This is just the eigenvalue equation forH 1 which we can write out in (pseudo)matrix form
H 1
α 1
α 2
α 3
α 4
=E(1)
α 1
α 2
α 3
α 4
Now, in fact, most of the matrix elements ofH 1 are zero. We will show that because [Lz,z] = 0,
that all the matrix elements between states of unequalmare zero. Another way of saying this is
that the operator z doesn’t “change”m. Here is a little proof.
〈Ylm|[Lz,z]|Yl′m′〉= 0 = (m−m′)〈Ylm|z|Yl′m′〉This implies that〈Ylm|z|Yl′m′〉= 0 unlessm=m′.
Lets define the one remaining nonzero (real) matrix element to beγ.
γ=eE〈φ 200 |z|φ 210 〉The equation (labeled with the basis states to define the order) is.
φ 200
φ 211
φ 210
φ 21 − 1
0 0 γ 0
0 0 0 0
γ 0 0 0
0 0 0 0
α 1
α 2
α 3
α 4
=E(1)
α 1
α 2
α 3
α 4
We can see by inspection that the eigenfunctions of this operator areφ 211 ,φ 21 − 1 , and√^12 (φ 200 ±φ 210 )
with eigenvalues (ofH 1 ) of 0, 0, and±γ.
What remains is to computeγ. RecallY 00 =√^14 πandY 10 =
√
3
4 πcosθ.γ = eE∫
(2a 0 )−^3 /^22(
1 −
r
2 a 0)
e−r/^2 a^0 Y 00 z(2a 0 )−^3 /^21
√
3
(
r
a 0)
e−r/^2 a^0 Y 10 d^3 r= 2eE(2a 0 )−^31
√
3
∫
r^3 d^3 r(
1 −
r
2 a 0)(
r
a 0)
e−r/a^0∫
1
√
4 πcosθY 10 dΩ= 2eE(2)−^31
√
3
1
√
3
∫∞
0(
r^4
a^40−
r^5
2 a^50)
e−r/a^0 dr=
a 0 eE
12[∫∞
0x^4 e−xdx−1
2
∫∞
0x^5 e−xdx]
=
a 0 eE
12[
4 · 3 · 2 · 1 −
5 · 4 · 3 · 2 · 1
2
]
=
a 0 eE
12(−36)
= − 3 eEa 0 ⇒ E(1)=∓ 3 eEa 0This is first order in the electric field, as we would expect in first order(degenerate) perturbation
theory.
If the states are not exactly degenerate, we have to leave in the diagonal terms ofH 0. Assume that
the energies of the two (mixed) states areE 0 ±∆, where ∆ comes from some other perturbation,
like the hydrogen fine structure. (Theφ 211 andφ21(−1)are still not mixed by the electric field.)
(
E 0 −∆ γ
γ E 0 + ∆
)(
α 1
α 2)
=E
(
α 1
α 2