220
3 Quantum Mechanics – II
(J x ) 12 =< 1|J x |2 >=
1
2
,
1
2
|J x |
1
2
, −
1
2
=
1
2
1
2
−
1
2
+ 1
1
2
+
1
2
1/2
=
1
/ 2
(J x ) 21 =
1
2
, −
1
2
|J |
1
2
,
1
2
=
1
/ 2
because the second delta factor survives
(J x ) 11 =< 1|J x |1 >=
1
2
,
1
2
|J x |
1
2
,
1
2
= 0 because of delta factors.
Similarly, (J x ) 22 = 0
Thus J x =
2
0 1
1 0
; J y =
2
0 −i
i 0
J z =
2
1 0
0 −1
(9)
These three matrices are known as Pauli matrices.
(b) J
2
= J
2
x + J
2
y + J
2
z
Using the matrices given in (6), squaring them and adding we get
J
2
=
3
/ 4
2
1 0
0 1
3.85 (a) For j = 1, m = 1, 0 and −1, the three base states are denoted by
|1 > ,|2 > and |3 >. In the | j, m > notation |1 >= |1, 1 >, |2 >=
|1, 0 >, |3 >= |1, −1 >
J z |1 >= m|1 >= |1 >
J z |2 >= 0.|2 >= 0
J z |3 >= −|3 >
(J z ) 11 =< 1|J z |1 >=< 1, 1|J z |1, 1 >=
(J z ) 22 =< 2|J z |2 >=< 1, 0|J z |1, 0 >= 0
(J z ) 33 =< 3|J z |3 >=< 1, −1|J z |1, −1 >= −
(J z ) 12 = (J z ) 21 = (J z ) 13 = (J z ) 31 = (J z ) 23 = (J z ) 32 = 0
because of δ – factor δ mm
J z =
⎛
⎝
1 0 0
0 0 0
0 0 −1
⎞
⎠
For the calculation of J x and J y , we need to work out J + and J − .
J + |1 >= 0
J + |2 >= [( j − m)( j + m + 1)]
1/2
|1 >