# Spin-boson model: Overview

In two earlier posts I gave some background information on superradiance and stimulated emission. With that, I’m ready to properly introduce the spin-boson model” theory of cold fusion, as advocated by Peter Hagelstein at MIT.

As of this writing (2014), I gather that the model is still under construction in many respects. Hagelstein and coworkers have published tons of papers about the model (at least a dozen just in 2011-13), but each of them is working out the details of some little ingredient in the model. They have been kinda noncommittal so far in some crucial details of how everything fits together and how it relates to the real palladium-deuterium system. (I don’t mean that as a criticism. It is perfectly normal and appropriate to publish papers on incremental progress and insights, without yet knowing all the answers.) Anyway, I don’t mind, because there is already plenty to blog about!

Since there is a lot here that I’m still confused about, I have few if any firm opinions. This post is just a chronicle of my ongoing progress in trying to understand this topic.

Overview

In its simplest manifestation, two deuterons fuse, and the energy goes directly into creating a billion phonons in a single phonon mode, i.e. exciting a very strong vibration of the atoms at a particular frequency and wavelength.

Note on terminology: In “spin-boson”, the “boson” part refers to this phonon mode. The “spin” part actually has nothing to do with spin in physics, but is instead invoking the mathematical analogy described in my earlier post about superradiance. Specifically, we have a lot of two-level systems, each of which is either a pair of deuterons sitting close to each other, or a helium-4 nucleus, and the word “spin” refers to the collective behavior of that system.

That’s the simplest model. There are some elaborations and modifications, like where there are losses in the system, or where energy transfers between different nuclear species, some of which may have more than two levels, and so on. But let’s stick with the simplest model for now.

Why this should strike you as crazy

direct transfer of energy from a deuterium fusion into a billion phonons: It sounds crazy! Following blog philosophy, we do not dismiss crazy-sounding theories, but it is certainly worth saying why it sounds so crazy, if it’s not obvious to you.

Quite generally, in physics, when different “degrees of freedom” (i.e. ways to store energy) exchange energy with each other, it is usually in a few-body process (involving 2 or 3 or 4 particles / quanta, rather than a billion), and it happens when the degrees of freedom have similar frequencies (in classical physics) or quantized energy (in quantum mechanics), whereas here there is a factor of a billion difference between the deuteron fusion energy quantum (24 MeV) and the phonon energy quantum (maybe 24 meV).

So for example, in your red laser pointer, an electron excitation with energy of 2eV turns into a red photon with energy of 2eV. It is extremely difficult to coax that same excited electron to instead emit two identical 1eV infrared photons. (It is possible with great effort; this is an obscure thing called a “2-photon laser”.) It is probably impossible (in practice) to get it to emit three identical 0.67eV photons, and it is utterly unthinkable that it would emit millions or billions of identical microwave photons.

(Multi-photon transitions can be “resonantly enhanced”, if there are less-excited states that the electron can [virtually] occupy as stepping stones. Resonant enhancement makes the process faster and more likely. Indeed, resonant enhancement is always used for those “2-photon lasers” I mentioned. But the spin-boson model for cold fusion is supposed to work without resonant enhancement. For example, there is no state where 19% of those billion phonons have been created, and the D+D is 19% of the way fused into Helium-4. The lack of resonant enhancement makes the claim all the more crazy-sounding.)

Hagelstein likes to bring up the example of high harmonic generation as a real (uncontroversial) example where thousands of low-energy photons combine their energy into one high-energy photon. However, we are looking for the reverse process, let’s call it “high subharmonic generation”, which seems far more exotic and unlikely.

Superradiance and stimulated emission to the rescue?

What can stimulated emission do for us? Well, it may be very unlikely that the first phonon would be emitted, but thanks to stimulated emission, the second identical phonon can be emitted more easily, the third even more easily, and so on all the way to the billionth phonon. This basic idea seems reasonable, but raises a few questions:

Q: If the phonon mode starts empty, isn’t it extremely unlikely for the first few phonons to be [virtually] emitted into the mode, having no help from stimulated emission? Or are there already tons of real phonons in the mode? If so, how did they get there? [Most phonon-creation processes tend to create phonons in many different modes, not just one.]

