|Ψ〉=|ψSpace〉⊗|χSpin〉Where the Symbol⊗denotes a so called tensor product. The Schroedinger equation reads
(Hˆ 1 ⊗‖ˆ+Sˆ⊗ˆ‖)|ψSpace〉⊗|χSpin〉=E|ψSpace〉⊗|χSpin〉
Hˆ 1 |ψ〉⊗|χ〉+|ψ〉⊗Sˆ|χ〉=E|ψ〉⊗|χ〉 (15)From eqn.(14), we see that the Spin part is essentially theSzoperator. It follows that the Spin part
of the state vector is|χ〉=|±z〉. Thus, we have
Sˆ|χ〉=−^2 gμB
~BzSˆz|±z〉=∓gμBBz|±z〉 (16)where we usedSˆz|±z〉=±~ 2. Using eqn.(16), we can rewrite the Schroedinger equation, eqn.(15),
resulting in
Hˆ 1 |ψ〉⊗|±z〉∓|ψ〉⊗gμBBz|±z〉=E|ψ〉⊗|±z〉
Hˆ 1 |ψ〉⊗|±z〉= (E±gμBBz)|ψ〉⊗|±z〉≡E′|ψ〉⊗|±z〉 (17)Where we defineE′=E±gμBBz. In order to obtain a differential equation in space variables, we
multiply eqn.(17) by〈r|⊗〈σ|from the left.
(〈r|⊗〈σ|)(Hˆ 1 |ψ〉⊗|±z〉) =E′(〈r|⊗〈σ|)(|ψ〉⊗|±z〉)
〈r|Hˆ 1 |ψ〉〈σ|±z〉=E′〈r|ψ〉〈σ|±z〉 (18)Canceling the Spin wave function〈σ|±z〉on both sides of eqn.(18) and plugging in the definition
Hˆ 1 =^1
2 m(ˆp−qA(ˆr))
(^2) from eqn.(15) yields
1
2 m
(~∇+qA(r))^2 ψ(r,t) =E′ψ(r,t) (19)For a magnetic fieldB= (0, 0 ,Bz)we can choose the vector potentialA(r) =Bzxy(this is called
Landau gauge). Using the Landau gauge and multiplying out the square leaves us with
1
2 m(−~^2 (
∂^2
∂x^2+
∂^2
∂y^2+
∂^2
∂z^2
) + 2i~qBzx∂
∂y
+q^2 Bz^2 x^2 )ψ=E′ψ (20)In eqn.(20) onlyxappears explicitly, which motivates the ansatz
ψ(r,t) =ei(kzz+kyy)φ(x) (21)After plugging this ansatz into eqn.(20), we get
1
2 m(−~^2 φ′′+~^2 (ky^2 +kz^2 )φ− 2 ~qBzkyxφ+q^2 Bz^2 x^2 φ) =E′φ (22)