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A Scale invariant inverse square potential

  1. Mar 10, 2017 #1

    hilbert2

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    Yesterday, I was thinking about a problem I had encountered many years before, the central force problem with a ##V(r) \propto r^{-2}## potential...

    If we have a Hamiltonian operator

    ##H = -\frac{\hbar^2}{2m}\nabla^2 - \frac{A}{r^2}##

    and do a coordinate transformation ##\mathbf{r} \rightarrow \lambda \mathbf{r}##, it's easy to see that if ##\psi (x,y,z)## is an eigenfunction of that ##H##, then also any scaled function ##\psi (\lambda x, \lambda y, \lambda z)## is, but with a different eigenvalue and normalization.

    In the classical mechanical case, you can make the scaling ##\mathbf{x} \rightarrow \lambda \mathbf{x}## and ##\mathbf{p} \rightarrow \mathbf{p}/\lambda## to turn one possible phase space trajectory of the orbiting point mass in this potential into another possible trajectory.

    Questions:

    1. Does the quantum inverse square potential system really have a continuum spectrum of eigenfunctions, as it seems here? Why is this different from the situation with a hydrogen atom?

    2. How could I explain, as simply as possible, why a scaling ##x \rightarrow \lambda x## is requires a simultaneous scaling ##p \rightarrow p/\lambda## in the classical mechanical case? In the quantum problem this is obvious because the momentum operator ##p_x## has a differentiation with respect to x in it, but it seems to be more difficult to explain in classical mechanical terms.

    p.s. don't confuse this with inverse square force...
     
  2. jcsd
  3. Mar 10, 2017 #2

    strangerep

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    I'm not sure why you'd think it wouldn't be different from the H-atom (or classical Kepler) case, since it's a different Hamiltonian.

    (Btw, classical Kepler does have a dilation-like symmetry, but it's non-uniform between space and time. See Kepler's 3rd law. :oldbiggrin:)

    A symmetry must preserve something, else it's not a symmetry. In the classical case, one typically wants to preserve the Poisson bracket, so we consider only canonical transformations that do so.

    Alternatively, one could consider symmetries as preserving the Hamilton equations of motion, but then one must introduce a corresponding scale transformation for time.
     
    Last edited: Mar 10, 2017
  4. Mar 11, 2017 #3

    hilbert2

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    Thanks. Usually bound states form a discrete spectrum, I guess the 1/r2 potential is a "falling to center" problem where the only bound state is one where the orbiting point mass is perfectly localized in the origin...

    I'll probably write a wikipedia page about this when I understand it well enough.
     
  5. Mar 11, 2017 #4
    If [itex]\psi(\vec{x})[/itex] is a solution with an energy [itex]E[/itex], then [itex]\overline{\psi}(\vec{x})=\psi(\lambda\vec{x})[/itex] is a solution with an energy [itex]\overline{E}=\lambda^2 E[/itex].

    Edit: I'm removing my conclusions because they contained a mistake related to the sign of the energy.
     
    Last edited: Mar 11, 2017
  6. Mar 11, 2017 #5

    hilbert2

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    But there seems to be nothing in the scaling property that wouldn't allow ##\lambda## to be an imaginary number, in which case you could make wavefunctions with arbitrarily large negative energies.

    EDIT: I guess something like ##V(r) \propto -e^{A/r}## would lead to falling-to-center behavior because there's an essential singularity, but I'm not sure.
     
  7. Mar 11, 2017 #6
    There is a lot of problem in using imaginary values for [itex]\lambda[/itex], and there is no reason to assume that you could change the sign of the energy by some change of variable.
     
  8. Mar 11, 2017 #7

    strangerep

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    But things get quite weird if you allow imaginary ##\lambda##. Position and momenta become imaginary. And time (which, afaict, must scale as ##\lambda^2## to preserve Hamilton's equations) would reverse.

    It might be interesting to explore the larger dynamical symmetry group for this problem. I've already worked through it for classical Kepler, so maybe I can adapt my computations.
     
  9. Mar 12, 2017 #8
    It is a common belief that localized states come with discrete spectrum with energy levels below the background flat potential, and non-localized states with a continuum spectrum with energy levels above the baground flat potential. If this [itex]\frac{1}{r^2}[/itex] potential debunks that belief as false, or gets close debunking it by introducing some subtleties, it would be quite interesting.

