As the title says, the Morse potential (https://en.wikipedia.org/wiki/Morse_potential) is an exactly solvable model in quantum mechanics, just like the quantum harmonic oscillator. Extensions of it are used in modeling molecular vibrational states to great success as far as I know.

However, in solid state physics we seem to stop at using phonons (collections of harmonic oscillators) to describe the vibrations of solids. Why not use the Morse potential to model the interactions between atoms in a solid instead of as harmonic springs, if it's just as exactly solvable?

  • $\begingroup$ +1. Welcome to the site, and thank you for asking your question here! We hope to see much more of you in the future !!! $\endgroup$ – Nike Dattani Sep 3 '20 at 20:44
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    $\begingroup$ Thanks! I was encouraged to post this question on here by the Physics discord. $\endgroup$ – Slenderman Sep 4 '20 at 3:29
  • $\begingroup$ Which physics discord is that? Can you send me a link (please)? $\endgroup$ – Nike Dattani Sep 4 '20 at 4:00
  • $\begingroup$ This one: discord.gg/physics $\endgroup$ – Slenderman Sep 9 '20 at 1:39

Some people do:

In this paper there is a system coupled to a bath of Morse oscillators rather than a bath of harmonic oscillators, but it is not exactly solvable, they used a numerical approach called . When it is said that the Morse potential is "exactly solvable", what it means is that you can solve the vibrational Schroedinger equation for a Morse potential analytically (the end result is a formula for the vibrational energies and wavefunctions of the system). What about if there is a spin coupled to 500 Morse oscillators, do we have an exact solution for the overall energies and wavefunctions of this very complicated system? We do not even have an exact solution in the case where the oscillators are harmonic oscillators; this would be called the "spin-boson problem" and it's not exactly solvable except in specific cases. Finally, even if there was only one Morse oscillator, the solutions that you see in that Wikipedia page you linked, are not very simple or easy to use: For example the exact solutions for the quantum harmonic oscillator do not involve generalized Laguerre polynomials but for the Morse oscillator they do.

Most people don't:

In solid state physics, many people model a system coupled to phonons using models like the spin-boson model or generalizations of it. For example in this paper of mine we study how a qubit would undergo decoherence if a quantum computer were to be made with GaAs quantum dots. The qubit can be defined as follows: 0 = absence of an exciton, 1 = presence of an exciton, so it's a 2-level system, but it's coupled to all of the vibrations of the semiconductor lattice in which it lies. 2-level systems can be considered as "spin-1/2 particles", so what we have is a spin interacting with a bunch of vibrations. These vibrations are approximated to be harmonic oscillators, so we simply have the aforementioned "spin-boson problem" which has been studied for several decades (and still cannot be solved exactly most of the time). Now there's several reasons why we chose this simple spin-boson model with harmonic oscillators rather than using a Morse potential:

  • Neither case (harmonic oscillators or Morse oscillators) is exactly solvable when you have not only the oscillators (nuclear vibrations) but also the spin (or electronic/excitonic degrees of freedom), but at least for the harmonic case we have simple analytic expressions for things such as the Feynman-Vernon influence functional which describes the influence of the vibrations on the spin; for Morse oscillators we do not have such a simple influence functional. So the calculations are much easier in the harmonic case compared to the Morse oscillator case.
  • A Morse oscillator is actually harmonic at the very bottom, and only starts to deviate from a harmonic oscillator for much larger internuclear distances as the system starts to dissociate. If we were to model a qubit in a GaAs semiconductor at such a high temperature that the semiconductor is in the midst of breaking apart (the Ga and As atom are dissociating from one another), we may wish to use an anharmonic potential, but this is rarely (or never!) the case. Consider this: Which vibrational levels of these oscillators actually have a significant population in your system? Surely not the ones at the top of this Morse potential (see v=8 and 9 and observe that their outer turning points are at internuclear distances of around r=12 and 14 which is about triple the internuclear distance at equilibrium; do you picture your solid-state lattice having its internuclear distances so far away from equilibrium in the normal scenario in your interest?).
  • For this particular system, a spectral distribution function (which tells you how strongly each oscillator couples to the "spin") was determined from fitting the dynamics of the spin-boson model, and the empirically obtained parameters of the spectral function matched almost exactly the values of those parameters obtained from first-principles calculations, and the chances of this being pure luck are so extremely low that we believe that the simple spin-boson model with harmonic oscillators is an excellent approximation.
  • If we were not in a case where we know the spin-boson model with harmonic oscillators is an excellent approximation (as described in the last bullet point), we can stop and think about all the other multitude of approximations we are making (maybe the ignoring of spin-orbit coupling, maybe the ignoring of the difference between relativistic mass and non-relativistic mass, maybe the use of the Born-Oppenheimer separation between the electronic/excitonic and nuclear/phononic/bosonic degrees of freedom, or maybe we don't have the exact parameters for the exact solid in question so we are just assuming we can use the parameters from a similar solid that has been studied in more detail, etc.), and then realize that there's just so many approximations going on that Jon von Neumann's quote applies: "Why be precise when we don't know what we're talking about?" Will you use quadruple precision (33-36 digits) for solving a differential equation telling you what weather it will be tomorrow, when the coefficients in the differential equation have enormous error bars? Then don't use an anharmonic bath if you don't have a fairly precise description of all other relevant information (relativistic effects, spin-orbit coupling, etc.).

