# Spin-boson model: a·cP

I’ve been puzzling over this paper, “Including nuclear degrees of freedom in a lattice Hamiltonian” by Peter Hagelstein and Irfan Chaudhary. I think I’ve made progress in understanding it. I’ll summarize what I think is going on. But I could be totally wrong 😛

If I understand correctly, the authors are trying to calculate how a nucleus being pushed back and forth (because of phonons) has some probability to jump from its ground state to a nuclear excited state. Or—vice versa—how a nucleus can relax to its ground state while increasing the strength of a phonon oscillation. This paper is a step towards that goal.

Nuclear excited states (also called nuclear isomers) are important in nuclear physics, but generally irrelevant in normal condensed matter physics (with the exception of Mössbauer spectroscopy), because nuclei are always in their ground states in normal condensed matter physics. But the authors are arguing that these nuclear excited states are more relevant than you might expect!

The authors do some calculations that are supposed to be related to this, and the calculations seem to be correct (free of errors), but the paper does not make it terribly obvious (to me) what the calculation means physically. My goal here is to clarify! I’ll start by developing some terminology and intuitions, and then at the end I’ll use that language to describe what I think the paper is about.

Two bases for the states of an isolated nucleus…

If you take an isolated nucleus (in a vacuum), it has many different possible quantum states that fill out an infinite-dimensional complex vector space. As always, we can write out a basis for this vector space. There are two bases in particular that I think are relevant to the paper.

The energy eigenbasis consists of the following states:

• The state where the nucleus is in the ground state, moving at momentum p. (One state for each p.)
• The state where the nucleus is in the first excited state, moving at momentum p. (One state for each p.)
• The nucleus is in the second excited state …

Here, for example, the ground-state nucleus at momentum p is exactly equal to a Lorentz boost of the ground-state nucleus at momentum p=0. The p‘s can be any real 3-vector.

The convoluted basis consists of the following states:

• $\int d^3 \mathbf{r} ; |\text{nucleus at rest in the ground state at } \mathbf{r} \rangle e^{i\mathbf{p}\cdot \mathbf{r}/\hbar}$. (One state for each p.)
• $\int d^3 \mathbf{r} ; |\text{nucleus at rest in 1st excited state at } \mathbf{r} \rangle e^{i\mathbf{p}\cdot \mathbf{r}/\hbar}$. (One state for each p.)

I’m using the term “convoluted” because these states are basically convolutions of the nucleus-at-rest with a plane wave. But I do also find these states a bit “convoluted” in the sense of “weird and liable to cause confusion”.

Pedantic readers may complain that if a nucleus is at rest, then it doesn’t have a location! Actually that’s not a big deal. I am just describing the state in a loose way—I’ll be more precise in a minute. First let’s pause and think about this qualitatively.

What is the difference between these two bases? When a ground-state nucleus is moving fast (particularly if it approaches the speed of light), it changes appearance as well as speed. For example, the nucleus looks squeezed by length contraction. In the energy eigenbasis, that’s what we see.

In a convoluted state, on the other hand, we are creating states by taking the “core” of a nucleus at rest, and simply putting it in motion without correspondingly changing the appearance. No length contraction, for example. If the nucleus were asleep in bed, we would be dragging its bed down the street, without waking it up.

Notice the length contraction on the right. Time dilation would also have a distorting effect (not shown).

Convoluted basis take 2 (more precise version)

In introductory mechanics and quantum mechanics, you sometimes solve problems where you separate the relative motion from the center-of-mass motion, and you usually find that the two types of motion are independent. For example, a $\text{H}_2$ molecule can vibrate (relative motion) while also traveling across a room (center-of-mass motion).

In this case, the relative motion is the configuration of protons and neutrons and gluons in the nucleus, i.e. whether the nucleus is in the ground state or excited states or some superposition thereof. The center-of-mass motion is how fast the nucleus is traveling through space.

However, here, the relative and center-of-mass motion are not exactly independent. For example, as the nucleus approaches the speed of light, the ground-state configuration should be squished in a way that’s consistent with relativistic length contraction. The non-independence of relative and center-of-mass motion is why the two bases here are different. In the energy eigenstates, the relative motion is adjusted to take into account the center-of-mass motion. In the convoluted states, you use the relative motion for a nucleus at rest, no matter what the center-of-mass motion actually is.