    I checked how far I could get by old fashioned technical PDE solving.

    [tex]
    \Big(-\frac{\hbar^2}{2m}\nabla^2 - \frac{A}{\|\vec{x}\|^2}\Big)\psi(\vec{x}) = E\psi(\vec{x})
    [/tex]

    If we substitute attempt [itex]\psi(\vec{x})=\xi(\|\vec{x}\|)[/itex], the PDE will be solved for [itex]\vec{x}\neq 0[/itex] when

    [tex]
    -\frac{\hbar^2}{2m}\Big(\frac{2}{r}\xi'(r) + \xi''(r)\Big) - \frac{A}{r^2}\xi(r) = E\xi(r)
    [/tex]

    is solved for [itex]r>0[/itex]. If we substitute [itex]\xi(r)=\frac{f(r)}{r}[/itex], then some terms cancel, and equation

    [tex]
    \frac{2}{r}\xi'(r) + \xi''(r) = \frac{f''(r)}{r}
    [/tex]

    turns out to be true. The differential equation for [itex]f[/itex] can be written as

    [tex]
    f''(r) = \Big(\alpha + \frac{\beta}{r^2}\Big)f(r)
    [/tex]

    The case [itex]E<0[/itex] is the same as [itex]\alpha>0[/itex], and [itex]A>0[/itex] is the same as [itex]\beta<0[/itex], so these cases are of the most interest.

    Anyone having any idea about the solutions for [itex]f[/itex]?
     
    Last edited: Mar 12, 2017
  10. Mar 12, 2017 #9
    As usual, the special cases [itex]\alpha=0[/itex] and [itex]\beta=0[/itex] are easier. The [itex]\beta=0[/itex] has no relevance, but [itex]\alpha=0[/itex] is equivalent to [itex]E=0[/itex], and there is nothing wrong with this energy, so we get few special solutions this way. Returning to the original notation in three dimensions, we can state the result in form that if we set

    [tex]
    \psi(\vec{x}) = \|\vec{x}\|^{-\frac{1}{2} \pm \sqrt{\frac{1}{4} - \frac{2mA}{\hbar^2}}}
    [/tex]

    then this will be a solution to the PDE with energy [itex]E=0[/itex].

    Unfortunately the scaling cannot be used to generate more solutions out of these, because the energy level remains constant, and the normalization factor will cancel changes to the wave function. These are some special solutions, anyway, and they are better than having nothing.
     
  11. Mar 12, 2017 #10

    hilbert2

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    Yeah, that's true. For some reason I had an idea that central force scattering states would have a ##\psi (r) \propto e^{ikr}## like behavior and you could make those normalizable with ##\mathbf{x} \rightarrow i\mathbf{x}##, but it's not like that after all.
     
  12. Mar 12, 2017 #11

    hilbert2

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  13. Mar 12, 2017 #12
    It is possible to guess and foresee how the eigenstates of this Hamiltonian are going to behave. You can compare them to the plane waves on a flat potential. The plane waves come with continuum energy levels, and they are not localized, but it is still possible to write localized wave packets out of them with continuous linear combinations (integrals). In the same way the eigenstates of this Hamiltonian are not going to be localized. It could be that the eigenstates will look a little bit localized, because they might approach zero at [itex]\|\vec{x}\|\to\infty[/itex], but actually they are going to approach zero so slowly, that they will not really be localized states. Despite these eigenstates not being localized, it will be possible to write localized wave packets out of them with continuous linear combinations (integrals). Then the time evolution of many of these wave packets is going to be such that they get sucked into the origin like into "a black hole".
     
  14. Mar 12, 2017 #13

    hilbert2

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    I'll probably try to do a Crank-Nicolson or diffusion monte carlo integration of the radial Schrödinger equation in imaginary time some day, using a Gaussian initial state ##\psi (r,t_0 ) = A\exp (-Br^2 )##, to see whether all the probability density collapses to the origin when ##s = it \rightarrow \infty##. Just need to use a potential ##\frac{1}{r^2 + \delta}## where the delta is some very small number, to prevent division by zero.
     