Conclusion: The bottom line is that you would be making your life harder (more difficult equations that can only be solved less efficiently or with less accuracy with the same resources), and most of the time you would not be gaining a better understanding of the relevant physics.

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    $\begingroup$ Accepted for being such a thorough and interesting answer! I will check out the papers you mentioned. There are other use cases for anharmonic effects than just near melting temperature. For example, near a structural phase transition (or a charge-density-wave transition) the energies of the phonons along the direction of transition seem to 'soften' (go to zero). In this case the leading order corrections (e.g. anharmonic or Born Oppenheimer) should make sure the process is stable (or doesn't lead to divergences, mathematically). This is more the use case I'm interested in. $\endgroup$ – Slenderman Sep 4 '20 at 3:39
  • $\begingroup$ Thank you. The question is still so young, so there is plenty of time for others to add. I don't know as much about phase transitions, but as I read this, I'm reminded of yet another paragraph I could have put in there, which would be about the statement by Caldeira & Leggett in their paper (now with 4000+ citations), that all spin-bath models can be transformed into a spin-harmonic-oscillator model except in some "pathological cases" (their term for it, not mine). They were essentially arguing that the Hamiltonian with an anharmonic bath could be transformed into one with a harmonic one. $\endgroup$ – Nike Dattani Sep 4 '20 at 4:08
  • $\begingroup$ I would have to read their paper again, but my current understanding is that you can transform an anharmonic model into a harmonic one, but at the cost of introducing more degrees of freedom. In that case, however, isn't that mainly nothing but the idea that one can approximate any function with a superposition of gaussians? $\endgroup$ – Slenderman Sep 5 '20 at 4:58
  • $\begingroup$ I don't think it's as simple as approximating a c-function (classical function) by a sum of basis functions (whether its cos(x) functions in a Fourier series or x^n functions in a Taylor series or Gaussians, etc.). We are approximating operators which are NxN matrices in Hilbert space such that the low-lying eigenvectors and eigenvalues are approximately preserved, rather than just making the function's value being near the right number. I would also not call more degrees of freedom a "cost" in this case. Infinite # of harmonic oscillators is easier for QUAPI than having having just 1 of them. $\endgroup$ – Nike Dattani Sep 5 '20 at 18:17

Even in treating molecular vibrations, the Morse potential is not always the best, because:

  • There's cases where the potential is more "harmonic" than "Morse-like", for example in the asymmetric stretching of water. It is the same case in solid state physics: considering the most crude approximation that we fix all the atoms in a solid while looking into the motion of one atom in bulk.
  • The Morse potential is more expensive to evaluate since it has two more parameters than the harmonic potential when considering just two atoms, and it is even harder when we consider the polyatomic case.
  • $\begingroup$ +1. This is a great addition that I hadn't even considered when I wrote my answer: Morse potentials are more complicated than Harmonic ones, so if evaluating the function many times (as in MD calculations) you would be slowing things down, and also you have to parameterize the Morse potential: what value do you use for the anharmonicity or the depth of the potential? The Morse potential has 3 parameters while the harmonic oscillator only depends on $\omega$. And it's also a good point that there's some cases where the real situation ism ore harmonic-like than Morse-like :) $\endgroup$ – Nike Dattani Sep 4 '20 at 18:28

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