Gauge Field Notes
55.1

Problem 55.1

srednickiChapter 55

习题 55.1

来源: 第55章, PDF第338页

55.1 Use eqs. (55.13–55.20) and [Ai,Aj]=[Πi,Πj]=0[A_{i}, A_{j}] = [\Pi_{i}, \Pi_{j}] = 0 (at equal times) to verify eqs. (55.21–55.23).

Referenced Equations:

Equation (55.13):

k⋅ελ(k)=0,(55.13)\mathbf{k} \cdot \boldsymbol{\varepsilon}_\lambda(\mathbf{k}) = 0 , \tag{55.13}

Equation (55.14):

ελ′(k)⋅ελ∗(k)=δλ′λ,(55.14)\boldsymbol{\varepsilon}_{\lambda'}(\mathbf{k}) \cdot \boldsymbol{\varepsilon}_\lambda^*(\mathbf{k}) = \delta_{\lambda'\lambda} , \tag{55.14}

Equation (55.15):

∑λ=±εiλ∗(k)εjλ(k)=δij−kikjk2.(55.15)\sum_{\lambda=\pm} \varepsilon_{i\lambda}^*(\mathbf{k}) \varepsilon_{j\lambda}(\mathbf{k}) = \delta_{ij} - \frac{k_i k_j}{\mathbf{k}^2} . \tag{55.15}

Equation (55.16):

aλ(k)=+iελ(k)⋅∫d3x e−ikx∂↔0A(x),(55.16)a_\lambda(\mathbf{k}) = +i \boldsymbol{\varepsilon}_\lambda(\mathbf{k}) \cdot \int d^3x \ e^{-ikx} \overleftrightarrow{\partial}_0 \mathbf{A}(x) , \tag{55.16}

Equation (55.17):

aλ†(k)=−iελ∗(k)⋅∫d3x e+ikx∂↔0A(x),(55.17)a_\lambda^\dagger(\mathbf{k}) = -i \boldsymbol{\varepsilon}_\lambda^*(\mathbf{k}) \cdot \int d^3x \ e^{+ikx} \overleftrightarrow{\partial}_0 \mathbf{A}(x) , \tag{55.17}

Equation (55.18):

Πi=∂L∂A˙i=A˙i.(55.18)\Pi_i = \frac{\partial \mathcal{L}}{\partial \dot{A}_i} = \dot{A}_i . \tag{55.18}

Equation (55.19):

H=ΠiA˙i−L=12ΠiΠi+12∇jAi∇jAi−JiAi+Hcoul,(55.19)\begin{aligned} \mathcal{H} &= \Pi_i \dot{A}_i - \mathcal{L} \\ &= \frac{1}{2} \Pi_i \Pi_i + \frac{1}{2} \nabla_j A_i \nabla_j A_i - J_i A_i + \mathcal{H}_{\text{coul}} , \end{aligned} \tag{55.19}

Equation (55.20):

[Ai(x,t),Πj(y,t)]=i(δij−∇i∇j∇2)δ3(x−y)=i∫d3k(2π)3eik⋅(x−y)(δij−kikjk2).(55.20)\begin{aligned} [A_i(\mathbf{x}, t), \Pi_j(\mathbf{y}, t)] &= i \left( \delta_{ij} - \frac{\nabla_i \nabla_j}{\nabla^2} \right) \delta^3(\mathbf{x} - \mathbf{y}) \\ &= i \int \frac{d^3k}{(2\pi)^3} e^{i\mathbf{k} \cdot (\mathbf{x}-\mathbf{y})} \left( \delta_{ij} - \frac{k_i k_j}{\mathbf{k}^2} \right) . \end{aligned} \tag{55.20}

Equation (55.21):

[aλ(k),aλ′(k′)]=0 ,[aλ†(k),aλ′†(k′)]=0 ,[aλ(k),aλ′†(k′)]=(2π)32ω δ3(k′−k)δλλ′ .\begin{align} [a_{\lambda}(\mathbf{k}), a_{\lambda'}(\mathbf{k}')] &= 0 \, , \tag{55.21} \\ [a_{\lambda}^{\dagger}(\mathbf{k}), a_{\lambda'}^{\dagger}(\mathbf{k}')] &= 0 \, , \tag{55.22} \\ [a_{\lambda}(\mathbf{k}), a_{\lambda'}^{\dagger}(\mathbf{k}')] &= (2\pi)^{3} 2\omega \, \delta^{3}(\mathbf{k}' - \mathbf{k}) \delta_{\lambda\lambda'} \, . \tag{55.23} \\ \end{align}

