I'm currently self-studying quantum mechanics and have encountered a challenge regarding higher angular momentum wave functions $\phi(r)$ on whether the corresponding Schrödinger equation has a bound state or not. I've searched through textbooks and online resources, but I'm still confused.
To illustrate my question, I'm considering a 3D Hamiltonian with a simple negative radial potential. I'm using the reduced form of the Schrödinger equation where $u(r)= r \phi(r)$: $$-u^{\prime \prime}(r) + \frac{\ell(\ell+1)}{r^2} u(r) +V(r)u(r) = Eu(r) \, .$$ Here, the nonnegative integer $l$ represents the angular momentum and also,the potential $V$ is given by \begin{equation} \label{mixpot} V(r) = \left\{ \begin{aligned} & 0 \;\;\; &\text{if} \quad & 0\leq r \leq r_0\\ -&\frac{c}{r^a} &\text{if} \quad & r>r_0 \end{aligned}\;,\right. \end{equation} where $a\geq 2$. For instance, consider $\ell=1$, i.e. a $p$-wave. Then, we will have a strong repulsive (positive) potential $\frac{2}{r^2}$ near to the origin and a negtive potential decaying to zero at infinity. (The situation is somehow resembles to the Lennard-Jones potential.)
Given this situation, if naïvely we consider $E=-k^2$ as a candidate for a bound state, something peculiar will happen. The solution of ODE before the point $r_0$ with a natural Dirichlet boundary condition $u(0)=0$ will be: $$ u(r)=\frac{\sqrt{i}\sqrt{\frac{2}{\pi}} c_1(k r \cosh (k r)-\sinh (k r))}{k r \sqrt{ k }} \, , $$ where $c_1$ is an arbitrary constant. As it is clear, it has a real and complex value. For the rest $r > r_0$, $u(r)$ will be only a real function. (For example in the case $a=2$, it becomes a Bessel ODE.)
My question is: How should I match the solution at $r=r_0$? Should I set the complex part to zero, implying that the solution before $r_0$ implying that the only admissible solution is zero? (Because we have to put $c_1 = 0$) Or, was my initial assumption of $E=-k^2$ incorrect and I should consider $E=k^2$, i.e. a resonance?
I'm trying to understand the intuition behind this. It appears that a positive potential doesn't yield a bound state, but when it becomes negative at some point, it can. This situation is similar to the Lennard-Jones potential. Does the Lennard-Jones potential produce a bound state, and if so, how? I'd greatly appreciate any help and insights on this matter!

  • $\begingroup$ I do not understand how the quantity $k$ appears in the derivations. How does it relate to energy? Do you substitute something into the radial Schrod. eq? $\endgroup$ Nov 3, 2023 at 14:36
  • $\begingroup$ To develop some intuition, it seems natural to consider "3D quantum well": $V(r)=-V_0$ if $r<a$ and $V(r)=0$ if $r>a$. In 1D bound state alwas exists (Dirac delta potential is the simplest illustration), in 2D bound state has exponentially small energy, in 3D the situation is more complicated. $\endgroup$ Nov 3, 2023 at 14:38
  • $\begingroup$ I'm not sure I understand why you think matching the boundary condition forces $c_1=0$. One can absorb the $\sqrt{i}$ into the constant $c_1$, making a new constant $\tilde{c}_1$, which can be a real number. Can you clarify? $\endgroup$
    – march
    Nov 3, 2023 at 15:36
  • $\begingroup$ @ArtemAlexandrov Thanks for your comment. First, $E= -k^2$ to have a negative energy and then, get a bound state. Second, if we have a negative potentail for $r<a$, of course we easily have a bound state; like every other negative potentail. The whole point is absence of any negative potential for small $r$. $\endgroup$
    – Re_Born
    Nov 3, 2023 at 16:11
  • 1
    $\begingroup$ @Re_Born , 1) solve Schrod eq. in domain $r<r_0$, the solution is linear combination of 1st & 2nd kind Bessel functions (I mean $J_n$ and $Y_n$ functions), 2) solve Schrod equation in domain $r>r_0$. In domain $r<r_0$ you have 2 contants to be determined. The first constant is 0, because wavefunction should be finite at $r=0$. The second constant is defined by matching wavefunctions + their derivatives in $r<r_0$ and $r>r_0$ domain. I can not obtain the mentioned in your answer solution. $\endgroup$ Nov 4, 2023 at 15:30

