في ميكانيكا الكم قاعدة الجمع في ميكانيكا الكم (بالإنجليزي : Sum rule in quantum mechanics) تصف الانتقالات بين مستويات,كما تستخدم لوصف العديد من الخصائص للأنظمة الفيزيائية الصلبة والذرية,النووية,نواة الذرة .
استنتاج مجموعة من القواعد[1]
عدل
نفرض أن هاميلتونيان
H
^
{\displaystyle {\hat {H}}}
|
n
⟩
{\displaystyle |n\rangle }
القيم الذاتية
ϵ
n
{\displaystyle \epsilon _{n}}
H
^
|
n
⟩
=
ϵ
n
|
n
⟩
.
{\displaystyle {\hat {H}}|n\rangle =\epsilon _{n}|n\rangle .}
نحدد معكوس المؤثر الهرميتي
A
^
{\displaystyle {\hat {A}}}
C
^
(
0
)
≡
A
^
C
^
(
1
)
≡
[
H
^
,
A
^
]
=
H
^
A
^
−
A
^
H
^
C
^
(
k
)
≡
[
H
^
,
C
^
(
k
−
1
)
]
,
k
=
1
,
2
,
…
{\displaystyle {\begin{aligned}{\hat {C}}^{(0)}&\equiv {\hat {A}}\\{\hat {C}}^{(1)}&\equiv [{\hat {H}},{\hat {A}}]={\hat {H}}{\hat {A}}-{\hat {A}}{\hat {H}}\\{\hat {C}}^{(k)}&\equiv [{\hat {H}},{\hat {C}}^{(k-1)}],\ \ \ k=1,2,\ldots \end{aligned}}}
نجد أن
C
^
(
0
)
{\displaystyle {\hat {C}}^{(0)}}
A
^
{\displaystyle {\hat {A}}}
C
^
(
1
)
{\displaystyle {\hat {C}}^{(1)}}
مؤثر غير هيرميتي
(
C
^
(
1
)
)
†
=
(
H
^
A
^
)
†
−
(
A
^
H
^
)
†
=
A
^
H
^
−
H
^
A
^
=
−
C
^
(
1
)
.
{\displaystyle \left({\hat {C}}^{(1)}\right)^{\dagger }=({\hat {H}}{\hat {A}})^{\dagger }-({\hat {A}}{\hat {H}})^{\dagger }={\hat {A}}{\hat {H}}-{\hat {H}}{\hat {A}}=-{\hat {C}}^{(1)}.}
وبالاستقراء نجد
(
C
^
(
k
)
)
†
=
(
−
1
)
k
C
^
(
k
)
{\displaystyle \left({\hat {C}}^{(k)}\right)^{\dagger }=(-1)^{k}{\hat {C}}^{(k)}}
أيضا
⟨
m
|
C
^
(
k
)
|
n
⟩
=
(
E
m
−
E
n
)
k
⟨
m
|
A
^
|
n
⟩
.
{\displaystyle \langle m|{\hat {C}}^{(k)}|n\rangle =(E_{m}-E_{n})^{k}\langle m|{\hat {A}}|n\rangle .}
|
⟨
m
|
A
^
|
n
⟩
|
2
=
⟨
m
|
A
^
|
n
⟩
⟨
m
|
A
^
|
n
⟩
∗
=
⟨
m
|
A
^
|
n
⟩
⟨
n
|
A
^
|
m
⟩
.
{\displaystyle |\langle m|{\hat {A}}|n\rangle |^{2}=\langle m|{\hat {A}}|n\rangle \langle m|{\hat {A}}|n\rangle ^{\ast }=\langle m|{\hat {A}}|n\rangle \langle n|{\hat {A}}|m\rangle .}
وباستخدام هذه العلاقة
⟨
m
|
[
A
^
,
C
^
(
k
)
]
|
m
⟩
=
⟨
m
|
A
^
C
^
(
k
)
|
m
⟩
−
⟨
m
|
C
^
(
k
)
A
^
|
m
⟩
=
∑
n
⟨
m
|
A
^
|
n
⟩
⟨
n
|
C
^
(
k
)
|
m
⟩
−
⟨
m
|
C
^
(
k
)
|
n
⟩
⟨
n
|
A
^
|
m
⟩
=
∑
n
⟨
m
|
A
^
|
n
⟩
⟨
n
|
A
^
|
m
⟩
(
E
n
−
E
m
)
k
−
(
E
m
−
E
n
)
k
⟨
m
|
A
^
|
n
⟩
⟨
n
|
A
^
|
m
⟩
=
∑
n
(
1
−
(
−
1
)
k
)
(
E
n
−
E
m
)
k
|
⟨
m
|
A
^
|
n
⟩
|
2
.
