Hamiltonian 定义为

H(p,q)≡δLδq˙q˙−L(q,q˙)H(p,q) \equiv\frac{\delta L}{\delta\dot{q}}\dot{q}-L(q,\dot{q})

作用量表达为

S=∫t1t2dt⋅[12q˙2−V(q)]S = \int_{t_1}^{t_2}\text{d}t\cdot\left[\frac{1}{2}\dot{q}^2-V(q)\right]

在广义相对论中，如果假设 dτ2=e2dt2\text{d}\tau^2=e^2\text{d}t^2，那么

S=∫t1t2edt⋅[12(dqdt)2e−2−V(q)]S=\int_{t_1}^{t_2}e\text{d}t\cdot\left[\frac{1}{2}\left(\frac{\text{d}q}{\text{d}t}\right)^2e^{-2}-V(q)\right]

做变分变的就是度规，因此是对 ee 做变分.

对于一个标量场，其作用量为

S=∫d4x⋅12(−∂μϕ∂μϕ−m2ϕ2)=∫dt∫d3x⋅12(ϕ˙x2−∇ϕ⋅∇ϕ−m2ϕ2)\begin{aligned} S &= \int\text{d}^4x\cdot\frac{1}{2}(-\partial_\mu\phi\partial^\mu\phi-m^2\phi^2)\\\\ &= \int\text{d}t\int\text{d}^3x\cdot\frac{1}{2}(\dot{\phi}_x^2-\nabla\phi\cdot\nabla\phi-m^2\phi^2) \end{aligned}

两个可变分的条件：

• δϕx⃗(t1)=δϕx⃗(t2)=0\delta\phi_{\vec{x}}(t_1)=\delta\phi_{\vec{x}}(t_2)=0.
• ϕx⃗∣∣x⃗∣→∞→0\phi_{\vec{x}}|_{|\vec{x}|\to\infty}\to0

现在可以变分，得到

δS=0=∫dtd3x(ημν∂μ∂νϕ−m2ϕ)δϕ\delta S = 0 =\int\text{d}t\text{d}^3x(\eta_{\mu\nu}\partial^\mu\partial^\nu\phi-m^2\phi)\delta\phi

求此运动方程的第一步就是对角化，为此做 Fourier 变换，

φ(x,t)=∫dt(2π)3ϕr⃗(t)eik⃗⋅r⃗\varphi(x,t) = \int\frac{\text{d}t}{(2\pi)^3}\phi_{\vec{r}}(t)e^{\text{i}\vec{k}\cdot\vec{r}}

解得 φ=ak∗eiωt+bk∗e−iωt\varphi=a_k^*e^{\text{i}\omega t}+b_k^*e^{-\text{i}\omega t}. 如果除以一个 2ωk2\omega k，那么整个式子变成一个 Lorentz 变换下的标量. 代回去，算得

ϕ=∫dϕ(2π)3⋅2uk(bk∗e−iωt+ik⃗⋅y⃗+ak∗eiωt−ik⃗⋅y⃗)\phi = \int\frac{\text{d}\phi}{(2\pi)^3\cdot 2u_k}\left(b_k^*e^{-\text{i}\omega t+\text{i}\vec{k}\cdot\vec{y}}+a_k^*e^{\text{i}\omega t-\text{i}\vec{k}\cdot\vec{y}} \right)

如果 ϕ\phi 想要变为实数，那么必须要求 bk=akb_k=a_k. General 的运动方程为

S=∫d4x⋅L(ϕ,∂μϕ)δS=∫d4x⋅(δL∂ϕ+δqδ∂μϕδ∂μϕ)⟹δLδϕ−∂μ(δLδ∂μϕ)=0\begin{aligned} S &= \int\text{d}^4x\cdot\mathcal{L}(\phi,\partial_\mu\phi)\\\\ \delta S &= \int\text{d}^4x\cdot\left(\frac{\delta\mathcal{L}}{\partial\phi}+\frac{\delta q}{\delta\partial_\mu\phi}\delta\partial_\mu\phi\right)\Longrightarrow\frac{\delta\mathcal{L}}{\delta\phi} - \partial_\mu\left(\frac{\delta\mathcal{L}}{\delta\partial_\mu\phi}\right)=0 \end{aligned}

