Gauge Field Notes
12.1

Problem 12.1

schwarzChapter 12

习题 12.1

来源: 第12章, PDF第223页

12.1 In a causal theory, commutators of observables should vanish outside the lightcone, [ϕ(x),ϕ(y)]=0[\phi(x), \phi(y)] = 0 for (x−y)2<0(x - y)^2 < 0. For spinors, we found that with anticommutation relations {ψˉ(x),ψ(y)}=0\{\bar{\psi}(x), \psi(y)\} = 0 outside the lightcone. This implies that integer spin quantities constructed out of spinors are automatically causal, e.g. [ψˉψ(x),ψˉψ(y)]=0[\bar{\psi}\psi(x), \bar{\psi}\psi(y)] = 0. However, this is not a proof that spinors must anticommute. What would happen to [ψˉψ(x),ψˉψ(y)][\bar{\psi}\psi(x), \bar{\psi}\psi(y)] outside the lightcone if we used commutation relations for spinors? For simplicity, you can just look at ⟨0∣[ψˉψ(x),ψˉψ(y)]∣0⟩\langle 0 | [\bar{\psi}\psi(x), \bar{\psi}\psi(y)] | 0 \rangle.

习题 12.1 - 解答

先分析物理背景

自旋统计定理表明，半整数自旋的费米场必须使用反对易关系进行量子化。题目要求我们假设旋量场使用对易关系（即错误的玻色子统计）进行量子化，并考察由其构成的双线性可观测量 ψˉψ(x)\bar{\psi}\psi(x) 在类空间隔下的因果性（即对易子是否为零）。

下面计算真空期望值 ⟨0∣[ψˉψ(x),ψˉψ(y)]∣0⟩\langle 0 | [\bar{\psi}\psi(x), \bar{\psi}\psi(y)] | 0 \rangle

首先，写出旋量场在对易关系假设下的两点关联函数。设标量场的 Wightman 传播子为 D(x−y)=∫d3p(2π)32Epe−ip(x−y)D(x-y) = \int \frac{d^3p}{(2\pi)^3 2E_p} e^{-ip(x-y)}。 对于使用对易关系 [apr,aqs†]=(2π)3δ3(p−q)δrs[a_{\mathbf{p}}^r, a_{\mathbf{q}}^{s\dagger}] = (2\pi)^3 \delta^3(\mathbf{p}-\mathbf{q})\delta^{rs} 和 [bpr,bqs†]=(2π)3δ3(p−q)δrs[b_{\mathbf{p}}^r, b_{\mathbf{q}}^{s\dagger}] = (2\pi)^3 \delta^3(\mathbf{p}-\mathbf{q})\delta^{rs} 量子化的 Dirac 场，非零的缩并（传播子）为： Sαβ+(x−y)≡⟨0∣ψα(x)ψˉβ(y)∣0⟩=∫d3p(2π)32Ep(p̸+m)αβe−ip(x−y)=(iγμ∂μ+m)αβD(x−y)S^+_{\alpha\beta}(x-y) \equiv \langle 0 | \psi_\alpha(x) \bar{\psi}_\beta(y) | 0 \rangle = \int \frac{d^3p}{(2\pi)^3 2E_p} (\not{p}+m)_{\alpha\beta} e^{-ip(x-y)} = (i\gamma^\mu \partial_\mu + m)_{\alpha\beta} D(x-y) Sβα−(x−y)≡⟨0∣ψˉα(x)ψβ(y)∣0⟩=∫d3p(2π)32Ep(p̸−m)βαe−ip(x−y)=(iγμ∂μ−m)βαD(x−y)S^-_{\beta\alpha}(x-y) \equiv \langle 0 | \bar{\psi}_\alpha(x) \psi_\beta(y) | 0 \rangle = \int \frac{d^3p}{(2\pi)^3 2E_p} (\not{p}-m)_{\beta\alpha} e^{-ip(x-y)} = (i\gamma^\mu \partial_\mu - m)_{\beta\alpha} D(x-y)

分两步处理对易子的真空期望值： ⟨0∣[ψˉψ(x),ψˉψ(y)]∣0⟩=⟨0∣ψˉα(x)ψα(x)ψˉβ(y)ψβ(y)∣0⟩−⟨0∣ψˉβ(y)ψβ(y)ψˉα(x)ψα(x)∣0⟩\langle 0 | [\bar{\psi}\psi(x), \bar{\psi}\psi(y)] | 0 \rangle = \langle 0 | \bar{\psi}_\alpha(x) \psi_\alpha(x) \bar{\psi}_\beta(y) \psi_\beta(y) | 0 \rangle - \langle 0 | \bar{\psi}_\beta(y) \psi_\beta(y) \bar{\psi}_\alpha(x) \psi_\alpha(x) | 0 \rangle