习题 55.1 - 解答

首先，根据双向导数 f∂↔0g=fg˙−f˙gf \overleftrightarrow{\partial}_0 g = f \dot{g} - \dot{f} g 的定义，以及度规约定 kx=−ωt+k⋅xkx = -\omega t + \mathbf{k} \cdot \mathbf{x}，我们可以将式 (55.16) 和 (55.17) 中的产生与湮灭算符在固定时间 tt 展开。

对于 e−ikx=eiωt−ik⋅xe^{-ikx} = e^{i\omega t - i\mathbf{k} \cdot \mathbf{x}}，有 ∂0e−ikx=iωe−ikx\partial_0 e^{-ikx} = i\omega e^{-ikx}。代入式 (55.16) 得到：

aλ(k)=iεiλ(k)∫d3x[e−ikxA˙i(x)−(∂0e−ikx)Ai(x)]=iεiλ(k)∫d3x eiωt−ik⋅x[Πi(x)−iωAi(x)]\begin{aligned} a_\lambda(\mathbf{k}) &= i \varepsilon_{i\lambda}(\mathbf{k}) \int d^3x \left[ e^{-ikx} \dot{A}_i(x) - (\partial_0 e^{-ikx}) A_i(x) \right] \\ &= i \varepsilon_{i\lambda}(\mathbf{k}) \int d^3x \ e^{i\omega t - i\mathbf{k} \cdot \mathbf{x}} \left[ \Pi_i(x) - i\omega A_i(x) \right] \end{aligned}

同理，对于 eikx=e−iωt+ik⋅xe^{ikx} = e^{-i\omega t + i\mathbf{k} \cdot \mathbf{x}}，有 ∂0eikx=−iωeikx\partial_0 e^{ikx} = -i\omega e^{ikx}。代入式 (55.17) 得到：

aλ†(k)=−iεiλ∗(k)∫d3x[eikxA˙i(x)−(∂0eikx)Ai(x)]=−iεiλ∗(k)∫d3x e−iωt+ik⋅x[Πi(x)+iωAi(x)]\begin{aligned} a_\lambda^\dagger(\mathbf{k}) &= -i \varepsilon_{i\lambda}^*(\mathbf{k}) \int d^3x \left[ e^{ikx} \dot{A}_i(x) - (\partial_0 e^{ikx}) A_i(x) \right] \\ &= -i \varepsilon_{i\lambda}^*(\mathbf{k}) \int d^3x \ e^{-i\omega t + i\mathbf{k} \cdot \mathbf{x}} \left[ \Pi_i(x) + i\omega A_i(x) \right] \end{aligned}

根据题目给定的等时对易关系 [Ai(x,t),Aj(y,t)]=[Πi(x,t),Πj(y,t)]=0[A_i(\mathbf{x}, t), A_j(\mathbf{y}, t)] = [\Pi_i(\mathbf{x}, t), \Pi_j(\mathbf{y}, t)] = 0 以及式 (55.20)：

[Ai(x,t),Πj(y,t)]=i∫d3p(2π)3eip⋅(x−y)(δij−pipjp2)≡iΔij(x−y)[A_i(\mathbf{x}, t), \Pi_j(\mathbf{y}, t)] = i \int \frac{d^3p}{(2\pi)^3} e^{i\mathbf{p} \cdot (\mathbf{x}-\mathbf{y})} \left( \delta_{ij} - \frac{p_i p_j}{\mathbf{p}^2} \right) \equiv i \Delta_{ij}(\mathbf{x}-\mathbf{y})

由于 Δij(x−y)\Delta_{ij}(\mathbf{x}-\mathbf{y}) 对 i,ji,j 对称且是 x−y\mathbf{x}-\mathbf{y} 的偶函数，故 [Πi(x,t),Aj(y,t)]=−iΔij(x−y)[\Pi_i(\mathbf{x}, t), A_j(\mathbf{y}, t)] = -i \Delta_{ij}(\mathbf{x}-\mathbf{y})。

验证式 (55.21)

计算 [aλ(k),aλ′(k′)][a_\lambda(\mathbf{k}), a_{\lambda'}(\mathbf{k}')]：

[aλ(k),aλ′(k′)]=−εiλ(k)εjλ′(k′)∫d3xd3y eiωt−ik⋅xeiω′t−ik′⋅y[Πi(x)−iωAi(x),Πj(y)−iω′Aj(y)][a_\lambda(\mathbf{k}), a_{\lambda'}(\mathbf{k}')] = -\varepsilon_{i\lambda}(\mathbf{k}) \varepsilon_{j\lambda'}(\mathbf{k}') \int d^3x d^3y \ e^{i\omega t - i\mathbf{k} \cdot \mathbf{x}} e^{i\omega' t - i\mathbf{k}' \cdot \mathbf{y}} [\Pi_i(x) - i\omega A_i(x), \Pi_j(y) - i\omega' A_j(y)]

