We have the following differential equation:

$$\left({- \hbar ^2 \over 2 \mu } \frac{d^2}{dr^2} + {\hbar^2 \ell(\ell+1) \hbar^2 \over 2 \mu r^2} + V(r) \right)U(r)= EU(r)$$

in order to find the eigenvector $\psi (\mathbf{r})$ which it is the unique basis for $H$ , $L_z$ and $L$.

We know that the effective potential $${\hbar^2 \ell(\ell+1) \hbar^2 \over 2 \mu r^2}$$ represents the centrifugal force which pushes the particle outward. It is necessary to know the behaviour of $U(r)$ or $V(r)$ at the origin $r=0$.

In the lecture, they indicated $\lim_{r\to 0} U(r)=0$, but I can't find a way to get this result. Could you explain "what's going on", mathematically and physically? By physically I mean that when the radius is too small the radial function vanishes; what's behind this result?

  • $\begingroup$ This is the DE of what exactly? $\endgroup$ – Gert Nov 6 '16 at 16:52
  • $\begingroup$ In radial PDEs, it is common to have both bounded and unbounded solutions at the origin. The condition $\lim_{r \to 0}U(r)=0$ is likely telling you to choose the bounded solution at the origin (hence it is not a result but an assumption, or, if you prefer, a boundary condition). $\endgroup$ – user130529 Nov 6 '16 at 17:01
  • $\begingroup$ @Gert it's the D.E. related to the particle in a central potential, the H atom, just the radial distance part. $\endgroup$ – user108787 Nov 6 '16 at 17:01
  • $\begingroup$ Essentially a duplicate of physics.stackexchange.com/q/183045/2451 & physics.stackexchange.com/q/134719/2451 The reason is the same: The boundary condition at $r=0$ follows because the wave function should be normalizable, and the kinetic energy finite. $\endgroup$ – Qmechanic Nov 6 '16 at 17:09
  • $\begingroup$ The general solution to the horror above is a series solution, it's the Laguerre polynomials, but to get to what you have in your post to the end involves a few subsitutions, which should be in your text book. By physically I mean that when the radius is too small the radial function vanishes; what's behind this result? One of us is lost here, sorry, physically what do expect to happen if the function goes to 0 as r goes to 0? I might be misunderstanding you, but I would read the textbook line by line. Maybe redo the derivation, keeping very careful track of the substitutions. $\endgroup$ – user108787 Nov 6 '16 at 17:30

This derivation utilises the idea of 3D isotropic harmonic oscillator, and is not simply based, but is a total line by line replication of part of this article Wikipedia 3D isotropic harmonic oscillator.

The article above linked above is a page I would urge the OP to read, also I would encourage the OP to watch and take notes on the relevant YouTube videos by J. Binney (Oxford) and L. Susskind (Stanford).

The potential of a 3D isotropic harmonic oscillator is

${\displaystyle V(r)={\frac {1}{2}}m_{0}\omega ^{2}r^{2}.}$

In this article it is shown that an N-dimensional isotropic harmonic oscillator has the energies

$${\displaystyle E_{n}=\hbar \omega {\Bigl (}n+{\frac {N}{2}}{\Bigr )}\quad {\text{with}}\quad n=0,1,\ldots ,\infty ,}$$

i.e., n is a non-negative integral number; $ω$ is the (same) fundamental frequency of the $N$ modes of the oscillator. In this case $N= 3$, so that the radial Schrödinger equation becomes,

$${\displaystyle \left[-{\hbar ^{2} \over 2m_{0}}{d^{2} \over dr^{2}}+{\hbar ^{2}l(l+1) \over 2m_{0}r^{2}}+{\frac {1}{2}}m_{0}\omega ^{2}r^{2}-\hbar \omega {\bigl (}n+{\tfrac {3}{2}}{\bigr )}\right]u(r)=0.}$$


