系列文章：Srednicki QFT 共 97 篇
自由标量场可以分解为一组独立的谐振子，每个动量模式的频率由自由粒子的能量决定。把上一节的高斯积分用于全部模式，就能求出自由场的外源生成泛函；二次核的逆给出费曼传播子，源导数则生成场的时间序关联函数。
把场看成一组耦合振子
从自由标量场的哈密顿量出发：
H 0 = ∫ d 3 x H 0 , H 0 = 1 2 Π 2 + 1 2 ( ∇ φ ) 2 + 1 2 m 2 φ 2 . (8.1) H_0=\int d^3x\,\mathcal H_0,\qquad
\mathcal H_0=\frac12\Pi^2+
\frac12(\nabla\varphi)^2+\frac12m^2\varphi^2.
\tag{8.1} H 0 = ∫ d 3 x H 0 , H 0 = 2 1 Π 2 + 2 1 ( ∇ φ ) 2 + 2 1 m 2 φ 2 . ( 8.1 ) 沿用 c = ℏ = 1 c=\hbar=1 c = ℏ = 1 ( − , + , + , + ) (-,+,+,+) ( − , + , + , + ) m > 0 m>0 m > 0 q ( t ) q(t) q ( t ) φ ( x , t ) \varphi(\mathbf x,t) φ ( x , t ) Q ( t ) Q(t) Q ( t ) φ ( x , t ) \varphi(\mathbf x,t) φ ( x , t ) f ( t ) f(t) f ( t ) J ( x , t ) J(\mathbf x,t) J ( x , t ) φ \varphi φ φ \varphi φ
先把这个对应放在有限系统中看清。将空间置于有限周期盒，格距记为 a a a r , s r,s r , s
[ φ r , Π s ] = i a 3 δ r s . q r = a 3 / 2 φ r , p r = a 3 / 2 Π r ⟹ [ q r , p s ] = i δ r s . (8.2) [\varphi_r,\Pi_s]=\frac{i}{a^3}\delta_{rs}.
\qquad
q_r=a^{3/2}\varphi_r,\quad p_r=a^{3/2}\Pi_r
\quad\Longrightarrow\quad [q_r,p_s]=i\delta_{rs}.
\tag{8.2} [ φ r , Π s ] = a 3 i δ rs . q r = a 3/2 φ r , p r = a 3/2 Π r ⟹ [ q r , p s ] = i δ rs . ( 8.2 ) 其中 a − 3 a^{-3} a − 3 a 3 ∑ s δ r s / a 3 = 1 a^3\sum_s\delta_{rs}/a^3=1 a 3 ∑ s δ rs / a 3 = 1 q r , p r q_r,p_r q r , p r
H 0 = 1 2 ∑ r p r 2 + 1 2 ∑ r s q r K r s q s , K = m 2 I − ∇ a 2 . (8.3) H_0=\frac12\sum_r p_r^2+\frac12\sum_{rs}q_rK_{rs}q_s,
\qquad K=m^2I-\nabla_a^2.
\tag{8.3} H 0 = 2 1 r ∑ p r 2 + 2 1 rs ∑ q r K rs q s , K = m 2 I − ∇ a 2 . ( 8.3 ) 势能中的梯度平方耦合相邻格点；在傅里叶模式中，这种耦合转化为不同的振动频率。最近邻差分作用于 e i k ⋅ x r e^{i\mathbf k\cdot\mathbf x_r} e i k ⋅ x r [ e i k j a + e − i k j a − 2 ] / a 2 = − 4 sin 2 ( k j a / 2 ) / a 2 [e^{ik_ja}+e^{-ik_ja}-2]/a^2=-4\sin^2(k_ja/2)/a^2 [ e i k j a + e − i k j a − 2 ] / a 2 = − 4 sin 2 ( k j a /2 ) / a 2 K K K
ω k , a 2 = m 2 + 4 a 2 ∑ j = 1 3 sin 2 k j a 2 ⟶ k 2 + m 2 . (8.4) \omega_{\mathbf k,a}^2
=m^2+\frac4{a^2}\sum_{j=1}^{3}\sin^2\frac{k_ja}{2}
\ \longrightarrow\ \mathbf k^2+m^2.
\tag{8.4} ω k , a 2 = m 2 + a 2 4 j = 1 ∑ 3 sin 2 2 k j a ⟶ k 2 + m 2 . ( 8.4 ) 由于 K K K q , p q,p q , p
与坐标的重标度相配，外源的对应也要包含胞元体积：
∫ d 3 x J φ ⟶ ∑ r a 3 J r φ r = ∑ r F r q r , F r = a 3 / 2 J r . (8.5) \int d^3x\,J\varphi
\ \longrightarrow\ \sum_r a^3J_r\varphi_r
=\sum_rF_rq_r,\qquad F_r=a^{3/2}J_r.
