# 海森堡繪景

（重定向自海森堡绘景

## 概述

${\displaystyle {\frac {\partial }{\partial {t}}}A_{\mathcal {H}}={1 \over i\hbar }[A_{\mathcal {H}},\,H]}$

${\displaystyle {\frac {\partial }{\partial {t}}}A=[A,\,H]_{Poisson}}$

${\displaystyle [\ ,\,\ ]_{Poisson}\ \to \ {\frac {[\ ,\,\ ]}{i\hbar }}}$

## 理論導引

${\displaystyle |\psi (t)\rangle _{\mathcal {S}}=U(t,0)|\psi (0)\rangle _{\mathcal {S}}}$

${\displaystyle U(t,0)U^{\dagger }(t,0)=1=U^{\dagger }(t,0)U(t,0)}$

${\displaystyle U(t,0)=e^{-iHt/\hbar }}$

${\displaystyle HU(t,0)=U(t,0)H}$

${\displaystyle |\psi (t)\rangle _{\mathcal {H}}{\stackrel {def}{=}}|\psi (0)\rangle _{\mathcal {H}}=|\psi (0)\rangle _{\mathcal {S}}}$
${\displaystyle A_{\mathcal {H}}(t){\stackrel {def}{=}}U^{\dagger }(t,0)A_{\mathcal {S}}U(t,0)}$

${\displaystyle {\frac {\partial U(t,0)}{\partial t}}={1 \over i\hbar }HU(t,0)}$
${\displaystyle {\frac {\partial U^{\dagger }(t,0)}{\partial t}}=-{1 \over i\hbar }U^{\dagger }(t,0)H}$

{\displaystyle {\begin{aligned}{d \over dt}A_{\mathcal {H}}(t)&={\frac {\partial U^{\dagger }(t,0)}{\partial t}}A_{\mathcal {S}}U(t,0)+U^{\dagger }(t,0)A_{\mathcal {S}}{\frac {\partial U(t,0)}{\partial t}}\\&=-{1 \over i\hbar }U^{\dagger }HA_{\mathcal {S}}U+{1 \over i\hbar }U^{\dagger }A_{\mathcal {S}}HU\\&=-{1 \over i\hbar }U^{\dagger }HUU^{\dagger }A_{\mathcal {S}}U+{1 \over i\hbar }U^{\dagger }A_{\mathcal {S}}UU^{\dagger }HU\\&={1 \over i\hbar }[U^{\dagger }A_{\mathcal {S}}U,U^{\dagger }HU]\\\end{aligned}}}

${\displaystyle H_{\mathcal {H}}=U^{\dagger }H_{\mathcal {S}}U=H_{\mathcal {S}}=H}$

${\displaystyle {d \over dt}A_{\mathcal {H}}(t)={1 \over i\hbar }[A_{\mathcal {H}}(t),H]}$

## 期望值

${\displaystyle \langle A\rangle _{\mathcal {S}}={}_{\mathcal {S}}\langle \psi (t)|A_{\mathcal {S}}|\psi (t)\rangle _{\mathcal {S}}={}_{\mathcal {S}}\langle \psi (0)|U^{\dagger }(t,0)A_{\mathcal {S}}U(t,0)|\psi (0)\rangle _{\mathcal {S}}}$

${\displaystyle \langle A\rangle _{\mathcal {H}}={}_{\mathcal {H}}\langle \psi (t)|A_{\mathcal {H}}(t)|\psi (t)\rangle _{\mathcal {H}}={}_{\mathcal {H}}\langle \psi (0)|A_{\mathcal {H}}(t)|\psi (0)\rangle _{\mathcal {H}}}$

${\displaystyle |\psi (t)\rangle _{\mathcal {H}}{\stackrel {def}{=}}|\psi (0)\rangle _{\mathcal {H}}=|\psi (0)\rangle _{\mathcal {S}}}$
${\displaystyle A_{\mathcal {H}}(t){\stackrel {def}{=}}U^{\dagger }(t,0)A_{\mathcal {S}}U(t,0)}$

## 貝克-豪斯多夫引理

${\displaystyle {e^{B}Ae^{-B}}=A+[B,A]+{\frac {1}{2!}}[B,[B,A]]+{\frac {1}{3!}}[B,[B,[B,A]]]+\cdots }$

${\displaystyle A_{\mathcal {H}}(t)=A_{\mathcal {H}}(0)+{\frac {it}{\hbar }}[H,A_{\mathcal {H}}(0)]-{\frac {t^{2}}{2!\hbar ^{2}}}[H,[H,A_{\mathcal {H}}(0)]]-{\frac {it^{3}}{3!\hbar ^{3}}}[H,[H,[H,A_{\mathcal {H}}(0)]]]+\cdots }$

