PDF《Homework10 答案（2018-2019）》

有?x ・ ?y = ? h/2. (提示:证明[b, b? ] = 1,再用b,b? 表示x, p). 证明: [b, b+ ] = [λa + νa+ , λa+ + νa] = (λa + νa+ )(λa+ + νa) ? (λa+ + νa)(λa + νa+ ) = (λ2 ? ν2 )[a, a+ ] =

1 (33) 逆变换得到a = λb ? νb+ ,a+ = λb+ ? νb,代入x =

1 √ 2α (a + a+ )及p = i? hα √

2 (a+ ? a),有x=1√2α (λ ? ν)(b + b+ ) (34)

9 p = i? hα √

2 (λ + ν)(b+ ? b) (35) 对b的本征态|β >

,b|β >

= β|β >

<

x >

=

1 √ 2α (λ ? ν) <

β|b + b+ |β >

=

1 √ 2α (λ ? ν)(β + β? ) (36) <

x2 >

=

1 2α2 (λ ? ν)2 <

β|b2 + b+2 + bb+ + b+ b|β >

=

1 2α2 (λ ? ν)2 <

β|b2 + b+2 + 2b+ b + 1|β >

=

1 2α2 (λ ? ν)2 (β2 + β?2 + 2|β|2 + 1) (37) <

p >

= i? hα √

2 (λ + ν) <

β|b+ ? b|β >

= i? hα √

2 (λ + ν)(β? ? β) (38) <

p2 >

= ? ? h2 α2

2 (λ + ν)2 <

β|b2 + b+2 ? bb+ ? b+ b|β >

= ? ? h2 α2

2 (λ + ν)2 <

β|b2 + b+2 ? 2b+ b ? 1|β >

= ? ? h2 α2

2 (λ + ν)2 (β2 + β?2 ? 2|β|2 ? 1) (39) 则?x = √ <

x2 >

? <

x >

2 =

1 √ 2α (λ ? ν) (40) ?p = <

p2 >

? <

p >

2 = ? hα √

2 (λ + ν) (41) 所以 ?x ・ ?p = ? h

2 (42) 6. 某体系能量算符为H =

5 3 a? a +

2 3 (a2 + (a? )2 ), [a, a? ] = 1. 求体系的能谱(所有能量本征态) 和基态波函数. (提示:引入b = ua + va? , 使得[b, b? ] = 1, 将H表示成 b? b的函数) 解:令b = ua + va? , b? = ua? + va, 其中u, v为实数.计算b, b? 的对易关系 [b, b? ] = [ua + va? , ua? + va] = [ua, ua? ] + [va? , va] = u2 ? v2 (43)

10 为使[b, b? ] = 1, 要求 u2 ? v2 =

1 (44) 用b, b? 来表示a, a? a = ub ? vb? (45) a? = ub? ? vb (46) 代入哈密顿量中得到 H =

5 3 (ub? ? vb)(ub ? vb? ) +

2 3 (ub ? vb? )2 +

2 3 (ub? ? vb)2 =

5 3 (u2 b? b + v2 bb? ? uv(b? )2 ? vub2 ) +

2 3 (v2 (b? )2 + u2 b2 ? uvb? b ? uvbb? ) +

2 3 (u2 (b? )2 + v2 b2 ? uvb? b ? uvbb? ) = [

5 3 (u2 + v2 ) ?

8 3 uv]b? b + [

2 3 (u2 + v2 ) ?

5 3 uv][(b? )2 + b2 ] +

5 3 v2 ?

4 3 uv (47) 为了消除第二项,要求

2 3 (u2 + v2 ) ?

5 3 uv =

0 (48) 结合式44,得到u =

4 3 , v =

1 3 , 哈密顿量化简为 H = b? b ?

1 3 (49) 对应能量本征态为En = n ?

1 3 , n = 0, 1, 2.... 设基态波函数为|ψ = n=0 cn|n , n |cn|2 = 1, 则有

0 = b|ψ = (ua + va? )( n=0 cn|n ) = n=1 cnu √ n|n ?

1 + n=0 cnv √ n + 1|n +

1 = c1u|0 + n=1 (ucn+1 √ n + 1|n + vcn?1 √ n)|n (50) 将u =

4 3 , v =

1 3 代入,我们得到 cn = 0, n = 1, 3, 5... 2cn+2 √ n +

2 + cn √ n +

1 = 0, n = 0, 2, 4...

11 解得 cn = ? ? ? ? ? 0, n = 1, 3, 5... c0 (n ? 1)!! n!!

1 2n/2 , n = 2, 4, 6... (51) |ψ = c0 n=0,2,4... (n ? 1)!! n!!

1 2n/2 |n (52) ........