Noether定理
最近看Atland的凝聚态场论, 对Noether定理的证明与讲解的部分感到有些含糊(毕竟这种东西应该放在理力里讲), 在此做一些总结
守恒律
考虑一个泛函\(\mathcal{J}\):
\[\mathcal{S}(q) = \int _{t_0}^{t_1} \mathcal{L}(t, q, \dot{q},\cdots, q^{(n)}){\rm d}t\tag{1}\]
若存在一个函数\(\phi(t, q, \dot{q},\cdots,q^{k})\)满足,
\[\frac{\rm d}{{\rm d} t}\phi(t, q, \dot{q},\cdots,q^{(k)}) = 0\tag{2}\]
则称上式为\(\mathcal{S}\)的\(k\) 阶守恒律, 相应的\(\phi\) 称为\(\mathcal{S}\) 的守恒量.
Example 1
考虑一个不显含时间的作用量: \[\mathcal{S}(q) = \int _{t_0}^{t_1}\mathcal{L}(q,\dot{q}){\rm d} t\] 显然的,我们可以找到其一阶守恒律 \[H = \dot{q} \frac{\partial \mathcal{L}}{\partial \dot{q}} - \mathcal{L}\]
对称性
对于一个给定的作用量：
\[ \mathcal{S}(q) = \int ^{t_1}_{t_0} \mathcal{L}(t,q,\dot{q}){\rm d} t \tag{3}\]
考虑两个单参变换,并要求他们是\(C^{\infty}\)的:
\[ X=\theta(t,q;\epsilon)\quad Y = \psi(t,q;\epsilon) \tag{4}\]
满足:
\[ \theta(t,q;0)=t\quad \psi(t,q;0) = q \tag{5}\]
Example 2
平移变换:\(X=x+\cos \theta \;\epsilon, Y=y+\sin \theta\; \epsilon\)
旋转变换:\(X = x\cos \epsilon + y \sin \epsilon,Y = -x\sin \epsilon + y\cos \epsilon\)
Jacobian矩阵为:
\[ J=\frac{\partial (X,Y)}{\partial(t,q)} = \begin{pmatrix}\frac{\partial \theta}{\partial t} & \frac{\partial \theta}{\partial q}\\ \frac{\partial \psi}{\partial t} & \frac{\partial \psi}{\partial q}\end{pmatrix} = \begin{pmatrix}\theta_{t}& \theta_{q}\\ \psi_{t} & \psi_{q}\end{pmatrix} \tag{6}\]
行列式为:
\[ |J| = \theta_{t} \psi_{q} - \theta_{t}\psi_{q} \tag{7}\]
当\(\epsilon=0\) 时,我们有:
\[ |J| = 1 \tag{8}\]
由于\(\theta\) 与\(\psi\) 均是\(C^{\infty}\)的, \(|J|\) 也是连续的,因此在\(\epsilon=0\) 的一个足够小的邻域内\(|J|\neq 0\),即, 我们可以找到一组逆映射将\((X,Y)\) 映回\((t,q)\):
\[ t = \Theta(X,Y;\epsilon)\quad q = \Psi(X,Y;\epsilon) \tag{9}\]
现在考虑一个给定的轨迹\(q(t)\),变换后的轨迹由一组参数方程给定:
\[ X_{\epsilon} = \theta(t,q;\epsilon)\quad Y_{\epsilon} = \psi(t,q;\epsilon) \tag{10}\]
现在我们可以定义变分不变性,作用量\(\mathcal{S}\) 的拉氏量\(\mathcal{L}\) 称为在区间\([t_0,t_1]\) 上变分不变,当其满足,对\(\forall [a,b]\in[t_0,t_1]\), \(\exist\delta\in \mathbb{R}\),使得\(\forall \epsilon< \delta\):
\[ \int^{a}_{b}\mathcal{L}(t,q,\dot{q}) {\rm d} t = \int^{a_{\epsilon}}_{b_{\epsilon}}\mathcal{L}(X_{\epsilon},Y_{\epsilon} ,\dot{Y_{\epsilon}}){\rm d} X \tag{11}\]
Example 3
\[ \mathcal{L}(t,q,\dot q) = \dot q^{2}(x) + q^{2}(x) \] 考虑时间平移变换: \[ X = t+\epsilon,\quad Y=q \] 我们有 \[ \dot Y_{\epsilon} = \dot q \] 则,对于给定的一个区间\([t_0,t_1]\), 我们有: \[ \int^{t_1}_{t_0}(\dot q^{2} + q^{2}) {\rm d}t = \int ^{t_1+\epsilon}_{t_0+\epsilon}(\dot Y_{\epsilon} + Y^{2}_{\epsilon})=\int^{t_1+\epsilon}_{t_0+\epsilon}(\dot q + q(t-\epsilon)){\rm d}t \] 成立,即上式的拉氏量在时间平移变换下是不变的
进一步的,在讨论Noether定理前,我们介绍一种描述变换的通用方式,即1 :
\[ \begin{aligned}X\approx t + \epsilon \zeta \\ Y\approx q+\epsilon \eta\end{aligned} \tag{12} \]
其中, \(\epsilon,\eta\) 为任意光滑函数, 称为变换的生成函数. 具体的, 与(4)式比较,我们有:
\[ \begin{aligned}\zeta(t,q) = \left.\frac{\partial \theta}{\partial \epsilon}\right|_{(t,q;0)} \\ \epsilon(t,q) = \left.\frac{\partial \psi}{\partial \epsilon}\right|_{(t,q;0)} \end{aligned} \tag{13}\]
Noether 定理
Noether定理:若\(\mathcal{L}(t,q,\dot q)\)在\([t_0,t_1]\)上对于生成函数为\(\zeta,\eta\) 的变换具有变分不变性, 则:
\[ \eta\frac{\partial \mathcal{L}}{\partial \dot q} + \zeta\left(\mathcal{L} - \frac{\partial \mathcal{L}}{\partial \dot q}\dot q\right) = \text{const} \tag{14}\]
在作用量\(\mathcal{S}\) 的任意极值轨道上成立.