Q: Do the phonons live in the mode long enough to be helpful? Or do they scatter / decay?

Superradiance can be equally useful. You can have a pair of deuterons every few lattice sites (occupying a vacancy in an ordered defect phase — long story), maybe 1 D-D pair per cubic nanometer. So if the phonon extends over a reasonable, macroscopic, volume, it seems reasonable to expect trillions or quadrillions of deuteron pairs that are collectively (superradiantly) emitting these phonons. Since superradiance can speed up a phonon-emission or -absorption process by up to the square of the number of D-D pairs, this seems like a promising way to turn an extremely-improbable process into a frequent one. Again, the basic principle seems reasonable, but raises some questions:

Q: How the first few fusions [virtually] happen? Remember, as discussed here, superradiance requires macroscopic numbers of both excited states (unfused D+D) and ground states (fused helium-4). With only excited states, there is no superradiant enhancement.

Q: Does quantum decoherence (a.k.a. “dephasing”) prevent superradiance from occurring? Remember, as discussed here, superradiance requires a coherent superposition (with a specific, fixed phase relation) between states like “D+D over here and He4 over there” versus “He4 over here and D+D over there”. Interactions with other particles will destroy this coherence. Does this happen so fast that superradiance is entirely irrelevant? (My guess is “yes”, but I’m not certain.)

The other interaction factor, V

The so-called “dimensionless coupling constant” for interactions between the nuclear fusion and phonons is (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 (see previous post for definitions). $\Delta E$ 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.

Ordinarily I would have expected V to be the product of the transition (electric) dipole moment for the fusion process, multiplied by the electric field that arises in the phonon oscillation. (Actually, since D+D and He4 are mirror-symmetric, I would expect V to be the product of a transition quadrupole moment times the phonon’s electric field gradient.) Anyway, it’s a very very small energy.

However, in recent papers (especially this one) the authors have argued that V is mainly due to a different, much larger interaction, which they call “a · cP”. (You are not expected to know what that means.) So far I can’t figure this out:

Q: What is this a · cP interaction (intuitively)? Is it correct? Is it really different from (and much larger than) an electric field times a transition dipole moment?

The Coulomb barrier is also part of V: It makes V much much smaller than it would otherwise be. I’ll discuss that below.

Quantum transitions: Perturbation theory vs. Strong coupling

In an introductory quantum mechanics course, we learn a simple formula to calculate how fast a transition happens in quantum mechanics. It’s called Fermi’s golden rule:

$T_{i \rightarrow f}= \frac{2 \pi} {\hbar} \left| \langle f|H'|i \rangle \right|^{2} \rho$

(transition rate from the initial to the final state is proportional to the interaction matrix element squared, times the density of final states).

That’s actually “first-order Fermi’s golden rule”, in that it is the first order of perturbation theory. If this rate is zero, we can try the second-order Fermi’s golden rule:

$T_{i \rightarrow f}^{(2)}= \frac{2 \pi}{\hbar} \left| \sum_m \frac{\langle f|H'|m \rangle \langle m | H' | i \rangle }{E_i - E_m} \right|^{2} \rho$

in other words, the interaction uses another state m as a stepping stone between i and f. The “energy denomenator” $E_i - E_m$ describes the fact that a transition to the intermediate state would violate energy conservation. The system “doesn’t like” to do that. The larger the energy-conservation violation, the less likely it is that m would be used as an intermediate step.

(The “resonant enhancement”, mentioned above, is the situation where there is an intermediate step m that has very little energy-conservation violation.)

In the problem at hand, i.e. cold fusion, the transition rate from second-order Fermi’s golden rule is still zero. We actually have to go to $R \equiv (E_n/E_\omega) \approx$1-billionth-order Fermi’s golden rule!!

