D.15 The hy­dro­gen ra­dial wave func­tions

This will be child’s play for har­monic os­cil­la­tor, {D.12}, and spher­i­cal har­mon­ics, {D.14}, vet­er­ans. If you re­place the an­gu­lar terms in (4.33) by $l(l+1)\hbar^2$, and then di­vide the en­tire equa­tion by $\hbar^2$, you get

\begin{displaymath}
- \frac{1}{R} \frac{{\rm d}}{{\rm d}r}\left(r^2\frac{{\rm d...
...}{4\pi\epsilon_0\hbar^2} r
= \frac{2m_{\rm e}}{\hbar^2} r^2 E
\end{displaymath}

Since $l(l+1)$ is nondi­men­sion­al, all terms in this equa­tion must be. In par­tic­u­lar, the ra­tio in the third term must be the rec­i­p­ro­cal of a con­stant with the di­men­sions of length; so, de­fine the con­stant to be the Bohr ra­dius $a_0$. It is con­ve­nient to also de­fine a cor­re­spond­ingly nondi­men­sion­al­ized ra­dial co­or­di­nate as $\rho$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r$$\raisebox{.5pt}{$/$}$$a_0$. The fi­nal term in the equa­tion must be nondi­men­sion­al too, and that means that the en­ergy $E$ must take the form $(\hbar^2/2{m_{\rm e}}a_0^2)\epsilon$, where $\epsilon$ is a nondi­men­sion­al en­ergy. In terms of these scaled co­or­di­nates you get

\begin{displaymath}
- \frac{1}{R} \frac{{\rm d}}{{\rm d}\rho}\left(\rho^2\frac{...
...}R}{{\rm d}\rho}\right)
+ l(l+1)
- 2 \rho
= \rho^2 \epsilon
\end{displaymath}

or writ­ten out

\begin{displaymath}
- \rho^2 R'' - 2\rho R' +[l(l+1)-2\rho-\epsilon\rho^2]R = 0
\end{displaymath}

where the primes de­note de­riv­a­tives with re­spect to $\rho$.

Sim­i­lar to the case of the har­monic os­cil­la­tor, you must have so­lu­tions that be­come zero at large dis­tances $\rho$ from the nu­cleus: $\int\vert\psi\vert^2{\,\rm d}^3{\skew0\vec r}$ gives the prob­a­bil­ity of find­ing the par­ti­cle in­te­grated over all pos­si­ble po­si­tions, and if $\psi$ does not be­come zero suf­fi­ciently rapidly at large $\rho$, this in­te­gral would be­come in­fi­nite, rather than one (cer­tainty) as it should. Now the ODE above be­comes for large $\rho$ ap­prox­i­mately $R''+\epsilon{R}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, which has so­lu­tions of the rough form $\cos(\sqrt{\epsilon}\rho+\alpha)$ for pos­i­tive $\epsilon$ that do not have the re­quired de­cay to zero. Zero scaled en­ergy $\epsilon$ is still too much, as can be checked by solv­ing in terms of Bessel func­tions, so you must have that $\epsilon$ is neg­a­tive. In clas­si­cal terms, the earth can only hold onto the moon since the moon’s to­tal en­ergy is less than the po­ten­tial en­ergy far from the earth; if it was not, the moon would es­cape.

Any­way, for bound states, you must have the scaled en­ergy $\epsilon$ neg­a­tive. In that case, the so­lu­tion at large $\rho$ takes the ap­prox­i­mate form $R$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ $e^{\pm\sqrt{-\epsilon}\rho}$. Only the neg­a­tive sign is ac­cept­able. You can make things a lot eas­ier for your­self if you peek at the fi­nal so­lu­tion and rewrite $\epsilon$ as be­ing $\vphantom{0}\raisebox{1.5pt}{$-$}$1$\raisebox{.5pt}{$/$}$$n^2$ (that is not re­ally cheat­ing, since you are not at this time claim­ing that $n$ is an in­te­ger, just a pos­i­tive num­ber.) In that case, the ac­cept­able ex­po­nen­tial be­hav­ior at large dis­tance takes the form $e^{-\frac12\xi}$ where $\xi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $2\rho$$\raisebox{.5pt}{$/$}$$n$. Split off this ex­po­nen­tial part by writ­ing $R$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e^{-\frac12\xi}\overline{R}$ where $\overline{R}(\xi)$ must re­main bounded at large $\xi$. Sub­sti­tut­ing these new vari­ables, the ODE be­comes

\begin{displaymath}
-\xi^2\overline{R}\,\strut''
+\xi(\xi-2)\overline{R}\,\strut'
+[l(l+1)-(n-1)\xi]\overline{R}=0
\end{displaymath}

where the primes in­di­cate de­riv­a­tives with re­spect to $\xi$.

