PDF《Homework10 答案（2018-2019）》

Homework10 答案(2018-2019) 1.

对于一维谐振子,(a) 处于能量本征态|n .利用代数方法求动能期望值和势能期望值,及 其标准差.(b) 初态ψ(0) =

1 √

2 ψ0 +

1 √

2 ψ1, 在S-pic和H-pic下分别计算 p(t) . 解:(a)利用a, a? 表示? x, ? p ? x = ( ? h 2mω )

1 2 (a + a? ) (1) ? p = i( m? hω

2 )

1 2 (a? ? a) (2) 把式(1),(2)代入动能项 ? T和势能项 ? V 得到 ? T = ? p2 2m = ? ? hω

4 (a ? a? )(a ? a? ) = ? ? hω

4 (a2 + (a? )2 ? a? a ? aa? ) = ? ? hω

4 (a2 + (a? )2 ? 2a? a ? 1) = ? ? hω

4 (a2 + (a? )2 ? 2? n ? 1) (3) ? V =

1 2 mω2 ? x2 = ? hω

4 (a + a? )(a + a? ) = ? hω

4 (a2 + (a? )2 + a? a + aa? ) = ? hω

4 (a2 + (a? )2 + 2a? a + 1) = ? hω

4 (a2 + (a? )2 + 2? n + 1) (4) 又<

n|a2 |n >

=<

n|(a? )2 |n >

=0.因此, n| ? T|n = ? ? hω

4 n|a2 + (a? )2 ? 2? n ? 1|n

1 2 = ? ? hω

4 n| ? 2? n ? 1|n = ? hω(2n + 1)

4 (5) n| ? V |n = ? hω

4 n|a2 + (a? )2 + 2? n + 1|n = ? hω

4 n|2? n + 1|n = ? hω(2n + 1)

4 (6) n| ? T2 |n = ( ? hω

4 )2 n|(a2 + (a? )2 ? 2? n ? 1)2 |n = ( ? hω

4 )2 n|a2 (a? )2 + (a? )2 a2 + (2? n + 1)2 |n = ( (2n + 1)? hω

4 )2 + ( ? hω

4 )2 n|a(aa? )a? + a? (a? a)a|n = ( (2n + 1)? hω

4 )2 + ( ? hω

4 )2 n|a(a? a + 1)a? + a? ? na|n = ( (2n + 1)? hω

4 )2 + ( ? hω

4 )2 n|a? na? + aa? + a? ? na|n = ( (2n + 1)? hω

4 )2 + ( ? hω

4 )2 n|(n + 1)2 + n +

1 + (n ? 1)n|n = ( (2n + 1)? hω

4 )2 + ( ? hω

4 )2 n|2n2 + 2n + 2|n = ( ? hω

4 )2 (6n2 + 6n + 3) (7) 同理, n| ? V

2 |n = ( ? hω

4 )2 n|(a2 + (a? )2 + 2? n + 1)2 |n = ( ? hω

4 )2 (2n + 1)2 + ( ? hω

4 )2 n|a2 (a? )2 + (a? )2 a2 |n = ( ? hω

4 )2 (6n2 + 6n + 3) (8) 由以上结果得到其标准偏差 ?T = √ <

T2 >

? <

T >

2 = ? hω

4 √ 2n2 + 2n +

2 (9) ?V = √ <

V

2 >

? <

V >

2 = ? hω

4 √ 2n2 + 2n +

2 (10)

3 (b) 在Schrodinger picture下,我们很容易可以得到t时刻的波函数 |ψ(t) =

1 √

2 e? i ? h E0t |0 +

1 √

2 e? i ? h E1t |1 =

1 √

2 e?i

1 2 ωt |0 +

1 √

2 e?i

3 2 ωt |1 (11) 由t时刻的波函数我们可以计算 <

p >

= ψ(t)|p|ψ(t) =

1 2 0|p|0 +

1 2 1|p|1 +

1 2 eiωt 1|p|0 +

1 2 e?iωt 0|p|1 =

1 2 eiωt 1|p|0 +

1 2 e?iωt 0|p|1 = i( m? hω

2 )

1 2

1 2 (eiωt 1|(a? ? a)|0 + e?iωt 0|(a? ? a)|1 ) = i( m? hω

2 )

1 2

1 2 (eiωt ? e?iωt ) = ?( m? hω

2 )

1 2 sinωt (12) 在Heisenberg picture下,由Heisenberg 运动方程得 d? p(t) dt =

1 i? h [? p(t), ? H(t)] = ?mω2 ? x(t) (13) d? x(t) dt =

1 i? h [? x(t), ? H(t)] =

1 m ? p(t) (14) 我们可以得到关于p(t)的二阶微分方程 d2 ? p(t) dt2 = ?ω2 ? p(t) (15) dp(t) dt |t=0 = ?mω2 ? x(0) (16) 解得 ? p(t) = ?mω2 ? x(0)sinωt + ? p(0)cosωt (17) ? p(t)期望值为 ? p(t) = ψ(0)|? p(t)|ψ(0)

4 = ?mω2 sinωt ψ(0)|x(0)|ψ(0) + cosωt ψ(0)|p(0)|ψ(0) = ?mω2 sinωt( ? h 2mω )

1 2 +

0 = ?( m? hω

2 )

1 2 sinωt (18) 2. 设谐振子初态为相干态|α . 请问 (a) 计算量子数算符a? a的期望值和标准偏差. (b)计算动 量的期望值随时间变化情况和标准偏差.在什么情况下此标准偏差可以忽略? 解: α|a? a|α = α|α? α|α = |α|2 (19) α|(a? a)2 |α = α|a? aa? a|α = α|α? (a? a + 1)α|α = |α|2 + α|α? α? αα|α = |α|4 + |α|2 (20) ?(a? a) = α|(a?a)2|α ? α|a?a|α

2 = |α| (21) (b) t时刻相干态为 |α(t) = e? |α|2

2 e?i ω

2 t ∞ n=0 αn √ n! e?iωnt |n a|α(t) = e?iωt α|α(t) α(t)|a? = α(t)|eiωt α? (22) 得到 α(t)|p|α(t) = ?i( m? hω