  15. Mar 12, 2017 #14

    hilbert2

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    Write "DSolve[D[f[r],r,r]==(a+b/r^2)*f[r],f,r]" in Wolfram Alpha or Mathematica, and you get a solution with BesselJ and BesselY functions in it. The BesselY functions are singular at the origin, so they're usually not valid forms of a wave function, but that could be an artifact of the falling-to-center property.
     
  16. Mar 12, 2017 #15

    Orodruin

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    First of all, let ##\hbar = 1##. The Hamiltonian is then of the form
    $$
    H = -\frac{1}{2m} \nabla^2 - \frac{A'}{r^2} \quad \Longrightarrow \quad 2m H = -\nabla^2 - \frac{A}{r^2}
    $$
    where ##A = 2mA'## has been introduced for brevity of notation. We are looking at the problem of finding eigenvalues to this operator, i.e., ##2m H \psi = \lambda \psi##. Using variable separation leads to ##\psi = R(r) Y_{\ell}^m (\theta, \varphi)##, where ##Y_{\ell}^m## is a spherical harmonic. Insertion into the eigenvalue equation then gives us
    $$
    - R''(r) - \frac{2R'(r)}{r} + \frac{\ell(\ell + 1)}{r^2} R(r) - \frac{A}{r^2} R(r) - \lambda R(r) = 0
    $$
    or, more compactly,
    $$
    - R''(r) - \frac{2R'(r)}{r} + \frac{A_\ell}{r^2} R(r) - \lambda R(r) = 0,
    $$
    where ##A_\ell = \ell(\ell + 1)-A##. Using the already mentioned substitution ##R(r) = f(r)/r^{1/2}##, we find that
    $$
    R'(r) = \frac{f'(r)}{r^{1/2}} - \frac{1}{2}\frac{f(r)}{r^{3/2}}, \quad
    R''(r)
    = \frac{f''(r)}{r^{1/2}} - \frac{f'(r)}{r^{3/2}} + \frac{3}{4} \frac{f(r)}{r^{5/2}}.
    $$
    Inserted into the differential equation for ##R(r)##, this leads to
    $$
    r^2 f''(r) - r f'(r) + \frac{3}{4} f(r) + 2r f'(r) - f(r) + A_\ell f(r) + \lambda r^2 f(r)
    =
    r^2 f''(r) + r f(r) - \underbrace{\left(A_\ell + \frac{1}{4}\right)}_{\equiv B^2} f(r) + \lambda r^2 f(r) = 0.
    $$
    For real ##B##, this is Bessel's differential equation with the general solution
    $$
    f(r) = c_1 J_B(\sqrt{\lambda} r) + c_2 Y_B(\sqrt{\lambda} r)
    $$
    for positive ##\lambda##. The Bessel functions of the second kind ##Y_B## are singular at the origin and we are left with ##f(r) \propto J_B(\sqrt{\lambda} r)##. Now, for negative ##\lambda##, we would instead have
    $$
    f(r) = c_1 I_B(\sqrt{-\lambda} r) + c_2 K_B(\sqrt{-\lambda} r).
    $$
    Here, ##I_B## grows unbounded as ##r\to \infty## and ##K_B## is singular at the origin, indicating that neither of these functions can be used to find normalised states. The conclusion is that there are no normalisable bound states as long as
    $$
    B^2 = A_\ell + \frac 14 = \ell(\ell + 1) - A + \frac 14 = \left(\ell + \frac 12\right)^2 - A \geq 0,
    $$
    which is always satisfied if ##A \leq 1/4##. I dare not say what happens when ##A > 1/4##.
     
  17. Mar 12, 2017 #16

    strangerep

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    Thanks for mentioning that! (I skimmed it a long time ago, but had completely forgotten about it.)

    Summarizing...

    It shows the importance of constructing a bona-fide self-adjoint quantum Hamiltonian on the entire domain of interest. Indeed, the superficial scaling property of this Hamiltonian leads to eigenstates with different energies, but which are not orthogonal -- so we know something is seriously wrong. (Eigenstates of a self-adjoint operator with distinct eigenvalues ought to be orthogonal.)

    The choice of a self-adjoint extension to ##H## at ##r=0## breaks the ordinary (continuous) scaling invariance, but some choices allow a discrete scaling symmetry.

    Apparently, in the quantum case, it "bounces" with a phase change at the singularity.
     
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