\tag{8.5} ∫ d 3 x J φ ⟶ r ∑ a 3 J r φ r = r ∑ F r q r , F r = a 3/2 J r . ( 8.5 ) 在某一时刻指定全部 q r q_r q r N s N_s N s N t + 1 N_t+1 N t + 1
( 1 2 π i δ t ) ( N t + 1 ) N s / 2 ∏ j = 1 N t ∏ r = 1 N s d q r j . (8.6) \left(\frac1{2\pi i\delta t}\right)^{(N_t+1)N_s/2}
\prod_{j=1}^{N_t}\prod_{r=1}^{N_s}dq_{rj}.
\tag{8.6} ( 2 πi δ t 1 ) ( N t + 1 ) N s /2 j = 1 ∏ N t r = 1 ∏ N s d q r j . ( 8.6 ) 转回场值时，还有 ∏ d q = a 3 N t N s / 2 ∏ d φ \prod dq=a^{3N_tN_s/2}\prod d\varphi ∏ d q = a 3 N t N s /2 ∏ d φ D φ ∝ ∏ x d φ ( x ) \mathcal D\varphi\propto\prod_x d\varphi(x) D φ ∝ ∏ x d φ ( x )
真空泛函与质量的无穷小虚部
与谐振子一样，场论积分也要保留选择真空的收敛处方。令 α = 1 − i ϵ 0 \alpha=1-i\epsilon_0 α = 1 − i ϵ 0 H 0 H_0 H 0 α \alpha α
L ϵ 0 + J φ = 1 2 α φ ˙ 2 − α 2 [ ( ∇ φ ) 2 + m 2 φ 2 ] + J φ . (8.7) \mathcal L_{\epsilon_0}+J\varphi
=\frac1{2\alpha}\dot\varphi^2
-\frac{\alpha}{2}\big[(\nabla\varphi)^2+m^2\varphi^2\big]
+J\varphi.
\tag{8.7} L ϵ 0 + J φ = 2 α 1 φ ˙ 2 − 2 α [ ( ∇ φ ) 2 + m 2 φ 2 ] + J φ . ( 8.7 ) 对于固定的 k \mathbf k k ω = ω k = k 2 + m 2 \omega=\omega_{\mathbf k}=\sqrt{\mathbf k^2+m^2} ω = ω k = k 2 + m 2 ( k 0 ) 2 / α − α ω k 2 (k^0)^2/\alpha-\alpha\omega_{\mathbf k}^2 ( k 0 ) 2 / α − α ω k 2 k 0 = ± α ω k k^0=\pm\alpha\omega_{\mathbf k} k 0 = ± α ω k
同一边界处方还可以用质量平方的微小负虚部表示。为与无量纲的 ϵ 0 \epsilon_0 ϵ 0 m 2 → m 2 − i η m^2\to m^2-i\eta m 2 → m 2 − i η η > 0 \eta>0 η > 0
D η = − ∂ 2 + m 2 − i η , D η ( k ) = k 2 + m 2 − i η , k 2 = − ( k 0 ) 2 + k 2 . (8.8) D_\eta=-\partial^2+m^2-i\eta,\qquad
D_\eta(k)=k^2+m^2-i\eta,
\qquad k^2=-(k^0)^2+\mathbf k^2.
\tag{8.8} D η = − ∂ 2 + m 2 − i η , D η ( k ) = k 2 + m 2 − i η , k 2 = − ( k 0 ) 2 + k 2 . ( 8.8 ) 这项质量虚部是对实轴边界值的简洁表示。精确二次核的相反数为 α ω k 2 − ( k 0 ) 2 / α \alpha\omega_{\mathbf k}^2-(k^0)^2/\alpha α ω k 2 − ( k 0 ) 2 / α − i ϵ 0 [ ( k 0 ) 2 + ω k 2 ] -i\epsilon_0[(k^0)^2+\omega_{\mathbf k}^2] − i ϵ 0 [( k 0 ) 2 + ω k 2 ] 2 ω k 2 2\omega_{\mathbf k}^2 2 ω k 2 D η ( k ) D_\eta(k) D η ( k ) η \eta η ϵ 0 \epsilon_0 ϵ 0
采用这一记法，自由场泛函可写为
Z 0 [ J ] = N η ∫ D φ e i S 0 [ φ , J ] , Z 0 [ 0 ] = 1 , S 0 [ φ , J ] = ∫ d 4 x [ − 1 2 ∂ μ φ ∂ μ φ − 1 2 ( m 2 − i η ) φ 2 + J φ ] . (8.9) \begin{aligned}
Z_0[J]&=\mathcal N_\eta\int\mathcal D\varphi\,
e^{iS_0[\varphi,J]},\qquad Z_0[0]=1,\\
S_0[\varphi,J]&=\int d^4x\left[
-\frac12\partial^\mu\varphi\,\partial_\mu\varphi
-\frac12(m^2-i\eta)\varphi^2+J\varphi\right].