To be more quantitative, let’s go to the limit of extremely small center-of-mass velocities. In this case, we really can separate the relative and center-of-mass motion. So here are the energy eigenstates, written as many-body wavefunctions:

$\Psi_{\mathbf{p},i}(\mathbf{r}_1,\ldots,\mathbf{r}_N) = e^{i\mathbf{p} \cdot \mathbf{R} / \hbar} \Psi_{\text{relative}}^{(i)}(\mathbf{r}_1-\mathbf{R}, \ldots, \mathbf{r}_N-\mathbf{R})$

where $\mathbf{r}_1, \ldots, \mathbf{r}_N$ are the locations of all the nucleons, and R is their center of mass, and p and i are two parameters characterizing the state. The relative wavefunction $\Psi_{\text{relative}}^{(i)}$ is one of a ladder of discrete possibilities. $i=0$ is the ground state, $i=1$ is the first excited state, etc. Note that $\Psi_{\text{relative}}^{(i)}$ is independent of p.

Again, these separable states are only energy eigenstates when p is very small compared to the speed of light.

The convoluted basis is defined as states like that but where p can be any value, not just very small values. When p is large, these won’t be energy eigenstates anymore, but they are still perfectly good quantum states.

Finally, what I think is going on in the paper

I think that all the math in the paper amounts to derivations of formulas that relate these two bases to each other (i.e., how do you write a convoluted state in the energy eigenbasis, and/or how do you write an energy eigenstate in the convoluted basis). At low nucleus velocity, the eigenstates are approximately equal to the corresponding convoluted states, but not exactly, and I think they are calculating what the discrepencies are.

Assuming that I’m right about this, what relevance would this have to cold fusion? Here is a critical question:

Q: If I have a ground-state nucleus at rest, and I gradually accelerate it to momentum p (by pushing with an electric field as part of a phonon oscillation), what is its final state? Is it: (A) The energy eigenstate for a ground-state nucleus at p? Or (B) The convoluted state for a ground-state nucleus at p?

My answer to this question is

A: Strictly speaking, it’s not exactly either, but for all intents and purposes, it’s (A).

The reason is the adiabatic theorem. The nuclear state energy differences are something like $10^{20}\text{Hz}$, which is much much much faster than the phonon frequency of $10^{13}\text{Hz}$. The process is not exactly adiabatic, but really really really close to adiabatic. A ground-state nucleus just stays in one of the energy eigenstates for a ground state in motion. The convoluted states never occur, there’s no reason to ever even mention them.

But think about this: What if someone (incorrectly) thought that the answer to the question were (B)? What would they conclude? That if you put a ground-state nucleus into motion (as part of a phonon oscillation), and then you measured the energy, you would sometimes find that it is in an excited state! Now, this sounds an awful lot like the kind of weird nuclear-phonon interaction that the authors are describing in the paper, and indeed they are doing just the right calculation for it. Hmm.

So, I’m concerned about the paper. The authors don’t say it explicitly, but I have an impression that they would give the wrong answer (B) to this question. Why else would they be doing these calculations related to convoluted states? (See update at bottom.) This is just speculation, I apologize for putting words in people’s mouths. Here’s a quote from the paper if it helps:

The origin of this effect is that the nuclear states of the composite nucleus transform under a boost in the many-particle Dirac model, which implies a mixing with other states. In the case of constant P the eigenvalue problem is seeking to create a version of the boosted wavefunction out of the rest frame states. However, with the composite momentum is dynamical (as occurs in a lattice), the model tries to develop boosted wavefunctions for a composite momentum that keeps changing magnitude and direction, which requires in this picture a dynamical admixture of different rest frame states

Regardless of what the authors do or don’t think, I conclude that there are two reasons that this kind of motion-induced nuclear transition would be rare. First, the nucleus is moving much slower than the speed of light, so there is very little difference between convoluted states and energy eigenstates, e.g. very little squishing due to length contraction. Second, the adiabatic theorem means that the transition rate will be a grillion times smaller still. This paper only addresses the first aspect, not the second. Maybe the authors are working on a follow-up paper using the Landau-Zener formula or whatever to fill in the second part. But I’m concerned that they’ve simply forgotten about the second part altogether (see update below). In which case, they would be dramatically overestimating the likelihood that a nuclear fusion process can transfer its energy to phonons.

It’s also quite possible that I’m misunderstanding the paper altogether!

[Update July 21: PH tells me that they have in fact calculated just how much or how little the adiabatic theorem is violated—even though they don’t discuss it in this particular paper—and claims that they have been taking this into account all along! According to their calculations, he says, there is enough non-adiabaticity to get the job done, after taking into account the other ingredients in the lossy-spin-boson model.]