耦合簇方法提供了一種近似求解不含時薛定諤方程 的方法：
H
^
|
Ψ
⟩
=
E
|
Ψ
⟩
{\displaystyle {\hat {H}}\vert {\Psi }\rangle =E\vert {\Psi }\rangle }
這裏
H
^
{\displaystyle {\hat {H}}}
|
Ψ
⟩
{\displaystyle \vert {\Psi }\rangle }
E 來表示。耦合簇理論的其它變體，如運動方程耦合簇方法 多參考態耦合簇方法 [ 4] [ 5]
體系的基態波函數可以用下面的擬設 來表出：
|
Ψ
⟩
=
e
T
^
|
Φ
0
⟩
{\displaystyle \vert {\Psi }\rangle =e^{\hat {T}}\vert {\Phi _{0}}\rangle }
式中
|
Φ
0
⟩
{\displaystyle \vert {\Phi _{0}}\rangle }
哈特里－福克 基態波函數，
T
^
{\displaystyle {\hat {T}}}
|
Φ
0
⟩
{\displaystyle \vert {\Phi _{0}}\rangle }
斯萊特行列式 的線性組合。（詳情見下文）
在擬設的選取上，CC 方法比起其它的方法例如組態相互作用方法 （CI）有優勢。這是因為這一擬設具有大小廣延性 。CC 方法的大小一致性取決於參考波函數的大小一致性。CC 方法的一個主要缺陷是，它不是變分 的。
簇算符由下式給出：
T
^
=
T
^
1
+
T
^
2
+
T
^
3
+
⋯
{\displaystyle {\hat {T}}={\hat {T}}_{1}+{\hat {T}}_{2}+{\hat {T}}_{3}+\cdots }
其中
T
^
1
{\displaystyle {\hat {T}}_{1}}
T
^
2
{\displaystyle {\hat {T}}_{2}}
正則量子化 表達為下列形式[ 6]
T
^
1
=
∑
i
∑
a
t
i
a
a
^
a
†
a
^
i
,
{\displaystyle {\hat {T}}_{1}=\sum _{i}\sum _{a}t_{i}^{a}{\hat {a}}_{a}^{\dagger }{\hat {a}}_{i},}
T
^
2
=
1
4
∑
i
,
j
∑
a
,
b
t
i
j
a
b
a
^
a
†
a
^
b
†
a
^
j
a
^
i
,
{\displaystyle {\hat {T}}_{2}={\frac {1}{4}}\sum _{i,j}\sum _{a,b}t_{ij}^{ab}{\hat {a}}_{a}^{\dagger }{\hat {a}}_{b}^{\dagger }{\hat {a}}_{j}{\hat {a}}_{i},}
余類推。
在上面的式子中，
a
^
†
{\displaystyle {\hat {a}}^{\dagger }}
a
^
{\displaystyle {\hat {a}}}
產生及湮沒算符 。下標 i, j 表示佔據軌道，而 a, b 表示空軌道。在耦合簇算符中的產生和湮沒算符按照正規序 排列。單粒子激發算符
T
^
1
{\displaystyle {\hat {T}}_{1}}
T
^
2
{\displaystyle {\hat {T}}_{2}}
|
Φ
0
⟩
{\displaystyle \vert {\Phi _{0}}\rangle }
t
i
a
{\displaystyle t_{i}^{a}}
t
i
j
a
b
{\displaystyle t_{ij}^{ab}}
考慮到簇算符
T
^
{\displaystyle {\hat {T}}}
e
T
^
{\displaystyle e^{\hat {T}}}
泰勒級數 ：
e
T
^
=
1
+
T
^
+
T
^
2
2
!
+
⋯
=
1
+
T
^
1
+
T
^
2
+
T
^
1
2
2
+
T
^
1
T
^
2
+
T
^
2
2
2
+
⋯
{\displaystyle e^{\hat {T}}=1+{\hat {T}}+{\frac {{\hat {T}}^{2}}{2!}}+\cdots =1+{\hat {T}}_{1}+{\hat {T}}_{2}+{\frac {{\hat {T}}_{1}^{2}}{2}}+{\hat {T}}_{1}{\hat {T}}_{2}+{\frac {{\hat {T}}_{2}^{2}}{2}}+\cdots }
事實上，這一級數是有限的，因為分子軌道的數目與激發的數目都是有限的。為了簡化求解係數
t
{\displaystyle t}
T
^
{\displaystyle {\hat {T}}}
T
^
5
{\displaystyle {\hat {T}}_{5}}
T
^
6
{\displaystyle {\hat {T}}_{6}}
n
{\displaystyle n}
T
^
=
T
^
1
+
.