\end{aligned}
\tag{8.9} Z 0 [ J ] S 0 [ φ , J ] = N η ∫ D φ e i S 0 [ φ , J ] , Z 0 [ 0 ] = 1 , = ∫ d 4 x [ − 2 1 ∂ μ φ ∂ μ φ − 2 1 ( m 2 − i η ) φ 2 + J φ ] . ( 8.9 ) 常数 N η \mathcal N_\eta N η + i η φ 2 / 2 +i\eta\varphi^2/2 + i η φ 2 /2 e − η ∫ d 4 x φ 2 / 2 e^{-\eta\int d^4x\,\varphi^2/2} e − η ∫ d 4 x φ 2 /2
四维 傅里叶 变换与完成平方
与上一节对时间的变换相仿，现在对全部时空坐标作傅里叶变换，使自由二次作用量在四动量中对角化。定义四维傅里叶变换为
φ ~ ( k ) = ∫ d 4 x e − i k x φ ( x ) , φ ( x ) = ∫ d 4 k ( 2 π ) 4 e i k x φ ~ ( k ) , k x = − k 0 t + k ⋅ x . (8.10) \widetilde\varphi(k)=\int d^4x\,e^{-ikx}\varphi(x),
\qquad
\varphi(x)=\int\frac{d^4k}{(2\pi)^4}e^{ikx}\widetilde\varphi(k),
\qquad kx=-k^0t+\mathbf k\cdot\mathbf x.
\tag{8.10} φ ( k ) = ∫ d 4 x e − ik x φ ( x ) , φ ( x ) = ∫ ( 2 π ) 4 d 4 k e ik x φ ( k ) , k x = − k 0 t + k ⋅ x . ( 8.10 ) 源 J J J k 0 k^0 k 0 ω k \omega_{\mathbf k} ω k ∂ μ e i k x = i k μ e i k x \partial_\mu e^{ikx}=ik_\mu e^{ikx} ∂ μ e ik x = i k μ e ik x
∫ d 4 x e i ( k + k ′ ) x = ( 2 π ) 4 δ 4 ( k + k ′ ) . (8.11) \int d^4x\,e^{i(k+k')x}=(2\pi)^4\delta^4(k+k').
\tag{8.11} ∫ d 4 x e i ( k + k ′ ) x = ( 2 π ) 4 δ 4 ( k + k ′ ) . ( 8.11 ) 先看动能项的符号。将两份场的傅里叶积分代入，两个 i i i 1 2 k μ k μ ′ \frac12 k^\mu k'_\mu 2 1 k μ k μ ′ k ′ = − k k'=-k k ′ = − k − 1 2 k 2 -\frac12 k^2 − 2 1 k 2
S 0 = 1 2 ∫ d 4 k ( 2 π ) 4 [ − φ ~ ( k ) D η ( k ) φ ~ ( − k ) + J ~ ( k ) φ ~ ( − k ) + J ~ ( − k ) φ ~ ( k ) ] . (8.12) \begin{aligned}
S_0=\frac12\int\frac{d^4k}{(2\pi)^4}
\big[&
-\widetilde\varphi(k)D_\eta(k)\widetilde\varphi(-k)\\
&+\widetilde J(k)\widetilde\varphi(-k)
+\widetilde J(-k)\widetilde\varphi(k)\big].
\end{aligned}
\tag{8.12} S 0 = 2 1 ∫ ( 2 π ) 4 d 4 k [ − φ ( k ) D η ( k ) φ ( − k ) + J ( k ) φ ( − k ) + J ( − k ) φ ( k ) ] . ( 8.12 ) 两份源项经 k ↔ − k k\leftrightarrow-k k ↔ − k 1 / 2 1/2 1/2 S 0 = − 1 2 ∫ φ D η φ + ∫ J φ S_0=-\frac12\int\varphi D_\eta\varphi+\int J\varphi S 0 = − 2 1 ∫ φ D η φ + ∫ J φ
第7节的二次核 A = E 2 − ω 2 + i 0 A=E^2-\omega^2+i0 A = E 2 − ω 2 + i 0 − D -D − D
χ ~ ( k ) = φ ~ ( k ) − J ~ ( k ) D η ( k ) . (8.13) \widetilde\chi(k)=\widetilde\varphi(k)
-\frac{\widetilde J(k)}{D_\eta(k)}.
