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电磁场中的带电粒子

2025/7/27 物理 量子力学 电动力学

带电粒子的哈密顿量

首先有麦克斯韦方程组

{∇⋅E→=ρϵ0∇×E→=−∂B→∂t∇⋅B→=0∇×B→=μ0J→+μ0ϵ0∂E→∂t\left\{\begin{matrix}\nabla \cdot \overrightarrow{E} =\frac{\rho}{\epsilon _0} \\ \nabla \times \overrightarrow{E} = - \frac{\partial \overrightarrow{B} }{\partial t} \\ \nabla \cdot \overrightarrow{B} = 0 \\ \nabla \times \overrightarrow{B} =\mu _0 \overrightarrow{J} + \mu _0 \epsilon _0 \frac{\partial \overrightarrow{E} }{\partial t} \end{matrix}\right.

由旋度无散可知 B→\overrightarrow{B} 可以写成某个矢量 A→\overrightarrow{A} 的旋度:

B→=∇×A→\overrightarrow{B}=\nabla \times \overrightarrow{A}

将其带回到方程中可得

∇×E→=−∂∂t(∇×A→)=−∇×∂A→∂t\nabla \times \overrightarrow{E}=- \frac{\partial}{\partial t}(\nabla \times \overrightarrow{A})=-\nabla \times \frac{\partial \overrightarrow{A}}{\partial t}

即

∇×(E→+∂A→∂t)=0\nabla \times (\overrightarrow{E}+\frac{\partial \overrightarrow{A}}{\partial t})=0

由梯度无旋可知 E→+∂A→∂t\overrightarrow{E}+\frac{\partial \overrightarrow{A}}{\partial t} 可以写成某个标量 φ\varphi 的梯度, 即

E→+∂A→∂t=−∇φ\overrightarrow{E}+\frac{\partial \overrightarrow{A}}{\partial t}=-\nabla \varphi

即

E→=−∇φ−∂A→∂t\overrightarrow{E}=-\nabla \varphi-\frac{\partial \overrightarrow{A}}{\partial t}

于是得到了 B→\overrightarrow{B} 与 E→\overrightarrow{E} 和 矢势 A→\overrightarrow{A} 与标势 φ\varphi 的关系:

{B→=∇×A→E→=−∇φ−∂A→∂t\left\{\begin{matrix}\overrightarrow{B}=\nabla \times \overrightarrow{A} \\ \overrightarrow{E}=-\nabla \varphi-\frac{\partial \overrightarrow{A}}{\partial t} \end{matrix}\right.

那么粒子所受的洛伦兹力可以写成:

F→=q(E→+v→×B→)=q[−∇φ−∂A→∂t+v→×(∇×A→)]\begin{aligned} \overrightarrow{F} &= q(\overrightarrow{E}+\overrightarrow{v} \times \overrightarrow{B}) \\ &= q[-\nabla \varphi-\frac{\partial \overrightarrow{A}}{\partial t}+\overrightarrow{v} \times (\nabla \times \overrightarrow{A})] \end{aligned}

又有广义力的表达式

Qi=−∂U∂qi+ddt∂U∂qα˙Q_i=-\frac{\partial U}{\partial q_i} + \frac{d}{dt} \frac{\partial U}{\partial \dot{q_{\alpha}}}

其中 UU 为广义势能, qiq_i 为广义坐标, 于是现在将洛伦兹力写成标量的形式并与广义力做对比, 现写出 xx 方向上的标量, 有

(∇φ)x=∂φ∂x(\nabla \varphi)_x =\frac{\partial \varphi}{\partial x}

(∂A→∂t)x=∂Ax∂t(\frac{\partial \overrightarrow{A}}{\partial t})_x =\frac{\partial A_x }{\partial t}

[v→×(∇×A→)]x=[v→×∣ex→ey→ez→∂∂x∂∂y∂∂zAxAyAz∣]x=[v→×[(∂Az∂y−∂Ay∂z)ex→+(∂Ax∂z−∂Az∂x)ey→+(∂Ay∂x−∂Ax∂y)ez→]]x=[∣ex→ey→ez→vxvyvz∂Az∂y−∂Ay∂z∂Ax∂z−∂Az∂x∂Ay∂x−∂Ax∂y∣]x=vy(∂Ay∂x−∂Ax∂y)−vz(∂Ax∂z−∂Az∂x)\begin{aligned} [\overrightarrow{v} \times (\nabla \times \overrightarrow{A})]_x &= [\overrightarrow{v} \times \begin{vmatrix} \overrightarrow{e_x} & \overrightarrow{e_y} & \overrightarrow{e_z} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_x & A_y & A_z \end{vmatrix} ]_x \\ &= [ \overrightarrow{v} \times [(\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}) \overrightarrow{e_x} + (\frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x}) \overrightarrow{e_y}+(\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}) \overrightarrow{e_z}]]_x \\ &= [\begin{vmatrix} \overrightarrow{e_x} & \overrightarrow{e_y} & \overrightarrow{e_z} \\ v_x & v_y & v_z \\ \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} & \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x} &\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \end{vmatrix} ]_x \\ &= v_y(\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y})-v_z(\frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x}) \end{aligned}

