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  Spectrum of 1D quantum systems with convex V(x)

+ 3 like - 0 dislike

The authors of this unreviewed article discuss the observation that the spectrum $E_n$ of the eigenvalues of the Hamiltonian $\hat{H}$ of a 1D quantum system is always of the form

$E_n = c_0 + c_1 n^{c_2},\hspace{20pt}c_j \in \mathbb{N}$,

when the potential energy is a monomial $V(x) = Ax^m$ with $m$ an even integer. When $m = 2$, it's clear that the constant $c_2$ has value $1$ because then it's a harmonic oscillator.

After calculating the first 10 or so energies for many different kinds of functions $V(x)$, it seems to me that actually the spectrum $E_n$ is of this form for any convex function $V(x)$, i.e. a potential with a graph that is turning upwards at any point $x$. This also holds for functions with a discontinuous derivative, such as $V(x) = A|x|$ and $V(x) = Ae^{|x|}$. When $V(x)$ is not convex, such as for a double well potential with a barrier in between, this does not hold. For instance, the most significant property of the 1D double well is that the spacing between the two lowest energy states is disproportionately small compared to other energy spacings, and therefore that power function doesn't fit the eigenvalues.

Has this been observed by anyone before, and does there exist a mathematical proof for it?

asked Aug 30, 2022 in Q&A by hilbert2 (15 points) [ no revision ]
retagged Aug 30, 2022

I think this is quite unlikely to hold exactly for all $V(x)=ax^2+bx^4$ with positive $a,b$. What is your evidence for it?

1 Answer

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I don't have any other evidence than for example this image where I tried to fit that power function to the calculated energy eigenvalues of that kind of system. I didn't find any convex function where this wouldn't seem to hold, but it's of course possible that it is only approximately true and you can see a clear difference if you construct some kind of extreme case.

I also wrote about this in my Wordpress blog (compphysdiary.wordpress.com and compphysdiary.wordpress.com/posts-page) but for some reason Google doesn't want to index that site and it sometimes automatically turns "private" so that I have to toggle that setting for anyone else to be able to visit that site, for whatever reason.

Edit: and the Schrödinger equation in that was non-dimensionalized by setting h-bar and particle mass to value 1. Just had to mention this to be sure.

answered Aug 30, 2022 by anonymous [ no revision ]

You could as well claim that the energy always has a cubic dependence, since thie would equally well fit data like yours. To establish a power law computationally you need many more data points up to large $n$ so that different laws start to differ substantially. And you'd need to work with much more than drwawing accuracy and show that your fit matches the accuracy with which you compute the spectrum.

That's a quite appropriate criticism, especially in this case where the most important component of $V(x)$ is quadratic and the $E_n$ increases close to linear fashion. However, a cubic spectrum $E_n \propto n^3$ isn't possible, because the energy levels of a 1D potential well can't asymptotically increase faster than $E_n \propto n^2$ which is the case with a "particle in a box". Reference for this: https://arxiv.org/abs/0811.1389 (see Subsection II.B.).

If there's some simple way to accurately estimate the values of parameters $c_0, c_1, c_2$ based on the power series expansion of $V(x)$, then this would be a very useful approximation at least for low quantum numbers, I think.

Edit: sorry for posting as "anonymous", I forget to log in because unlike Physics Forums this site allows easy posting as a guest.

There are still many ways to achieve a less steep increase for large n, such as a cubic/linear ansatz, or a linear combnation of $n^e$ with a list of several exponents. Thus your evidence indicates nothing at all. Try to fit a few hundred energies, and the limitations of your formula will become apparent.

Yes, I just attempted to fit that empirical formula to the calculated energies of a half-linear half-harmonic oscillator where $V(x) = Ax^2$ for $x\leq 0$ and $V(x) = Bx$ for $x>0$, found in the Table 1 of the document in this link, and I can already see a small but visible error in the fit with only 10 first eigenenergies.

Looks like I was too quick to draw conclusions from what these spectra seemed to be. If there's a way to show that, for convex $V(x)$, there's some upper bound for the max error of the optimal fit of this type with first $N$ eigenvalues included, then it could be useful. But I'm not trying to prove that myself, it's more of pure mathematics.

Edit: It could also be possible to prove the converse statement, i.e. if the spectrum of a 1D potential well is $E_n = c_0 + c_1 n^{c_2}$, then this implies that the $V(x)$ is a convex function. The harmonic oscillator and the particle in a box are the two known examples where the spectrum is that kind of a power function, and both are "convex" (the box potential is not a real function but is convex in the practical sense here). The equation (5a) from reference https://arxiv.org/pdf/0811.1389.pdf and standard properties of determinants would probably be used in the proof.

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