A.17 The hydrogen radial wave functions

This will be child’s play for harmonic oscillator, {A.12}, and spherical harmonics, {A.15}, veterans. If you replace the angular terms in (3.15) by $l(l+1)\hbar^2$, and then divide the entire equation by $\hbar^2$, you get

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

Since $l(l+1)$ is nondimensional, all terms in this equation must be. In particular, the ratio in the third term must be the inverse 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 $\rho=r/a_0$. The final term in the equation must be nondimensional too, and that means that the energy $E$ must take the form $(\hbar^2/2m_e a_0^2)\epsilon$, 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 $R''+\epsilon{R}=0$, which has solutions of the rough form $\cos(\sqrt{\epsilon}\rho+\phi)$ 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 $R\approx{e}^{\pm\sqrt{-\epsilon}\rho}$. 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 $-1/n^2$ (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 $\xi=2\rho/n$. Split off this exponential part by writing $R=e^{-\frac12\xi}\overline{R}$ 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 $\overline{R}=\xi^l\overline{\overline{R}}$, 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 $\overline{\overline{R}}=\sum{c}_p\xi^p$, 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 $p=q$ 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 $\epsilon=-1/n^2$ with $n>l$ has been obtained.

With $n$ identified, you can identify the ODE as Laguerre's associated differential equation, e.g. [20, 30.26], the $(2l+1)$-th derivative of Laguerre's differential equation, e.g. [20, 30.1], and the polynomial solutions as the associated Laguerre polynomials $L_{n+l}^{2l+1}$, e.g. [20, 30.27], the $(2l+1)$-th derivatives of the Laguerre's polynomials $L_{n+l}$, e.g. [20, 30.2]. To normalize the wave function use an integral from a table book, e.g. [20, 30.46].

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

$$\psi_{nlm} = -\frac{2}{n^2} \sqrt{\frac{(n-l-1)!}{[(n+l)... ...ft(\frac{2\rho}n\right) e^{-\rho/n} Y_l^m(\theta,\phi) %$$ (A.30)

The properties of the associated Laguerre polynomials $L_{n+l}^{2l+1}(2\rho/n)$ are in table books like [20, pp. 169-172], and the spherical harmonics were given earlier in section 3.1.3 and in note {A.15}, (A.28).

Do keep in mind that different references have contradictory definitions of the associated Laguerre polynomials. This book follows the notations of [20, 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 [11, 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 (A.28) with the uncertain definition of the Laguerre polynomials in the formulae (A.30) 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 (A.30), it is seen that

$$\psi_{nlm} \propto r^l \quad\mbox{for}\quad r \to 0 %$$ (A.31)

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 $L_n(0)=n!$, [20, 30.19], while the differential equation for these function implies that $L_n'(0)=-nL_n(0)$. Therefore:

$$\psi_{n00}(0) = \frac{1}{\sqrt{n^3 \pi a_0^3}} %$$ (A.32)