在量子力学 里，相互作用绘景 （interaction picture），是在薛定谔绘景 与海森堡绘景 之间的一种表述，为纪念物理学者保罗·狄拉克 而又命名为狄拉克绘景 。在这绘景里，描述量子系统的态矢量 与表达可观察量 的算符 都会随着时间 流易而演化。有些实际案例会涉及到因相互作用而使得量子态与可观察量发生改变，这类案例通常会使用狄拉克绘景。
保罗·狄拉克 狄拉克绘景与薛定谔绘景 、海森堡绘景 不同。在薛定谔绘景里，描述量子系统的态矢量 随着时间流易而演化。在海森堡绘景里，表达可观察量 的算符 会随着时间流易而演化。
这三种绘景殊途同归，所获得的结果完全一致。这是必然的，因为它们都是在表达同样的物理行为。[ 1] :80-84 [ 2] [ 3]
为了便利分析，位于下标的符号
H
{\displaystyle {}_{\mathcal {H}}}
I
{\displaystyle {}_{\mathcal {I}}}
S
{\displaystyle {}_{\mathcal {S}}}
通过对于基底 的一种幺正变换
在量子力学里，对于大多数案例的哈密顿量 ，通常无法找到薛定谔方程 的精确解，只有少数案例可以找到精确解。因此，为了要能够解析其它没有精确解的案例，必须将薛定谔绘景里的哈密顿量
H
S
{\displaystyle H_{\mathcal {S}}\,\!}
[ 1] :337-339
H
S
=
H
0
,
S
+
H
1
,
S
{\displaystyle H_{\mathcal {S}}=H_{0,\,{\mathcal {S}}}+H_{1,\,{\mathcal {S}}}\,\!}
其中，
H
0
,
S
{\displaystyle H_{0,\,{\mathcal {S}}}\,\!}
H
1
,
S
{\displaystyle H_{1,\,{\mathcal {S}}}\,\!}
摄动 。
假若哈密顿量
H
S
{\displaystyle H_{\mathcal {S}}\,\!}
外电场 作用的量子系统，其哈密顿量会含时），则通常会将显性 含时部分放在
H
1
,
S
{\displaystyle H_{1,\,{\mathcal {S}}}\,\!}
H
0
,
S
{\displaystyle H_{0,\,{\mathcal {S}}}\,\!}
U
(
t
)
{\displaystyle U(t)\,\!}
U
(
t
)
=
e
−
i
H
0
,
S
t
/
ℏ
{\displaystyle U(t)=e^{-iH_{0,\,{\mathcal {S}}}\,t/\hbar }\,\!}
其中，
t
{\displaystyle t\,\!}
假若对于某些案例，
H
0
,
S
{\displaystyle H_{0,\,{\mathcal {S}}}\,\!}
时间演化算符 的公式会变得较为复杂：[ 1] :70-71
U
(
t
)
=
e
−
i
ℏ
∫
0
t
H
0
,
S
(
t
′
)
d
t
′
{\displaystyle U(t)=e^{-{\frac {i}{\hbar }}\int \limits _{0}^{t}H_{0,\,{\mathcal {S}}}(t^{'})\,dt^{'}}\,\!}
本条目以下内容假设
H
0
,
S
{\displaystyle H_{0,\,{\mathcal {S}}}\,\!}
在狄拉克绘景里，态矢量
|
ψ
(
t
)
⟩
I
{\displaystyle |\psi (t)\rangle _{\mathcal {I}}\,\!}
|
ψ
(
t
)
⟩
I
=
d
e
f
e
i
H
0
,
S
t
/
ℏ
|
ψ
(
t
)
⟩
S
{\displaystyle |\psi (t)\rangle _{\mathcal {I}}{\stackrel {def}{=}}e^{iH_{0,\,{\mathcal {S}}}\,t/\hbar }|\psi (t)\rangle _{\mathcal {S}}\,\!}
其中，
|
ψ
(
t
)
⟩
S
{\displaystyle |\psi (t)\rangle _{\mathcal {S}}\,\!}
由于在薛定谔绘景里， 态矢量
|
ψ
(
t
)
⟩
S
{\displaystyle |\psi (t)\rangle _{\mathcal {S}}\,\!}
|
ψ
(
t
)
⟩
S
=
e
−
i
H
S
t
/
ℏ
|
ψ
(
0
)
⟩
S
{\displaystyle |\psi (t)\rangle _{\mathcal {S}}=e^{-iH_{\mathcal {S}}\,t/\hbar }|\psi (0)\rangle _{\mathcal {S}}\,\!}
所以，在
H
0
,
S
,
H
S
{\displaystyle H_{0,{\mathcal {S}}},H_{\mathcal {S}}}
|
ψ
(
t
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⟩
I
=
e