即

Fx=q[−∂φ∂x−∂Ax∂t+vy(∂Ay∂x−∂Ax∂y)−vz(∂Ax∂z−∂Az∂x)]F_x =q[-\frac{\partial \varphi}{\partial x} -\frac{\partial A_x }{\partial t} +v_y(\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y})-v_z(\frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x})]

为了进一步化简和对比, 则由

dAxdt=∂Ax∂t+∂Ax∂x∂x∂t+∂Ax∂y∂y∂t+∂Ax∂z∂z∂t=∂Ax∂t+∂Ax∂xvx+∂Ax∂xvy+∂Ax∂xvz\begin{aligned} \frac{d A_x}{dt} &= \frac{\partial A_x}{\partial t}+\frac{\partial A_x}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial A_x}{\partial y}\frac{\partial y}{\partial t} + \frac{\partial A_x}{\partial z}\frac{\partial z}{\partial t} \\ &= \frac{\partial A_x}{\partial t}+ \frac{\partial A_x}{\partial x} v_x +\frac{\partial A_x}{\partial x} v_y +\frac{\partial A_x}{\partial x} v_z \end{aligned}

可将 FxF_x 写为

Fx=q[−∂φ∂x−dAxdt+∂Ax∂xvx+∂Ax∂xvy+∂Ax∂xvz]=q[−∂φ∂x−dAxdt+v→⋅∂A→∂x]=q[−∂φ∂x−dAxdt+∂∂x(A→⋅v→)]v不显含t=q[−∂∂x(φ−A→⋅v→)−dAxdt]\begin{aligned} F_x &= q[-\frac{\partial \varphi}{\partial x} - \frac{d A_x}{dt} + \frac{\partial A_x}{\partial x} v_x +\frac{\partial A_x}{\partial x} v_y +\frac{\partial A_x}{\partial x} v_z] \\ &= q[-\frac{\partial \varphi}{\partial x} - \frac{d A_x}{dt} + \overrightarrow{v} \cdot \frac{\partial \overrightarrow{A}}{\partial x} ] \\ &= q[-\frac{\partial \varphi}{\partial x} - \frac{d A_x}{dt} + \frac{\partial}{\partial x}(\overrightarrow{A} \cdot \overrightarrow{v} ) ] \qquad v不显含t \\ &= q[-\frac{\partial}{\partial x}(\varphi -\overrightarrow{A}\cdot \overrightarrow{v})-\frac{d A_x}{dt}] \end{aligned}

为了进一步凑出广义力的形式, 由

dAxdt=ddt[∂∂vx(A→⋅v→)]A不显含v=ddt∂∂vx(−φ+A→⋅v→)\begin{aligned} \frac{d A_x}{dt} &=\frac{d}{dt}[\frac{\partial}{\partial v_x}(\overrightarrow{A}\cdot \overrightarrow{v})] \qquad A不显含v \\ &=\frac{d}{dt}\frac{\partial}{\partial v_x}(-\varphi +\overrightarrow{A}\cdot \overrightarrow{v}) \end{aligned}

可得

Fx=q[−∂∂x(φ−A→⋅v→)+ddt∂∂vx(−φ+A→⋅v→)]F_x =q[-\frac{\partial}{\partial x}(\varphi -\overrightarrow{A}\cdot \overrightarrow{v})+\frac{d}{dt}\frac{\partial}{\partial v_x}(-\varphi +\overrightarrow{A}\cdot \overrightarrow{v})]

于是对比广义力 QiQ_i 的式子, 可得广义势能为

U=q(φ−A→⋅v→)U=q(\varphi -\overrightarrow{A}\cdot \overrightarrow{v})

于是拉格朗日量为

L=T−U=12mv2−q(φ−A→⋅v→)L=T-U=\frac{1}{2}mv^2-q(\varphi -\overrightarrow{A}\cdot \overrightarrow{v})

则广义动量, 或这里可称之为正则动量为

Pi=∂L∂vi=mvi+qAiP_i =\frac{\partial L}{\partial v_i}=mv_i +qA_i

即机械动量为

mvi=Pi−qAimv_i=P_i - qA_i

则哈密顿量为

H=T+qφ=mv22m+qφ=(P→−qA→)22m+qφH=T+q\varphi =\frac{mv^2}{2m}+q\varphi =\frac{(\overrightarrow{P}-q\overrightarrow{A})^2}{2m}+ q\varphi