被积函数中的对易子为：

[Πi(x)−iωAi(x),Πj(y)−iω′Aj(y)]=−iω′[Πi(x),Aj(y)]−iω[Ai(x),Πj(y)]=−iω′(−iΔij(x−y))−iω(iΔij(x−y))=(ω−ω′)Δij(x−y)\begin{aligned} [\Pi_i(x) - i\omega A_i(x), \Pi_j(y) - i\omega' A_j(y)] &= -i\omega' [\Pi_i(x), A_j(y)] - i\omega [A_i(x), \Pi_j(y)] \\ &= -i\omega' (-i \Delta_{ij}(\mathbf{x}-\mathbf{y})) - i\omega (i \Delta_{ij}(\mathbf{x}-\mathbf{y})) \\ &= (\omega - \omega') \Delta_{ij}(\mathbf{x}-\mathbf{y}) \end{aligned}

代回积分中，并对空间坐标 x\mathbf{x} 和 y\mathbf{y} 积分：

[aλ(k),aλ′(k′)]=−εiλ(k)εjλ′(k′)ei(ω+ω′)t∫d3p(2π)3(δij−pipjp2)(ω−ω′)∫d3x ei(p−k)⋅x∫d3y e−i(p+k′)⋅y=−εiλ(k)εjλ′(k′)ei(ω+ω′)t∫d3p(2π)3(δij−pipjp2)(ω−ω′)(2π)3δ3(p−k)(2π)3δ3(p+k′)=−εiλ(k)εjλ′(k′)ei(ω+ω′)t(2π)3δ3(k+k′)(δij−kikjk2)(ω−ω′)\begin{aligned} [a_\lambda(\mathbf{k}), a_{\lambda'}(\mathbf{k}')] &= -\varepsilon_{i\lambda}(\mathbf{k}) \varepsilon_{j\lambda'}(\mathbf{k}') e^{i(\omega+\omega')t} \int \frac{d^3p}{(2\pi)^3} \left( \delta_{ij} - \frac{p_i p_j}{\mathbf{p}^2} \right) (\omega - \omega') \int d^3x \ e^{i(\mathbf{p}-\mathbf{k}) \cdot \mathbf{x}} \int d^3y \ e^{-i(\mathbf{p}+\mathbf{k}') \cdot \mathbf{y}} \\ &= -\varepsilon_{i\lambda}(\mathbf{k}) \varepsilon_{j\lambda'}(\mathbf{k}') e^{i(\omega+\omega')t} \int \frac{d^3p}{(2\pi)^3} \left( \delta_{ij} - \frac{p_i p_j}{\mathbf{p}^2} \right) (\omega - \omega') (2\pi)^3 \delta^3(\mathbf{p}-\mathbf{k}) (2\pi)^3 \delta^3(\mathbf{p}+\mathbf{k}') \\ &= -\varepsilon_{i\lambda}(\mathbf{k}) \varepsilon_{j\lambda'}(\mathbf{k}') e^{i(\omega+\omega')t} (2\pi)^3 \delta^3(\mathbf{k}+\mathbf{k}') \left( \delta_{ij} - \frac{k_i k_j}{\mathbf{k}^2} \right) (\omega - \omega') \end{aligned}

由于 δ3(k+k′)\delta^3(\mathbf{k}+\mathbf{k}') 强制要求 k′=−k\mathbf{k}' = -\mathbf{k}，这蕴含着 ω′=∣k′∣=∣−k∣=ω\omega' = |\mathbf{k}'| = |-\mathbf{k}| = \omega。因此因子 (ω−ω′)=0(\omega - \omega') = 0，得到：

[aλ(k),aλ′(k′)]=0(55.21)\boxed{ [a_\lambda(\mathbf{k}), a_{\lambda'}(\mathbf{k}')] = 0 } \tag{55.21}

验证式 (55.22)

计算 [aλ†(k),aλ′†(k′)][a_\lambda^\dagger(\mathbf{k}), a_{\lambda'}^\dagger(\mathbf{k}')]：