\tag{8.13} χ ( k ) = φ ( k ) − D η ( k ) J ( k ) . ( 8.13 ) 将 φ ~ = χ ~ + D η − 1 J ~ \widetilde\varphi=\widetilde\chi+D_\eta^{-1}\widetilde J φ = χ + D η − 1 J （8.12） ，负二次项产生的两份负交叉项被显式源项分别消去；剩余源平方项的系数为 − 1 + 1 + 1 = 1 -1+1+1=1 − 1 + 1 + 1 = 1
S 0 = 1 2 ∫ d 4 k ( 2 π ) 4 [ J ~ ( k ) J ~ ( − k ) D η ( k ) − χ ~ ( k ) D η ( k ) χ ~ ( − k ) ] . (8.14) S_0=\frac12\int\frac{d^4k}{(2\pi)^4}
\left[
\frac{\widetilde J(k)\widetilde J(-k)}{D_\eta(k)}
-\widetilde\chi(k)D_\eta(k)\widetilde\chi(-k)
\right].
\tag{8.14} S 0 = 2 1 ∫ ( 2 π ) 4 d 4 k [ D η ( k ) J ( k ) J ( − k ) − χ ( k ) D η ( k ) χ ( − k ) ] . ( 8.14 ) 原来的场变量已经与外源分开。要在积分中使用这个复平移，可直接沿用式（7.15） 的恒等式，并把本题的核和测度代入。若四维离散胞元体积为 v 4 v_4 v 4 A = − v 4 D η A=-v_4D_\eta A = − v 4 D η v 4 J v_4J v 4 J v 4 η I v_4\eta I v 4 η I
I ( v 4 J ) I ( 0 ) = exp [ − i 2 ( v 4 J ) T ( − v 4 D η ) − 1 ( v 4 J ) ] = exp [ i 2 v 4 J T D η − 1 J ] . (8.15) \frac{I(v_4J)}{I(0)}
=\exp\!\left[
-\frac i2(v_4J)^T(-v_4D_\eta)^{-1}(v_4J)\right]
=\exp\!\left[\frac i2v_4J^TD_\eta^{-1}J\right].
\tag{8.15} I ( 0 ) I ( v 4 J ) = exp [ − 2 i ( v 4 J ) T ( − v 4 D η ) − 1 ( v 4 J ) ] = exp [ 2 i v 4 J T D η − 1 J ] . ( 8.15 ) 从矩阵逆转为连续积分核时，还要计入求和的体积因子：离散核是 ( Δ η ) r s = ( D η − 1 ) r s / v 4 (\Delta_\eta)_{rs}=(D_\eta^{-1})_{rs}/v_4 ( Δ η ) rs = ( D η − 1 ) rs / v 4 exp [ ( i / 2 ) v 4 2 ∑ r s J r ( Δ η ) r s J s ] \exp[(i/2)v_4^2\sum_{rs}J_r(\Delta_\eta)_{rs}J_s] exp [( i /2 ) v 4 2 ∑ rs J r ( Δ η ) rs J s ]
将外源因子提出后，取相同的连续极限与真空边界极限，得到自由生成泛函：
Z 0 [ J ] = exp [ i 2 ∫ d 4 k ( 2 π ) 4 J ~ ( k ) J ~ ( − k ) k 2 + m 2 − i 0 ] = exp [ i 2 ∫ d 4 x d 4 y J ( x ) Δ ( x − y ) J ( y ) ] , (8.16) \begin{aligned}
Z_0[J]
&=\exp\!\left[
\frac i2\int\frac{d^4k}{(2\pi)^4}
\frac{\widetilde J(k)\widetilde J(-k)}{k^2+m^2-i0}\right]\\
&=\exp\!\left[
\frac i2\int d^4x\,d^4y\,
J(x)\Delta(x-y)J(y)\right],
\end{aligned}
\tag{8.16} Z 0 [ J ] = exp [ 2 i ∫ ( 2 π ) 4 d 4 k k 2 + m 2 − i 0 J ( k ) J ( − k ) ] = exp [ 2 i ∫ d 4 x d 4 y J ( x ) Δ ( x − y ) J ( y ) ] , ( 8.16 ) 这里定义Feynman传播子（Feynman propagator）：
Δ ( z ) = lim η ↓ 0 Δ η ( z ) , Δ η ( z ) = ∫ d 4 k ( 2 π ) 4 e i k z k 2 + m 2 − i η . (8.17) \Delta(z)=\lim_{\eta\downarrow0}\Delta_\eta(z),
\qquad
\Delta_\eta(z)=\int\frac{d^4k}{(2\pi)^4}
\frac{e^{ikz}}{k^2+m^2-i\eta}.