{\displaystyle {\begin{aligned}\langle m|[{\hat {A}},{\hat {C}}^{(k)}]|m\rangle &=\langle m|{\hat {A}}{\hat {C}}^{(k)}|m\rangle -\langle m|{\hat {C}}^{(k)}{\hat {A}}|m\rangle \\&=\sum _{n}\langle m|{\hat {A}}|n\rangle \langle n|{\hat {C}}^{(k)}|m\rangle -\langle m|{\hat {C}}^{(k)}|n\rangle \langle n|{\hat {A}}|m\rangle \\&=\sum _{n}\langle m|{\hat {A}}|n\rangle \langle n|{\hat {A}}|m\rangle (E_{n}-E_{m})^{k}-(E_{m}-E_{n})^{k}\langle m|{\hat {A}}|n\rangle \langle n|{\hat {A}}|m\rangle \\&=\sum _{n}(1-(-1)^{k})(E_{n}-E_{m})^{k}|\langle m|{\hat {A}}|n\rangle |^{2}.\end{aligned}}}
يمكن كتابة النتيجة كالتالي :
⟨
m
|
[
A
^
,
C
^
(
k
)
]
|
m
⟩
=
{
0
,
if
k
is even
2
∑
n
(
E
n
−
E
m
)
k
|
⟨
m
|
A
^
|
n
⟩
|
2
,
if
k
is odd
.
{\displaystyle \langle m|[{\hat {A}},{\hat {C}}^{(k)}]|m\rangle ={\begin{cases}0,&{\mbox{if }}k{\mbox{ is even}}\\2\sum _{n}(E_{n}-E_{m})^{k}|\langle m|{\hat {A}}|n\rangle |^{2},&{\mbox{if }}k{\mbox{ is odd}}.\end{cases}}}
وبوضع
k
=
1
{\displaystyle k=1}
⟨
m
|
[
A
^
,
[
H
^
,
A
^
]
]
|
m
⟩
=
2
∑
n
(
E
n
−
E
m
)
|
⟨
m
|
A
^
|
n
⟩
|
2
.
{\displaystyle \langle m|[{\hat {A}},[{\hat {H}},{\hat {A}}]]|m\rangle =2\sum _{n}(E_{n}-E_{m})|\langle m|{\hat {A}}|n\rangle |^{2}.}
انظر أيضا
عدل
المراجع
عدل
![{\displaystyle {\begin{aligned}{\hat {C}}^{(0)}&\equiv {\hat {A}}\\{\hat {C}}^{(1)}&\equiv [{\hat {H}},{\hat {A}}]={\hat {H}}{\hat {A}}-{\hat {A}}{\hat {H}}\\{\hat {C}}^{(k)}&\equiv [{\hat {H}},{\hat {C}}^{(k-1)}],\ \ \ k=1,2,\ldots \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2e38c4e3192a59e4492e20010700727a6e2b1f18)
![{\displaystyle {\begin{aligned}\langle m|[{\hat {A}},{\hat {C}}^{(k)}]|m\rangle &=\langle m|{\hat {A}}{\hat {C}}^{(k)}|m\rangle -\langle m|{\hat {C}}^{(k)}{\hat {A}}|m\rangle \\&=\sum _{n}\langle m|{\hat {A}}|n\rangle \langle n|{\hat {C}}^{(k)}|m\rangle -\langle m|{\hat {C}}^{(k)}|n\rangle \langle n|{\hat {A}}|m\rangle \\&=\sum _{n}\langle m|{\hat {A}}|n\rangle \langle n|{\hat {A}}|m\rangle (E_{n}-E_{m})^{k}-(E_{m}-E_{n})^{k}\langle m|{\hat {A}}|n\rangle \langle n|{\hat {A}}|m\rangle \\&=\sum _{n}(1-(-1)^{k})(E_{n}-E_{m})^{k}|\langle m|{\hat {A}}|n\rangle |^{2}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0856da3cfd137892a729122ae07b0d76030fd99)
![{\displaystyle \langle m|[{\hat {A}},{\hat {C}}^{(k)}]|m\rangle ={\begin{cases}0,&{\mbox{if }}k{\mbox{ is even}}\\2\sum _{n}(E_{n}-E_{m})^{k}|\langle m|{\hat {A}}|n\rangle |^{2},&{\mbox{if }}k{\mbox{ is odd}}.\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9170c4d8b9085011605cca49e3810a2e64e4a525)
![{\displaystyle \langle m|[{\hat {A}},[{\hat {H}},{\hat {A}}]]|m\rangle =2\sum _{n}(E_{n}-E_{m})|\langle m|{\hat {A}}|n\rangle |^{2}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/36320be103235cab1fcbb667f753d87d88f9507c)
