22-2
More Spin Algebra
The final result derived in the homework above, 2, is a proper eigenvalue of
S
2
corresponding to
S
= 1. That means that
α
(1)
α
(2) is an eigenfunction of
S
2
. Since
S
= 1, it
must be the case that
M
S
can equal
−
1, 0, or 1. Recall that evaluation of
M
S
is
straightforward. One simply adds one-half for each
α
electron and subtracts one-half for
each
β
electron. So, for the case of
α
(1)
α
(2), we have
M
S
= 1. The other two cases,
−
1
and 0, will be degenerate in energy in the absence of a magnetic field, so this state is
triply degenerate. Such states are called “triplet” states, and indicated by a superscript 3
to the left of the wave function.
Technically, we did not prove that the expectation value of
S
2
over the
full
wave
function
Ψ
III
is 2, we only showed it for the spin part of the wave function. It is a trivial
matter, however, to show that the spatial portion of the wave function is normalized (it’s
obviously a 2
x
2 determinant of orthonormal spatial functions preceded by the square
root of 2!, so that’s that). So, we should indicate the triplet character of
Ψ
III
by writing
3
Ψ
III
.
Some quick consideration of the symmetric nature of the
β
(1)
β
(2) case should
convince one that
1
( )
!
2
( )
S
2
!
1
( )
!
2
( )
=
2
(22-1)
and that this is the
M
S
=
−
1 component of the triplet,
3
Ψ
IV
(the homework to prove this
will also, no doubt, be convincing…)
We now turn to
Ψ
V
. Evaluation of
S
2
proceeds in the fashion to which we should,
by now, be accustomed
!
V
S
2
!
V
=
1
2
1
( )
"
1
( )
b
2
( )
#
2
( )
S
2
a
1
( )
"
1
( )
b
2
( )
#
2
( )
$
a
1
( )
"
1
( )
b
2
( )
#
2
( )
S
2
a
2
( )
"
2
( )
b
1
( )
#
1
( )
$
a
2
( )
"
2
( )
b
1
( )
#
1
( )
S
2
a
1
( )
"
1
( )
b
2
( )
#
2
( )
+
a
2
( )
"
2
( )
b
1
( )
#
1
( )
S
2
a
2
( )
"
2
( )
b
1
( )
#
1
( )
%
&
’
’
’
’
’
’
’
’
’
’
(
)
*
*
*
*
*
*
*
*
*
*
(22-2)
Note
that
the
second
and
third
integrals
on
the
r.h.s.
are
zero
because
of
the
orthonormality of the spatial orbitals
a
and
b
, whose products appear over the same
electronic coordinate in those integrals. The spatial functions integrate to one in the first
and fourth integrals, and the remaining spin expectation values are just those of eqs. 21-
14 and 21-15. Thus, the expectation value of eq. 22-2 is 1/2(1
−
0
−
0 + 1) = 1. There is
no integer value of
S
for which
S
(
S
+ 1) = 1, so evidently
Ψ
V
is