.
.
+
T
^
n
{\displaystyle {\hat {T}}={\hat {T}}_{1}+...+{\hat {T}}_{n}}
那麼由於耦合算符具有指數形式，高於
n
{\displaystyle n}
T
^
n
{\displaystyle {\hat {T}}_{n}}
n
{\displaystyle n}
耦合簇方程就是展開係數
t
{\displaystyle t}
H
^
e
T
^
|
Ψ
0
⟩
=
E
e
T
^
|
Ψ
0
⟩
{\displaystyle {\hat {H}}e^{\hat {T}}\vert {\Psi _{0}}\rangle =Ee^{\hat {T}}\vert {\Psi _{0}}\rangle }
假設現在共有
q
{\displaystyle q}
t
{\displaystyle t}
q
{\displaystyle q}
t
{\displaystyle t}
t
i
j
k
.
.
.
a
b
c
.
.
.
{\displaystyle t_{ijk...}^{abc...}}
|
Φ
0
⟩
{\displaystyle \vert {\Phi _{0}}\rangle }
i
,
j
,
k
,
⋯
{\displaystyle i,j,k,\cdots }
a
,
b
,
c
,
⋯
{\displaystyle a,b,c,\cdots }
q
{\displaystyle q}
⟨
Ψ
∗
|
H
^
e
T
^
|
Ψ
0
⟩
=
E
⟨
Ψ
∗
|
e
T
^
|
Ψ
0
⟩
{\displaystyle \langle {\Psi ^{*}}\vert {\hat {H}}e^{\hat {T}}\vert {\Psi _{0}}\rangle =E\langle {\Psi ^{*}}\vert e^{\hat {T}}\vert {\Psi _{0}}\rangle }
式中
|
Ψ
∗
⟩
{\displaystyle \vert {\Psi ^{*}}\rangle }
t
{\displaystyle t}
e
−
T
^
{\displaystyle e^{-{\hat {T}}}}
Ψ
0
{\displaystyle \Psi _{0}}
Ψ
∗
{\displaystyle \Psi ^{*}}
⟨
Ψ
0
|
e
−
T
^
H
^
e
T
^
|
Ψ
0
⟩
=
E
{\displaystyle \langle {\Psi _{0}}\vert e^{-{\hat {T}}}{\hat {H}}e^{\hat {T}}\vert {\Psi _{0}}\rangle =E}
⟨
Ψ
∗
|
e
−
T
^
H
^
e
T
^
|
Ψ
0
⟩
=
E
⟨
Ψ
∗
|
e
−
T
^
e
T
^
|
Ψ
0
⟩
=
0
{\displaystyle \langle {\Psi ^{*}}\vert e^{-{\hat {T}}}{\hat {H}}e^{\hat {T}}\vert {\Psi _{0}}\rangle =E\langle {\Psi ^{*}}\vert e^{-{\hat {T}}}e^{\hat {T}}\vert {\Psi _{0}}\rangle =0}
第一式提供了求解 CC 能量的方法，第二式則是用來求解
t
{\displaystyle t}
⟨
Ψ
0
|
e
−
(
T
^
1
+
T
^
2
)
H
^
e
(
T
^
1
+
T
^
2
)
|
Ψ
0
⟩
=
E
{\displaystyle \langle {\Psi _{0}}\vert e^{-({\hat {T}}_{1}+{\hat {T}}_{2})}{\hat {H}}e^{({\hat {T}}_{1}+{\hat {T}}_{2})}\vert {\Psi _{0}}\rangle =E}
⟨
Ψ
S
|
e
−
(
T
^
1
+
T
^
2
)
H
^
e
(
T
^
1
+
T
^
2
)
|
Ψ
0
⟩
=
0
{\displaystyle \langle {\Psi _{S}}\vert e^{-({\hat {T}}_{1}+{\hat {T}}_{2})}{\hat {H}}e^{({\hat {T}}_{1}+{\hat {T}}_{2})}\vert {\Psi _{0}}\rangle =0}
⟨
Ψ
D
|
e
−
(
T
^
1
+
T
^
2
)
H
^
e
(
T
^
1
+
T
^
2
)
|
Ψ
0
⟩
=
0
{\displaystyle \langle {\Psi _{D}}\vert e^{-({\hat {T}}_{1}+{\hat {T}}_{2})}{\hat {H}}e^{({\hat {T}}_{1}+{\hat {T}}_{2})}\vert {\Psi _{0}}\rangle =0}
上式中經相似變換後的哈密頓量（用
H
¯