\tag{8.17} Δ ( z ) = η ↓ 0 lim Δ η ( z ) , Δ η ( z ) = ∫ ( 2 π ) 4 d 4 k k 2 + m 2 − i η e ik z . ( 8.17 ) 将双源指数从动量表示改写为坐标表示时，两份源的傅里叶相位最初给出 e − i k ( x − y ) e^{-ik(x-y)} e − ik ( x − y ) k → − k k\to-k k → − k [ φ ] = 1 [\varphi]=1 [ φ ] = 1 [ J ] = 3 [J]=3 [ J ] = 3 [ Δ ] = 2 [\Delta]=2 [ Δ ] = 2 3 + 3 + 2 − 8 = 0 3+3+2-8=0 3 + 3 + 2 − 8 = 0
作完能量积分
为了看出传播子怎样联系两个时空点，令 z = x − y = ( τ , r ) z=x-y=(\tau,\mathbf r) z = x − y = ( τ , r ) （8.17） 中固定三动量并作 k 0 k^0 k 0 E = k 0 E=k^0 E = k 0 ω = ω k \omega=\omega_{\mathbf k} ω = ω k Ω k = ω k 2 − i η \Omega_{\mathbf k}=\sqrt{\omega_{\mathbf k}^2-i\eta} Ω k = ω k 2 − i η τ > 0 \tau>0 τ > 0 − i [ − e − i Ω k τ / ( 2 Ω k ) ] -i[-e^{-i\Omega_{\mathbf k}\tau}/(2\Omega_{\mathbf k})] − i [ − e − i Ω k τ / ( 2 Ω k )] τ < 0 \tau<0 τ < 0 i e i Ω k τ / ( 2 Ω k ) i e^{i\Omega_{\mathbf k}\tau}/(2\Omega_{\mathbf k}) i e i Ω k τ / ( 2 Ω k ) η \eta η
∫ d k 0 2 π e − i k 0 τ − ( k 0 ) 2 + ω k 2 − i 0 = i 2 ω k e − i ω k ∣ τ ∣ . (8.18) \int\frac{dk^0}{2\pi}
\frac{e^{-ik^0\tau}}{-(k^0)^2+\omega_{\mathbf k}^2-i0}
=\frac{i}{2\omega_{\mathbf k}}e^{-i\omega_{\mathbf k}|\tau|}.
\tag{8.18} ∫ 2 π d k 0 − ( k 0 ) 2 + ω k 2 − i 0 e − i k 0 τ = 2 ω k i e − i ω k ∣ τ ∣ . ( 8.18 ) 圆弧积分与式（7.20） 、（7.21） 一样消失：所选半平面中的指数不增长，二次分母又提供 R − 2 R^{-2} R − 2
Δ ( z ) = i ∫ d 3 k ( 2 π ) 3 2 ω k e i k ⋅ r − i ω k ∣ τ ∣ . (8.19) \Delta(z)
=i\int\frac{d^3k}{(2\pi)^3\,2\omega_{\mathbf k}}\,
e^{i\mathbf k\cdot\mathbf r-i\omega_{\mathbf k}|\tau|}.
\tag{8.19} Δ ( z ) = i ∫ ( 2 π ) 3 2 ω k d 3 k e i k ⋅ r − i ω k ∣ τ ∣ . ( 8.19 ) 因此，在壳测度 d k ~ d\widetilde k d k 2 ω k 2\omega_{\mathbf k} 2 ω k
为了把两支的时间次序显式写出，对负时间支将空间积分变量换成 − k -\mathbf k − k ω k \omega_{\mathbf k} ω k
Δ ( z ) = i θ ( τ ) ∫ d k ~ e i k z + i θ ( − τ ) ∫ d k ~ e − i k z , k 0 = ω k . (8.20) \Delta(z)
=i\theta(\tau)\int d\widetilde k\,e^{ikz}
+i\theta(-\tau)\int d\widetilde k\,e^{-ikz},
\qquad k^0=\omega_{\mathbf k}.
