Physlets run in a Java-enabled browser, except Chrome, on the latest Windows & Mac operating systems. If Physlets do not run, click here for help on updating Java and setting Java security.
Section 13.7: The Coulomb Potential for the Idealized Hydrogen Atom
Please wait for the animation to completely load.
Now consider the radial part of the Schrödinger equation in Eq. (13.21) written as
[−(ħ2/2μ)(1/r2) d/dr (r2 d/dr) + l(l + 1)ħ2/(2μr2) + V(r)] R(r) = ER(r) . (13.26)
We want to rewrite these terms, especially the derivative, into a more standard form. The substitution, R(r) = u(r)/r, simplifies the derivative term, in the above equation to yield
[−(ħ2/2μ) (d2/dr2) + l(l + 1)ħ2/(2μr2) + V(r)] u(r) = Eu(r) .
When we find solutions to this equation, we must keep in mind that we are solving for u(r), not R(r), and that we must divide u(r) by r to give the true radial wave function, R(r). Before looking at a particular V(r), we look at the general equation and interpret terms. We see what we can interpret as an effective potential:
Veff = l(l + 1)ħ2/(2μr2) + V(r) ,
where l(l + 1)ħ2/(2μr2) is the potential associated with the so-called the centrifugal barrier. Now consider the following Coulomb potential, V = −e2/r, which describes the potential energy function for an electron in the proximity of a proton: the potential responsible for the basic structure of the hydrogen atom.4 When we insert this Coulomb potential in the radial differential equation, we have a differential equation that describes the electron:
[−(ħ2/2μe) (d2/dr2) + l(l + 1)ħ2/(2μer2) − e2/r] u(r) = Eu(r) . (13.27)
where μe is the electron's mass and e is the charge of the electron. In Animation 1 the effective potential for the Coulomb problem Veff = l(l + 1)ħ2/(2μer2) + V(r) is shown for l = 0, 1, 2. Notice that as l gets bigger, the centrifugal barrier increases as well. To get this equation into standard form, divide by −ħ2/2μe, which yields
[ (d2/dr2) − l(l + 1)/r2 + 2μee2/(ħ2r)] u(r) = (−2μeE/ħ2) u(r) .
We now define κ2 = (−2μeE/ħ2) (which is real since E < 0) and the dimensionless quantities ρ = κr and ρ0 = 2μee2/(κ ħ2). Making these substitutions yields
[ (d2/dρ2) − l(l + 1)/ρ2 +ρ0/ρ − 1] u = 0 .
We begin our analysis of the solutions of this differential equation by considering the two special limiting cases:
Case I: ρ→ 0 (r→ 0). In this case the centrifugal barrier dominates in Eq. (13.46)
[ (d2/dρ2) − l(l + 1)/ρ2] u = 0 . (13.28)
We find that the general solution to this equation is u = Aρl+1+Bρ−l and therefore the normalizable piece is just u ∝ ρl+1 ∝ rl+1.
Case II: ρ → ∞ (r → ∞). In this case the centrifugal term, l(l + 1)/ρ2, and the potential energy, ρ0/ρ, vanish from Eq. (13.28) at large ρ. This leaves
[ (d2/dρ2) − 1] u = 0 . (13.48)
which for E < 0 gives the normalizable solution u = exp(−ρ) = exp(−κr).
Now that we have an idea of what the bound states should look like asymptotically, we can find the entire solution. After much algebra we first find that
E = μee4/(2n2ħ2) = −R (1/n2) ,
where R = μee4/(2ħ2) is the Rydberg and is 13.6 eV. This result describes the energy levels for the Coulomb problem, and hence, the basic energy level structure for the hydrogen atom. We now simplify ρ. We use ρ = κr and the definition of κ to find
ρ = μee2/(nħ2) r = (r/na0) ,
where a0 = ħ2/μe2 is the Bohr radius. We can make use of further substitutions, this time yielding the radial wave functions
Rnl(r) = Anl e−r/na0 [(r/na0)l+1/r] vn(r/na0) , (13.29)
where Anl = [(2/(na0)3(n − l − 1)!/(2n[(n + l)!]3)] is the normalization for the radial energy eigenfunction. In addition, vn( r/na0) = L2l+1n−l−1(2r/na0) are the associated Laguerre polynomials. The unnormalized radial wave functions, above without Anl, are shown in Animation 2. In the animation, distances are given in terms of Bohr radii, a0. You may enter values of n and l and see the radial energy eigenfunction that results.
We find that the entire energy eigenfunction, properly normalized, is simply the product of the radial and angular solutions:
ψnlm = Rnl(r) Ylm(θ,φ) .
Note that given that there are n2 states per n value, and that the energy just depends on n, the solutions have an n2 energy degeneracy.
When you get a good-looking graph, right-click on it to clone the graph and resize it for a better view.
4In Chapter 14 we discuss corrections to the Coulomb potential which are responsible for the remaining structure in the hydrogen spectral lines. We also generalize the Coulomb potential to include hydrogenic atoms, those with one electron and Z protons.