Anharmonic oscillators with closed-form solutions

[hilbert2]

TL;DR Summary

Some quantum oscillators with higher than quadratic V(x) seem to have an exact solution. Does this have any applications in QFT?

There are some articles from the 1980s where the authors discuss 1D quantum oscillators where $V(x)$ has higher than quadratic terms in it but an exact solution can still be found. One example is in this link:

https://iopscience.iop.org/article/10.1088/0305-4470/14/9/001

Has anyone tried to form a quantum field where the normal modes have a behavior similar to this kind of oscillator, and write the ground state of this interacting field in terms of the ground state and creation operators of the non-interacting case? It would likely not have any practical use, but it would just be a QFT system where something can be solved non-perturbatively.

 

[Keith_McClary]

That one seems to be on the half line.

Here is one in 2D:
https://arxiv.org/pdf/quant-ph/0003100.pdf

 

[A. Neumaier]

Has anyone tried to form a quantum field where the normal modes have a behavior similar to this kind of oscillator, and write the ground state of this interacting field in terms of the ground state and creation operators of the non-interacting case? It would likely not have any practical use, but it would just be a QFT system where something can be solved non-perturbatively.

There are quite a number of exactly solvable relativistic QFTs in 2 spacetime dimensions. The most important one is perhaps the Schwinger model, which is QED in 2 instead of 4 dimensions.

Quantum wires are described by nonrelativistic QFTs, some of which are exactly solvable and explain (in simplified models) the fractional quantum Hall effect.

 

[hilbert2]

That one seems to be on the half line.

Here is one in 2D:
https://arxiv.org/pdf/quant-ph/0003100.pdf

Oh, I didn't notice that right away. A system defined only on one half-line of the position axis wouldn't be very useful, because the degrees of freedom of the corresponding field would have only positive values then.

There are quite a number of exactly solvable relativistic QFTs in 2 spacetime dimensions. The most important one is perhaps the Schwinger model, which is QED in 2 instead of 4 dimensions.

Quantum wires are described by nonrelativistic QFTs, some of which are exactly solvable and explain (in simplified models) the fractional quantum Hall effect.

Thanks for the reply, I didn't remember the Schwinger model. A solvable scalar field system where the Lagrangian density contains a 6th or higher power of the field function would be in some way interesting because higher than phi-fourth interactions are not renormalizable (if I understood D. Tong's lecture notes correctly) and with an exact solution you could still see how the system behaves.

 

[A. Neumaier]

A solvable scalar field system where the Lagrangian density contains a 6th or higher power of the field function would be in some way interesting because higher than phi-fourth interactions are not renormalizable (if I understood D. Tong's lecture notes correctly) and with an exact solution you could still see how the system behaves.

In 2D all bosonic fields with polynomial interactions are renormalizable, and existence has been proved rigorously if the interactions are bounded below. See the book by Glimm and Jaffe.

But for field theories, sextic interactions are not exactly solvable; at least I don't know of an example. Typically, the exactly solvable theories have very special quartic interactions.

The 2D Gross-Neveu Model is nonrenormalizable and exactly solvable, I believe.

 

[Demystifier]

The 2D Gross-Neveu Model is nonrenormalizable and exactly solvable, I believe.

How can a theory be both nonrenormalizable and exactly solvable at the same time? Does it mean that it is ill defined perturbatively but well defined non-perturbatively?

 

[A. Neumaier]

How can a theory be both nonrenormalizable and exactly solvable at the same time? Does it mean that it is ill defined perturbatively but well defined non-perturbatively?

Yes if you adhere to the old belief that nonrenormalizable theories are perturbatively undefined.

But the modern view is that nonrenormalizable theories are as well-defined perturbatively as renormalizable ones, except that they depend on a countable (rather than finite) number of parameters. This is not a real defect since analytic functions also depend on in finitely many parameters but as long as the power series converge fast enough a few known parameters produce accurate results.

Thus the most correct statement is that nonrenormalizable and exactly solvable theories are underspecified perturbatively but well defined non-perturbatively.

 

[hilbert2]

I believe that the way how the authors of the linked articles have found the exact solutions for those oscillators has been by rearranging the nondimensional ($\hbar = m = 1$) Schrödinger equation

$\displaystyle -\frac{1}{2}\frac{d^2 \psi (x)}{dx^2} + V(x)\psi (x)=E\psi(x)$

to the inverse problem equation

$\displaystyle V(x)=E+\frac{1}{2}\frac{\psi ''(x)}{\psi (x)}\hspace{30pt}$ (*)

and then by noting that for instance in the case $\psi (x) = Ae^{-ax^2 -bx^4}$ the resulting potential energy is

$\displaystyle V(x) = E + (2ax + 4bx^3)^2 + 2a + 12bx^2$,

which is a sixth-degree polynomial. So, the ground state of the system with $V(x)$ having the form of that polynomial is like a Gaussian but with higher than second powers of $x$ in the exponent. However, if the ground state is like that, then none of the excited state wave functions have the simple form of the same exponential multiplied with polynomials (like in the harmonic oscillator case, where those multipliers are Hermite polynomials). Actually, I'm able to prove this in the following way (tell me if there's a logical error somewhere):

Suppose the unnormalized ground state wave function of an anharmonic oscillator is $\psi_0 (x) = e^{P_1 (x)}$, where $P_1 (x)$ is some polynomial of $x$ containing only terms of order divisible by 2. Then, according to (*), the 1D potential energy for which this is one of the energy eigenfunctions is

$\displaystyle V(x) = E_0 + P_{1}^{''} (x) + [P_{1}^{'} (x)]^2$.

