# Radial equation for hydrogen atom

Radial equation for hydrogen atom refers to one of equations that appear at the separation of variables for the Stationary Schroedinger equation for the Hydrogen atom in spherical coordinates.

## Coordinates and the equation

$$x=\cos(\theta)\cos(\phi)$$
$$y=\cos(\theta)\sin(\phi)$$
$$z=\sin(\phi)$$

The wave function appears as $$\psi=R(r )\Theta(\theta)\Phi(\phi)$$ in the spherical system of coordinates $$r, \theta, \phi$$; then the Angular part of the equation gives

$$\Phi(\phi)=\exp(\pm \mathrm i m \phi)$$

$$\Theta(\theta)=\displaystyle P_{\ell,m}(\sin(\theta)) = \mathrm{LegendreP}_{\ell,m}(\sin(\theta))$$,
where LegendreP is associated Legendre function,

and for $$R$$, the separation gives the following equation

$$\displaystyle \frac{1}{r^2} \partial_r (r^2 R') - \frac{\ell(\ell\!+\!1)}{r^2} R + \left( \frac{2\mu} {\hbar^2} \frac{e^2}{r} + \frac{2\mu} {\hbar^2} E \right) R = 0$$

Here $$R=R(r )$$, $$\mu$$ is effective mass of electron, $$\hbar$$ is Planck constant, $$e$$ is charge of electron, $$\ell$$ is non–negative integer parameter, that comes from the equations for the angular part of the wave function. Parameter $$E$$ appears as eigenvalue of the Hamiltonian and, therefore, is interpreted as energy of the stationary state.

Define the Bohr radius $$\displaystyle ~a\!=\!\frac{\hbar^2}{\mu\, e^2}$$

Let $$R(r )=F(x)$$, where $$x$$ is new dimensionless variable that has nothing to do with variable $$x$$ used in the initial Cartesian system of coordinates. Then, for $$F$$, the equation becomes

$$\displaystyle \frac{1}{x^2} \partial_x (x^2 F') - \frac{\ell(\ell\!+\!1)}{x^2} F + \left( \frac{1}{x} + \frac{\hbar^2}{\mu e^4} E \right) F = 0$$

Parameter $$\displaystyle -\varepsilon=\frac{\hbar^2}{\mu e^4} E$$ can be interpreted as dimension-less energy.

$$\displaystyle F''+ \frac{2}{x} F' - \frac{\ell(\ell\!+\!1)}{x^2} F + \left( \frac{1}{x} - \varepsilon \right) F = 0$$

Seach for the solution $$F(x)=\exp(-kx) f(x)$$, that takes into account the asymptotic behaviour of the solution at $$x\gg 1$$, assuming, that $$k^2=\varepsilon$$. The substitution gives:

$$\displaystyle f'' -2k f' +\frac{2}{x} f' +\frac{1\!-\!2k}{x}f - \frac{\ell(\ell\!+\!1)}{x^2} f= 0$$

for $$\ell \!=\!0$$, this equation has, among other, solution $$f(x)\!=\!\rm const$$ at $$k\!=\!1/2$$.

The Mathematica code

DSolve[f''[x] - 1/n f'[x] + 2/x f'[x] + (1 - 1/n)/x f[x] - (m (m + 1))/x^2 f[x] == 0, {f[x]}, x]

among other solution, suggests for f[x] the following:

x^m LaguerreL[-1-m+n, 1+2m, x/n]

This expression suggests values $$\displaystyle k=\frac{1}{2n}$$; then, for positive integer $$n$$, the solution $$f$$ is polynomial, easily expressed through the Associated Laguerre polynomial [1]

## Solution

The physically-meaningful solution can be written as [2]:

$$\displaystyle R(r )=\frac{N}{an} \left( \frac{r}{an}\right)^\ell \exp\!\left(- \frac{r}{an} \right) ~ L_{n-\ell-1}^{~2 \ell +1} \left( \frac{2r}{an} \right)$$

where $$n$$ is positive integer parameter called also principal quantum number, $$N$$ is normalisation factor (that may depend on the quantum numbers) and

$$L_{n-\ell-1}^{~2 \ell +1}(t)=\mathrm{LaguerreL}(n\!-\!\ell\!-\!1,2 \ell \!+\!1,t)$$

is the LaguerreL polynomial.

## Wave function

Often, the angular part of the wave function is denoted with

$$\displaystyle Y_n^m (\theta,\phi) = \sqrt{\frac{2n\!+\!1}{4\pi} \, \frac{(n\!-\!m)!}{(n\!+\!m)!} } ~ \exp(\mathrm i m \phi) ~ P_\ell^{\,n}(\sin \theta)$$

Then, the wave function can be written as follows:

$$\displaystyle \psi_{n,\ell,m}= \sqrt{ \left( \frac{2}{na} \right)^3 \frac{(n\!-\!\ell\!-\!1)!}{2 n (n\!+\!1)!} } ~ \exp\left(\frac{-r}{na}\right) ~ L_{n-\ell-1}^{~ 2\ell+1}\!\left(\frac{2r}{na}\right) ~ Y_n^m (\theta,\phi)$$

## Jim Branson

Expression of the Radial wave function through the polynomials LaguerreL looks as a patch, due to the expressions in the subscript and superscript of the polynomial $$L$$. Namely for evaluation of the radial wave function, the polynomials can be represented in a little bit more direct way.

