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A A need for an exactly solvable model

  1. Dec 27, 2016 #1

    Demystifier

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    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
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  3. Dec 27, 2016 #2

    hilbert2

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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## ?
     
    Last edited: Dec 27, 2016
  4. Dec 27, 2016 #3

    mfb

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    That would surprise me - you don't reproduce the two lines of zero potential with such a linear transformation and then u4+v4.
     
  5. Dec 27, 2016 #4

    hilbert2

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    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...
     
  6. Dec 27, 2016 #5

    Vanadium 50

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    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.
     
  7. Dec 27, 2016 #6

    hilbert2

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    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
  8. Dec 27, 2016 #7
    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.
     
  9. Dec 27, 2016 #8

    mfb

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    @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).
     
  10. Dec 27, 2016 #9
    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
     
  11. Dec 28, 2016 #10

    Demystifier

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

    I know, but I wanted to check out.
     
  12. Dec 28, 2016 #11

    Demystifier

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    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).
     
  13. Dec 28, 2016 #12

    Demystifier

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    For my purposes, this Hamiltonian is better viewed as two particles moving in 1 dimension. The constraint x=y would make no physical sense.
     
  14. Dec 28, 2016 #13

    Demystifier

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    Thank you all for your tries. I guess I will need to go with the approximative perturbative approach (which, of course, is straightforward).
     
  15. Dec 28, 2016 #14

    hilbert2

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    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.
     
  16. Dec 28, 2016 #15
    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
  17. Dec 28, 2016 #16

    hilbert2

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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)##.
     
  18. Dec 28, 2016 #17

    Demystifier

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    But mathematics does not depend on that. The constraint x=y is unjustified in both cases.
     
  19. Dec 28, 2016 #18

    Demystifier

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    No. The force is a physical measurable quantity.
     
  20. Dec 28, 2016 #19

    hilbert2

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    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
     
  21. Dec 28, 2016 #20

    Demystifier

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