若将其量子化则有

H^=(P^−qA→)22m+qφ\hat{H}=\frac{(\hat{P}-q\overrightarrow{A})^2}{2m}+ q\varphi

其中正则动量算符 P^\hat{P} 为

P^=−iℏ∇\hat{P}=-i \hbar \nabla

若使用 GaussGauss 单位制则哈密顿算符为

H^=(P^−qcA→)22m+qφ\hat{H}=\frac{(\hat{P}-\frac{q}{c} \overrightarrow{A})^2}{2m}+ q\varphi



规范不变性

由前面得到的 B→\overrightarrow{B} 与 E→\overrightarrow{E} 和 矢势 A→\overrightarrow{A} 与标势 φ\varphi 的关系:

{B→=∇×A→E→=−∇φ−∂A→∂t\left\{\begin{matrix}\overrightarrow{B}=\nabla \times \overrightarrow{A} \\ \overrightarrow{E}=-\nabla \varphi-\frac{\partial \overrightarrow{A}}{\partial t} \end{matrix}\right.

易知由给定的一组 A→\overrightarrow{A} 与 φ\varphi 可以唯一确定 B→\overrightarrow{B} 与 E→\overrightarrow{E} , 但给定的一组 B→\overrightarrow{B} 与 E→\overrightarrow{E} 并不能唯一确定 A→\overrightarrow{A} 与 φ\varphi , 换言之, 可以有多组 A→\overrightarrow{A} 与 φ\varphi 对应同一组 B→\overrightarrow{B} 与 E→\overrightarrow{E} , 并且满足规范不变性:

{A′→=A→(r→,t)+∇Λ(r→,t)φ′(r→,t)=φ(r→,t)−∂Λ(r→,t)∂t\left\{\begin{matrix} \overrightarrow{A^{'}}=\overrightarrow{A} (\overrightarrow{r},t) +\nabla \Lambda (\overrightarrow{r},t) \\ \varphi ^{'} (\overrightarrow{r},t)=\varphi (\overrightarrow{r},t)-\frac{\partial \Lambda (\overrightarrow{r},t)}{\partial t} \end{matrix}\right.

显然经过上面两式的变换后 A′→\overrightarrow{A^{'}} , φ′\varphi ^{'} 和原先的 A→\overrightarrow{A} , φ\varphi 对应的是同样的 B→\overrightarrow{B} 与 E→\overrightarrow{E} , 证明:

B′→=∇×A′→=∇×(A→+∇Λ)=∇×A→梯度无旋=B→\begin{aligned} \overrightarrow{B^{'}} &=\nabla \times \overrightarrow{A^{'}} = \nabla \times (\overrightarrow{A}+\nabla \Lambda) \\ &= \nabla \times \overrightarrow{A} \qquad 梯度无旋 \\ &= \overrightarrow{B} \end{aligned}

E′→=−∇φ′−∂A→∂t=−∇φ+∇∂Λ∂t−∂A→∂t−∂∂t∇Λ=−∇φ−∂A→∂t∇与∂∂t的位置可互换=E→\begin{aligned} \overrightarrow{E^{'}} &= -\nabla \varphi ^{'} - \frac{\partial \overrightarrow{A}}{\partial t} \\ &= -\nabla \varphi + \nabla \frac{\partial \Lambda}{\partial t} -\frac{\partial \overrightarrow{A}}{\partial t} - \frac{\partial }{\partial t} \nabla \Lambda \\ &= - \nabla \varphi - \frac{\partial \overrightarrow{A}}{\partial t} \qquad \nabla 与 \frac{\partial}{\partial t}的位置可互换 \\ &= \overrightarrow{E} \end{aligned}

即经过规范变换后的 A′→\overrightarrow{A^{'}} , φ′\varphi ^{'} 不会改变 B→\overrightarrow{B} 与 E→\overrightarrow{E} .

对于波函数 ψ\psi 同样拥有规范变换

ψ′=eiqΛ/ℏψ\psi ^{'}=e^{iq \Lambda / \hbar}\psi

前面得到过电磁场中的带电粒子哈密顿量为:

H^=(P^−qA→)22m+qφ\hat{H}=\frac{(\hat{P}-q\overrightarrow{A})^2}{2m}+q\varphi

对于薛定谔方程

iℏ∂ψ∂t=H^ψi\hbar \frac{\partial \psi}{\partial t}=\hat{H}\psi

则左式

iℏ∂ψ∂t=iℏ∂∂t(e−iqΛ/ℏψ′)=e−iqΛ/ℏ(iℏ∂ψ′∂t+q∂Λ∂tψ)\begin{aligned} i\hbar \frac{\partial \psi}{\partial t} &= i\hbar \frac{\partial}{\partial t} (e^{-iq \Lambda / \hbar}\psi ^{'}) \\ &= e^{-iq \Lambda / \hbar} (i\hbar \frac{\partial \psi ^{'}}{\partial t} +q \frac{\partial \Lambda}{\partial t} \psi ) \end{aligned}

