文档介绍：第二章
,令~ = 1,有:
Z
xˆ|p > = xˆ|x >< x|p > dx
Z
= dx x eip·x|x >
Z
∂
= −i dx eip·x|x >
∂p
Z
∂
= −i dx |x >< x|p >
∂p
∂
= −i |p > (1)
∂p
同理
∂
< p | xˆ= i < p |
∂p
< p | pˆ= p < p |
pˆ| p > = p | p > (2)
2. 取~n = ~ez,有
cos θ− sin θ 0
g(~ez, θ) = sin θ cos θ 0 ,
0 0 1
0 −1 0
∂g(~ez, θ)
|θ=0 = 1 0 0 (3)
∂θ
0 0 0
又有:
[x, Jz] = [x, xpy − ypx] = −i~y
[y, Jz] = [y, xpy − ypx] = i~x
[z, Jz] = 0 (4)
1
所以,
0 −1 0 x
[~r, Jz] = i~ 1 0 0 y
0 0 0 z
∂g(~e , θ)
= i~ z | ~r (5)
∂θθ=0
. 参考教材 P50,51 习题 1, 2
e = c = ~ = 1 ,
1 X 1
H = (p − A )2 + φ−~σ· H~
2m j j 2m
j
1 X 1
= (p − A )2 + φ−~σ· (∇× A~)
2m j j 2m
j
1 X X ε
= (p − A )2 + φ− ijk σ(∂ A )
2m j j 2m i j k
j ijk
1 X X iε
= (p − A )2 + φ− ijk σ(p A − A p )
2m j j 2m i j k k j
j ijk
1 X X iε
= (p − A )2 + φ− ijk σ(−p A − A p )
2m j j 2m i k j k j
j ijk
1 X X iε
= δ(p − A )(p − A ) + φ− ijk σ(p p − p A − A p + A A )
2m jk j j k k 2m i k j k j k j k j
jk ijk
1 X X
= φ+ (δ+ iεσ)(p − A )(p − A )
2m jk ijk i k k j j
jk i
1 X
= φ+ (σσ)(p − A )(p − A )
2m j k k k j j
jk
1 X X
= φ+ σ(p − A ) σ(p − A )
2m j j j k k k
j k
1 X
= φ+ ( σ(p − A ))2 (6)
2m j j j
j
+
6.(a) [a, a ] = 1/2[βx,ˆ−i/~βpˆ] + 1/2[