## 自由粒子範例

${\displaystyle H={\frac {p^{2}}{2m}}}$

${\displaystyle {d \over dt}p(t)={1 \over i\hbar }[p,H]=0}$

${\displaystyle p(t)=p(0)}$

${\displaystyle {d \over dt}x(t)={1 \over i\hbar }[x,H]={\frac {p}{m}}}$

${\displaystyle x(t)=x(0)+{\frac {p(0)}{m}}t}$

${\displaystyle x(t)=e^{iHt/\hbar }x(0)e^{-iHt/\hbar }}$

${\displaystyle x(t)=x(0)+{\frac {it}{\hbar }}[H,x(0)]-{\frac {t^{2}}{2!\hbar ^{2}}}[H,[H,x(0)]]-{\frac {it^{3}}{3!\hbar ^{3}}}[H,[H,[H,x(0)]]]+\cdots }$

${\displaystyle [H,x(0)]={\frac {-i\hbar p(0)}{m}}}$
${\displaystyle [H,p(0)]=0}$

${\displaystyle x(t)=x(0)+{\frac {p(0)}{m}}t}$

${\displaystyle [x(t),x(0)]=\left[{\frac {p(0)t}{m}},x(0)\right]={\frac {-i\hbar t}{m}}}$

## 諧振子範例

${\displaystyle H={\frac {p^{2}}{2m}}+{\frac {m\omega ^{2}x^{2}}{2}}}$

${\displaystyle {d \over dt}p(t)={1 \over i\hbar }[p(t),H]=-m\omega ^{2}x(t)}$
${\displaystyle {d \over dt}x(t)={1 \over i\hbar }[x(t),H]={\frac {p(t)}{m}}}$

${\displaystyle {d^{2} \over dt^{2}}p(t)={-m\omega ^{2} \over i\hbar }[x(t),H]=-\omega ^{2}p(t)}$
${\displaystyle {d^{2} \over dt^{2}}x(t)={1 \over im\hbar }[p(t),H]=-\omega ^{2}x(t)}$

${\displaystyle p(0)=p_{0}}$
${\displaystyle x(0)=x_{0}}$

${\displaystyle {\dot {p}}(0)=-m\omega ^{2}x_{0}}$
${\displaystyle {\dot {x}}(0)={\frac {p_{0}}{m}}}$

${\displaystyle p(t)=p_{0}\cos(\omega t)-m\omega \!x_{0}\sin(\omega t)}$
${\displaystyle x(t)=x_{0}\cos(\omega t)+{\frac {p_{0}}{m\omega }}\sin(\omega t)}$

${\displaystyle [p(t_{1}),p(t_{2})]=i\hbar m\omega \sin(\omega t_{2}-\omega t_{1})}$
${\displaystyle [x(t_{1}),x(t_{2})]={\frac {i\hbar }{m\omega }}\sin(\omega t_{2}-\omega t_{1})}$
${\displaystyle [x(t_{1}),p(t_{2})]=i\hbar \cos(\omega t_{2}-\omega t_{1})}$

${\displaystyle x(t)=e^{iHt/\hbar }x(0)e^{-iHt/\hbar }}$

${\displaystyle x(t)=x(0)+{\frac {it}{\hbar }}[H,x(0)]-{\frac {t^{2}}{2!\hbar ^{2}}}[H,[H,x(0)]]-{\frac {it^{3}}{3!\hbar ^{3}}}[H,[H,[H,x(0)]]]+\cdots }$

${\displaystyle [H,x(0)]={\frac {-i\hbar p(0)}{m}}}$
${\displaystyle [H,p(0)]=i\hbar m\omega ^{2}x(0)}$

{\displaystyle {\begin{aligned}x(t)&=x(0)+{\frac {p(0)}{m\omega }}\omega t-x(0){\frac {\omega ^{2}t^{2}}{2!}}-{\frac {p(0)}{m\omega }}{\frac {\omega ^{3}t^{3}}{3!}}+\cdots \\&=x(0)\cos(\omega t)+{\frac {p(0)}{m\omega }}\sin(\omega t)\\\end{aligned}}}

