# A need for an exactly solvable model

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• Demystifier
In summary: I may be misunderstanding what you want? ... It looks like a Hamiltonian for a single particle in 2 dimensions, labelled x and y. But you refer to "a free particle with position ##y## and a harmonic oscillator with position ##x##" which sounds like two separate particles labelled x and y. Also you have two different masses, m and M. Well, I'll guess you're talking about one particle which is free in y and subject to harmonic potential in x? And assume m=M.Then you could use f(y) = k/y^2. Then the third term is (x/y)^2, times a constant. Now constrain the particle to move on a line through the
Demystifier
Gold Member
Consider a Hamiltonian of the form
$$H=\frac{p_x^2}{2m}+\frac{p_y^2}{2M}+\frac{k(y)x^2}{2}$$
where ##k(y)## is some free function which I can choose at will. For ##k(y)=k=constant##, the Hamiltonian is a trivial combination of a free particle with position ##y## and a harmonic oscillator with position ##x##. However, I need the exact solution (of the Schrodinger equation) for some nontrivial case, when ##k(y)## is not a constant. Does anybody know such a function ##k(y)## for which the system can be solved exactly? A reference would be very welcome.

(I need it for a research paper on which I am working, so coautorship is also possible, in which case one can send me a private message. The research is related to the Casimir effect.)

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Here's an article about solutions for a quartic potential oscillator system where ##V(x) \propto x^4##: https://arxiv.org/pdf/physics/0603165.pdf . I think there's an exact solution that can be written as some kind of a series expansion or iteration, but there's no compact expression for the solutions in terms of familiar special functions.

There is probably some way to make a change of variables ##x = au+bv##, ##y = cu + dv## that leads to ##kx^2 y^2 = \lambda (u^4 + v^4)## so that you can think of an oscillator with a ##kx^2 y^2## potential as a combination of two independent quartic oscillators.

EDIT: Also, a system with a potential energy ##V(x)## that is a sixth-order polynomial in ##x## is exactly solvable up to some n:th excited state if the coefficients of the polynomial are chosen in the right way. Maybe try to do a change of variables on ##V(x,y)=kx^2 y^4## ?

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Demystifier
hilbert2 said:
There is probably some way to make a change of variables ##x = au+bv##, ##y = cu + dv## that leads to ##kx^2 y^2 = \lambda (u^4 + v^4)## so that you can think of an oscillator with a ##kx^2 y^2## potential as a combination of two independent quartic oscillators.
That would surprise me - you don't reproduce the two lines of zero potential with such a linear transformation and then u4+v4.

bhobba
mfb said:
That would surprise me - you don't reproduce the two lines of zero potential with such a linear transformation and then u4+v4.

Yeah, good point... We'd need a potential of the form ##u^4 - v^4## so that the lines ##u=v## and ##u=-v## would correspond to zero potential energy, and then there would be no bound states...

Can you cheat? Have k(y) be [some other solvable potential]/x2?

As you know, there are very few exactly solvable potentials in QM. I don't think any are of the form you want.

Does the equation ##f(y)x^2 = g(ax+by) + h(cx+dy)## hold for any nontrivial functions ##f##, ##g## and ##h## even if the matrix ##
\left[ {\begin{array}{cc}
a & b \\ c & d \ \end{array} } \right]## is not required to be orthogonal?

EDIT: Also, you could form a potential ##V(x,y)=kf(y)x^2##, where ##f(y)## is a kind of a particle-in-a-box potential:

##f(y)=1##, when ##|y|<L/2## and ##f(y)=\infty##, when ##|y| \geq L/2##, and ##k## is a constant with dimensions of a Hooke's law constant.

Then the ##y## coordinate of the particle would be constrained between ##-L/2## and ##L/2##, except possibly in the special situation where ##x=0## (the relevance of which is unclear to me). The box potential is kind of a limit of the anharmonic potential ##V(x)=k\left(\frac{2x}{L}\right)^{2n}## when ##n \rightarrow \infty##...

EDIT 2: And if we have a system with ##V(x,y)=kx^{2}(\sin{(2\pi y / \lambda)}+1)##, where the wavelength ##\lambda## is very small compared to the spring constant and the y-component of the momentum of the particle is very large, then the system would probably behave as if the x and y degrees of freedom wouldn't "see" each other at all.

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I may be misunderstanding what you want? ... It looks like a Hamiltonian for a single particle in 2 dimensions, labelled x and y. But you refer to "a free particle with position ##y## and a harmonic oscillator with position ##x##" which sounds like two separate particles labelled x and y. Also you have two different masses, m and M. Well, I'll guess you're talking about one particle which is free in y and subject to harmonic potential in x? And assume m=M.

