Download E618: Pertubation theory for Helium atom

1
E618: Pertubation theory for Helium atom
Submitted by: Amir Waxman
The problem:
The Helium Atom includes 2 electrons with half spin. In this problem, the interaction between the electrons should
be treated as a pertubation
(1) Write the Hamiltonian that describes the electrons in the atom.
(2) Find the stable states of the atom, neglecting the interaction between the electrons.
(3) What is the Ionization energy of the Helium atom in the zeroth order of the pertubation
theory?
(4) What is the first order correction for the ionization energy?
(5) What is the leading order correction for the bound excited states?
The solution:
(1) The unpertubed hamiltonian of the helium atom includes the interactins between the two electrons and the nuclei.
the interaction between the electrons is refered to as a pertubation. the total hamiltonian looks like that:
H=
P2 2
α
α
e2
P1 2
+
−
−
+
2M
2M
r1
r2
|r1 − r2 |
while α = ze2 and z = 2 for Helium.
(2) first let us define the base we are working in:
base = |ν1 l1 m1 σ1 i ⊗ |ν2 l2 m2 σ2 i
These are the same quntum numbers that describe the electron in the hydrogen atom, only here we have two electrons.
the eigen functions are thus, the same like in those of hydrogen:
ϕνlm (r, Ω) = hrΩ|lmνi = Rlν (r)Y lm (Ω)
The energies of the unpertubed hamiltonian will be:
El1m1ν1,l2m2ν2 = −M α2
1
2
(l1 + ν1 )
+
!
1
2
(l2 + ν2 )
The stable states are those with the energies between Eground = −M α2 (1 + 1) andE∞ = −M α2 (1 + 0) as above E∞ ,
the atom can be ionized (one of the electrons is taken to infinity). all the states below E∞ are stable, and all the
states above it belong to the continum. Thus, the stable states are those with at least one electron in the ground
state (l + ν = 1) if we have, for example, two electrons in the excited state, this is not a stable state, as the energy is
greater than E∞ .
(3) The ionzation energy in the zeroth order will thus be:
Eionization = E∞ − Eground = M α2
(4) The first order correction of the energy is:
∆E = hΨ|V |Ψi
And in the ground state:
Ψ = [|l = 0, ν = 1i ⊗ |l = 0, ν = 1i] ⊗ [singlet]
When singlet = √12 (| ↑↓i − | ↓↑i) is the unsymetric spin state. The reason we took only the singlet state (and not
the triplet) is that the ”‘general”’ wave function has to be unsymetric to the exchange of the electrons (fermions),
2
and as the orbital function is symetric, the spin function has to be unsymetric.The pertubation is:V =
correction to the energy will thus be calculated by the integral on the Radial functions:
Z
e2
1
dr1 dr2 [R01 (r1 )R01 (r2 )]∗
R01 (r1 )R01 (r2 )
∆E = 2
π
|r1 − r2 |
1
z
∆E = ( )2 ( )6 e2
π
a0
Z
zr1
dr1 r12 e− a0
Z
zr2
dr2 r22 e− a0
Z
dΩ1 dΩ2
=
2
2r1 r2
The
1
|r1 − r2 |
when we use spherical coordinatesand,and the eigen function are those of the hydrogen atom. a0 =
Bhor. first, we have to solve the integral over Ω
Z
Z 1
1
1
dΩ1 dΩ2
= 4π · 2π
dcos(θ) p 2
2 − 2r r cos(θ)
|r1 − r2 |
+
r
r
−1
1 2
2
1
= 8π 2
e2
|r1 −r2 |
1
me2
is radius
q
r12 + r22 − 2r1 r2 cos(θ)
8π 2
(r1 + r2 − |r1 − r2 |)
r1 r2
= 16π 2 ·
1
r>
∆E = 16
z 6 e2
a60
Z
∞
dr1 r12 e−
2zr1
a0
0
Z
0
∞
dr2 r22 e−
2zr2
a0
1
r>
we divide the integral over r1 into two integrals. one for r1 < r2 , and one for r1 > r2 . we then get
Z r2
Z
Z ∞
2zr2
r2 2zr1
r2 2zr1
z 6 e2 ∞
dr2 r22 e− a0 ·
dr1 1 e− a0 +
dr1 1 e− a0
∆E = 16 6
a0 0
r2
r1
0
r2
8z 5 e2
=− 5
a0
Z
4z 4 e2
a40
Z
=−
=−
∞
dr2 r22 e−
2zr2
a0
dr2 r22 e−
4zr2
a0
0
∞
0
2zr
2zr
a20
a0
a20
a0
− a 2
− a 2
0
0
+
· e
(r2 +
)−
−e
(r2 + )
z
2zr2
2zr2
2z
+
4zr2
a0
a0 2zr2
r2 e − a0 − e − a0
z
z
4z 4 e2
−a0 3 a0 −a0 2 a0 −a0 2
·
2(
)
+
(
)
−
(
)
a40
4z
z 4z
z 2z
∆E =
5 ze2
5
= M α2 .
8 a0
8
(5) In the first excited state of helium, one of the electrons has n = l + ν = 2 the other one got n = 1. The n = 2
electron can have both l = 1 and l = 0 states. first we deal the case in which both of the electrons got l = 0. The
symetrical and unsymetrycal states will be:
1
Ψ+ = √ (|l = 0, ν = 2i ⊗ |l = 0, ν = 1i + |l = 0, ν = 1i ⊗ |l = 0, ν = 2i) ⊗ [singlet](degeneration = 1)
2
3
1
Ψ− = √ (|l = 0, ν = 2i ⊗ |l = 0, ν = 1i − |l = 0, ν = 1i ⊗ |l = 0, ν = 2i) ⊗ [triplet](degeneration = 3)
2
and one can notice we attached a singlet (unsymetric) to the symetric orbital function and a triplet to the unsymetric
orbital function in order to keep the general wave function unsymetric. in the case that 1 electron has l = 1 and the
other has l = 0 we get:
1
Ψ+ = √ (|l = 1, m, ν = 1i ⊗ |l = 0, ν = 1i + |l = 0, ν = 1i ⊗ |l = 1, m, ν = 1i) ⊗ [singlet](degeneration = 3)
2
1
Ψ− = √ (|l = 1, m, ν = 1i ⊗ |l = 0, ν = 1i − |l = 0, ν = 1i ⊗ |l = 1, m, ν = 1i) ⊗ [triplet](degeneration = 9)
2
we will now want to calculate the change in the energy due to pertubation:
∆E = hΨ|V |Ψi
and thus:
Z
∆E =
∗
dr1 dr2 ϕ100 (r1 , Ω1 )ϕνlm (r2 , Ω2 ) ± ϕ100 (r2 , Ω2 )ϕνlm (r1 , Ω1 )
1
ϕ100 (r1 , Ω1 )ϕνlm (r2 , Ω2 )±
|r1 −2 |
±ϕ100 (r2 , Ω2 )ϕνlm (r1 , Ω1 ) = J ± K
when ϕνlm (r, Ω) was defined in section 2 and ϕ100 (r, Ω) goes for the electron in the ground state. J and K are the
self and exchange integral that were discussed in class. note that this is just a general form and we get different J
and K values for different quantum numbers. All together we get a 16 × 16 diagonal matrix. the reason that we got
a diagonal matrix relates to our choice of base ( two particles base ) where between two states, atleast one of the
quantum numbers is different. in a one particle base we would have got off diagonal elements.