## 各種繪景比較摘要

 演化 海森堡繪景 交互作用繪景 薛丁格繪景 右矢 常定 ${\displaystyle |\psi (t)\rangle _{\mathcal {I}}=e^{iH_{0}t/\hbar }|\psi (t)\rangle _{\mathcal {S}}}$ ${\displaystyle |\psi (t)\rangle _{\mathcal {S}}=e^{-iHt/\hbar }|\psi (0)\rangle _{\mathcal {S}}}$ 可觀察量 ${\displaystyle A_{\mathcal {H}}(t)=e^{iHt/\hbar }A_{\mathcal {S}}e^{-iHt/\hbar }}$ ${\displaystyle A_{\mathcal {I}}(t)=e^{iH_{0}t/\hbar }A_{\mathcal {S}}e^{-iH_{0}t/\hbar }}$ 常定 密度算符 常定 ${\displaystyle \rho _{\mathcal {I}}(t)=e^{iH_{0}t/\hbar }\rho _{S}(t)e^{-iH_{0}/\hbar }}$ ${\displaystyle \rho _{\mathcal {S}}(t)=e^{-iHt/\hbar }\rho _{\mathcal {S}}(0)e^{iHt/\hbar }}$

## 註釋

1. ^ 在薛丁格繪景裏，假若勢能與時間有關，${\displaystyle V=V(t)}$，則哈密頓算符${\displaystyle H={\frac {P^{2}}{2m}}+V(t)}$也與時間有關。
2. ^ ${\displaystyle U(t,0)}$是无穷维矩阵时， ${\displaystyle U(t,0)U^{\dagger }(t,0)}$${\displaystyle U^{\dagger }(t,0)U(t,0)}$未必相等；甚至${\displaystyle U(t,0)}$ 的行列指标可以分别是可数无穷维与连续不可数无穷维。
3. ^ 在薛丁格繪景裏，假若哈密頓算符含時，${\displaystyle H_{\mathcal {S}}=H_{\mathcal {S}}(t)}$，則時間演化算符會比較複雜。更詳盡內容，請查閱條目時間演化算符
4. ^ 處於不同時間${\displaystyle t_{1}}$${\displaystyle t_{2}}$的哈密頓算符${\displaystyle H(t_{1})}$${\displaystyle H(t_{2})}$可能會不相互對易：
${\displaystyle [H(t_{1}),H(t_{2})]\neq [H(t_{2}),H(t_{1})]}$
對於這案例，時間演化算符必須用戴森級數英语Dyson series來表示，時間演化算符與哈密頓算符也會不相互對易。[2]:69-71
5. ^ 假若算符${\displaystyle A_{\mathcal {S}}}$含時：
${\displaystyle A_{\mathcal {S}}=A_{\mathcal {S}}(t)}$
則算符${\displaystyle A_{\mathcal {H}}(t)}$對於時間的導數是
{\displaystyle {\begin{aligned}{d \over dt}A_{\mathcal {H}}(t)&={\frac {\partial U^{\dagger }(t,0)}{\partial t}}A_{\mathcal {S}}U(t,0)+U^{\dagger }(t,0)A_{\mathcal {S}}{\frac {\partial U(t,0)}{\partial t}}+U^{\dagger }(t,0){\frac {\partial A_{\mathcal {S}}}{\partial t}}U(t,0)\\&={1 \over i\hbar }[U^{\dagger }A_{\mathcal {S}}U,U^{\dagger }HU]+U^{\dagger }(t,0){\frac {\partial A_{\mathcal {S}}}{\partial t}}U(t,0)\\\end{aligned}}}
6. ^ 假若處於不同時間${\displaystyle t_{1}}$${\displaystyle t_{2}}$的哈密頓算符${\displaystyle H(t_{1})}$${\displaystyle H(t_{2})}$不相互對易：
${\displaystyle [H(t_{1}),H(t_{2})]\neq [H(t_{2}),H(t_{1})]}$
則哈密頓量在兩種繪景裏的形式可能不相同。[2]:69-71

## 參考文獻

• Cohen-Tannoudji, Claude; Bernard Diu, Frank Laloe. Quantum Mechanics (Volume One). Paris: Wiley. 1977: 312–314. ISBN 047116433X.
1. ^ Robert D. Klauber. Student Friendly Quantum Field Theory: Basic Principles and Quantum Electrodynamics (PDF). Sandtrove Press. 2013. ISBN 978-0-9845139-3-2.
2. Sakurai, J. J.; Napolitano, Jim, Modern Quantum Mechanics 2nd, Addison-Wesley, 2010, ISBN 978-0805382914
3. ^ Parker, C.B. McGraw Hill Encyclopaedia of Physics 2nd. Mc Graw Hill. 1994: 786, 1261. ISBN 0-07-051400-3.
4. ^ Y. Peleg, R. Pnini, E. Zaarur, E. Hecht. Quantum mechanics. Schuam's outline series 2nd. McGraw Hill. 2010: 70. ISBN 9-780071-623582.
5. ^ Goldstein, Herbert, Classical Mechanics 3rd, United States of America: Addison Wesley, 1980, ISBN 0201657023 （英语）