The general Schroedinger Equation is: $$\left[-\frac{\hbar^2}{2m}\triangle +V(r,\vartheta,\varphi)\right]\psi_{nlm}=E\psi_{nlm}$$ When considering free waves, i.e. $V(r,\vartheta,\varphi)=0$ and a separation ansatz $\psi_{nlm}=R_{nl}(r)\cdot Y_{lm}(\vartheta,\varphi)$, one can simplify this to the Radial Equation: $$R_{nl}^{\prime\prime}+\frac{2}{r}R_{nl}^{\prime}-\frac{l(l+1)}{r^2}R_{nl}=-k^2R_{nl}$$ with $k\equiv \frac{\sqrt{2mE}}{\hbar}$.

On the one hand, the solution to this is usually given as: $$R_{nl}(r)=\left[\frac{(n-l-1)!(2k)^3}{2n((n+l)!)^3}\right]^{\frac{1}{2}}(2k r)^l e^{-k r}L_{n+l}^{2l+1}(2k r)$$ with the Laguerre Polynoms $$L_r^s(x)=\frac{d^s}{dx^s}e^x \frac{d^r}{dx^r}e^{-x}x^r$$

On the other hand, now with $\rho\equiv k\cdot r$ this becomes the Bessel Differential Equation: $$\left[\rho^2 \partial^2_{\rho}+2\rho\partial_{\rho}-l(l+1)+\rho^2\right]R_{nl}(r)=0$$ with the solution: $$R_{nl}(r)=A\cdot j_l(\rho)$$ The $j_l$ are the spherical Bessel functions and $A$ is a constant.

Are these two solutions for $R_{nl}(r)$ equivalent? If yes, how can I show this?

I tried with the following representation of the spherical Bessel function, but no success. $$j_l(\rho)=\sqrt{\pi}\sum_{s=0}^{\infty}\frac{(-1)^s\cdot \rho^{2s+l}}{s!\cdot(s+l+\frac{1}{2})!\cdot2^{2s+l+1}}$$

Even for the simplest case $n=1,l=0$, I have on the one hand: $$R_{10}(r)\propto e^{-r}$$ and from the Bessel solution: $$R_{10}(r)\propto \frac{\sin kr}{kr}$$

What am I doing wrong?


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  • 1
    $\begingroup$ I only remember seeing laguerre polynomials when there is a coulomb potential. Are you sure the radial wavefunctions can be expressed in terms of laguerre polynomials when there is no coulomb potential? I would think the spherical bessel functions make more sense, but I haven't done the math. For example, I wouldn't expect the exponential decay in $r$ (as seen in the laguerre solution) if there is no confining potential. $\endgroup$ – Brian Moths Jun 29 '15 at 19:52
  • $\begingroup$ oh right, the solution with Laguerre Polynoms is only for a Coulomb potential.. and why should free waves decay exponentially?^^ this explains why I was confused $\endgroup$ – Andy Jun 29 '15 at 20:10
  • $\begingroup$ What do you mean? You know what the wave function means, right? and it's not really a wave function, and I presume that something in your solution is wrong. $\endgroup$ – iharob Jun 29 '15 at 20:13
  • $\begingroup$ You can only separate if the potential is a purely radial potential, while the potential in your first equation is completely general, it's merely written in spherical coordinates. There is no more a general solution to the Schroedinger equation than there is to Newton's equation. $\endgroup$ – CuriousOne Jun 29 '15 at 20:45

With this transformation $\rho = kr$ you can't possibly change the form of the equation, because it's a scale transformation that does nothing special to the derivatives, so I think that you have computed the derivatives wrong.

  • $\begingroup$ So, the first $R_{10}(r)\propto e^{-r}$ is wrong? But e.g. here this is state: quantummechanics.ucsd.edu/ph130a/130_notes/node233.html .. or is the Bessel solution wrong? $\endgroup$ – Andy Jun 29 '15 at 19:55
  • $\begingroup$ I think you are confusing the Schroedinger equation for spherical symetry, with the Hydrohen Atom solution to the Schroedinger equation, the difference is $V(r)$. $\endgroup$ – iharob Jun 29 '15 at 20:00
  • $\begingroup$ ah.. so the solution with Laguerre Polynoms is only valid for a radial potential? $\endgroup$ – Andy Jun 29 '15 at 20:07
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    $\begingroup$ Not necessarily, but certainly for a coulomb potential of the type $\frac{k}{r}$, where $k$ is a constant. Do you want a fun radial potential for which the solution is not Laguerre polynomials, read Yukawa's Potential. $\endgroup$ – iharob Jun 29 '15 at 20:08
  • $\begingroup$ well yes! There seem to be only approximate solutions possible: arxiv.org/abs/1210.5886 $\endgroup$ – Andy Jun 29 '15 at 20:26

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