[aλ†(k),aλ′†(k′)]=−εiλ∗(k)εjλ′∗(k′)∫d3xd3y e−iωt+ik⋅xe−iω′t+ik′⋅y[Πi(x)+iωAi(x),Πj(y)+iω′Aj(y)][a_\lambda^\dagger(\mathbf{k}), a_{\lambda'}^\dagger(\mathbf{k}')] = -\varepsilon_{i\lambda}^*(\mathbf{k}) \varepsilon_{j\lambda'}^*(\mathbf{k}') \int d^3x d^3y \ e^{-i\omega t + i\mathbf{k} \cdot \mathbf{x}} e^{-i\omega' t + i\mathbf{k}' \cdot \mathbf{y}} [\Pi_i(x) + i\omega A_i(x), \Pi_j(y) + i\omega' A_j(y)]

被积函数中的对易子为：

[Πi(x)+iωAi(x),Πj(y)+iω′Aj(y)]=iω′[Πi(x),Aj(y)]+iω[Ai(x),Πj(y)]=iω′(−iΔij(x−y))+iω(iΔij(x−y))=(ω′−ω)Δij(x−y)\begin{aligned} [\Pi_i(x) + i\omega A_i(x), \Pi_j(y) + i\omega' A_j(y)] &= i\omega' [\Pi_i(x), A_j(y)] + i\omega [A_i(x), \Pi_j(y)] \\ &= i\omega' (-i \Delta_{ij}(\mathbf{x}-\mathbf{y})) + i\omega (i \Delta_{ij}(\mathbf{x}-\mathbf{y})) \\ &= (\omega' - \omega) \Delta_{ij}(\mathbf{x}-\mathbf{y}) \end{aligned}

同理，对空间坐标积分会产生 (2π)3δ3(p+k)(2π)3δ3(p−k′)(2\pi)^3 \delta^3(\mathbf{p}+\mathbf{k}) (2\pi)^3 \delta^3(\mathbf{p}-\mathbf{k}')，从而给出整体的动量守恒因子 δ3(k+k′)\delta^3(\mathbf{k}+\mathbf{k}')。这同样要求 ω′=ω\omega' = \omega，使得因子 (ω′−ω)=0(\omega' - \omega) = 0，得到：

[aλ†(k),aλ′†(k′)]=0(55.22)\boxed{ [a_\lambda^\dagger(\mathbf{k}), a_{\lambda'}^\dagger(\mathbf{k}')] = 0 } \tag{55.22}

验证式 (55.23)

计算 [aλ(k),aλ′†(k′)][a_\lambda(\mathbf{k}), a_{\lambda'}^\dagger(\mathbf{k}')]：

[aλ(k),aλ′†(k′)]=εiλ(k)εjλ′∗(k′)∫d3xd3y eiωt−ik⋅xe−iω′t+ik′⋅y[Πi(x)−iωAi(x),Πj(y)+iω′Aj(y)][a_\lambda(\mathbf{k}), a_{\lambda'}^\dagger(\mathbf{k}')] = \varepsilon_{i\lambda}(\mathbf{k}) \varepsilon_{j\lambda'}^*(\mathbf{k}') \int d^3x d^3y \ e^{i\omega t - i\mathbf{k} \cdot \mathbf{x}} e^{-i\omega' t + i\mathbf{k}' \cdot \mathbf{y}} [\Pi_i(x) - i\omega A_i(x), \Pi_j(y) + i\omega' A_j(y)]

被积函数中的对易子为：

[Πi(x)−iωAi(x),Πj(y)+iω′Aj(y)]=iω′[Πi(x),Aj(y)]−iω[Ai(x),Πj(y)]=iω′(−iΔij(x−y))−iω(iΔij(x−y))=(ω+ω′)Δij(x−y)\begin{aligned} [\Pi_i(x) - i\omega A_i(x), \Pi_j(y) + i\omega' A_j(y)] &= i\omega' [\Pi_i(x), A_j(y)] - i\omega [A_i(x), \Pi_j(y)] \\ &= i\omega' (-i \Delta_{ij}(\mathbf{x}-\mathbf{y})) - i\omega (i \Delta_{ij}(\mathbf{x}-\mathbf{y})) \\ &= (\omega + \omega') \Delta_{ij}(\mathbf{x}-\mathbf{y}) \end{aligned}

