220 3 Quantum Mechanics – II
(Jx) 12 =< 1 |Jx| 2 >=〈
1
2
,
1
2
|Jx|1
2
,−
1
2
〉
=
1
2
[(
1
2
−
1
2
+ 1
)(
1
2
+
1
2
)] 1 / 2
=^1 / 2
(Jx) 21 =〈
1
2
,−
1
2
|J|
1
2
,
1
2
〉
=^1 / 2
because the second delta factor survives
(Jx) 11 =< 1 |Jx| 1 >=〈 1
2 ,
1
2 |Jx|1
2 ,1
2〉
=0 because of delta factors.
Similarly, (Jx) 22 = 0
ThusJx= 2(
01
10
)
;Jy= 2(
0 −i
i 0)
Jz=2
(
10
0 − 1
)
(9)
These three matrices are known as Pauli matrices.
(b)J^2 =Jx^2 +J^2 y+Jz^2
Using the matrices given in (6), squaring them and adding we getJ^2 =^3 / 4 ^2(
10
01
)
3.85 (a) For j = 1 ,m = 1 ,0 and−1, the three base states are denoted by
| 1 >,| 2 >and| 3 >.Inthe|j,m>notation| 1 >=| 1 , 1 >,| 2 >=
| 1 , 0 >,| 3 >=| 1 ,− 1 >
Jz| 1 >=m| 1 >=| 1 >
Jz| 2 >= 0 .| 2 >= 0
Jz| 3 >=−| 3 >
(Jz) 11 =< 1 |Jz| 1 >=< 1 , 1 |Jz| 1 , 1 >=
(Jz) 22 =< 2 |Jz| 2 >=< 1 , 0 |Jz| 1 , 0 >= 0
(Jz) 33 =< 3 |Jz| 3 >=< 1 ,− 1 |Jz| 1 ,− 1 >=−
(Jz) 12 =(Jz) 21 =(Jz) 13 =(Jz) 31 =(Jz) 23 =(Jz) 32 = 0
because ofδ– factorδmm′Jz=⎛
⎝
10 0
00 0
00 − 1
⎞
⎠
For the calculation ofJxandJy, we need to work outJ+andJ−.
J+| 1 >= 0
J+| 2 >=[(j−m)(j+m+1)]^1 /^2 | 1 >