# A A need for an exactly solvable model

1. Dec 27, 2016

### Demystifier

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.)

Last edited: Dec 27, 2016
2. Dec 27, 2016

### hilbert2

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 $V(x,y)=kx^2 y^4$ ?

Last edited: Dec 27, 2016
3. Dec 27, 2016

### Staff: Mentor

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

4. Dec 27, 2016

### hilbert2

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...

5. Dec 27, 2016

Staff Emeritus
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.

6. Dec 27, 2016

### hilbert2

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.

Last edited: Dec 27, 2016
7. Dec 27, 2016

### secur

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.

8. Dec 27, 2016

### Staff: Mentor

@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).

9. Dec 27, 2016

### secur

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.

Obviously

Ok ... that's not usually necessarily the case

10. Dec 28, 2016

### Demystifier

No, because k(y) should not depend on x.

I know, but I wanted to check out.

11. Dec 28, 2016

### Demystifier

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).

12. Dec 28, 2016

### Demystifier

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

13. Dec 28, 2016

### Demystifier

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

14. Dec 28, 2016

### hilbert2

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.

15. Dec 28, 2016

### secur

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.

Last edited: Dec 28, 2016
16. Dec 28, 2016

### hilbert2

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 $V(x,y)$.

17. Dec 28, 2016

### Demystifier

But mathematics does not depend on that. The constraint x=y is unjustified in both cases.

18. Dec 28, 2016

### Demystifier

No. The force is a physical measurable quantity.

19. Dec 28, 2016

### hilbert2

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

20. Dec 28, 2016

### Demystifier

Interesting, but the result is numerical, not analytical. It's probably not very useful for my purposes.