保羅·狄拉克 在量子力學 裏，相互作用繪景 （interaction picture），是在薛丁格繪景 與海森堡繪景 之間的一種表述，為紀念物理學者保羅·狄拉克 而又命名為狄拉克繪景 。在這繪景裏，描述量子系統的態向量 與表達可觀察量 的算符 都會隨著時間 流易而演化。有些實際案例會涉及到因相互作用而使得量子態與可觀察量發生改變，這類案例通常會使用狄拉克繪景。
狄拉克繪景與薛丁格繪景 、海森堡繪景 不同。在薛丁格繪景裏，描述量子系統的態向量 隨著時間流易而演化。在海森堡繪景裏，表達可觀察量 的算符 會隨著時間流易而演化。
這三種繪景殊途同歸，所獲得的結果完全一致。這是必然的，因為它們都是在表達同樣的物理行為。[ 1] :80-84 [ 2] [ 3]
為了便利分析，位於下標的符號
H
{\displaystyle {}_{\mathcal {H}}}
I
{\displaystyle {}_{\mathcal {I}}}
S
{\displaystyle {}_{\mathcal {S}}}
通過對於基底 的一種幺正變換
在量子力學裏，對於大多數案例的哈密頓量 ，通常無法找到薛丁格方程式 的精確解，只有少數案例可以找到精確解。因此，為了要能夠解析其它沒有精確解的案例，必須將薛丁格繪景裏的哈密頓量
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}}}\,\!}
微擾 。
假若哈密頓量
H
S
{\displaystyle H_{\mathcal {S}}\,\!}
外電場 作用的量子系統，其哈密頓量會含時），則通常會將顯性 含時部分放在
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\,\!}
假若對於某些案例，
H
0
,
S
{\displaystyle H_{0,\,{\mathcal {S}}}\,\!}
時間演化算符 的公式會變得較為複雜：[ 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^{'}}\,\!}
本條目以下內容假設
H
0
,
S
{\displaystyle H_{0,\,{\mathcal {S}}}\,\!}
在狄拉克繪景裏，態向量
|
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}}\,\!}
由於在薛丁格繪景裏， 態向量
|
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
)
⟩
I
=
e
−
i
H
1
,
S
t
/
ℏ
|
ψ
(
0
)
⟩
S
{\displaystyle |\psi (t)\rangle _{\mathcal {I}}=e^{-iH_{1,\,{\mathcal {S}}}\,t/\hbar }|\psi (0)\rangle _{\mathcal {S}}\,\!}
在狄拉克繪景裏的算符
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)\,\!}
（請注意，
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 }\,\!}
對易 ，不論在薛丁格繪景裏或在狄拉克繪景裏，
H
0
,
S
{\displaystyle H_{0,\,{\mathcal {S}}}\,\!}
H
0
,
I
{\displaystyle H_{0,\,{\mathcal {I}}}\,\!}
[ 註 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}\,\!}
哈密頓算符的微擾成分
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 }\,\!}
除非對易關係式
[
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}}}\,\!}
與算符類似，在薛丁格繪景裏的密度矩陣 也可以變換到在狄拉克繪景裏。設定
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
(
t
)
⟩
I
I
⟨
ψ
n
(
t
)
|
=
∑
n
p
n
e
i
H
0
,
S
t
/
ℏ
|
ψ
n
(
t
)
⟩
S
S
⟨
ψ
n
(
t
)
|
e
−
i
H
0
,
S
t
/
ℏ
=
e
i
H
0
,
S
t
/
ℏ
ρ
S
(
t
)
e
−
i
H
0
,
S
t
/
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}}\,\!}
。 本文以下內容，算符
H
0
,
S
{\displaystyle H_{0,\,{\mathcal {S}}}\,\!}
H
0
,
I
{\displaystyle H_{0,\,{\mathcal {I}}}\,\!}
H
0
{\displaystyle H_{0}\,\!}
[ 1] :337-339
從態向量的定義式，可以得到態向量對於時間的導數是
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}}}
將算符的定義式代入，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}}\,\!}
這是施溫格-朝永振一郎方程式 [ 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}}\,\!}
。 這與在海森堡繪景裏，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]\,\!}
應用施溫格-朝永振一郎方程式於密度矩陣，則可得到
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]\,\!}
應用狄拉克繪景的目的是促使
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)\,\!}
時變微擾理論 來計算
H
1
,
S
(
t
)
{\displaystyle H_{1,\,{\mathcal {S}}}(t)\,\!}
費米黃金定則 的導引裏[ 1] :359–363 ，或在推導戴森級數 [ 1] :355–357 ，通常都會用到狄拉克繪景。
各種繪景隨著時間流易會呈現出不同的演化：[ 1] :86-89, 337-339
演化
海森堡繪景
交互作用繪景
薛丁格繪景
右矢
常定
|
ψ
(
t
)
⟩
I
=
e
i
H
0
t
/
ℏ
|
ψ
(
t
)
⟩
S
{\displaystyle |\psi (t)\rangle _{\mathcal {I}}=e^{iH_{0}t/\hbar }|\psi (t)\rangle _{\mathcal {S}}}
|
ψ
(
t
)
⟩
S
=
e
−
i
H
t
/
ℏ
|
ψ
(
0
)
⟩
S
{\displaystyle |\psi (t)\rangle _{\mathcal {S}}=e^{-iHt/\hbar }|\psi (0)\rangle _{\mathcal {S}}}
可觀察量
A
H
(
t
)
=
e
i
H
t
/
ℏ
A
S
e
−
i
H
t
/
ℏ
{\displaystyle A_{\mathcal {H}}(t)=e^{iHt/\hbar }A_{\mathcal {S}}e^{-iHt/\hbar }}
A
I
(
t
)
=
e
i
H
0
t
/
ℏ
A
S
e
−
i
H
0
t
/
ℏ
{\displaystyle A_{\mathcal {I}}(t)=e^{iH_{0}t/\hbar }A_{\mathcal {S}}e^{-iH_{0}t/\hbar }}
常定
密度算符
常定
ρ
I
(
t
)
=
e
i
H
0
t
/
ℏ
ρ
S
(
t
)
e
−
i
H
0
t
/
ℏ
{\displaystyle \rho _{\mathcal {I}}(t)=e^{iH_{0}t/\hbar }\rho _{S}(t)e^{-iH_{0}t/\hbar }}
ρ
S
(
t
)
=
e
−
i
H
t
/
ℏ
ρ
S
(
0
)
e
i
H
t
/
ℏ
{\displaystyle \rho _{\mathcal {S}}(t)=e^{-iHt/\hbar }\rho _{\mathcal {S}}(0)e^{iHt/\hbar }}
^ 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 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\,\!}](https://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}}}](https://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}}\,\!}](https://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]\,\!}](https://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]\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a6473d13d1b0428fea8ae8de051965a7a671aad2)
