Appendix A — Quick review for hydrogenic atoms

As discussed elsewhere, I don’t include a solution of the hydrogen atom in these notes; it’s a common enough textbook example and a very special case. (Maybe I’ll be convinced to add one to this appendix later on…) Actually, Sakurai’s presentation of the hydrogen solution is also fairly brief. If you really want to see the details I recommend chapter 13 of Shankar.

However, there is lots of interesting physics that relies on knowing the form of the solutions to hydrogen, so here I include a short review and a collection of formulas and constants.

Strictly speaking, we consider the solution for a “hydrogenic” atom, i.e. a bound state of a single electron in the electric field generated by a nucleus with total charge Z: V(r) = -\frac{Ze^2}{r}. Clearly to be a bound state, the electron energy must be less than zero. Since the Coulomb potential is a central potential (spherically symmetric), the solution to the Schrödinger equation splits into radial and angular components, \psi_{nlm}(r, \theta, \phi) = R_{nl}(r) Y_l^m(\theta, \phi). Solving the radial equation for the hydrogen atom is a difficult task, and as we’ve seen in other examples the solutions are written only in terms of special functions, either the associated Laguerre polynomials L_p^q(\rho) or the confluent hypergeometric functions, which are written as F(a;c;\rho). Here is the general formula in Sakurai’s notation:

R_{nl}(r) = \frac{1}{(2l+1)!} \left( \frac{2Zr}{na_0}\right)^l e^{-Zr / (na_0)} \\ \times \sqrt{ \left( \frac{2Z}{na_0} \right)^3 \frac{(n+l)!}{2n(n-l-1)!} } F\left(-n+l+1; 2l+2; \frac{2Zr}{na_0} \right)

and here it is again conveniently in Mathematica code:

R[n_, l_] := 1/(2*l + 1)!*((2*Z*r)/(n*a0))^l*Exp[(-Z*r)/(n*a0)]*
Sqrt[((2*Z)/(n*a0))^3*(n + l)!/(2 n*(n - l - 1)!)]*
Hypergeometric1F1[-n + l + 1, 2 l + 2, (2*Z*r)/(n*a0)];

The constant length scale a_0 is known as the Bohr radius, a_0 = \frac{\hbar^2}{m_e e^2}. As we observed generally for a particle in a central potential, the wavefunctions will be peaked away from the origin except for l=0; in particular only l=0 gives R_{n0}(0) \neq 0. The energy eigenstates are labelled by the principal quantum number, n, which is a linear combination of a radial quantum number q and the angular momentum quantum number l: n = q + l + 1. This imposes the restrictions that l \leq n-1 and n \geq 1; for the n=1 state, only l=0 is allowed. The energy eigenvalues are E_n = -\frac{1}{2} mc^2 \frac{Z^2 \alpha^2}{n^2} where \alpha is the fine-structure constant, \alpha = \frac{e^2}{\hbar c}. The value of the ground-state n=1 energy for the hydrogen atom (Z=1) is sometimes used as a unit of energy, known as the Rydberg: we see that \textrm{Ry} = \frac{me^4}{2\hbar^2} = \frac{e^2}{2a_0}, and so E_n = -\frac{Z^2}{n^2}\ \textrm{Ry}. I’ll emphasize that the energy eigenvalues of this solution depend only on n, not l or m. The E_n energy level is thus made up of \sum_l (2l+1) = n^2 degenerate states.

The ground state radial wavefunction is relatively simple, and we should have it handy for some of the examples we’re about to work: it is equal to R_{10}(r) = 2 \sqrt{\frac{Z^3}{a_0^3}} e^{-Zr/a_0}.