代回积分中，并对空间坐标 x\mathbf{x} 和 y\mathbf{y} 积分：

[aλ(k),aλ′†(k′)]=εiλ(k)εjλ′∗(k′)ei(ω−ω′)t∫d3p(2π)3(δij−pipjp2)(ω+ω′)∫d3x ei(p−k)⋅x∫d3y e−i(p−k′)⋅y=εiλ(k)εjλ′∗(k′)ei(ω−ω′)t∫d3p(2π)3(δij−pipjp2)(ω+ω′)(2π)3δ3(p−k)(2π)3δ3(p−k′)=εiλ(k)εjλ′∗(k′)ei(ω−ω′)t(2π)3δ3(k−k′)(δij−kikjk2)(ω+ω′)\begin{aligned} [a_\lambda(\mathbf{k}), a_{\lambda'}^\dagger(\mathbf{k}')] &= \varepsilon_{i\lambda}(\mathbf{k}) \varepsilon_{j\lambda'}^*(\mathbf{k}') e^{i(\omega-\omega')t} \int \frac{d^3p}{(2\pi)^3} \left( \delta_{ij} - \frac{p_i p_j}{\mathbf{p}^2} \right) (\omega + \omega') \int d^3x \ e^{i(\mathbf{p}-\mathbf{k}) \cdot \mathbf{x}} \int d^3y \ e^{-i(\mathbf{p}-\mathbf{k}') \cdot \mathbf{y}} \\ &= \varepsilon_{i\lambda}(\mathbf{k}) \varepsilon_{j\lambda'}^*(\mathbf{k}') e^{i(\omega-\omega')t} \int \frac{d^3p}{(2\pi)^3} \left( \delta_{ij} - \frac{p_i p_j}{\mathbf{p}^2} \right) (\omega + \omega') (2\pi)^3 \delta^3(\mathbf{p}-\mathbf{k}) (2\pi)^3 \delta^3(\mathbf{p}-\mathbf{k}') \\ &= \varepsilon_{i\lambda}(\mathbf{k}) \varepsilon_{j\lambda'}^*(\mathbf{k}') e^{i(\omega-\omega')t} (2\pi)^3 \delta^3(\mathbf{k}-\mathbf{k}') \left( \delta_{ij} - \frac{k_i k_j}{\mathbf{k}^2} \right) (\omega + \omega') \end{aligned}

此时 δ3(k−k′)\delta^3(\mathbf{k}-\mathbf{k}') 强制要求 k′=k\mathbf{k}' = \mathbf{k}，这蕴含着 ω′=ω\omega' = \omega。因此时间依赖项 ei(ω−ω′)t=1e^{i(\omega-\omega')t} = 1，且 (ω+ω′)=2ω(\omega + \omega') = 2\omega。表达式化简为：

[aλ(k),aλ′†(k′)]=(2π)32ωδ3(k−k′)[εiλ(k)εjλ′∗(k)(δij−kikjk2)][a_\lambda(\mathbf{k}), a_{\lambda'}^\dagger(\mathbf{k}')] = (2\pi)^3 2\omega \delta^3(\mathbf{k}-\mathbf{k}') \left[ \varepsilon_{i\lambda}(\mathbf{k}) \varepsilon_{j\lambda'}^*(\mathbf{k}) \left( \delta_{ij} - \frac{k_i k_j}{\mathbf{k}^2} \right) \right]

展开方括号内的极化矢量缩并：

εiλ(k)εjλ′∗(k)(δij−kikjk2)=ελ(k)⋅ελ′∗(k)−(k⋅ελ(k))(k⋅ελ′∗(k))k2\varepsilon_{i\lambda}(\mathbf{k}) \varepsilon_{j\lambda'}^*(\mathbf{k}) \left( \delta_{ij} - \frac{k_i k_j}{\mathbf{k}^2} \right) = \boldsymbol{\varepsilon}_\lambda(\mathbf{k}) \cdot \boldsymbol{\varepsilon}_{\lambda'}^*(\mathbf{k}) - \frac{(\mathbf{k} \cdot \boldsymbol{\varepsilon}_\lambda(\mathbf{k})) (\mathbf{k} \cdot \boldsymbol{\varepsilon}_{\lambda'}^*(\mathbf{k}))}{\mathbf{k}^2}

根据式 (55.13) 的横向条件 k⋅ελ(k)=0\mathbf{k} \cdot \boldsymbol{\varepsilon}_\lambda(\mathbf{k}) = 0，第二项为零。 根据式 (55.14) 的正交归一条件 ελ′(k)⋅ελ∗(k)=δλ′λ\boldsymbol{\varepsilon}_{\lambda'}(\mathbf{k}) \cdot \boldsymbol{\varepsilon}_\lambda^*(\mathbf{k}) = \delta_{\lambda'\lambda}，对其取复共轭即得 ελ′∗(k)⋅ελ(k)=δλλ′\boldsymbol{\varepsilon}_{\lambda'}^*(\mathbf{k}) \cdot \boldsymbol{\varepsilon}_\lambda(\mathbf{k}) = \delta_{\lambda\lambda'}。 代入后最终得到：

[aλ(k),aλ′†(k′)]=(2π)32ωδ3(k−k′)δλλ′(55.23)\boxed{ [a_\lambda(\mathbf{k}), a_{\lambda'}^\dagger(\mathbf{k}')] = (2\pi)^3 2\omega \delta^3(\mathbf{k}-\mathbf{k}') \delta_{\lambda\lambda'} } \tag{55.23}
55.2