{\displaystyle {\bar {H}}}
BCH 公式
H
¯
=
e
−
T
^
H
^
e
T
^
=
H
^
+
[
H
^
,
T
^
]
+
1
2
[
[
H
^
,
T
^
]
,
T
^
]
+
⋯
{\displaystyle {\bar {H}}=e^{-{\hat {T}}}{\hat {H}}e^{\hat {T}}={\hat {H}}+\left[{\hat {H}},{\hat {T}}\right]+{\frac {1}{2}}\left[\left[{\hat {H}},{\hat {T}}\right],{\hat {T}}\right]+\cdots }
H
¯
{\displaystyle {\bar {H}}}
傳統上耦合簇方法依照
T
^
{\displaystyle {\hat {T}}}
T
^
n
{\displaystyle {\hat {T}}_{n}}
S - 單激發 (在英語的 CC 術語裏面簡稱 singles )
D - 雙激發 (doubles )
T - 三激發 (triples )
Q - 四激發 (quadruples ) 例如，CCSDT 方法裏面簇算符
T
^
{\displaystyle {\hat {T}}}
T
=
T
^
1
+
T
^
2
+
T
^
3
.
{\displaystyle T={\hat {T}}_{1}+{\hat {T}}_{2}+{\hat {T}}_{3}.}
在圓括號裏面的項則表示它們是通過微擾理論 求得的。例如 CCSD(T) 表示：
耦合簇方法
包含完整的單激發和雙激發
三激發則採用微擾理論而不是迭代求解
^
Kümmel, H. G. A biography of the coupled cluster method. Xian, R. F.; Brandes, T.; Gernoth, K. A.; Walet, N. R. (編). Recent progress in many-body theories Proceedings of the 11th international conference. Singapore: World Scientific Publishing. 2002: 334–348. ISBN 978-981-02-4888-8
^ Cramer, Christopher J. Essentials of Computational Chemistry . Chichester: John Wiley & Sons, Ltd. 2002: 191 –232. ISBN 0-471-48552-7 ^
Shavitt, Isaiah; Bartlett, Rodney J. Many-Body Methods in Chemistry and Physics: MBPT and Coupled-Cluster Theory. Cambridge University Press. 2009. ISBN 978-0-521-81832-2
^ Koch, Henrik; Jo̸rgensen, Poul. Coupled cluster response functions. The Journal of Chemical Physics. 1990, 93 : 3333. Bibcode:1990JChPh..93.3333K doi:10.1063/1.458814 ^ Stanton, John F.; Bartlett, Rodney J. The equation of motion coupled-cluster method. A systematic biorthogonal approach to molecular excitation energies, transition probabilities, and excited state properties. The Journal of Chemical Physics. 1993, 98 : 7029. Bibcode:1993JChPh..98.7029S doi:10.1063/1.464746 ^ The Cluster Operator . [2012-06-24 ] . （原始內容 存檔於2012-06-16）.
![{\displaystyle {\bar {H}}=e^{-{\hat {T}}}{\hat {H}}e^{\hat {T}}={\hat {H}}+\left[{\hat {H}},{\hat {T}}\right]+{\frac {1}{2}}\left[\left[{\hat {H}},{\hat {T}}\right],{\hat {T}}\right]+\cdots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a38e833503154cfb18fed144ff4be1eb5f3e98cc)