Jim Branson suggests another (and perhaps equivalent) representation of the solution, that seems to be more suitable namely for the Radial equation for Hydrogen atom [3]:

Let $$\displaystyle \rho=\sqrt{\frac{-8\mu E}{\hbar^2}\,}\,r$$ , and let $$~\displaystyle \lambda = \frac{Z e^2}{\hbar} \sqrt{\frac{-\mu}{2 E}\,}$$

Then, the radial equation can be written as follows:

$$\displaystyle \frac{\mathrm d^2 R}{\mathrm d \rho^2} + \frac{2}{\rho} \, \frac{\mathrm d R}{\mathrm d \rho} - \frac{\ell(\ell\!+\!1)}{\rho^2} R + \left(\frac{\lambda}{\rho}-\frac{1}{4}\right) R=0$$

It is convenient to denote dependence of $$R$$ on $$\rho$$ with some name; let it be first latter of the last name of Jim Branson, id est, $$R(r )=B(\rho)$$. The equation for $$B$$ can be written as follows:

$$\displaystyle B''(\rho) + \frac{2}{\rho} B'(\rho) - \frac{\ell(\ell\!+\!1)}{\rho^2} B(\rho) + \frac{\lambda}{\rho} B(\rho)-\frac{1}{4} B(\rho)=0$$

Then, the only positive integer values of $$\lambda$$ happen to be allowed. These values determine energy $$E$$ and, therefore, relation between $$r$$ and $$\rho$$. Through this $$\rho$$, the solution can be written as follows:

$$\displaystyle B_{n,\ell}=\rho^\ell \sum_{k=0}^\infty a_k \rho^k \exp(-\rho/2)$$

where

$$\displaystyle a_{k+1}=\frac{k+\ell+1-n}{(k+1)(k+2\ell+2)} a_k$$

Here, $$a$$ are dimensionless coefficients of expansion; they should not be confused with the Bohr radius $$a$$. The last expression can be written in a little bit more compact form, replacing $$k\!+\!1$$ to $$k$$:

$$\displaystyle a_{k}=\frac{k+\ell - n}{k\, (k+2\ell+1)} a_{k-1}$$

Practically, the summation above stops, as the nominator of the fraction becomes zero; the last term corresponds to $$k=n-1-\ell$$; and the only positive integer $$n$$ correspond to the normalisable eigenstates of the Hamiltonian. The Radial part of the Wave function can be expressed with

$$\displaystyle R_{n,\ell}(r )=B_{n,\ell}\left(\sqrt{\frac{-8\mu E}{\hbar^2}\,}r \right) =F_{n,\ell}\left(\frac{2\mu Z e^2}{\hbar^2} r \right)$$

where

$$\displaystyle F_{n,\ell}(x)=B_{n,\ell} (x/n)$$

and this $$F$$ satisfies the dimensionless radial Schroedinger equation

$$\displaystyle {F_{n,\ell}}''(x) + \frac{2}{x} {F_{n,\ell}}'(x) - \frac{\ell(\ell\!+\!1)}{x^2} F_{n,\ell}(x) + \frac{1}{x} F_{n,\ell}(\rho)-\frac{1}{4 n^2} F_{n,\ell}(x)=0$$

where $$-\frac{1}{4 n^2}$$ plays role of the dimensionless energy. Тhe dimensional energy

$$\displaystyle E=- \frac{(Ze^2)^2 \mu}{2\hbar^2 n^2}$$

The radial wave functions are orthogonal;

$$\displaystyle \int_0^\infty \, F_{n,\ell} (x)\, F_{m,\ell} (x)\, x^2 \, \mathrm d x = \Nu_{n,\ell} \delta _{n,m}~ ~$$ where $$~\delta~$$ is Kronecker symbol.

## References

1. http://mathworld.wolfram.com/AssociatedLaguerrePolynomial.html
2. https://www.physics.drexel.edu/~tim/open/hydrofin/hyd.pdf TIMOTHY JONES. ELEMENTARY QUANTUM MECHANICAL MODEL OF THE HYDROGEN ATOM. 2009-02-11.
3. http://quantummechanics.ucsd.edu/ph130a/130_notes/node236.html Jim Branson. Solution of Hydrogen Radial Equation. 2013-04-22

http://quantummechanics.ucsd.edu/ph130a/130_notes/node233.html The Radial Wavefunction Solutions. Defining the Bohr radius, $$\displaystyle a_0={\hbar\over\alpha mc}$$