1 Answer 1


First: check the discussion of bound state here

We have a quantum particle in the potential $$V(r)=\begin{cases}0,\quad r<r_0, \\ -c/r^a,\quad r>r_0\end{cases}$$ with $a\geq 2$. Such potential admits the separation of variables, so we use the ansatz $f(r)=ru(r)$ for the radial part of Schrodinger equation, $$-\frac{\hbar^2}{2m}\frac{d^2}{dr^2}\left(ru\right)+\left(\frac{l(l+1)}{2mr^2}+V(r)\right)(ru)=ru(r)E.$$ Denoting $k^2=2mE/\hbar^2$, we find $$\frac{d^2u}{dr^2}+\frac{2}{r}\frac{du}{dr}+\left(k^2-\frac{l(l+1)}{r^2}-V_{\text{eff}}(r)\right)u(r)=0,$$ where we denote $V_{\text{eff}}=2mV(r)/\hbar^2$. In this equation we perform change of variables, $z=kr$, so $$\frac{d^2u}{dr^2}=k^2\frac{d^2u}{dz^2},$$ which gives us $$k^2\frac{d^2u}{dz^2}+\frac{2k}{z}\frac{du}{dz}+\left(1-\frac{l(l+1)}{z^2}+\frac{V_{\text{eff}}}{z^2}\right)u=0.$$ Now we solve in the domain $r<r_0$, where $V_{\text{eff}}\equiv 0$. The solution consists of two spherical Bessel functions, $$u(r)=Aj_l(z)+By_l(z).$$ It is known that the function $y_l(z)$ is divergent at zero, so $\boxed{u_{<}(z)=j_l(z)}$. Now we solve at $r>r_0$ domain. The equation becomes $$\frac{d^2u}{dr^2}+\frac{2}{r}\frac{du}{dr}+\left(k^2-\frac{l(l+1)}{r^2}+\frac{c}{r^a}\right)u=0,$$ where we absorb some constants into the definition of $c$. For simplicity, let us consider the special case $a=2$. In this case, we still have spherical Bessel functions, $$u_{>}(z)=Cj_n(z)+By_n(z),\quad n=\frac{1}{2}\left(-1+\sqrt{(2l+1)^2-4c}\right).$$ Now we match functions $u_{<}(z)$ and $u_{>}(z)$ at $r=r_0$, $$u_{>}(kr_0)=u_{<}(kr_0),$$ $$\left.u_{>}'(kr)\right|_{r_0}=\left.u_{<}'(kr)\right|_{r_0}.$$ The first equation gives us $$j_l(z_0)=Aj_n(z_0)+By_n(z_0),$$ the second equation gives us $$\frac{lj_l(z_0)}{z_0}-j_{l+1}(z_0)=\frac{n(Aj_n(z_0)+By_n(z_0))}{z}-(Aj_{n+1}(z_0)+By_{n+1}(z_0))$$ where we denote $z_0\equiv kr_0$. It the system of two linear equations for two variables, $A$ and $B$. After some tedious derivations, we find $$A=\frac{j_l(z_0)y_n(z_0)-nj_l(z_0)y_n(z_0)-z_0j_l(z_0)y_{n+1}(z_0)-z_0j_{l+1}(z_0)y_n(z_0)}{z_0j_n(z_0)y_{n+1}(z_0)-z_0j_{n+1}(z_0)y_n(z_0)},$$ $$B=-\frac{lj_l(z_0)j_{n}(z_0)-nj_l(z_0)j_n(z_0)+z_0j_l(z_0)j_{n+1}(z_0)-z_0j_{l+1}(z_0)j_{n}(z_0)}{z_0j_n(z_0)y_{n+1}(z_0)-z_0j_{n+1}(z_0)y_n(z_0)}.$$ Having defined the coefficients, we take the equation $j_l(z_0)=Aj_n(z_0)+By_n(z_0)$ and set $l=1$. It gives $$j_l(z_0)=\frac{\sin z_0}{z_0^2}-\frac{\cos z_0}{z_0}.$$ At this stage I can not add more. In my view, you should solve the equation $j_l(z_0)=Aj_n(z_0)+By_n(z_0)$ for $z_0$, i.e. for $k$.


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