The particle of mass $m$ moves in potential $$V(x) = \dfrac{\alpha \left( \left( 2 \alpha +1 \right)x^2-a^2 \right)}{m \left( a^2 + x^2\right)^2},$$ and $\alpha > 1/4$.

Find the energy and the wave function of the ground state of the system. How many related states exist for a given system? And what happens when $0<\alpha \leq1/4$.

I do not know what to do with such an equation. Maybe there are some other ways to find the ground state? But then how to find the number of bound states? I'm completely at a loss.

With the following replacement $z = x^2 + a^2$ the equation turns into

$$ 2\Psi_{zz} (z -a^2) + \Psi_z + \left( mE - \dfrac{\alpha \left( \left( 2 \alpha +1 \right)z-a^2 2 (\alpha + 1) \right)}{z^2} \right) \Psi = 0.$$

for $x \rightarrow \infty$, our equation becomes simply

$$ \Psi_{xx} + \left(2mE - 2\alpha \left( 2\alpha +1 \right) x^{-2} \right) \Psi = 0 $$ Solution for this

$$ \Psi (x) \rightarrow C_1 \sqrt{x} J_{ \frac{1}{2} \sqrt{4 B + 1}} \left( \sqrt{A} x \right) + C_2 \sqrt{x} Y_{ \frac{1}{2} \sqrt{4 B + 1}} \left( \sqrt{A} x \right), $$ where $B = 2\alpha \left( 2\alpha +1 \right) $ and $ A = 2 m E$. $J, Y$ is bessel function of first and second kind. For $ x \rightarrow 0 $. The Bessel functions of the second kind tend to $-\infty$, so $C_2 = 0$.

For $x \rightarrow 0$ $$ \Psi (x) \rightarrow C_1 e^{iA^{1/2}x} + C_2 e^{-iA^{1/2}x},$$ where $A = 2mE + 2 \alpha^2 a^{-2}$.

But what do these asymptotics give us?

  • $\begingroup$ Have you tried just solving the equation? $\endgroup$ May 20 '18 at 16:26
  • $\begingroup$ @ZeroTheHero, I tried to make various substitutions, but this equation does not turn into something good. Although it is solved for $a = 0$. And the solutions are the spherical functions of Bessel. $\endgroup$
    – Ann
    May 20 '18 at 16:28
  • 2
    $\begingroup$ maybe then you could expand your question to include some of the steps you took in searching for a solution. As it stands now you're just asking people to do the work for you. $\endgroup$ May 20 '18 at 16:30
  • $\begingroup$ @ZeroTheHero. But they are also defined only for positive values of x. $\endgroup$
    – Ann
    May 20 '18 at 16:31
  • 1
    $\begingroup$ Hint: find $A,\,B$ such that $V(x)=A+\frac{B}{a^2+x^2}$. $\endgroup$
    – J.G.
    May 20 '18 at 16:32

I have never seen this potential and would like to know if there is any physical context to it, or if it's just a random potential known to have a solution.

I don't know what the answer is but I would recommend you proceed in parallel to the way the solution is found in the hydrogen atom.

If you go back to the way the Schrodinger equation is solved, one interpolates between the large $x$ and small $x$ behaviour using an unknown function. In the case of the hydrogen atom, for instance, one uses $$ \chi(r/a)= (r/a)^{\ell+1}L(r/a) e^{-\sqrt{\nu}(r/a)} \tag{1} $$ where $(r/a)^{\ell+1}$ captures the small $r$ behaviour, and $ e^{-\sqrt{\nu}(r/a)}$ captures the large $r$ behaviour of the differential equation. Plugging (1) into the Schrodinger equation yields a differential equation for $L$ which is then solved.

The key observation here is that, for the ground state, the function $L$ is actually constant, i.e. the ground state wave function $\chi_0$ is basically $\chi(r/a)= (r/a)^{\ell+1} e^{-\sqrt{\nu}(r/a)} $.

As a result I would recommend you find the equivalent to Eq.(1) for your problem and try to set $L=1$ again (you might want to review the hydrogen case or any similar problem with a hard boundary at $x=0$). If guess function is indeed a solution you should get the ground state energy as a result of algebraic cancellations.

Edit: Your potential for $a=1, \alpha=1, m=1$ looks like enter image description here

so you may expect that the ground state solution will have no nodes, i.e. will not cross the axis; thus the interpolating function for the ground state will probably be constant. Maybe reviewing the solution to the 3d harmonic oscillator in spherical or cylindrical coordinates would be more physically appropriate than the hydrogen atom.


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