Then you could use f(y) = k/y^2. Then the third term is (x/y)^2, times a constant. Now constrain the particle to move on a line through the origin, for instance x=y. Then the third term is constant. Rotate 45 degrees (assuming simplest case x=y) and the system becomes a particle in 1 dimension with constant potential (modulo constant terms) for which the solution is well known.

@secur: The particle can have different effective masses along different axes, nothing wrong with that (you can also re-scale the axes to make the masses agree if you want). If the potential for x and y is independent as discussed in post 1 you can solve for the motion of the same particle in both dimensions by separately solving the motion in each dimension.

You cannot constrain the motion to x=y or something like that, that makes the whole problem pointless.The box potential approach should have an analytic solution, although it is not a proper mathematical function k(y).

bhobba
mfb said:
The particle can have different effective masses along different axes, nothing wrong with that (you can also re-scale the axes to make the masses agree if you want).
True ... I suppose that's what he means? Different effective masses happens with relativistic velocities (and some other odd, "virtual" cases), but he's using Schrodinger's and non-relativistic Hamiltonian. ... Not that it matters.

mfb said:
If the potential for x and y is independent as discussed in post 1 you can solve for the motion of the same particle in both dimensions by separately solving the motion in each dimension.
Obviously

mfb said:
You cannot constrain the motion to x=y or something like that, that makes the whole problem pointless.
Ok ... that's not usually necessarily the case

Can you cheat? Have k(y) be [some other solvable potential]/x2?
No, because k(y) should not depend on x.

As you know, there are very few exactly solvable potentials in QM. I don't think any are of the form you want.
I know, but I wanted to check out.

hilbert2 said:
EDIT: Also, you could form a potential ##V(x,y)=kf(y)x^2##, where ##f(y)## is a kind of a particle-in-a-box potential:
Nice try, but it is also important that the derivative with respect to y is neither zero nor infinite. This derivative is related to a force, and I need a finite force (which can mimick Casimir force).

secur said:
I may be misunderstanding what you want? ... It looks like a Hamiltonian for a single particle in 2 dimensions, labelled x and y. But you refer to "a free particle with position ##y## and a harmonic oscillator with position ##x##" which sounds like two separate particles labelled x and y. Also you have two different masses, m and M. Well, I'll guess you're talking about one particle which is free in y and subject to harmonic potential in x? And assume m=M.

Then you could use f(y) = k/y^2. Then the third term is (x/y)^2, times a constant. Now constrain the particle to move on a line through the origin, for instance x=y. Then the third term is constant. Rotate 45 degrees (assuming simplest case x=y) and the system becomes a particle in 1 dimension with constant potential (modulo constant terms) for which the solution is well known.
For my purposes, this Hamiltonian is better viewed as two particles moving in 1 dimension. The constraint x=y would make no physical sense.

Thank you all for your tries. I guess I will need to go with the approximative perturbative approach (which, of course, is straightforward).

Demystifier said:
Nice try, but it is also important that the derivative with respect to y is neither zero nor infinite. This derivative is related to a force, and I need a finite force (which can mimick Casimir force).

Is it sufficient if the force ##\frac{\partial V}{\partial y}## exists in the sense of a distribution (generalized function)? You could make piecewise constant functions ##k(y)## that have finite potential steps, or something that has Dirac delta functions ##\delta (y-y_0 )## in it, and it could be solved analytically.

Demystifier said:
For my purposes, this Hamiltonian is better viewed as two particles moving in 1 dimension. The constraint x=y would make no physical sense.

Ok, but it's worth noting that (it appears) most others also thought it was one, 2-d, particle. So maybe now that it's clear, give them another chance to think about the right problem.

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If it's enough that the ground state wavefunction of the system is known exactly, you could try forming functions ##\psi (x,y)## that are not of simple product form ##\psi (x,y) = \phi (x) \theta (y)## - e.g. something like ##\psi (x,y) = A e^{-Bx^2-Cx^2 y^2}##, and solving the quantum inverse problem ##V(x,y)=\frac{\hbar^2}{2m}\frac{\nabla^2 \psi}{\psi}##. The trial function ##\psi (x,y)## should be normalizable and have no nodes, so that you could be certain that it's the ground state wavefunction of the system described by the resultant potential function ##V(x,y)##.

secur said:
Ok, but it's worth noting that (it appears) most others also thought it was one, 2-d, particle. So maybe now that it's clear, give them another chance to think about the right problem.
But mathematics does not depend on that. The constraint x=y is unjustified in both cases.

mfb
hilbert2 said:
Is it sufficient if the force ##\frac{\partial V}{\partial y}## exists in the sense of a distribution (generalized function)?
No. The force is a physical measurable quantity.