而对于 H^ψ\hat{H}\psi 有

(P^−qA→)ψ=(−iℏ∇−qA→)e−iqΛ/ℏψ′=−iℏ∇(e−iqΛ/ℏψ′)−qA→e−iqΛ/ℏψ′=−iℏ(q∇Λiℏe−iqΛ/ℏψ′+e−iqΛ/ℏ∇ψ′)−qA→e−iqΛ/ℏψ′=−iℏe−iqΛ/ℏ∇ψ′−q∇Λe−iqΛ/ℏψ′−qA→e−iqΛ/ℏψ′=−iℏe−iqΛ/ℏ∇ψ′−q(A→+∇Λ)e−iqΛ/ℏψ′=e−iqΛ/ℏ[−iℏ∇−q(A→+∇Λ)]ψ′\begin{aligned} (\hat{P}-q \overrightarrow{A})\psi &= (-i\hbar \nabla -q \overrightarrow{A} )e^{-iq \Lambda / \hbar}\psi ^{'} \\ &= -i\hbar \nabla (e^{-iq \Lambda / \hbar}\psi ^{'}) - q \overrightarrow{A} e^{-iq \Lambda / \hbar}\psi ^{'} \\ &=-i\hbar (\frac{q \nabla \Lambda}{i \hbar}e^{-iq \Lambda / \hbar}\psi ^{'}+e^{-iq \Lambda / \hbar} \nabla \psi ^{'}) - q \overrightarrow{A} e^{-iq \Lambda / \hbar}\psi ^{'} \\ &=-i\hbar e^{-iq \Lambda / \hbar} \nabla \psi ^{'} -q\nabla \Lambda e^{-iq \Lambda / \hbar} \psi ^{'} - q \overrightarrow{A} e^{-iq \Lambda / \hbar}\psi ^{'} \\ &= -i\hbar e^{-iq \Lambda / \hbar} \nabla \psi ^{'} -q(\overrightarrow{A}+\nabla \Lambda) e^{-iq \Lambda / \hbar} \psi ^{'} \\ &= e^{-iq \Lambda / \hbar}[-i\hbar \nabla -q(\overrightarrow{A}+\nabla \Lambda) ]\psi ^{'} \end{aligned}

于是则有

e−iqΛ/ℏ(iℏ∂ψ′∂t+q∂Λ∂tψ)=e−iqΛ/ℏ{12m[−iℏ∇−q(A→+∇Λ)]2+qφ}ψ′iℏ∂ψ′∂t+q∂Λ∂tψ={12m[−iℏ∇−q(A→+∇Λ)]2+qφ}ψ′iℏ∂ψ′∂t={12m[−iℏ∇−q(A→+∇Λ)]2+q(φ−∂Λ∂t)}ψ′iℏ∂ψ′∂t={12m[P^−qA′→]2+qφ′}ψ′iℏ∂ψ′∂t=H′^ψ′\begin{aligned} e^{-iq \Lambda / \hbar} (i\hbar \frac{\partial \psi ^{'}}{\partial t} +q \frac{\partial \Lambda}{\partial t} \psi )&= e^{-iq \Lambda / \hbar}\left \{ \frac{1}{2m}[-i\hbar \nabla -q(\overrightarrow{A}+\nabla \Lambda) ]^2 +q \varphi \right \} \psi ^{'} \\ i\hbar \frac{\partial \psi ^{'}}{\partial t} +q \frac{\partial \Lambda}{\partial t} \psi &= \left \{ \frac{1}{2m}[-i\hbar \nabla -q(\overrightarrow{A}+\nabla \Lambda) ]^2 +q \varphi \right \} \psi ^{'} \\ i\hbar \frac{\partial \psi ^{'}}{\partial t} &= \left \{ \frac{1}{2m}[-i\hbar \nabla -q(\overrightarrow{A}+\nabla \Lambda) ]^2 +q( \varphi - \frac{\partial \Lambda}{\partial t}) \right \} \psi ^{'} \\ i\hbar \frac{\partial \psi ^{'}}{\partial t} &= \left \{ \frac{1}{2m}[\hat{P}-q \overrightarrow{A^{'}}]^2+q \varphi ^{'} \right \} \psi ^{'} \\ i\hbar \frac{\partial \psi ^{'}}{\partial t} &= \hat{H^{'}} \psi ^{'} \end{aligned}

即经过规范变换后的 ψ′\psi ^{'} 也满足薛定谔方程



Aharonov-Bohm 效应


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爱の规范不变性