\[ \mathcal{S} =\int ^{t_1}_{t_0}\mathcal{L}(t,q,\dot q){\rm d} t \tag{15}\]
(13)式也可以写为:
\[ \eta p - \zeta \mathcal{H} = \text{const} \tag{16}\]
\(p = \frac{\partial \mathcal{L}}{\partial \dot q}\) 为相对于\(q\) 的广义动量, \(\mathcal{H} = \mathcal{L} - p\dot q\)为哈密顿量.
Proof. 令 \[ \widetilde{\mathcal{S}}(q) = \int^{b}_{a}\mathcal{L}(t,q,\dot q){\rm d} t \] 由定义(11), \(\forall [a,b]\in [t_0, t_1]\), 有: \[ \widetilde{S}(Y_{\epsilon}) - \widetilde{S}(q) = \int^{b_{\epsilon}}_{a_{\epsilon}}\mathcal{L}(X,Y_{\epsilon},\dot Y_{\epsilon}) {\rm d} X - \int^{b}_{a}\mathcal{L}(t,q,\dot q){\rm d} t = 0\] 对于足够小的\(\epsilon\)成立, 我们有: \[ X = t+\epsilon \zeta + O(\epsilon^{2}) = t + \epsilon X_{0} \] \[ Y_{\epsilon} = q+ \epsilon \eta + O(\epsilon^{2}) = q + \epsilon Y_{0} \] 或者, 上式给定了\(\delta t = \zeta, \delta q = \eta\).
对于上述变分问题, 其取极值时, 我们有: \[ \left.p \delta q - \mathcal{H} \delta t\right|^{x_1}_{x_2} = 0\] 代入: \[ \eta p - \zeta \mathcal{H}|^{b}_{a} = 0 \] 由于\(a,b\)为满足 \(t_0\leq a < b\leq t_1\)的任意实数, 因此上式成立的唯一可能为: \[ \eta p - \zeta \mathcal{H} = \text{const} \] Q.E.D
进一步的, 对于\(n\) 个自由度的体系,Noether定理可以进行推广,拉氏量的形式变为\(\mathcal{L}(t,{\bf q},\dot {\bf q})\), \({\bf q} = (q_1,\cdots, q_{n})\), 给定由\(\zeta,\eta_{k}\)生成的变换,其中:
\[ Y_{k} = q_{k} + \epsilon \eta_{k} \tag{17}\]
广义动量与哈氏量为2 : \[ \begin{aligned}&p_{k} = \frac{\partial \mathcal{L}}{\partial \dot q_{k}} \\ &\mathcal{H} = p_{k}\dot q_{k} - \mathcal{L}\end{aligned}\tag{18}\] 则守恒量为:
\[ p_{k} \eta_{k} - \mathcal{H} \zeta = \text{const} \]
接下来,我们展示几个具体的例子:
时间平移不变性和能量守恒
考虑时间变换\(T=t+\epsilon, Q_{k} = q_{k}\) 我们有:\[\begin{aligned}\zeta = 1, \eta_{k} = 0\end{aligned}\]
则相应的守恒量为:
\[ \mathcal{H} = \text{const} \]
即, 能量守恒对应于时间平移不变性.
空间平移不变性与动量守恒
考虑空间平移变换:\[ T = t, Q_{k} = q_{k} + \epsilon \eta_{k} \]
则守恒量为:
\[ p_{k} \eta_{k} = \text{const} \]
即, 动量守恒对应于空间平移不变性.
空间旋转不变性与角动量守恒
由于三维旋转较为复杂, 我们作为例子仅考虑绕\(z\) 轴的一个旋转.\[T = t\] \[Q_{1} = q_{1}\cos \epsilon + q_{2} \sin \epsilon\] \[Q_{2} = -q_{1}\sin \epsilon + q_{2} \cos \epsilon\] \[Q_{3} = q_{3}\]
取\(\epsilon\to 0\) 的极限(即,无穷小旋转),我们有:
\[ \zeta=0, \eta_{1} = q_2, \eta_{2} = -q_1, \eta_3 = 0\]
对应的守恒量为:
\[ p_1 q_2 - p_2 q_1 = \text{const} \]
这就是角动量的\(z\) 分量,即空间旋转对称性对应于角动量守恒.