If you do a power se­ries so­lu­tion of this ODE, you see that it must start with ei­ther power $\xi^l$ or with power $\xi^{-l-1}$. The lat­ter is not ac­cept­able, since it would cor­re­spond to an in­fi­nite ex­pec­ta­tion value of en­ergy. You could now ex­pand the so­lu­tion fur­ther in pow­ers of $\xi$, but the prob­lem is that tab­u­lated poly­no­mi­als usu­ally do not start with a power $l$ but with power zero or one. So you would not eas­ily rec­og­nize the poly­no­mial you get. There­fore it is best to split off the lead­ing power by defin­ing $\overline{R}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\xi^l\overline{\overline{R}}$, which turns the ODE into

\begin{displaymath}
\xi \overline{\overline{R}}\,\strut''
+ [2(l+1) - \xi]\ove...
...e{\overline{R}}\,\strut'
+ [n-l-1]\overline{\overline{R}} = 0
\end{displaymath}

Sub­sti­tut­ing in a power se­ries $\overline{\overline{R}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sum{c}_p\xi^p$, you get

\begin{displaymath}
\sum p[p+2l+1]c_p\xi^{p-1} = \sum [p+l+1-n]c_p\xi^p
\end{displaymath}

The ac­cept­able low­est power $p$ of $\xi$ is now zero. Again the se­ries must ter­mi­nate, oth­er­wise the so­lu­tion would be­have as $e^{\xi}$ at large dis­tance, which is un­ac­cept­able. Ter­mi­na­tion at a high­est power $p$ $\vphantom0\raisebox{1.5pt}{$=$}$ $q$ re­quires that $n$ equals $q+l+1$. Since $q$ and $l$ are in­te­gers, so must be $n$, and since the fi­nal power $q$ is at least zero, $n$ is at least $l+1$. The cor­rect scaled en­ergy $\epsilon$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$1$\raisebox{.5pt}{$/$}$$n^2$ with $n$ $\raisebox{.3pt}{$>$}$ $l$ has been ob­tained.

With $n$ iden­ti­fied, you can iden­tify the ODE as La­guerre's as­so­ci­ated dif­fer­en­tial equa­tion, e.g. [40, 30.26], the $(2l+1)$-​th de­riv­a­tive of La­guerre's dif­fer­en­tial equa­tion, e.g. [40, 30.1], and the poly­no­mial so­lu­tions as the as­so­ci­ated La­guerre poly­no­mi­als $L_{n+l}^{2l+1}$, e.g. [40, 30.27], the $(2l+1)$-​th de­riv­a­tives of the La­guerre's poly­no­mi­als $L_{n+l}$, e.g. [40, 30.2]. To nor­mal­ize the wave func­tion use an in­te­gral from a ta­ble book, e.g. [40, 30.46].

Putting it all to­gether, the generic ex­pres­sion for hy­dro­gen eigen­func­tions are, drums please:

\begin{displaymath}
\psi_{nlm} = -\frac{2}{n^2}
\sqrt{\frac{(n-l-1)!}{[(n+l)!a...
...}\left(\frac{2\rho}n\right)
e^{-\rho/n}
Y_l^m(\theta,\phi) %
\end{displaymath} (D.8)

The prop­er­ties of the as­so­ci­ated La­guerre poly­no­mi­als $L_{n+l}^{2l+1}(2\rho/n)$ are in ta­ble books like [40, pp. 169-172], and the spher­i­cal har­mon­ics were given ear­lier in chap­ter 4.2.3 and in de­riva­tion {D.14}, (D.5).

Do keep in mind that dif­fer­ent ref­er­ences have con­tra­dic­tory de­f­i­n­i­tions of the as­so­ci­ated La­guerre poly­no­mi­als. This book fol­lows the no­ta­tions of [40, pp. 169-172], who de­fine

\begin{displaymath}
L_n(x)=e^x\frac{{\rm d}^n}{{\rm d}x^n}\left(x^n e^{-x}\right),
\quad
L_n^m=\frac{{\rm d}^m}{{\rm d}x^m} L_n(x).
\end{displaymath}

In other words, $L_n^m$ is sim­ply the $m$-​th de­riv­a­tive of $L_n$, which cer­tainly tends to sim­plify things. Ac­cord­ing to [25, p. 152], the most nearly stan­dard no­ta­tion de­fines

\begin{displaymath}
L_n^m=(-1)^m\frac{{\rm d}^m}{{\rm d}x^m} L_{n+m}(x).
\end{displaymath}

Com­bine the messy de­f­i­n­i­tion of the spher­i­cal har­mon­ics (D.5) with the un­cer­tain de­f­i­n­i­tion of the La­guerre poly­no­mi­als in the for­mu­lae (D.8) for the hy­dro­gen en­ergy eigen­func­tions $\psi_{nlm}$ above, and there is of course al­ways a pos­si­bil­ity of get­ting an eigen­func­tion wrong if you are not care­ful.

Some­times the value of the wave func­tions at the ori­gin is needed. Now from the above so­lu­tion (D.8), it is seen that

\begin{displaymath}
\psi_{nlm} \propto r^l \quad\mbox{for}\quad r \to 0 %
\end{displaymath} (D.9)

so only the eigen­func­tions $\psi_{n00}$ are nonzero at the ori­gin. To find the value re­quires $L_n^1(0)$ where $L_n^1$ is the de­riv­a­tive of the La­guerre poly­no­mial $L_n$. Skim­ming through ta­ble books, you can find that $L_n(0)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n!$, [40, 30.19], while the dif­fer­en­tial equa­tion for these func­tion im­plies that $L_n'(0)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-nL_n(0)$. There­fore:
\begin{displaymath}
\psi_{n00}(0) = \frac{1}{\sqrt{n^3 \pi a_0^3}} %
\end{displaymath} (D.10)