对于 SS 来说，在 ϕ\phi 变化时 SS 不变，也就是 SS 和时空本身无关 (因为标量场的变化是 ϕ→ϕ+dϕ\phi\to\phi+\text{d}\phi). 因此

0=δS=∫d4x(δLδϕ−∂μ∂Lδ∂μϕ)δϕ+∫d4x∂μ(δLδ∂μϕδϕ)0=\delta S=\int\text{d}^4x\left(\frac{\delta\mathcal{L}}{\delta\phi}-\partial_\mu\frac{\partial\mathcal{L}}{\delta\partial_\mu\phi}\right)\delta\phi + \int\text{d}^4x\partial_\mu\left(\frac{\delta\mathcal{L}}{\delta\partial_\mu\phi}\delta\phi\right)

也就是存在一个守恒流，定义为 Jμ=δLδ∂μϕΔϕJ^\mu = \displaystyle{\frac{\delta\mathcal{L}}{\delta\partial_\mu\phi}\Delta\phi}. 这便是 Noether 定理.

一个例子，考虑 complex scalar，

L=−∂μϕ∂μϕ∗−m2ϕ∗ϕ\mathcal{L} = -\partial_\mu\phi\partial^\mu\phi^*-m^2\phi^*\phi

它们的变换分别是 ϕ→ϕeiα\phi\to\phi e^{\text{i}\alpha}，ϕ∗→ϕ∗e−iα\phi^* \to\phi^*e^{-\text{i}\alpha}，变分分别是 δϕ=iαϕ\delta\phi=\text{i}\alpha\phi，ϕ∗=−iαϕ∗\phi^*=-\text{i}\alpha\phi^*，所以守恒流就是

Jμ=−∂μϕ∗iϕ+∂μϕiϕ∗J^\mu = -\partial^\mu\phi^*\text{i}\phi+\partial^\mu\phi\text{i}\phi^*

验证发现 ∂μJμ\partial_\mu J^\mu 确实为零，上式守恒.

来看这种对称性意味着什么. Spacetime Related Symmetry：在某个平移中，

ϕ′(x′)=ϕ(x)\phi'(x')=\phi(x)

因此，δϕ(x)=∂μϕ(x)δxμ\delta\phi(x) = \partial_\mu\phi(x)\delta x^\mu (同时我们默认有 x′=x+δxx'=x+\delta x，因为这是一个空间平移)，

S=∫d4x⋅L(ϕ,∂μϕ)\begin{aligned} S &= \int\text{d}^4x\cdot\mathcal{L}(\phi,\partial_\mu\phi) \end{aligned}

对于空间的平移，一个巨大的面积 SμS^\mu 变化，会导致边缘上有一个 Lagrangian 的变化；同时面内部会有另一个变化. 总的 δS\delta S 受到两部分贡献，得到

δS=∫dSμδxμL+∫d4x(δLδϕ(x)δϕ(x)+δLδ∂μϕ(x)δ∂μϕ(x))=∫dSμδxμL+∫d4x[(δLδϕ(x)−∂μδLδ∂μϕ(x))‾0δx]+∫d4x⋅∂μ[δLδ∂μϕδϕ(x)]\begin{aligned} \delta S &= \int\text{d}S_\mu\delta x^\mu\mathcal{L}+\int\text{d}^4x\left(\frac{\delta\mathcal{L}}{\delta\phi(x)}\delta\phi(x)+\frac{\delta\mathcal{L}}{\delta\partial_\mu\phi(x)}\delta\partial_\mu\phi(x)\right)\\\\ &= \int\text{d}S_\mu\delta x^\mu\mathcal{L} + \int\text{d}^4x\left[\underset{0}{\underline{\left(\frac{\delta\mathcal{L}}{\delta\phi(x)}-\partial_\mu\frac{\delta\mathcal{L}}{\delta\partial_\mu\phi(x)}\right)}}\delta x\right] + \int\text{d}^4x\cdot\partial_\mu\left[\frac{\delta\mathcal{L}}{\delta\partial_\mu\phi}\delta\phi(x)\right] \end{aligned}

