down. So how helpful is this cooperative effect really?

In particular, I can easily say “Assume there are a million nuclei, or a billion nuclei, or a billion billion nuclei…” After all, even a billion billion nuclei is not that many nuclei. It would sure be nice if a billion billion nuclei made the reaction a billion billion times more probable. Isn’t that the idea of superradiance? Alas, it’s not the case. Let’s explore why…

The Hamiltonian. In the simplest version of the spin-boson model, the Hamiltonian is (ref):

$H = (\Delta E/\hbar) S_z + \hbar \omega_0 a^\dagger a + (V/\hbar)(S_+ + S_-)(a^\dagger + a)$

The first term says that each time a D-D fuses into helium-4, the system’s energy goes down by $\Delta E$ = 24MeV. The second term says that the phonon energy is the number of phonons ($a^\dagger a$) times the energy of a phonon ($\hbar \omega_0 \approx$ maybe 10meV). The third term is at the heart of the spin-boson model. It says that there is an interaction in which the D-D either fuses or unfuses, while simultaneously a phonon is either created or destroyed (all four combinations are possible). (The $S_\pm$ and $S_z$ and $a$ and $a^\dagger$ notation is explained in these two earlier posts.)

Think the first two terms as the “unperturbed Hamiltonian” and the third term as the perturbation, to which we can apply perturbation theory. Well, in that case, the fusion would be a billionth-order-perturbation-theory process, which means it will not happen at all (see my earlier post). It can only possibly happen if perturbation theory breaks down, i.e. “strong coupling”.

So a question to keep in mind is: “Would strong coupling really occur?”

There is a so-called “dimensionless coupling constant$g$, with the property that when $g<<1$ the phonons and nucleons are weakly coupled to each other, and you can use perturbation theory (and the process will never happen);  and when $g >>1$, they are strongly coupled and you can’t use perturbation theory (and it’s at least possible that the process will happen). Here is the equation for $g$ (ref):

$g=\frac{V \sqrt{n} \sqrt{S^2 - m^2}}{\Delta E}$

The $\sqrt{n}$ is from stimulated emission (n is the number of phonons in the mode), and the $\sqrt{S^2-m^2}$ is from superradiance (as described in my earlier post, S is the number of D-D pairs, and m is the number of pairs that have already (perhaps virtually) fused into helium-4). $\Delta E$ is the fusion energy, 24MeV. What remains is V, the baseline interaction that couples (i.e. allows energy to transfer between) the nuclear fusion and the phonons.

I’m mainly skeptical about how the process gets started, let’s say the first 10 steps into virtual states. I picked the number 10 because it’s very small compared to the billion steps required for the whole process (each step creates one phonon), but big enough that if the first 10 steps do not have strong coupling, the reaction probably won’t happen at all. I mean, you rarely see 10th-order-perturbation-theory processes occur, as opposed to 1st- or 2nd-order. The reason is, when you multiply together 10 numbers much less than 1, you get a number very very close to 0. I will also assume that the number of D-D pairs participating is much more than 10. I mean, I was talking about a billion billion of them a few paragraphs ago.

Reminder: My suspicion is that the first few (virtual) steps of the transition are astronomically unlikely to occur, since they benefit from neither superradiance nor stimulated emission. So, maybe there are strongly-coupled states, but I don’t see how you can get to any of them! This post concerns those crucial early steps. (Blue boxes = real states. Pink boxes = virtual states.)

So we have $(S-m) < 10$ and $S>>10$, so therefore $\sqrt{S^2-m^2} \approx \sqrt{S} \sqrt{S-m}$.

The letter V is already taken, so let’s call the phonon volume $mathcal{V}$. Maybe this is the size of a little micro-grain of palladium.

What’s the effect of increasing the phonon volume? One thing is, $V$ gets smaller. $V$ describes the interaction between the phonon and any one particular D-D pair. If the phonon is more spread out, it is less likely to be found close to that D-D pair. More details: If the volume of the phonon quadruples, then the normalized amplitude for the phonon wavefunction goes down by half, so the matrix element V between that phonon and one lattice site also goes down by half. So if we define $V = V_0 / \sqrt{\mathcal{V}}$, then $V_0$ will be a material parameter, independent of $\mathcal{V}$.

Another effect is, larger volume proportionally increases the number of D-D pairs participating. Let’s define a density $\rho_S = S/\mathcal{V}$.

Does a larger volume increase n, the number of phonons? I say “no”. Remember, I’m investigating the first 10 steps of perturbation theory, so we’re talking about the phonons that are already in the crystal independent of any nuclear fusion, plus up to 10 more.

If the crystal is in thermodynamic equilibrium, then the case is clear-cut: The equilibrium number of phonons per mode does not depend on the mode volume, only the mode frequency. A larger volume has more phonons, I admit, but that is because it has more modes, not because it has more phonons per mode.

Similarly, if you create phonons by knocking atoms around (the authors remind us that there are ions getting electrochemically dragged through the crystal), this is a local interaction, so it creates phonons in all modes equally. (Well, in principle, if you knock atoms around, there can be stimulated emission of phonons, but I believe there are too many modes and too many phonons for any one mode to get more than its share of phonons.) A bigger volume may have more atoms knocking around in it, but it will also have more modes sharing those phonons.

One more thing: Phonons can switch modes by scattering. Is it possible for phonons to start in many modes and wind up all gathering in one mode? Well, it’s not inconceivable. There is something called “stimulated scattering” (analogous to stimulated emission) which allows huge numbers of bosons to scatter into a single mode. That’s the basis for polariton lasers. But normally, entropy prevents that from happening, and I’ve never heard of it for phonons. (I don’t know the details here.)

Anyway, it seems pretty safe to assume that n is independent of volume.

Finally, how does the coupling constant depend on volume?

$g=\frac{V \sqrt{n} \sqrt{S^2 - m^2}}{\Delta E}= \frac{V_0 \sqrt{n}\sqrt{S-m} \sqrt{\rho_S}}{\Delta E}$

There is no dependence on $\mathcal{V}$ here. Under these circumstances, the coupling constant is independent of phonon mode volume. If the first ten steps do not reach strong coupling with 40 D-D pairs cooperating, then they will not reach strong coupling with 40 billion billion D-D pairs cooperating!

When I was first learning about this model, I kinda had the impression that you could have a pretty small coupling constant $V<<\Delta E$, but make for the difference by having trillions of D-D pairs cooperating in the process.  But now I understand that V has to be huge, definitely in the MeV range (divided by the square-root of the number of unit cells), for the lossy spin-boson model to have any chance at working. (For the first 10 steps, remember that $S-m<10$ and $n \lesssim 15$ or so.)

I suspect that Dr Hagelstein and coauthors would generally agree with this post, because I vaguely remember reading something where they said that V has to be comparable to ΔE. I bet I am rediscovering a fact that they have known for decades. Update: Dr. Hagelstein disagrees with this blog post, see here

So $V_0$ has to be more than 1MeV per square-root-unit-cell. That seems far-fetched; I would have guessed that it would be, oh, a million times smaller than that. If $V_0$ also includes the Coulomb barrier, that would make it much much much much smaller than it would otherwise be, and then it would be pretty much inconceivable that $V_0$ would be so large. I guess this is why the authors have been recently emphasizing more complicated “donor/receiver” models where additional high-energy nuclear transitions are involved, as a stepping stone between the D-D reaction and the phonons. (See this paper and references therein.) The stepping stones do not have the Coulomb barrier, so at least there is a chance of strong coupling.