(
~r×B~) 2
= riBjεijkrmBnεmnk= (riBjriBj−riBjrjBi)= r^2 B^2 −(
~r·B~) 2
− 0
So, plugging these two equations in, we get
− ̄h^2
2 μ∇^2 ψ+
e
2 μcB~·Lψ~ + e2
8 mc^2[
r^2 B^2 −(
~r·B~) 2 ]
ψ= (E+eφ)ψ.We see that there are two new terms due to the magnetic field. The first one is the magnetic moment
term we have already used and the second will be negligible in atoms.
20.5.4 The Size of the B field Terms in Atoms
In the equation
− ̄h^2
2 μ∇^2 ψ+
e
2 μcB~·Lψ~ + e2
8 mc^2[
r^2 B^2 −(
~r·B~) 2 ]
ψ= (E+eφ)ψ.the second term divided by (e^2 /a 0 )
e
2 μc
B~·~L/(e^2 /a 0 ) ∼ e
2 μc
B(m ̄h)/(e^2 /a 0 )= mαeBa 0
2/
(
e^2 /a 0)
=mαa^20
2 eB
= mB(
0. 5 × 10 −^8 cm) 2
(2)(137) (4. 8 × 10 −^10 )
=mB
5 × 109 gauss
(
α=e^2
̄hca 0 =̄h
αmc)
Divide the third term by the second:
B^2 a^20 e2
8 mc^2
e
2 μcB ̄h=αa^20
4 eB=
(
0. 5 × 10 −^8
) 2
(4)(137) (4. 8 × 10 −^10 )
=
B
1010 gauss20.5.5 Energy States of Electrons in a Plasma I
−
̄h^2
2 me∇^2 ψ+
eB
2 mecLzψ+
e^2 B^2
8 mec^2(
x^2 +y^2)
ψ=EψFor uniformB~field, cylindrical symmetry⇒applycylindrical coordinatesρ,φ,z. Then
∇^2 =
∂^2
∂z^2+
∂^2
∂ρ^2+
1
ρ∂
∂ρ+
1
ρ^2∂^2
∂φ^2