# How to show that $\lim_{r\to0} U(r)=0$ in the radial differential equation?

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?

• This is the DE of what exactly? – Gert Nov 6 '16 at 16:52
• 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). – user130529 Nov 6 '16 at 17:01
• @Gert it's the D.E. related to the particle in a central potential, the H atom, just the radial distance part. – user108787 Nov 6 '16 at 17:01
• 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. – Qmechanic Nov 6 '16 at 17:09
• 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. – 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,

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

Introducing

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

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${\displaystyle \gamma =2\nu \,}$. Derivation 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

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

• 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 – user108787 Nov 7 '16 at 19:15