Problem 55.2

srednickiChapter 55

习题 55.2

来源: 第55章, PDF第338页

55.2 Use eqs. (55.11), (55.14), (55.19), and (55.21–55.23) to verify eq. (55.24).

Referenced Equations:

Equation (55.11):

A(x)=∑λ=±∫dk~[ελ∗(k)aλ(k)eikx+ελ(k)aλ†(k)e−ikx],(55.11)\mathbf{A}(x) = \sum_{\lambda=\pm} \int \widetilde{dk} \left[ \boldsymbol{\varepsilon}_\lambda^*(\mathbf{k}) a_\lambda(\mathbf{k}) e^{ikx} + \boldsymbol{\varepsilon}_\lambda(\mathbf{k}) a_\lambda^\dagger(\mathbf{k}) e^{-ikx} \right] , \tag{55.11}

Equation (55.14):

ελ′(k)⋅ελ∗(k)=δλ′λ,(55.14)\boldsymbol{\varepsilon}_{\lambda'}(\mathbf{k}) \cdot \boldsymbol{\varepsilon}_\lambda^*(\mathbf{k}) = \delta_{\lambda'\lambda} , \tag{55.14}

Equation (55.19):

H=ΠiA˙i−L=12ΠiΠi+12∇jAi∇jAi−JiAi+Hcoul,(55.19)\begin{aligned} \mathcal{H} &= \Pi_i \dot{A}_i - \mathcal{L} \\ &= \frac{1}{2} \Pi_i \Pi_i + \frac{1}{2} \nabla_j A_i \nabla_j A_i - J_i A_i + \mathcal{H}_{\text{coul}} , \end{aligned} \tag{55.19}

Equation (55.21):

[aλ(k),aλ′(k′)]=0 ,[aλ†(k),aλ′†(k′)]=0 ,[aλ(k),aλ′†(k′)]=(2π)32ω δ3(k′−k)δλλ′ .\begin{align} [a_{\lambda}(\mathbf{k}), a_{\lambda'}(\mathbf{k}')] &= 0 \, , \tag{55.21} \\ [a_{\lambda}^{\dagger}(\mathbf{k}), a_{\lambda'}^{\dagger}(\mathbf{k}')] &= 0 \, , \tag{55.22} \\ [a_{\lambda}(\mathbf{k}), a_{\lambda'}^{\dagger}(\mathbf{k}')] &= (2\pi)^{3} 2\omega \, \delta^{3}(\mathbf{k}' - \mathbf{k}) \delta_{\lambda\lambda'} \, . \tag{55.23} \\ \end{align}

Equation (55.24):

H=∑λ=±∫dk~ ω aλ†(k)aλ(k)+2E0V−∫d3x J(x)⋅A(x)+Hcoul ,(55.24)H = \sum_{\lambda=\pm} \int \widetilde{dk} \, \omega \, a_{\lambda}^{\dagger}(\mathbf{k}) a_{\lambda}(\mathbf{k}) + 2\mathcal{E}_{0}V - \int d^{3}x \, \mathbf{J}(x) \cdot \mathbf{A}(x) + H_{\text{coul}} \, , \tag{55.24}

习题 55.2 - 解答

根据题意，我们需要利用给定的公式推导出电磁场与外部电流相互作用的总哈密顿量 HH（公式 55.24）。总哈密顿量是哈密顿密度 H\mathcal{H} 在全空间的积分：

H=∫d3x HH = \int d^3x \, \mathcal{H}

将公式 (55.19) 代入，总哈密顿量可以分为自由电磁场部分 H0H_0、相互作用部分和库仑部分：

H=12∫d3x(ΠiΠi+∇jAi∇jAi)⏟H0−∫d3x J(x)⋅A(x)+HcoulH = \underbrace{\frac{1}{2} \int d^3x \left( \Pi_i \Pi_i + \nabla_j A_i \nabla_j A_i \right)}_{H_0} - \int d^3x \, \mathbf{J}(x) \cdot \mathbf{A}(x) + H_{\text{coul}}

其中 Hcoul=∫d3x HcoulH_{\text{coul}} = \int d^3x \, \mathcal{H}_{\text{coul}}。接下来我们集中计算自由场哈密顿量 H0H_0。

首先，由公式 (55.11) 求出共轭动量 Π(x)=A˙(x)\mathbf{\Pi}(x) = \dot{\mathbf{A}}(x) 和空间导数 ∇jA(x)\nabla_j \mathbf{A}(x)。注意到 kx=k⋅x−ωtkx = \mathbf{k} \cdot \mathbf{x} - \omega t，对其求导可得：

