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

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

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

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

or written out

$$- \rho^2 R'' - 2\rho R' +[l(l+1)-2\rho-\epsilon\rho^2]R = 0$$

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 $\vphantom0\raisebox{1.5pt}{$-$}$1$\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

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

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

$$\xi \overline{\overline{R}}\,\strut'' + [2(l+1) - \xi]\o... ...\overline{R}}\,\strut' + [n-l-1]\overline{\overline{R}} = 0$$

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

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

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 $\epsilon$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$1$\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. [39, 30.26], the $(2l+1)$-th derivative of Laguerre's differen­tial equation, e.g. [39, 30.1], and the polynomial solutions as the associated Laguerre polynomials $L_{n+l}^{2l+1}$, e.g. [39, 30.27], the $(2l+1)$-th derivatives of the Laguerre's polynomials $L_{n+l}$, e.g. [39, 30.2]. To normalize the wave function use an integral from a table book, e.g. [39, 30.46].

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

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

The properties of the associated Laguerre polynomials $L_{n+l}^{2l+1}(2\rho/n)$ are in table books like [39, 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 [39, pp. 169-172], who define

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

In other words, $L_n^m$ is simply the $m$-th derivative of $L_n$, which certainly tends to simplify things. According to [24, p. 152], the “most nearly standard” notation defines

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

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

$$\psi_{nlm} \propto r^l \quad\mbox{for}\quad r \to 0 %$$ (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 $L_n(0)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n!$, [39, 30.19], while the differen­tial equation for these function implies that $L_n'(0)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-nL_n(0)$. Therefore:

$$\psi_{n00}(0) = \frac{1}{\sqrt{n^3 \pi a_0^3}} %$$ (D.10)