The One-Electron Atom
121
and also
S x S y + S y S x = –
2
2
(
)
2
i S
S
+
−
−
= 0
(4.84)
Together with the commutation relation S x S y – S y S x =
z
i S
satisfied by all
angular momentum operators [see Eq. (3.163)], this implies
S x S y = –S y S x = 2
z
i S
(4.85)
By symmetry,
S y S z = –S z S y = 2
x
i S
(4.86)
S z S x = –S x S z = 2
y
i S
For writing down the Schrödinger equation for a free, spin 1/2 particle of
mass m, it is proposed that the kinetic energy be written as
E =
2
2
m
(p ⋅ S) (p ⋅ S)
(4.87)
This expression is equivalent to
1
2m
p
2
for the free particle, as can be
shown by using Eqs. (4.80), (4.85) and (4.86). On using the operator expressions
for E and p, it leads to the Schrödinger equation for a free particle,
i t
∂ψ
∂
= –
2
m
(∆ ⋅ S) (∆ ⋅ S)ψ
(4.88)
The interaction with the electrostatic potential φ, can be introduced by the
prescription that
E → E – qφ
(4.89)
where q is the charge of the particle. However, since both
tot
1
,






E
c
p
and
1
,


φ




c
A
transform as relativistic 4-vectors, requirements of relativistic
covariance imply that the prescription in Eq. (4.89) should be accompanied by
the replacement
p → p – qA
(4.90)
The fact that only the kinetic energy appears in Eq. (4.87) does not alter the
essential results since the addition of a constant to the energy only redefines the