Π(x)=∑λ=±∫dk~(−iω)[ελ∗(k)aλ(k)eikx−ελ(k)aλ†(k)e−ikx]\mathbf{\Pi}(x) = \sum_{\lambda=\pm} \int \widetilde{dk} (-i\omega) \left[ \boldsymbol{\varepsilon}_\lambda^*(\mathbf{k}) a_\lambda(\mathbf{k}) e^{ikx} - \boldsymbol{\varepsilon}_\lambda(\mathbf{k}) a_\lambda^\dagger(\mathbf{k}) e^{-ikx} \right]
∇jA(x)=∑λ=±∫dk~(ikj)[ελ∗(k)aλ(k)eikx−ελ(k)aλ†(k)e−ikx]\nabla_j \mathbf{A}(x) = \sum_{\lambda=\pm} \int \widetilde{dk} (ik_j) \left[ \boldsymbol{\varepsilon}_\lambda^*(\mathbf{k}) a_\lambda(\mathbf{k}) e^{ikx} - \boldsymbol{\varepsilon}_\lambda(\mathbf{k}) a_\lambda^\dagger(\mathbf{k}) e^{-ikx} \right]

将它们代入 H0H_0 并展开平方项。在计算 ∫d3x Π2\int d^3x \, \mathbf{\Pi}^2 和 ∫d3x (∇jA)2\int d^3x \, (\nabla_j \mathbf{A})^2 时，空间积分 ∫d3x\int d^3x 会产生动量守恒的狄拉克 δ\delta 函数：

  1. 对于包含 aλ(k)aλ′(k′)a_\lambda(\mathbf{k}) a_{\lambda'}(\mathbf{k}') 和 aλ†(k)aλ′†(k′)a_\lambda^\dagger(\mathbf{k}) a_{\lambda'}^\dagger(\mathbf{k}') 的项，空间积分给出 (2π)3δ3(k+k′)(2\pi)^3 \delta^3(\mathbf{k} + \mathbf{k}')，这意味着 k′=−k\mathbf{k}' = -\mathbf{k}，从而 ω′=ω\omega' = \omega。

    • 在 Π2\mathbf{\Pi}^2 中，这些项的系数包含 (−iω)(−iω′)=−ω2(-i\omega)(-i\omega') = -\omega^2。
    • 在 (∇jA)2(\nabla_j \mathbf{A})^2 中，系数包含 (ikj)(ikj′)=−k⋅k′=k2=ω2(ik_j)(ik'_j) = -\mathbf{k} \cdot \mathbf{k}' = \mathbf{k}^2 = \omega^2。
    • 两者相加为 −ω2+ω2=0-\omega^2 + \omega^2 = 0，因此这些项完全抵消。

  2. 对于包含 aλ(k)aλ′†(k′)a_\lambda(\mathbf{k}) a_{\lambda'}^\dagger(\mathbf{k}') 和 aλ†(k)aλ′(k′)a_\lambda^\dagger(\mathbf{k}) a_{\lambda'}(\mathbf{k}') 的交叉项，空间积分给出 (2π)3δ3(k−k′)(2\pi)^3 \delta^3(\mathbf{k} - \mathbf{k}')，这意味着 k′=k\mathbf{k}' = \mathbf{k}，从而 ω′=ω\omega' = \omega。

    • 在 Π2\mathbf{\Pi}^2 中，这些项的系数包含 −(−iω)(iω′)=ω2-(-i\omega)(i\omega') = \omega^2。
    • 在 (∇jA)2(\nabla_j \mathbf{A})^2 中，系数包含 −(ikj)(−ikj′)=k⋅k′=ω2-(ik_j)(-ik'_j) = \mathbf{k} \cdot \mathbf{k}' = \omega^2。
    • 两者相加为 ω2+ω2=2ω2\omega^2 + \omega^2 = 2\omega^2。

利用上述结果，并对 k′\mathbf{k}' 积分（利用 ∫dk′~(2π)3δ3(k−k′)=12ω\int \widetilde{dk'} (2\pi)^3 \delta^3(\mathbf{k} - \mathbf{k}') = \frac{1}{2\omega}），自由场哈密顿量简化为：