\tag{8.20} Δ ( z ) = i θ ( τ ) ∫ d k e ik z + i θ ( − τ ) ∫ d k e − ik z , k 0 = ω k . ( 8.20 ) 第一支将正频率模式从较早的 y 0 y^0 y 0 x 0 x^0 x 0
外源作用后的响应由推迟Green函数描述。将两个能量极点都移到下半平面，定义
Δ r e t ( τ , r ) = lim ϵ ↓ 0 ∫ d 3 k d E ( 2 π ) 4 e i k ⋅ r − i E τ ω k 2 − ( E + i ϵ ) 2 . \Delta_{\rm ret}(\tau,\mathbf r)
=\lim_{\epsilon\downarrow0}\int\frac{d^3k\,dE}{(2\pi)^4}
\frac{e^{i\mathbf k\cdot\mathbf r-iE\tau}}
{\omega_{\mathbf k}^2-(E+i\epsilon)^2}. Δ ret ( τ , r ) = ϵ ↓ 0 lim ∫ ( 2 π ) 4 d 3 k d E ω k 2 − ( E + i ϵ ) 2 e i k ⋅ r − i E τ . 当 τ < 0 \tau<0 τ < 0 τ > 0 \tau>0 τ > 0 E = ω k − i ϵ E=\omega_{\mathbf k}-i\epsilon E = ω k − i ϵ E = − ω k − i ϵ E=-\omega_{\mathbf k}-i\epsilon E = − ω k − i ϵ
− i [ − e − i ω k τ 2 ω k + e i ω k τ 2 ω k ] = sin ( ω k τ ) ω k , Δ r e t ( τ , r ) = θ ( τ ) ∫ d 3 k ( 2 π ) 3 e i k ⋅ r sin ( ω k τ ) ω k . \begin{aligned}
-i\left[-\frac{e^{-i\omega_{\mathbf k}\tau}}{2\omega_{\mathbf k}}
+\frac{e^{i\omega_{\mathbf k}\tau}}{2\omega_{\mathbf k}}\right]
&=\frac{\sin(\omega_{\mathbf k}\tau)}{\omega_{\mathbf k}},\\
\Delta_{\rm ret}(\tau,\mathbf r)
&=\theta(\tau)\int\frac{d^3k}{(2\pi)^3}
e^{i\mathbf k\cdot\mathbf r}
\frac{\sin(\omega_{\mathbf k}\tau)}{\omega_{\mathbf k}}.
\end{aligned} − i [ − 2 ω k e − i ω k τ + 2 ω k e i ω k τ ] Δ ret ( τ , r ) = ω k sin ( ω k τ ) , = θ ( τ ) ∫ ( 2 π ) 3 d 3 k e i k ⋅ r ω k sin ( ω k τ ) . 它在零时刻连续，时间导数从零跳到 δ 3 ( r ) \delta^3(\mathbf r) δ 3 ( r ) D Δ r e t = δ 4 D\Delta_{\rm ret}=\delta^4 D Δ ret = δ 4 φ J ( x ) = ∫ d 4 y Δ r e t ( x − y ) J ( y ) \varphi_J(x)=\int d^4y\,\Delta_{\rm ret}(x-y)J(y) φ J ( x ) = ∫ d 4 y Δ ret ( x − y ) J ( y ) ω k 2 − ( E − i 0 ) 2 \omega_{\mathbf k}^2-(E-i0)^2 ω k 2 − ( E − i 0 ) 2
Δ a d v ( τ , r ) = − θ ( − τ ) ∫ d 3 k ( 2 π ) 3 e i k ⋅ r sin ( ω k τ ) ω k . \Delta_{\rm adv}(\tau,\mathbf r)
=-\theta(-\tau)\int\frac{d^3k}{(2\pi)^3}
e^{i\mathbf k\cdot\mathbf r}\frac{\sin(\omega_{\mathbf k}\tau)}{\omega_{\mathbf k}}. Δ adv ( τ , r ) = − θ ( − τ ) ∫ ( 2 π ) 3 d 3 k e i k ⋅ r ω k sin ( ω k τ ) .
Green 方程中的接触项
传播子是自由二次核的逆，因此应满足相应的Green方程。先利用 ∂ 2 e i k z = − k 2 e i k z \partial^2e^{ikz}=-k^2e^{ikz} ∂ 2 e ik z = − k 2 e ik z η \eta η
D η Δ η ( z ) = ∫ d 4 k ( 2 π ) 4 e i k z = δ 4 ( z ) . (8.21) D_\eta\Delta_\eta(z)
=\int\frac{d^4k}{(2\pi)^4}e^{ikz}
=\delta^4(z).