$T_{i \rightarrow f}^{(R)} = \frac{2\pi}{\hbar} \left| \sum_{m_1,m_2,\ldots, m_{R-1}} \frac{\langle f | H' | m_{R-1} \rangle}{E_i-E_{m_{R-1}}} \frac{\langle m_{R-1} | H' | m_{R-2} \rangle}{E_i-E_{m_{R-2}}} \cdots \frac{\langle m_2 | H' | m_1 \rangle}{E_i-E_{m_1}} \langle m_1 | H' | i \rangle \right|^2 \rho$

I think it’s helpful (pedagogically) to write down that equation, but it turns out that we can’t use it. As a matter of fact, the billionth-order-Fermi’s-golden-rule equation cannot possibly be relevant for any real-world process:

• If each of the billion terms is somewhere between 0 and 1 (in absolute value), then their product is guaranteed to be ~0. You could wait a grillion years, the transition would never occur.
• If any two consecutive terms (out of the billion terms) has a product greater than 1 (in absolute value), then you shouldn’t be using Fermi’s golden rule (FGR). The reason is: The total transition rate is the sum of the rate for 1st-order FGR, 2nd-order FGR, etc. In this situation, that sum is a divergent series. For example, if the terms for $m_{300}$ and $m_{301}$ have a product larger than 1, then whatever transition rate you have for start→1→2→…→end in Fermi’s golden rule, you have a larger transition rate for the higher-order Fermi’s-golden-rule process start→1→2→…→300→301→300→301→…→300→301→302→…→end. This is a sign that FGR is not applicable.

The latter situation is called “strong coupling”. It’s the situation where perturbation theory (including Fermi’s golden rule) cannot be used because it would involve summing a divergent series. In fact, calculating transition rates of strongly-coupled systems is generally a quite difficult math problem. That’s why you can find papers like this where Hagelstein uses a computer to numerically calculate a transition rate. He would have saved so much time and effort by using perturbation theory (Fermi’s golden rule) if that was an option, but he couldn’t, because he was assuming strong-coupling.

I think the appropriate way to think about it is that some groups of states are strongly coupled to each other, meaning that there are direct transitions between them, and that the corresponding interaction energy is bigger than the energy difference between the states. The corresponding term in Fermi’s golden rule would be bigger than 1. Imagine that the states in such a group are all rooms in a single building; you can easily walk from one to the other. Then the remaining pairs of states are weakly coupled to each other. Imagine that each of them is in a different city; it’s hard and slow to travel between them.

Fermi’s golden rule is the right way to calculate how long it takes to travel from one city to another, but not from one room to another.

So we can think of two aspects to the problem:

Q: Can the system enter the strong-coupling regime in the first place?

Q: Once we’re in the strong-coupling regime, does the transition happen at a reasonable rate?

The authors seem to have written more about the second question than the first. For the second question, they suggest that the transition is intolerably slow in the straightforward spin-boson model, but pretty fast in the “lossy spin-boson model”. I will explain what that means in another post. Anyway, so far I’ve been thinking more about the first question. It seems to me that the answer is “no”, but I’m not 100% sure.

My impression 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 is one of my questions / concerns about the model. (Blue boxes = real states. Pink boxes = virtual states.)

Coulomb barrier

The Coulomb barrier can be penetrated by quantum tunneling. Everyone agrees on that. The problem is that it’s extremely unlikely—or, to put it another way, you need to wait a long time before it happens (roughly 5 grillion bajillion years).

If we write out the two-body wavefunction for a D+D system, there is a probability amplitude a for finding the two D’s close enough to fuse, on account of quantum-tunneling through the Coulomb barrier. This quantity a is very very small (but not zero). The rate that fusions occur (using Fermi’s golden rule) is proportional to $|a|^2$, which is very very very very small! In the spin-boson model, it is argued that the rate that fusions occur is proportional to $|a|$ rather than $|a|^2$. If this is correct, then that makes a huge difference! It starts to make it seem plausible that the Coulomb barrier isn’t such a big deal after all.

Q: Why is the rate proportional to $|a|$ rather than $|a|^2$? It seems plausible, but I can’t find where the details are spelled out. It’s presumably related to the fact that the strong-coupling calculation is quite different from Fermi’s golden rule. For example, in the formula for Rabi frequency, you don’t square the amplitude, if I’m not mistaken.

Summary