# Solution of $p$-wave time independent Schrödinger equation with a simple negative potential

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 \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. 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!

• 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? Nov 3, 2023 at 14:36
• 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. Nov 3, 2023 at 14:38
• 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? Nov 3, 2023 at 15:36
• @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 $r$. Nov 3, 2023 at 16:11
• @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. Nov 4, 2023 at 15:30

First: check the discussion of bound state here

We have a quantum particle in the potential $$V(r)=\begin{cases}0,\quad rr_0\end{cases}$$ 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 $$r, where $$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$$.