Lagrangian and Hamiltonian

经典作用量

$$S[q_n]=\int L(q_n,\dot{q}_n)\dd t$$

作用量对$q_n$变分为零给出 Euler-Lagrange方程

$$\frac{\dd}{\dd t}\frac{\partial L}{\partial \dot{q}_n}-\frac{\partial L}{\partial q_n}=0$$

定义动量 $\displaystyle p_n\equiv\frac{\partial L}{\partial \dot{q}_n}$，由此定义Hamiltonian

$$H(q_n,p_n)\equiv\sum_{n}p_n\dot{q}_n-L(q_n,\dot{q}_n)$$

定义泊松括号

$$\{f,g\}\equiv\frac{\partial f}{\partial q_n}\frac{\partial g}{\partial p_n}-\frac{\partial f}{\partial p_n}\frac{\partial g}{\partial q_n}$$

对易关系

$$\{q_m,p_n\}=\delta_{mn}$$

与Euler-Lagrange方程等价 Hamilton方程

$$\dot{q}_n=\{q_n,H\}=\frac{\partial H}{\partial p_n}\qquad \dot{p}_n=\{p_n,H\}=-\frac{\partial H}{\partial q_n}$$

量子化

$$[q_m,p_n]=i\hbar\delta_{mn}$$

Heisenberg方程

$$\dot{q}_n=\frac{1}{i\hbar}[q_n,H]$$

场形式化

将位置弥散为位置密度 $\phi(\boldsymbol{x})$ 可得 经典场作用量

$$S=\int\text{d}^4x\; \mathcal{L}(\phi_a,\partial_\mu\phi_a)$$

作用量对$\phi_a$变分为零给出 Euler-Lagrange方程

$$\partial_\mu\left(\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi_a)}\right)-\frac{\partial \mathcal{L}}{\partial \phi_a}=0$$

定义动量密度 $\displaystyle \pi(\boldsymbol{x})\equiv\frac{\partial \mathcal{L}}{\partial \dot\phi(\boldsymbol{x})}$，由此定义Hamiltonian密度

$$\mathcal{H}\equiv\pi(\boldsymbol{x})\dot{\phi}(\boldsymbol{x})-\mathcal{L}$$

$$H\equiv \int\text{d}^3 x\;[\pi(\boldsymbol{x})\dot{\phi}(\boldsymbol{x})-\mathcal{L}]$$

自由标量场Lagrange密度

$$\mathcal{L}=\frac{1}{2}\partial_\mu\phi\;\partial^\mu\phi-\frac{1}{2}m^2\phi^2$$

Euler-Lagrange方程给出 Klein-Gordon方程

$$(\partial^\mu\partial_\mu+m^2)\phi=0$$

Hamiltonian

$$H=\int\text{d}^3x\;\left[\frac{1}{2}\pi^2+\frac{1}{2}(\nabla\phi)^2+\frac{1}{2}m^2\phi^2\right]$$

Symmetry and Noether’s theorem

场作用量应满足 Lorentz不变性，在变换 $x^\mu\to (x’)^\mu=\Lambda_{\;\nu}^{\mu}x^\nu$ 下

$$g_{\mu\nu}\to g_{\mu\nu}\Lambda_{\;\rho}^\mu\Lambda_{\;\sigma}^\nu$$

$$\phi(\boldsymbol{x})\to\phi(\Lambda^{-1}\boldsymbol{x}’)$$

$$A_\mu(\boldsymbol{x})\to \Lambda A_\mu(\Lambda^{-1}\boldsymbol{x}’)$$

Noether’s theorem$\quad$ 每个连续对称性对应一个Noether current $j^\mu(\boldsymbol{x})$

$$\partial_\mu j^\mu=0$$

对称性保持作用量不变，说明对称变换前后Lagrangian至多相差一个散度

$$\phi(\boldsymbol{x})\to\phi(\boldsymbol{x})+\alpha\Delta\phi(\boldsymbol{x})$$

$$\mathcal{L}(\boldsymbol{x})\to \mathcal{L}(\boldsymbol{x})+\alpha\partial_\mu \mathcal{J}^\mu(\boldsymbol{x})$$

同时直接计算Lagrangian的变化可得

$$\begin{aligned}\alpha\Delta\mathcal{L}&=\frac{\partial\mathcal{L}}{\partial\phi}(\alpha\Delta\phi)+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\partial_\mu(\alpha\Delta\phi)\\&=\alpha\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi\right)\end{aligned}$$

因此若$\Delta\phi$为对称变换，即可得Noether current的表达形式

$$j^\mu=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi-\mathcal{J}^\mu$$

守恒流 $j^\mu$ 同时对应了守恒荷 $Q$，假设 $|x|\to\infty$ 时 $\boldsymbol{j}$ 以足够快的速度趋于零

$$Q=\int_{\mathbb{R}^3}j^0\text{d}^3x$$

$$\frac{\dd Q}{\dd t}=\int_{\mathbb{R}^3}\frac{\dd j^0}{\dd t}\text{d}^3x=-\int_{\mathbb{R}^3}\nabla\cdot\boldsymbol{j}\;\text{d}^3x=0$$

能动张量$\quad$对时空平移对称应用Noether定理

$$x^\mu\to x^\mu-a^\mu$$

$$\phi(x)\to\phi(x+a)=\phi(x)+a^\mu\partial_\mu\phi(x)$$

$$\mathcal{L}\to \mathcal{L}+a^\mu\partial_\mu\mathcal{L}=\mathcal{L}+a^\nu\partial_\mu(\delta_{\;\nu}^\mu\mathcal{L})$$

可得对应守恒流，其中 $T^{0\mu}$ 为能量密度和能流，$T^{i\mu}$ 为动量密度和动量流密度

$$T^\mu_{\;\;\nu}=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\partial_\nu\phi-\delta_{\;\nu}^\mu\mathcal{L}$$

然而如此定义的能动张量并不关于 $\mu,\nu$ 对称，可以附加合适的散度项将其对称化，其中 $S^{\mu\nu\rho}$ 关于前两个指标反对称，从而保证对称后的能动张量散度仍然为零

$$\Theta^{\mu\nu}=T^{\mu\nu}+\partial_\rho S^{\rho\mu\nu}$$

定义能动张量的另一种更加本质的方式如下

$$S=\int_\text{d}^4x\;\sqrt{-g}\mathcal{L}(g_{\mu\nu},\phi,\partial_\mu\phi)$$

$$T^{\mu\nu}=-\frac{2}{\sqrt{-g}}\frac{\partial(\sqrt{-g}\mathcal{L})}{\partial g_{\mu\nu}}$$

对于Maxwell理论，其能动张量如下

$$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\qquad F_{\mu\nu}=\partial_\mu A_\nu-\partial_\mu A_\nu$$

$$T^{\mu\nu}=-F^{\mu\sigma}\partial^\nu A_\sigma-g^{\mu\nu}\mathcal{L}$$

$$S^{\rho\mu\nu}=F^{\mu\rho}A^\nu\qquad\Theta^{\mu\nu}=-F^{\mu\rho}F^\nu_{\;\;\rho}+\frac{1}{4}g^{\mu\nu}F_{\rho\lambda}F^{\rho\lambda}$$

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