PDF《Homework10 答案（2018-2019）》

2 )

1 2 α(t)|(a ? a? )|α(t) = ?i( m? hω

2 )

1 2 (e?iωt α ? eiωt α? ) = ?i( m? hω

2 )

1 2 |α|(ei(θ?ωt) ? ei(ωt?θ) ), with α = |α|eiθ = 2( m? hω

2 )

1 2 |α|sin(θ ? ωt) (23) α(t)|p2 |α(t) = ?( m? hω

2 ) α(t)|(a ? a? )2 |α(t)

5 = ?( m? hω

2 ) α(t)|a2 + (a? )2 ? 2a? a ? 1|α(t) = ?( m? hω

2 )(e?i2ωt α2 + ei2ωt (α? )2 ? 2|α|2 ? 1) = ( m? hω

2 )(1 + 2|α|2 ? 2cos(2θ ? 2ωt)|α|2 ) = m? hω

2 + [2( m? hω

2 )

1 2 |α|sin(θ ? ωt)]2 (24) 则?p = ( m? hω

2 )

1 2 (25) 那么 ?p p =

1 2|α|sin(θ ? ωt) (26) 当|α| 1时,相对误差很小,此标准偏差可以忽略. 3. 粒子在对数形式中心力场V (r) = ?V0ln(r/r0), V0 <

0中运动.利用Hellman和维里定理证 明: (a)各束缚本征态的动能期望值相同;

(b)能级间距不随粒子质量变. 解: (a)由维里定理得

2 T n = r ・ ?V n = ?V0 r ・ ( r r2 ) n = ?V0 (27) 即各束缚态动能期望值相同. (b) 由Hellman定理 T n = p2 2m n ? V0

2 = ?m ?H ?m n ? V0

2 = ?m ?En ?m En = V0

2 lnm + en (28) 其中en是和m无关的量,因此能级间隔?E = ?e和m无关.

6 4. 考虑轨道角动量.限定l = 1, 取基矢为|1,

1 , |1,

0 , |1, ?1 . (a)利用代数方法求Lx, Ly矩阵 表示及本征值和本征矢量. (b)将L2 , Lx的共同本征函数表示成Ylm的线性叠加. 解:令|1,

1 = ? ? ? ?

1 0

0 ? ? ? ? , |1,

0 = ? ? ? ?

0 1

0 ? ? ? ? , |1, ?1 = ? ? ? ?

0 0

1 ? ? ? ? (29) 利用 L±|l, m = ? h l(l + 1) ? m(m ± 1)|l, m ±

1 (30) L± = Lx ± iLy (31) 则Lx =

1 2 (L+ + L?) Ly =

1 2i (L+ + L?) (32) 可以分别计算Lx, Ly的矩阵元 (Lx)11 = 1, 1|Lx|1,

1 , (Lx)12 = 1, 1|Lx|1,

0 , (Lx)13 = 1, 1|Lx|1, ?1 (Lx)21 = (Lx)? 12, (Lx)22 = 1, ?1|Lx|1, ?1 , (Lx)23 = 1, 0|Lx|1, ?1 (Lx)31 = (Lx)? 13, (Lx)32 = (Lx)? 23, (Lx)33 = 1, ?1|Lx|1, ?1 (Ly)11 = 1, 1|Ly|1,

1 , (Ly)12 = 1, 1|Ly|1,

0 , (Ly)13 = 1, 1|Ly|1, ?1 (Ly)21 = (Ly)? 12, (Ly)22 = 1, ?1|Ly|1, ?1 , (Ly)23 = 1, 0|Ly|1, ?1 (Ly)31 = (Ly)? 13, (Ly)32 = (Ly)? 23, (Ly)33 = 1, ?1|Ly|1, ?1 容易求得对角上的矩阵元均为0,非对角元 (Lx)12 =

1 2 1, 1|L+ + L?|1,

0 = √

2 2 ? h (Lx)13 =

1 2 1, 1|L+ + L?|1, ?1 =

0 (Lx)21 = √

2 2 ? h (Lx)23 =

1 2 1, 0|L+ + L?|1, ?1 = √

2 2 ? h (Lx)31 =

0 (Lx)32 = √

2 2 ? h (Ly)12 =

1 2i 1, 1|L+ ? L?|1,

0 = ?i √

2 2 ? h

7 (Ly)13 =

1 2i 1, 1|L+ ? L?|1, ?1 =

0 (Ly)21 = i √

2 2 ? h (Ly)23 =

1 2i 1, 0|L+ ? L?|1, ?1 = ?i √

2 2 ? h (Ly)31 =

0 (Ly)32 = i √

2 2 ? h 最后求得 Lx = √

2 2 ? h ? ? ? ?

0 1

0 1

0 1

0 1

0 ? ? ? ? Ly = √

2 2 ? h ? ? ? ?

0 ?i

0 i

0 ?i

0 i

0 ? ? ? ? 设Lx的本征矢为φx = ? ? ? ? a b c ? ? ? ?, 解Lx的本征方程 √

2 2 ? h ? ? ? ?

0 1

0 1

0 1

0 1

0 ? ? ? ? ? ? ? ? a b c ? ? ? ? = λ ? ? ? ? a b c ? ? ? ? 得到 λx1 = ? h, φx1 =

1 2 ? ? ? ?

1 √

2 1 ? ? ? ? λx2 = 0, φx2 =

1 √

2 ? ? ? ?

1 0 ?1 ? ? ? ? λx3 = ?? h, φx3 =

1 2 ? ? ? ?

1 ? √

2 1 ? ? ? ?

8 同理求得Ly的本征值和本征矢为 λy1 = ? h, φy1 =

1 2 ? ? ? ?

1 i √

2 ?1 ? ? ? ? λy2 = 0, φy2 =

1 √

2 ? ? ? ?

1 0

1 ? ? ? ? λy3 = ?? h, φy3 =

1 2 ? ? ? ?

1 ?i √

2 ?1 ? ? ? ? (b) 由(a)结果可知 φx1 =

1 2 |1,

1 + √

2 2 |1,

0 +

1 2 |1, ?1 φx2 =

1 √

2 |1,

1 ?

1 √

2 |1, ?1 φx3 =

1 2 |1,

1 ? √

2 2 |1,

0 +

1 2 |1, ?1 5.考虑谐振子升降算符a, a? 的线性变换 b = λa + νa? λ, ν为实数,满足λ2 ? ν2 = 1.证明:对于b的任意本征态|β >