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
)
⟩
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
(
t
)
⟩
I
I
⟨
ψ
n
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t
)
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=
∑
n
p
n
e
i
H
0
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S
t
/
ℏ
|
ψ
n
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t
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⟩
S
S
⟨
ψ
n
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t
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|
e
−
i
H
0
,
S
t
/
ℏ
=
e
i
H
0
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t
/
ℏ
ρ
S
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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}}\,\!}
。 
![{\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)