−
i
H
1
,
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t
/
ℏ
|
ψ
(
0
)
⟩
S
{\displaystyle |\psi (t)\rangle _{\mathcal {I}}=e^{-iH_{1,\,{\mathcal {S}}}\,t/\hbar }|\psi (0)\rangle _{\mathcal {S}}\,\!}
在狄拉克绘景里的算符
A
I
(
t
)
{\displaystyle A_{\mathcal {I}}(t)\,\!}
A
I
(
t
)
=
e
i
H
0
,
S
t
/
ℏ
A
S
(
t
)
e
−
i
H
0
,
S
t
/
ℏ
{\displaystyle A_{\mathcal {I}}(t)=e^{iH_{0,\,{\mathcal {S}}}\,t/\hbar }A_{\mathcal {S}}(t)\,e^{-iH_{0,\,{\mathcal {S}}}\,t/\hbar }\,\!}
其中，
A
S
(
t
)
{\displaystyle A_{\mathcal {S}}(t)\,\!}
（请注意，
A
S
(
t
)
{\displaystyle A_{\mathcal {S}}(t)\,\!}
A
S
{\displaystyle A_{\mathcal {S}}\,\!}
H
S
(
t
)
{\displaystyle H_{\mathcal {S}}(t)\,\!}
假若
H
0
,
S
{\displaystyle H_{0,\,{\mathcal {S}}}\,\!}
H
0
,
S
{\displaystyle H_{0,\,{\mathcal {S}}}\,\!}
e
i
H
0
,
S
t
/
ℏ
{\displaystyle e^{iH_{0,\,{\mathcal {S}}}\,t/\hbar }\,\!}
对易 ，不论在薛定谔绘景里或在狄拉克绘景里，
H
0
,
S
{\displaystyle H_{0,\,{\mathcal {S}}}\,\!}
H
0
,
I
{\displaystyle H_{0,\,{\mathcal {I}}}\,\!}
[ 注 1]
H
0
,
I
(
t
)
=
e
i
H
0
,
S
t
/
ℏ
H
0
,
S
e
−
i
H
0
,
S
t
/
ℏ
=
H
0
,
S
{\displaystyle H_{0,\,{\mathcal {I}}}(t)=e^{iH_{0,\,{\mathcal {S}}}\,t/\hbar }H_{0,\,{\mathcal {S}}}\,e^{-iH_{0,\,{\mathcal {S}}}\,t/\hbar }=H_{0,\,{\mathcal {S}}}\,\!}
所以，算符
H
0
,
S
{\displaystyle H_{0,\,{\mathcal {S}}}\,\!}
H
0
,
I
{\displaystyle H_{0,\,{\mathcal {I}}}\,\!}
H
0
{\displaystyle H_{0}\,\!}
哈密顿算符的摄动成分
H
1
,
I
{\displaystyle H_{1,\,{\mathcal {I}}}\,\!}
H
1
,
I
(
t
)
=
e
i
H
0
,
S
t
/
ℏ
H
1
,
S
e
−
i
H
0
,
S
t
/
ℏ
{\displaystyle H_{1,\,{\mathcal {I}}}(t)=e^{iH_{0,\,{\mathcal {S}}}\,t/\hbar }H_{1,\,{\mathcal {S}}}\,e^{-iH_{0,\,{\mathcal {S}}}\,t/\hbar }\,\!}
除非对易关系式
[
H
1
,
S
,
H
0
,
S
]
=
0
{\displaystyle [H_{1,\,{\mathcal {S}}},H_{0,\,{\mathcal {S}}}]=0\,\!}
H
1
,
I
{\displaystyle H_{1,\,{\mathcal {I}}}\,\!}
与算符类似，在薛定谔绘景里的密度矩阵 也可以变换到在狄拉克绘景里。设定
ρ
I
{\displaystyle \rho _{\mathcal {I}}\,\!}
ρ
S
{\displaystyle \rho _{\mathcal {S}}\,\!}
|
ψ
n
⟩
{\displaystyle |\psi _{n}\rangle \,\!}
p
n
{\displaystyle p_{n}\,\!}
ρ
I
(
t
)
=
∑
n
p
n
|
ψ
n
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t
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⟩
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I
⟨
ψ
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p