\tag{8.21} D η Δ η ( z ) = ∫ ( 2 π ) 4 d 4 k e ik z = δ 4 ( z ) . ( 8.21 ) 分母被波算符抵消后只剩傅里叶表示的delta函数，取边界极限便有
( − ∂ 2 + m 2 ) Δ ( z ) = δ 4 ( z ) . (8.22) (-\partial^2+m^2)\Delta(z)=\delta^4(z).
\tag{8.22} ( − ∂ 2 + m 2 ) Δ ( z ) = δ 4 ( z ) . ( 8.22 ) 也可从式（8.19） 的时间表示求出同一个delta源。两支在 τ = 0 \tau=0 τ = 0
∂ τ Δ ( 0 + , r ) − ∂ τ Δ ( 0 − , r ) = 2 ∫ d k ~ ω k e i k ⋅ r = ∫ d 3 k ( 2 π ) 3 e i k ⋅ r = δ 3 ( r ) . (8.23) \begin{aligned}
\partial_\tau\Delta(0^+,\mathbf r)
-\partial_\tau\Delta(0^-,\mathbf r)
&=2\int d\widetilde k\,\omega_{\mathbf k}
e^{i\mathbf k\cdot\mathbf r}\\
&=\int\frac{d^3k}{(2\pi)^3}
e^{i\mathbf k\cdot\mathbf r}
=\delta^3(\mathbf r).
\end{aligned}
\tag{8.23} ∂ τ Δ ( 0 + , r ) − ∂ τ Δ ( 0 − , r ) = 2 ∫ d k ω k e i k ⋅ r = ∫ ( 2 π ) 3 d 3 k e i k ⋅ r = δ 3 ( r ) . ( 8.23 ) 这些等时关系按空间分布理解。离开 τ = 0 \tau=0 τ = 0 ( ∂ τ 2 + ω k 2 ) e − i ω k ∣ τ ∣ = 0 (\partial_\tau^2+\omega_{\mathbf k}^2)e^{-i\omega_{\mathbf k}|\tau|}=0 ( ∂ τ 2 + ω k 2 ) e − i ω k ∣ τ ∣ = 0 δ ( τ ) δ 3 ( r ) \delta(\tau)\delta^3(\mathbf r) δ ( τ ) δ 3 ( r )
也可以直接对式（8.20） 中的阶跃函数求导，但须将所产生的 δ \delta δ δ ′ \delta' δ ′ F ( τ ) F(\tau) F ( τ ) F ( 0 ) = 0 F(0)=0 F ( 0 ) = 0
F ( τ ) δ ′ ( τ ) = F ( 0 ) δ ′ ( τ ) − F ′ ( 0 ) δ ( τ ) . (8.24) F(\tau)\delta'(\tau)
=F(0)\delta'(\tau)-F'(0)\delta(\tau).
\tag{8.24} F ( τ ) δ ′ ( τ ) = F ( 0 ) δ ′ ( τ ) − F ′ ( 0 ) δ ( τ ) . ( 8.24 ) 所以二阶导数中暂时出现的 2 F ′ δ + F δ ′ 2F'\delta+F\delta' 2 F ′ δ + F δ ′ F ′ ( 0 ) δ F'(0)\delta F ′ ( 0 ) δ
剩余空间积分的 Bessel 表示
剩余空间积分可用Bessel函数表示。第4节已求出等时类空核，现在只需把时间依赖接到同一积分中。为控制收敛，先取正实欧几里得时间 s > 0 s>0 s > 0
C E ( s , r ) = ∫ d k ~ e i k ⋅ r − ω k s . (8.25) C_E(s,\mathbf r)=
\int d\widetilde k\,e^{i\mathbf k\cdot\mathbf r-\omega_{\mathbf k}s}.
\tag{8.25} C E ( s , r ) = ∫ d k e i k ⋅ r − ω k s . ( 8.25 ) 因 s s s k 4 k_4 k 4 k 4 = i ω k k_4=i\omega_{\mathbf k} k 4 = i ω k ∫ d k 4 e i k 4 s / [ 2 π ( k 4 2 + ω k 2 ) ] = e − ω k s / ( 2 ω k ) \int dk_4\,e^{ik_4s}/[2\pi(k_4^2+\omega_{\mathbf k}^2)]
=e^{-\omega_{\mathbf k}s}/(2\omega_{\mathbf k}) ∫ d k 4 e i k 4 s / [ 2 π ( k 4 2 + ω k 2 )] = e − ω k s / ( 2 ω k ) C E C_E C E 1 / ( k E 2 + m 2 ) = ∫ 0 ∞ d u e − u ( k E 2 + m 2 ) 1/(k_E^2+m^2)=\int_0^\infty du\,e^{-u(k_E^2+m^2)} 1/ ( k E 2 + m 2 ) = ∫ 0 ∞ d u e − u ( k E 2 + m 2 )
C E ( s , r ) = ∫ d 4 k E ( 2 π ) 4 e i k E X k E 2 + m 2 = 1 16 π 2 ∫ 0 ∞ d u u 2 exp [ − m 2 u − R 2 4 u ] , R = r 2 + s 2 > 0. (8.26) \begin{aligned}
C_E(s,\mathbf r)
&=\int\frac{d^4k_E}{(2\pi)^4}
\frac{e^{ik_EX}}{k_E^2+m^2}\\
&=\frac1{16\pi^2}\int_0^\infty\frac{du}{u^2}
\exp\!\left[-m^2u-\frac{R^2}{4u}\right],
\qquad R=\sqrt{r^2+s^2}>0 .
