21.8.3 Applying theS^2 Operator toχ 1 mandχ
We wish to verify that the states we have deduced are really eigenstates of theS^2 operator. We will
really compute this in the most brute force.
S^2 =
(
S~ 1 +S~ 2
) 2
=S 12 +S^22 + 2S~ 1 ·S~ 2
S^2 χ(1)+χ(2)+ = s 1 (s 1 + 1) ̄h^2 χ(1)+χ(2)+ +s 2 (s 2 + 1) ̄h^2 χ+(1)χ(2)+ + 2S~ 1 χ(1)+ ·S~ 2 χ(2)+=3
2
̄h^2 χ(1)+χ(2)+ + 2(
S(1)x Sx(2)+S(1)y Sy(2)+S(1)z S(2)z)
χ(1)+χ(2)+Sxχ+ =̄h
2(
0 1
1 0
)(
1
0
)
=
̄h
2(
0
1
)
=
̄h
2χ−Syχ+ =̄h
2(
0 −i
i 0)(
1
0
)
=
̄h
2(
0
i)
=īh
2χ−Sxχ− =̄h
2(
0 1
1 0
)(
0
1
)
=
̄h
2(
1
0
)
=
̄h
2χ+Syχ− =̄h
2(
0 −i
i 0)(
0
1
)
=
̄h
2(
−i
0)
=−īh
2χ+S^2 χ
(1)
+χ(2)
+ =3
2
̄h^2 χ
(1)
+χ(2)
+ +̄h^2
2[(
0
1
)
1(
0
1
)
2+
(
0
i)
1(
0
i)
2+
(
1
0
)
1(
1
0
)
2]
=
3
2
̄h^2 χ(1)+χ(2)+ +
̄h^2
2[(
0
1
)
1(
0
1
)
2−
(
0
1
)
1(
0
1
)
2+
(
1
0
)
1(
1
0
)
2]
=
3
2
̄h^2 χ(1)+χ(2)+ +̄h^2
2χ(1)+χ(2)+ = 2 ̄h^2 χ(1)+χ(2)+Note thats(s+ 1) = 2, so that the 2 ̄his what we expected to get. This confirms that we have an
s=1 state.
Now lets do theχ 00 state.
S^2 χ 00 =(
S 12 +S 22 + 2S~ 1 ·S~ 2
)
χ 00=(
S 12 +S 22 + 2
(
Sx(1)S(2)x +Sy(1)S(2)y +Sz(1)S(2)z))
χ 00=(
S 12 +S 22 + 2Sz(1)Sz(2)+ 2(
Sx(1)S(2)x +Sy(1)S(2)y))
χ 00=
(
3
4
+
3
4
− 2
1
4
)
̄h^2 χ 00 + 2(
S(1)x Sx(2)+S(1)y Sy(2))
χ 00= ̄h^2 χ 00 + 2(
S(1)x Sx(2)+S(1)y Sy(2)) 1
√
2
(
χ
(1)
+χ(2)
− −χ(1)
−χ(2)
+)
= ̄h^2 χ 00 +̄h^2
21
√
2
(
χ
(1)
−χ(2)
+ −χ(1)
+χ(2)
− +i(−i)χ(1)
−χ(2)
+ −(−i)iχ(1)
+χ(2)
−