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e
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n
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⟩
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⟨
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n
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−
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,
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t
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ℏ
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i
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/
ℏ
ρ
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−
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ℏ
{\displaystyle {\begin{aligned}\rho _{\mathcal {I}}(t)&=\sum _{n}p_{n}|\psi _{n}(t)\rangle _{\mathcal {I}}\,{}_{\mathcal {I}}\langle \psi _{n}(t)|\\&=\sum _{n}p_{n}\,e^{iH_{0,\,{\mathcal {S}}}\,t/\hbar }|\psi _{n}(t)\rangle _{\mathcal {S}}\,{}_{\mathcal {S}}\langle \psi _{n}(t)|e^{-iH_{0,\,{\mathcal {S}}}\,t/\hbar }\\&=e^{iH_{0,\,{\mathcal {S}}}\,t/\hbar }\rho _{\mathcal {S}}(t)\,e^{-iH_{0,\,{\mathcal {S}}}\,t/\hbar }\\\end{aligned}}\,\!}
。
从态矢量的定义式，可以得到态矢量对于时间的导数是
i
ℏ
d
d
t
|
ψ
(
t
)
⟩
I
=
e
i
H
0
t
/
ℏ
[
−
H
0
|
ψ
(
t
)
⟩
S
+
i
ℏ
d
d
t
|
ψ
(
t
)
⟩
S
]
=
e
i
H
0
t
/
ℏ
[
−
H
0
|
ψ
(
t
)
⟩
S
+
H
S
|
ψ
(
t
)
⟩
S
]
=
e
i
H
0
t
/
ℏ
H
1
,
S
|
ψ
(
t
)
⟩
S
=
e
i
H
0
t
/
ℏ
H
1
,
S
e
−
i
H
0
t
/
ℏ
|
ψ
(
t
)
⟩
I
{\displaystyle {\begin{aligned}i\hbar {\frac {d}{dt}}|\psi (t)\rangle _{\mathcal {I}}&=e^{iH_{0}\,t/\hbar }\left[-H_{0}|\psi (t)\rangle _{\mathcal {S}}+i\hbar {\frac {d}{dt}}|\psi (t)\rangle _{\mathcal {S}}\right]\\&=e^{iH_{0}\,t/\hbar }\left[-H_{0}|\psi (t)\rangle _{\mathcal {S}}+H_{\mathcal {S}}|\psi (t)\rangle _{\mathcal {S}}\right]\\&=e^{iH_{0}\,t/\hbar }H_{1,\,{\mathcal {S}}}|\psi (t)\rangle _{\mathcal {S}}\\&=e^{iH_{0}\,t/\hbar }H_{1,\,{\mathcal {S}}}\,e^{-iH_{0}\,t/\hbar }|\psi (t)\rangle _{\mathcal {I}}\\\end{aligned}}}
将算符的定义式代入，可以得到
i
ℏ
d
d
t
|
ψ
(
t
)
⟩
I
=
H
1
,
I
|
ψ
(
t
)
⟩
I
{\displaystyle i\hbar {\frac {d}{dt}}|\psi (t)\rangle _{\mathcal {I}}=H_{1,\,{\mathcal {I}}}|\psi (t)\rangle _{\mathcal {I}}\,\!}
这是施温格-朝永振一郎方程 [ 4] :153-155