H0=12∑λ,λ′∫dk~12ω(2ω2)[ελ∗(k)⋅ελ′(k)aλ(k)aλ′†(k)+ελ(k)⋅ελ′∗(k)aλ†(k)aλ′(k)]H_0 = \frac{1}{2} \sum_{\lambda, \lambda'} \int \widetilde{dk} \frac{1}{2\omega} (2\omega^2) \left[ \boldsymbol{\varepsilon}_\lambda^*(\mathbf{k}) \cdot \boldsymbol{\varepsilon}_{\lambda'}(\mathbf{k}) a_\lambda(\mathbf{k}) a_{\lambda'}^\dagger(\mathbf{k}) + \boldsymbol{\varepsilon}_\lambda(\mathbf{k}) \cdot \boldsymbol{\varepsilon}_{\lambda'}^*(\mathbf{k}) a_\lambda^\dagger(\mathbf{k}) a_{\lambda'}(\mathbf{k}) \right]

根据公式 (55.14) 的极化矢量正交归一性 ελ′(k)⋅ελ∗(k)=δλ′λ\boldsymbol{\varepsilon}_{\lambda'}(\mathbf{k}) \cdot \boldsymbol{\varepsilon}_\lambda^*(\mathbf{k}) = \delta_{\lambda'\lambda}，上式进一步化简为：

H0=12∑λ=±∫dk~ ω[aλ(k)aλ†(k)+aλ†(k)aλ(k)]H_0 = \frac{1}{2} \sum_{\lambda=\pm} \int \widetilde{dk} \, \omega \left[ a_\lambda(\mathbf{k}) a_\lambda^\dagger(\mathbf{k}) + a_\lambda^\dagger(\mathbf{k}) a_\lambda(\mathbf{k}) \right]

为了将哈密顿量正规序化（Normal Ordering），我们使用对易关系公式 (55.23)：

aλ(k)aλ†(k)=aλ†(k)aλ(k)+(2π)32ωδ3(0)a_\lambda(\mathbf{k}) a_\lambda^\dagger(\mathbf{k}) = a_\lambda^\dagger(\mathbf{k}) a_\lambda(\mathbf{k}) + (2\pi)^3 2\omega \delta^3(\mathbf{0})

将其代入 H0H_0 中，得到：

H0=∑λ=±∫dk~ ω aλ†(k)aλ(k)+12∑λ=±∫dk~ ω(2π)32ωδ3(0)H_0 = \sum_{\lambda=\pm} \int \widetilde{dk} \, \omega \, a_\lambda^\dagger(\mathbf{k}) a_\lambda(\mathbf{k}) + \frac{1}{2} \sum_{\lambda=\pm} \int \widetilde{dk} \, \omega (2\pi)^3 2\omega \delta^3(\mathbf{0})

第二项是真空零点能（Zero-point energy）。利用测度定义 dk~=d3k(2π)32ω\widetilde{dk} = \frac{d^3k}{(2\pi)^3 2\omega} 以及全空间体积 V=(2π)3δ3(0)V = (2\pi)^3 \delta^3(\mathbf{0})，我们可以计算该零点能项：

Ezp=12∑λ=±∫d3k(2π)32ωω(2π)32ωδ3(0)=∑λ=±V∫d3k(2π)312ωE_{\text{zp}} = \frac{1}{2} \sum_{\lambda=\pm} \int \frac{d^3k}{(2\pi)^3 2\omega} \omega (2\pi)^3 2\omega \delta^3(\mathbf{0}) = \sum_{\lambda=\pm} V \int \frac{d^3k}{(2\pi)^3} \frac{1}{2} \omega

由于单个标量场的零点能密度定义为 E0=∫d3k(2π)312ω\mathcal{E}_0 = \int \frac{d^3k}{(2\pi)^3} \frac{1}{2} \omega，且光子有两个独立的极化自由度（λ=±\lambda = \pm），因此总零点能为：

Ezp=2E0VE_{\text{zp}} = 2 \mathcal{E}_0 V

于是自由场哈密顿量最终写为：

H0=∑λ=±∫dk~ ω aλ†(k)aλ(k)+2E0VH_0 = \sum_{\lambda=\pm} \int \widetilde{dk} \, \omega \, a_\lambda^\dagger(\mathbf{k}) a_\lambda(\mathbf{k}) + 2 \mathcal{E}_0 V

最后，将 H0H_0 代回总哈密顿量 HH 的表达式中，即可得到公式 (55.24)：

H=∑λ=±∫dk~ ω aλ†(k)aλ(k)+2E0V−∫d3x J(x)⋅A(x)+Hcoul\boxed{ H = \sum_{\lambda=\pm} \int \widetilde{dk} \, \omega \, a_\lambda^\dagger(\mathbf{k}) a_\lambda(\mathbf{k}) + 2\mathcal{E}_0 V - \int d^3x \, \mathbf{J}(x) \cdot \mathbf{A}(x) + H_{\text{coul}} }