$${\displaystyle \gamma \equiv {\frac {m_{0}\omega }{\hbar }}}$$

and recalling that ${\displaystyle u(r)=rR(r)\,}$, we will show that the radial Schrödinger equation has the normalized solution,

$${\displaystyle R_{n,l}(r)=N_{nl}\,r^{l}\,e^{-{\frac {1}{2}}\gamma r^{2}}\;L_{{\frac {1}{2}}(n-l)}^{(l+{\frac {1}{2}})}(\gamma r^{2}),}$$

where the function ${\displaystyle L_{k}^{(\alpha )}(\gamma r^{2})}$ is a generalized Laguerre polynomial in $γr^2$ of order $k$ (i.e., the highest power of the polynomial is proportional to $γkr^2k$).

The normalization constant $N_{nl} is,

$${\displaystyle N_{nl}=\left[{\frac {2^{n+l+2}\,\gamma ^{l+{\frac {3}{2}}}}{\pi ^{\frac {1}{2}}}}\right]^{\frac {1}{2}}\left[{\frac {[{\frac {1}{2}}(n-l)]!\;[{\frac {1}{2}}(n+l)]!}{(n+l+1)!}}\right]^{\frac {1}{2}}.}$$

The eigenfunction $R_n,l(r)$ belongs to energy $E_n$ and is to be multiplied by the spherical harmonic ${\displaystyle Y_{lm}(\theta ,\phi )\,}$, where

$${\displaystyle l=n,n-2,\ldots ,l_{\min }\quad {\hbox{with}}\quad l_{\min }={\begin{cases}1&\mathrm {if} \;n\;\mathrm {odd} \\0&\mathrm {if} \;n\;\mathrm {even} \end{cases}}}$$

This is the same result as given in this article if we realize that ${\displaystyle \gamma =2\nu \,}$.


First we transform the radial equation by a few successive substitutions to the generalized Laguerre differential equation, which has known solutions: the generalized Laguerre functions. Then we normalize the generalized Laguerre functions to unity. This normalization is with the usual volume element $r^2 dr$.

First we scale the radial coordinate

$${\displaystyle y={\sqrt {\gamma }}r\quad {\hbox{with}}\quad \gamma \equiv {\frac {m_{0}\omega }{\hbar }},}$$

and then the equation becomes

$${\displaystyle \left[{d^{2} \over dy^{2}}-{l(l+1) \over y^{2}}-y^{2}+2n+3\right]v(y)=0}$$

with $${\displaystyle v(y)=u\left(y/{\sqrt {\gamma }}\right)}$$

Consideration of the limiting behaviour of $v(y)$ at the origin and at infinity suggests the following substitution for $v(y)$,

$${\displaystyle v(y)=y^{l+1}e^{-y^{2}/2}f(y).}$$

This substitution transforms the differential equation to

$${\displaystyle \left[{d^{2} \over dy^{2}}+2\left({\frac {l+1}{y}}-y\right){\frac {d}{dy}}+2n-2l\right]f(y)=0,}$$

where we divided through with ${\displaystyle y^{l+1}e^{-y^{2}/2}}$, which can be done so long as $y$ is not zero.

Transformation to Laguerre polynomials

If the substitution ${\displaystyle x=y^{2}\,}$ is used, ${\displaystyle y={\sqrt {x}}}$, and the differential operators become

$${\displaystyle {\frac {d}{dy}}={\frac {dx}{dy}}{\frac {d}{dx}}=2y{\frac {d}{dx}}=2{\sqrt {x}}{\frac {d}{dx}},{\text{ and }}}{\displaystyle {\frac {d^{2}}{dy^{2}}}={\frac {d}{dy}}\left(2y{\frac {d}{dx}}\right)=4x{\frac {d^{2}}{dx^{2}}}+2{\frac {d}{dx}}.}$$

The expression between the square brackets multiplying $f(y)$ becomes the differential equation characterizing the generalized Laguerre equation :

$${\displaystyle x{\frac {d^{2}g}{dx^{2}}}+{\Big (}(l+{\frac {1}{2}})+1-x{\Big )}{\frac {dg}{dx}}+{\frac {1}{2}}(n-l)g(x)=0}$$

with ${\displaystyle g(x)\equiv f({\sqrt {x}})\,\;}$.