The constant $E_0$ is the ground state energy. Now, if another eigenfunction of the same Hamiltonian is $\psi_n (x) = P_2 (x)e^{P_1 (x)}$ with $P_2$ a polynomial, the function $P_2$ should obey the differential equation (derived from (*))

$\displaystyle P_2^{''}(x) + 2P_1^{'}(x)P_2^{'}(x) = 2EP_2 (x),$

where $E$ is a constant that doesn't depend on $x$. Suppose the polynomial $P_1$ is of $m$:th order and $P_2$ is of $n$:th order. Then the term on the LHS that contains highest powers of $x$ is the $2P_1^{'}(x)P_2^{'}(x)$, where the order is $m-1 + n-1$. The highest power of $x$ on the RHS is of $n$:th order. But the terms $2P_1^{'}(x)P_2^{'}(x)$ and $2EP_2 (x)$ can't be of same order in $x$ unless $m=2$, as in the harmonic oscillator. Therefore, none of the excited states of this kind of anharmonic oscillators can't have the form of a simple finite-degree polynomial multiplying the ground state wave function.

Edit: The term $2P_1^{'}(x)P_2^{'}(x)$ is of order $m-1 + n-1$, not $(m-1)(n-1)$ as in the unedited post.

 

[A. Neumaier]

I believe that the way how the authors of the linked articles have found the exact solutions for those oscillators has been by rearranging the nondimensional ($\hbar = m = 1$) Schrödinger equation

$\displaystyle -\frac{1}{2}\frac{d^2 \psi (x)}{dx^2} + V(x)\psi (x)=E\psi(x)$

to the inverse problem equation

$\displaystyle V(x)=E+\frac{1}{2}\frac{\psi ''(x)}{\psi (x)}\hspace{30pt}$ (*)

and then by noting that for instance in the case $\psi (x) = Ae^{-ax^2 -bx^4}$ the resulting potential energy is

$\displaystyle V(x) = E + (2ax + 4bx^3)^2 + 2a + 12bx^2$,

which is a sixth-degree polynomial.

This approach only works for a single oscillator. But you asked about quantum fields, where one needs infinitely many of them!

 

[hilbert2]

This approach only works for a single oscillator. But you asked about quantum fields, where one needs infinitely many of them!

Yes, if the Fourier modes of the self-interacting field behave like anharmonic oscillators with different "spring constants", not all of them necessarily have an exact solution for the ground state energy and wave function.

Edit: It's possible that I'm considering field systems in a too simplified way here; if some scalar field equation with a nonlinear interaction has a Fourier transform

$\displaystyle \frac{\partial^2}{\partial t^2} \phi(\mathbf{p},t) +(|\mathbf{p}|^2+m^2 )\phi(\mathbf{p},t)+\beta \phi(\mathbf{p},t)^3 = 0$,

where $\beta$ is some parameter, then the original untransformed field equation is not necessarily an acceptable local equation of motion. Or is it?

 

[hilbert2]

Using the shooting method to calculate first 10 energy eigenvalues of some pure anharmonic oscillators

$\displaystyle -\frac{1}{2}\frac{d^2 \psi (x)}{dx^2}+\lambda x^{2m} \psi(x) = E\psi (x)$,

with particle mass and $\hbar$ set to value 1, I found out that a function like $E(n) = C_0 + C_1 n^{C_2}$ fits to the data points $E(n)$ ($n$th energy eigenvalue) if the quantum number $n$ is seen as a continuum variable. For multiplier $\lambda =2$ and exponents $m=1$ to $m=4$, the curve fits look like in this graph (note that the ground state of the harmonic oscillator case is also labelled $n=1$, not the usual $n=0$):

The fitting parameters $C_0$, $C_1$ and $C_2$ are shown in this table:

So this seems to be kind of an exact result, but not completely as you have to determine the parameters empirically for each oscillator.

I initially though I was the first to notice this feature of those energy levels, but yesterday I found this article preprint made by researchers in Iran in 2019 and describing the same thing:

https://arxiv.org/pdf/1910.02746.pdf

Is anyone else familiar with this power-law property of the pure anharmonic oscillator? The exponent $C_2$ clearly approaches $2$ when $m\rightarrow\infty$, because an anharmonic oscillator of infinite order is practically the same as a 1D particle-in-box system with sudden infinite walls around it.

Actually, that same trial function also fits to data points describing the spectrum of an oscillator with both a quadratic and quartic term in $V(x)$, but then the large variety of combinations of parameter values makes it difficult to find a general logic behind the results.