最后一项和第一项 combine 为一个守恒，

∂μ[δLδ∂μϕ(x)δϕ(x)+Lδxμ]=0⟹∂μ[(−δLδ∂μϕ(x)∂νϕ(x)+Lδμν)δxν]=0\partial_\mu\left[\frac{\delta\mathcal{L}}{\delta\partial_\mu\phi(x)}\delta\phi(x)+\mathcal{L}\delta x^\mu\right]=0\Longrightarrow\partial_\mu\left[\left(-\frac{\delta\mathcal{L}}{\delta\partial_\mu\phi(x)}\partial_\nu\phi(x)+\mathcal{L}\delta^\mu{}_\nu\right)\delta x^\nu\right]=0

这是空间变换.

对于一个一般的时空平移变换 (Lorentz 变换)，应该由 x′μ=Λμνxν≈xμ+εμxνx'^\mu=\Lambda^\mu{}_\nu x^\nu\approx x^\mu+\varepsilon^\mu x^\nu，也就是

δxμ=εμνxν\delta x^\mu = \varepsilon^\mu{}_\nu x^\nu

我们得到

Lδxμ−δLδ∂μϕ∂μϕδxμ=Lεμλxλ−δLδ∂μϕ∂νϕενλxλ=(Lεμλ−δLδ∂μϕ∂νϕενλ)xλ=(Lδμλ−δLδ∂μϕ∂λϕ)ελνxν=−Tμλελνxν=−Tμλελνxν\begin{aligned} \mathcal{L}\delta x^\mu - \frac{\delta \mathcal{L}}{\delta\partial_\mu\phi}\partial_\mu\phi\delta x^\mu&= \mathcal{L}\varepsilon^\mu{}_\lambda x^\lambda-\frac{\delta\mathcal{L}}{\delta\partial_\mu\phi}\partial_\nu\phi\varepsilon^\nu{}_\lambda x^\lambda\\\\ &= \left(\mathcal{L}\varepsilon^\mu{}_\lambda-\frac{\delta\mathcal{L}}{\delta\partial_\mu\phi}\partial_\nu\phi\varepsilon^\nu{}_\lambda\right)x^\lambda\\\\ &= \left(\mathcal{L}\delta^\mu{}_\lambda-\frac{\delta\mathcal{L}}{\delta\partial_\mu\phi}\partial_\lambda\phi\right)\varepsilon^\lambda{}_\nu x^\nu = -T^\mu{}_\lambda\varepsilon^\lambda{}_\nu x^\nu = -T^\mu{}_\lambda\varepsilon^{\lambda\nu}x_\nu \end{aligned}

因此得到一个守恒量

∂μ(Tμλxν)ελν=0⟹∂μ(Tμλxν−Tμνxλ)=0\partial_\mu(T^{\mu}{}_\lambda x_\nu)\varepsilon^{\lambda\nu} = 0\Longrightarrow\partial_\mu(T^\mu{}_\lambda x_\nu-T^\mu{}_\nu x_\lambda)=0

把这个守恒量叫作 MμλνM^\mu{}_{\lambda\nu}，那么守恒流可以在三维下积分一次，

Jij=∫d3x⋅M0ij=∫d3x(T0ixj−T0jxi)J_{ij} = \int\text{d}^3x\cdot M^0{}_{ij} = \int\text{d}^3x(T^0{}_ix_j-T^0{}_jx_i)

这是一个角动量！但是它仅仅是一个轨道角动量，也就是说，标量场没有自旋角动量.