假若算符
A
S
{\displaystyle A_{\mathcal {S}}\,\!}
A
I
(
t
)
{\displaystyle A_{\mathcal {I}}(t)\,\!}
i
ℏ
d
d
t
A
I
(
t
)
=
i
ℏ
d
d
t
(
e
i
H
0
t
/
ℏ
A
S
e
−
i
H
0
t
/
ℏ
)
=
−
H
0
e
i
H
0
t
/
ℏ
A
S
e
−
i
H
0
t
/
ℏ
+
e
i
H
0
t
/
ℏ
A
S
e
−
i
H
0
t
/
ℏ
H
0
=
A
I
(
t
)
H
0
−
H
0
A
I
(
t
)
=
[
A
I
(
t
)
,
H
0
]
{\displaystyle {\begin{aligned}i\hbar {\frac {d}{dt}}A_{\mathcal {I}}(t)&=i\hbar {\frac {d}{dt}}(e^{iH_{0}\,t/\hbar }A_{\mathcal {S}}\,e^{-iH_{0}\,t/\hbar })\\&=-H_{0}\,e^{iH_{0}\,t/\hbar }A_{\mathcal {S}}\,e^{-iH_{0}\,t/\hbar }+e^{iH_{0}\,t/\hbar }A_{\mathcal {S}}\,e^{-iH_{0}\,t/\hbar }H_{0}\\&=A_{\mathcal {I}}(t)H_{0}-H_{0}A_{\mathcal {I}}(t)\\&=\left[A_{\mathcal {I}}(t),\,H_{0}\right]\\\end{aligned}}\,\!}
。 这与在海森堡绘景里，算符
A
H
(
t
)
{\displaystyle A_{\mathcal {H}}(t)\,\!}
i
ℏ
d
d
t
A
H
(
t
)
=
[
A
H
(
t
)
,
H
]
{\displaystyle i\hbar {\frac {d}{dt}}A_{\mathcal {H}}(t)=\left[A_{\mathcal {H}}(t),\,H\right]\,\!}
应用施温格-朝永振一郎方程于密度矩阵，则可得到
i
ℏ
d
d
t
ρ
I
(
t
)
=
[
H
1
,
I
(
t
)
,
ρ
I
(
t
)
]
{\displaystyle i\hbar {\frac {d}{dt}}\rho _{\mathcal {I}}(t)=\left[H_{1,\,{\mathcal {I}}}(t),\rho _{\mathcal {I}}(t)\right]\,\!}
应用狄拉克绘景的目的是促使
H
0
{\displaystyle H_{0}\,\!}
H
1
,
I
(
t
)
{\displaystyle H_{1,\,{\mathcal {I}}}(t)\,\!}
H
1
,
I
(
t
)
{\displaystyle H_{1,\,{\mathcal {I}}}(t)\,\!}
假若
H
0
{\displaystyle H_{0}\,\!}
H
1
,
S
(
t
)
{\displaystyle H_{1,\,{\mathcal {S}}}(t)\,\!}
时变摄动理论 来计算
H
1
,
S
(
t
)
{\displaystyle H_{1,\,{\mathcal {S}}}(t)\,\!}
费米黄金定则 的导引里[ 1] :359–363 ，或在推导戴森级数 [ 1] :355–357 ，通常都会用到狄拉克绘景。
Sakurai, J. J.; Napolitano, Jim, Modern Quantum Mechanics 2nd, Addison-Wesley, 2010, ISBN 978-0805382914 Parker, C.B. McGraw Hill Encyclopaedia of Physics 2nd. Mc Graw Hill. 1994: 786, 1261. ISBN 0-07-051400-3 Y. Peleg, R. Pnini, E. Zaarur, E. Hecht. Quantum mechanics. Schuam's outline series 2nd. McGraw Hill. 2010: 70. ISBN 9-780071-623582 Ian J R Aitchison; Anthony J.G. Hey. Gauge Theories in Particle Physics: A Practical Introduction, Volume 1: From Relativistic Quantum Mechanics to QED, Fourth Edition. CRC Press. 17 December 2012. ISBN 978-1-4665-1302-0
Townsend, John S. A Modern Approach to Quantum Mechanics, 2nd ed.. Sausalito, CA: University Science Books. 2000. ISBN 1-891389-13-0 . 