Provided ${\displaystyle k\equiv (n-l)/2\,}$ is a non-negative integral number, the solutions of this equations are generalized (associated) Laguerre polynomials

$${\displaystyle g(x)=L_{k}^{(l+{\frac {1}{2}})}(x).}$$

From the conditions on $k$ follows: (i) ${\displaystyle n\geq l\,}$ and (ii) n and l are either both odd or both even. This leads to the condition on $l$ given above.

Recovery of the normalized radial wavefunction

Remembering that ${\displaystyle u(r)=rR(r)\,}$, we get the normalized radial solution

$${\displaystyle R_{n,l}(r)=N_{nl}\,r^{l}\,e^{-{\frac {1}{2}}\gamma r^{2}}\;L_{{\frac {1}{2}}(n-l)}^{(l+{\frac {1}{2}})}(\gamma r^{2}).}$$

The normalization condition for the radial wavefunction is

$${\displaystyle \int _{0}^{\infty }r^{2}|R(r)|^{2}\,dr=1.}$$

Substituting ${\displaystyle q=\gamma r^{2}\,\;}$, gives ${\displaystyle dq=2\gamma r\,dr\,\;}$ and the equation becomes

$${\displaystyle {\frac {N_{nl}^{2}}{2\gamma ^{l+{3 \over 2}}}}\int _{0}^{\infty }q^{l+{1 \over 2}}e^{-q}\left[L_{{\frac {1}{2}}(n-l)}^{(l+{\frac {1}{2}})}(q)\right]^{2}\,dq=1.}$$

By making use of the orthogonality properties of the generalized Laguerre polynomials, this equation simplifies to

$${\displaystyle {\frac {N_{nl}^{2}}{2\gamma ^{l+{3 \over 2}}}}\cdot {\frac {\Gamma [{\frac {1}{2}}(n+l+1)+1]}{[{\frac {1}{2}}(n-l)]!}}=1.}$$

Hence, the normalization constant can be expressed as

$${\displaystyle N_{nl}={\sqrt {\frac {2\,\gamma ^{l+{3 \over 2}}\,({\frac {n-l}{2}})!}{\Gamma ({\frac {n+l}{2}}+{\frac {3}{2}})}}}}$$

Other forms of the normalization constant can be derived by using properties of the gamma function, while noting that n and lare both of the same parity. This means that $n + l$ is always even, so that the gamma function becomes

$${\displaystyle \Gamma \left[{1 \over 2}+\left({\frac {n+l}{2}}+1\right)\right]={\frac {{\sqrt {\pi }}(n+l+1)!!}{2^{{\frac {n+l}{2}}+1}}}={\frac {{\sqrt {\pi }}(n+l+1)!}{2^{n+l+1}[{\frac {1}{2}}(n+l)]!}},}$$

where we used the definition of the double factorial. Hence, the normalization constant is also given by

$${\displaystyle N_{nl}=\left[{\frac {2^{n+l+2}\,\gamma ^{l+{3 \over 2}}\,[{1 \over 2}(n-l)]!\;[{1 \over 2}(n+l)]!}{\;\pi ^{1 \over 2}(n+l+1)!}}\right]^{1 \over 2}={\sqrt {2}}\left({\frac {\gamma }{\pi }}\right)^{1 \over 4}\,({2\gamma })^{\ell \over 2}\,{\sqrt {\frac {2\gamma (n-l)!!}{(n+l+1)!!}}}.}$$

  • $\begingroup$ No problem. There is 1 very good book that goes through 1st and 2nd year q.m material, it's schaumms and it's just worked example after worked example (so it's cheap : ). He doesn't teach much, he just shows you. Anyway, best of luck with your studies $\endgroup$ – user108787 Nov 7 '16 at 19:15

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