自旋来自于下面的情况：generally，

ϕr(x)→ϕr′(x′)+12εμνSμνϕs(x)\phi_r(x)\to \phi'_r(x')+\frac{1}{2}\varepsilon^{\mu\nu}S_{\mu\nu}\phi_s(x)

后面这一项是变化造成的场「内部」的变化，由 SμνS_{\mu\nu} 来描述.

自旋项带来的影响为

M(s)μλκ=δLδ∂μϕrSrs,λκφs(x)M^{(s)\mu}{}_{\lambda\kappa} = \frac{\delta\mathcal{L}}{\delta\partial_\mu\phi_r}S_{rs,\lambda\kappa}\varphi_s(x)

以有自旋的光子场为例，来计算这里的 SS.

Aμ′(x′)=∂xν∂x′μAν(x)=Aμ(x)−ενμAν(x)=Aμ(x)−ενληλμAν(x)=Aμ(x)−12ενλ(ηλμAν(x)−ηλνAμ(x))=Aμ(x)−12ενλ(ημλδκν−ημνδκλ)Aμ(x)\begin{aligned} A'_\mu(x') &= \frac{\partial x^\nu}{\partial x'^\mu}A_\nu(x)\\\\ &= A_\mu(x)-\varepsilon^\nu{}_\mu A_\nu(x)\\\\ &= A_\mu(x)-\varepsilon^{\nu\lambda}\eta_{\lambda\mu}A_\nu(x)\\\\ &= A_\mu(x)-\frac{1}{2}\varepsilon^{\nu\lambda}(\eta_{\lambda\mu}A_\nu(x)-\eta_{\lambda\nu}A_\mu(x))\\\\ &= A_\mu(x)-\frac{1}{2}\varepsilon^{\nu\lambda}(\eta_{\mu\lambda}\delta^\kappa{}_\nu-\eta_{\mu\nu}\delta^\kappa{}_\lambda)A_\mu(x) \end{aligned}

因此这里 SS 是

(Sνλ)μκ=−12(ημλδνκ−ημνδλκ)(S_{\nu\lambda})_\mu{}^\kappa = -\frac{1}{2}(\eta_{\mu\lambda}\delta_\nu{}^\kappa-\eta_{\mu\nu}\delta_\lambda{}^\kappa)

下面用 Noether 流计算电磁场能动张量：依然是 L=−14FμνFμν\mathcal{L} = -\displaystyle{\frac{1}{4}F_{\mu\nu}F^{\mu\nu}}，

Tμν=−δLEMδ∂μAρ∂μAρ+ημνLEM=Fμρ∂νAρ−14ημνF2T^{\mu\nu} = -\frac{\delta\mathcal{L}_{EM}}{\delta\partial_\mu A_\rho}\partial^\mu A_\rho + \eta^{\mu\nu}\mathcal{L}_{EM} = F^{\mu\rho}\partial^\nu A_\rho - \frac{1}{4}\eta^{\mu\nu}F^2

规范变换存在一个自由度，Aρ→Aρ+∂ρφA_\rho\to A_\rho+\partial_\rho\varphi，因此观测量不能含有这种内容.

Tμν=Fμρ(∂νAρ−∂ρAν+∂νAρ)−14ημνF2=FμρFνρ−14ημνF2+Fμρ∂ρAν\begin{aligned} T^{\mu\nu} &= F^{\mu\rho}(\partial^\nu A_\rho-\partial_\rho A^\nu+\partial_\nu A^\rho)-\frac{1}{4}\eta^{\mu\nu}F^2\\\\ &= F^{\mu\rho}F^\nu{}_\rho - \frac{1}{4}\eta^{\mu\nu}F^2+F^{\mu\rho}\partial_\rho A^\nu \end{aligned}