![{\displaystyle [H_{1,\,{\mathcal {S}}},H_{0,\,{\mathcal {S}}}]=0\,\!}](http://wikimedia.org/api/rest_v1/media/math/render/svg/cca226bf23cfabe53bb863dc7725ce9ef4d7fe1c)
![{\displaystyle {\begin{aligned}i\hbar {\frac {d}{dt}}|\psi (t)\rangle _{\mathcal {I}}&=e^{iH_{0}\,t/\hbar }\left[-H_{0}|\psi (t)\rangle _{\mathcal {S}}+i\hbar {\frac {d}{dt}}|\psi (t)\rangle _{\mathcal {S}}\right]\\&=e^{iH_{0}\,t/\hbar }\left[-H_{0}|\psi (t)\rangle _{\mathcal {S}}+H_{\mathcal {S}}|\psi (t)\rangle _{\mathcal {S}}\right]\\&=e^{iH_{0}\,t/\hbar }H_{1,\,{\mathcal {S}}}|\psi (t)\rangle _{\mathcal {S}}\\&=e^{iH_{0}\,t/\hbar }H_{1,\,{\mathcal {S}}}\,e^{-iH_{0}\,t/\hbar }|\psi (t)\rangle _{\mathcal {I}}\\\end{aligned}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/cb32c499da6b74f1729fbccb2f6fc463003a6be2)
![{\displaystyle {\begin{aligned}i\hbar {\frac {d}{dt}}A_{\mathcal {I}}(t)&=i\hbar {\frac {d}{dt}}(e^{iH_{0}\,t/\hbar }A_{\mathcal {S}}\,e^{-iH_{0}\,t/\hbar })\\&=-H_{0}\,e^{iH_{0}\,t/\hbar }A_{\mathcal {S}}\,e^{-iH_{0}\,t/\hbar }+e^{iH_{0}\,t/\hbar }A_{\mathcal {S}}\,e^{-iH_{0}\,t/\hbar }H_{0}\\&=A_{\mathcal {I}}(t)H_{0}-H_{0}A_{\mathcal {I}}(t)\\&=\left[A_{\mathcal {I}}(t),\,H_{0}\right]\\\end{aligned}}\,\!}](http://wikimedia.org/api/rest_v1/media/math/render/svg/b504216b6dd44224ea8cc426b5c48111e24afffc)
![{\displaystyle i\hbar {\frac {d}{dt}}A_{\mathcal {H}}(t)=\left[A_{\mathcal {H}}(t),\,H\right]\,\!}](http://wikimedia.org/api/rest_v1/media/math/render/svg/75f49b2e479c51c7f4fbb086f37bce4661df66de)
![{\displaystyle i\hbar {\frac {d}{dt}}\rho _{\mathcal {I}}(t)=\left[H_{1,\,{\mathcal {I}}}(t),\rho _{\mathcal {I}}(t)\right]\,\!}](http://wikimedia.org/api/rest_v1/media/math/render/svg/a6473d13d1b0428fea8ae8de051965a7a671aad2)