利用对易场的 Wick 定理（算符交换时不产生费米负号），第一项的连通图部分贡献为： V(x,y)≡⟨0∣ψˉα(x)ψα(x)ψˉβ(y)ψβ(y)∣0⟩conn=⟨0∣ψˉα(x)ψβ(y)∣0⟩⟨0∣ψα(x)ψˉβ(y)∣0⟩V(x,y) \equiv \langle 0 | \bar{\psi}_\alpha(x) \psi_\alpha(x) \bar{\psi}_\beta(y) \psi_\beta(y) | 0 \rangle_{\text{conn}} = \langle 0 | \bar{\psi}_\alpha(x) \psi_\beta(y) | 0 \rangle \langle 0 | \psi_\alpha(x) \bar{\psi}_\beta(y) | 0 \rangle =Sβα−(x−y)Sαβ+(x−y)=Tr[S−(x−y)S+(x−y)]= S^-_{\beta\alpha}(x-y) S^+_{\alpha\beta}(x-y) = \text{Tr}[S^-(x-y) S^+(x-y)] （注：非连通图部分 ⟨0∣ψˉψ∣0⟩2\langle 0|\bar{\psi}\psi|0 \rangle^2 在 x↔yx \leftrightarrow y 交换下显然对称，在对易子中会直接抵消，故只需关注连通部分）。

代入 S+S^+ 和 S−S^- 的表达式计算 Dirac 迹： Tr[S−(x−y)S+(x−y)]=Tr[(iγμ∂μ−m)D(x−y)(iγν∂ν+m)D(x−y)]\text{Tr}[S^-(x-y) S^+(x-y)] = \text{Tr}\left[ (i\gamma^\mu \partial_\mu - m)D(x-y) (i\gamma^\nu \partial_\nu + m)D(x-y) \right] 展开并利用 Dirac 矩阵的迹性质 Tr[γμ]=0\text{Tr}[\gamma^\mu] = 0 和 Tr[γμγν]=4ημν\text{Tr}[\gamma^\mu \gamma^\nu] = 4\eta^{\mu\nu}： V(x,y)=Tr[−γμγν(∂μD(x−y))(∂νD(x−y))−m2D(x−y)2]V(x,y) = \text{Tr}\left[ -\gamma^\mu \gamma^\nu (\partial_\mu D(x-y)) (\partial_\nu D(x-y)) - m^2 D(x-y)^2 \right] V(x,y)=−4[∂μD(x−y)∂μD(x−y)+m2D(x−y)2]V(x,y) = -4 \left[ \partial_\mu D(x-y) \partial^\mu D(x-y) + m^2 D(x-y)^2 \right]

分析类空间隔下的行为

当处于光锥外即 (x−y)2<0(x-y)^2 < 0 时，标量传播子 D(x−y)D(x-y) 仅依赖于洛伦兹不变量 (x−y)2(x-y)^2，因此它是一个关于坐标差 (x−y)(x-y) 的偶函数。 其一阶导数 ∂μD(x−y)\partial_\mu D(x-y) 是关于 (x−y)(x-y) 的奇函数，但其平方项 ∂μD(x−y)∂μD(x−y)\partial_\mu D(x-y) \partial^\mu D(x-y) 依然是偶函数。 因此，整个迹 V(x,y)V(x,y) 在类空间隔下是关于 (x−y)(x-y) 的偶函数，即： V(x,y)=V(y,x)for (x−y)2<0V(x,y) = V(y,x) \quad \text{for } (x-y)^2 < 0

将此结果代回对易子期望值中，得到： ⟨0∣[ψˉψ(x),ψˉψ(y)]∣0⟩=V(x,y)−V(y,x)=0\langle 0 | [\bar{\psi}\psi(x), \bar{\psi}\psi(y)] | 0 \rangle = V(x,y) - V(y,x) = 0

物理结论

即使我们错误地使用对易关系来量子化旋量场，由其构成的双线性可观测量在光锥外依然是对易的。这解释了题目中的论断：“这并不能证明旋量必须反对易”。自旋统计定理对于费米子的真正强制性要求，并不来源于可观测量的微观因果性，而是为了保证系统的哈密顿量具有能量下界（Energy Positivity）。

最终结果为： ⟨0∣[ψˉψ(x),ψˉψ(y)]∣0⟩=0for (x−y)2<0\boxed{ \langle 0 | [\bar{\psi}\psi(x), \bar{\psi}\psi(y)] | 0 \rangle = 0 \quad \text{for } (x-y)^2 < 0 }