Here's an article about solving the energy eigenvalues for 3D double-well potentials that are of type ##V(x,y,z;Z, \lambda )=-Z(x^2 + y^2 +z^2) + \lambda ( a_{xx} x^4 + a_{yy} y^4 + a_{zz} z^4 +a_{xy} x^2 y^2 + a_{yz} y^2 z^2 + a_{xz} x^2 z^2 )##, of which a 2D potential ##V(x,y) = a_{xy} y^2 x^2## is a special case where all the parameters except ##a_{xy}## are zero.

http://www.sciencedirect.com/science/article/pii/S0377042796000490

Demystifier
hilbert2 said:
Here's an article about solving the energy eigenvalues for 3D double-well potentials that are of type ##V(x,y,z;Z, \lambda )=-Z(x^2 + y^2 +z^2) + \lambda ( a_{xx} x^4 + a_{yy} y^4 + a_{zz} z^4 +a_{xy} x^2 y^2 + a_{yz} y^2 z^2 + a_{xz} x^2 z^2 )##, of which a 2D potential ##V(x,y) = a_{xy} y^2 x^2## is a special case where all the parameters except ##a_{xy}## are zero.

http://www.sciencedirect.com/science/article/pii/S0377042796000490
Interesting, but the result is numerical, not analytical. It's probably not very useful for my purposes.

If I try to plug a 2D power series trial function ##\psi (x,y) = \sum_{i=0}^\infty \sum_{j=0}^\infty c_{ij}x^i y^j## in the Schroedinger equation that has a ##V(x,y) \propto x^2 y^2 ## potential, I get a recurrence relation for a two index object ##c_{ij}## instead of a usual sequence ##c_i##... Apparently there is no closed form solution for some recurrence equations, so it's a bit unclear whether this kind of solution can be written on paper. The power series could probably be simplified by using the symmetry properties that the solution must have (V(x,y) is invariant in interchange of x and y, and in reflections ##x \rightarrow -x## and ##y \rightarrow -y##), and its predicted asymptotic behavior for ##x^2 y^2 \rightarrow \infty##.

A progress report: Today I have found a way to solve the equation in a non-perturbative semi-classical way for an arbitrary k(y). It turned out that the solution leads to a correction to the Casimir force that has not been known before. I still need to work out some details, but hopefully in a not so far future I will write a paper where the details will be revealed.
This, of course, is the job for the next year.

Mentz114, bhobba, mfb and 2 others
Demystifier said:
A progress report: Today I have found a way to solve the equation in a non-perturbative semi-classical way for an arbitrary k(y). It turned out that the solution leads to a correction to the Casimir force that has not been known before. I still need to work out some details, but hopefully in a not so far future I will write a paper where the details will be revealed.
This, of course, is the job for the next year.

Good work! Are there some special conditions when the semiclassical approximation works best, like a slowly varying function k(y) or a large mass associated with the y degree of freedom?

Henon Heiles hamiltonian comes to my mind.

Demystifier
hilbert2 said:
Are there some special conditions when the semiclassical approximation works best
For ##M\gg m##, ##y## can be treated semi-classicaly while ##x## is treated fully quantum mechanically. Think, for instance, of a textbook analysis of the hydrogen atom, where the nucleus is treated as a classical particle (classical source of Coulomb potential).

## 1. What is an exactly solvable model?

An exactly solvable model is a mathematical or physical model that can be solved analytically, meaning that all of its parameters and equations can be solved for in a closed form solution. This allows for a complete understanding of the system and its behavior.

## 2. Why is there a need for an exactly solvable model?

Exactly solvable models are useful for understanding complex systems and predicting their behavior. They also serve as a starting point for more realistic models and can provide insights into the underlying principles of a system.

## 3. How are exactly solvable models different from other models?

Unlike other models, exactly solvable models have closed form solutions, meaning that they can be solved for using mathematical techniques such as integration, differentiation, and algebra. This allows for a more complete understanding of the system and its behavior.

## 4. What types of systems can be modeled using exactly solvable models?

Exactly solvable models can be used to model a wide range of systems, including physical systems such as particles, fluids, and gases, as well as mathematical systems such as differential equations and statistical models.

## 5. What are the benefits of using an exactly solvable model?

Exactly solvable models have the advantage of providing a complete understanding of a system, as well as allowing for easy prediction of its behavior under different conditions. They also serve as a useful tool for developing more complex and realistic models.

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