Hydrogen Atom The Hamiltonian in spherical polar ... - Chemistry

Hydrogen Atom
The Hamiltonian in spherical polar coordinates is
H = − h2
2µ ( ∂2
∂r 2 + 2 r
∂
∂r ) + l 2
op
2µr 2 − Ze 2
r
where is the square of angular momentum operator l 2 op . Note that the
2
potential depends only on r and so H commutes with l op and l z . Thus
eigenfunctions of H can be written in the form R n,l (r)Y l,m (θ,φ ), where
R n,l (r) satisfies
[− h2
2µ ( ∂2
∂r 2 + 2 ∂ l(l +1)h2
) +
r ∂r 2µr 2 − Ze 2
− En ]R
r nl (r)= 0
To proceed, examine the solution for r large, and also need to examine
behavior as r goes to zero, because of the singular nature of Coulomb
potential. It is sometimes convenient to introduce a new function
u n,l (r)=rR n,l (r) in terms of which the above equation is
[− h2
2µ
d 2 l(l +1)h2
2 +
2µr 2 − Ze 2
− En ]u
r nl (r)= 0
dr
(Prove this for homework).
Re-write as
[ d2 l(l +1)
2 −
dr 2r 2 + 2µZe 2 2µE
rh 2 + n
h 2 ]u nl (r)=0

as usual, get the asymptotic solution. The differential equation is then
[ d2
dr 2 + 2µE n
h 2 ]u nl (r)=0
and recall that E n is negative. So the asymptotic solution is of the form
u nl (r)~exp(−κr)
κ =
− 2µE n
h 2 .
Near the origin, and for l > 0 the differential equation is
[ d2 l(l +1)
2 −
dr 2r 2 ]u nl (r)=0.
The general solution is u nl (r)= Ar l+1 + Br −l , but B must equal zero for an
acceptable solution. Thus, the general solution is of the form
u nl (r)=r l+1 e −κr v(r)
Introduce a new variable as follows ρ =κr, and then let
and the equation for v is
u nl (r)= ρ l+1 e −ρ v nl (ρ)
d 2 v +1 dv
2 + 2(l
dρ ρ 2 −1)
dρ + [V E
2(l +1)
− ]v = 0
ρ
where V is the Coulomb potential. Develop a series for v and examine the
asymptotic form, and see that the series diverges as e 2ρ unless it terminates

at some power N. Doing that gives the result that Ze2 κ
E
= 2(N + l +1) and rearranging
give
E n = − Z 2 e 4 µ
2h 2 2 ; n = (N + l +1).
n
Thus, as in the two-dimensional isotropic harmonic oscillator, there is a
degeneracy in the eigenvalues, i.e., for each n l varies from 0 to n-1, and so
there are n+1 values of l. In addtion, for each lthere are 2 l+1 values of m.
Recall - l≤ m ≤ l. So, the total degeneracy is
n −1
g n
= ∑ 2l +1 = 1+ 3 + 5 + ... + 2(n −1) +1 = n 2
l = 0