\end{aligned}
\tag{8.26} C E ( s , r ) = ∫ ( 2 π ) 4 d 4 k E k E 2 + m 2 e i k E X = 16 π 2 1 ∫ 0 ∞ u 2 d u exp [ − m 2 u − 4 u R 2 ] , R = r 2 + s 2 > 0. ( 8.26 ) 每个方向给出 ( 4 π u ) − 1 / 2 e − X j 2 / ( 4 u ) (4\pi u)^{-1/2}e^{-X_j^2/(4u)} ( 4 π u ) − 1/2 e − X j 2 / ( 4 u ) u ∈ [ ε , M ] u\in[\varepsilon,M] u ∈ [ ε , M ] u ↓ 0 u\downarrow0 u ↓ 0 e − R 2 / ( 4 u ) e^{-R^2/(4u)} e − R 2 / ( 4 u ) u → ∞ u\to\infty u → ∞
此时剩余的正是式（4.14） ，只须把 r r r R R R v = R 2 / ( 4 u ) v=R^2/(4u) v = R 2 / ( 4 u ) d u / u 2 = − 4 d v / R 2 du/u^2=-4\,dv/R^2 d u / u 2 = − 4 d v / R 2 （4.16） 已求出的 ∫ 0 ∞ d v e − v − ( m R ) 2 / ( 4 v ) = m R K 1 ( m R ) \int_0^\infty dv\,e^{-v-(mR)^2/(4v)}=mR K_1(mR) ∫ 0 ∞ d v e − v − ( m R ) 2 / ( 4 v ) = m R K 1 ( m R )
C E ( s , r ) = m 4 π 2 R K 1 ( m R ) . (8.27) C_E(s,\mathbf r)=\frac{m}{4\pi^2R}K_1(mR).
\tag{8.27} C E ( s , r ) = 4 π 2 R m K 1 ( m R ) . ( 8.27 ) 接下来将欧几里得时间延拓回实时间。在复 s s s （8.25） 因 e − ω k Re s e^{-\omega_{\mathbf k}\operatorname{Re}s} e − ω k Re s s s s s = δ + i ∣ τ ∣ s=\delta+i|\tau| s = δ + i ∣ τ ∣ δ > 0 \delta>0 δ > 0 （8.19） 得到
Δ ( z ) = i m 4 π 2 z 2 + i 0 K 1 ( m z 2 + i 0 ) , z 2 = r 2 − τ 2 . (8.28) \Delta(z)=
\frac{im}{4\pi^2\sqrt{z^2+i0}}\,
K_1\!\left(m\sqrt{z^2+i0}\right),
\qquad z^2=r^2-\tau^2 .
\tag{8.28} Δ ( z ) = 4 π 2 z 2 + i 0 im K 1 ( m z 2 + i 0 ) , z 2 = r 2 − τ 2 . ( 8.28 ) 这个延拓同时指定平方根的分支：类空区域取正实根，类时区域从负实轴上方趋近，光锥上则按相应的分布边界值理解。
在 z 2 > 0 z^2>0 z 2 > 0 i i i m → 0 m\to0 m → 0 （4.15） 中的积分趋于一，传播子简化为
Δ m = 0 ( z ) = i 4 π 2 ( z 2 + i 0 ) . (8.29) \Delta_{m=0}(z)=\frac{i}{4\pi^2(z^2+i0)}.
\tag{8.29} Δ m = 0 ( z ) = 4 π 2 ( z 2 + i 0 ) i . ( 8.29 ) 传播子在光锥上具有奇异性；相互作用展开中的同点传播子 Δ ( 0 ) \Delta(0) Δ ( 0 )
场关联函数与 Wick 配对
自由生成泛函一经求出，关联函数仍按第7节的方法产生。将每次位置源微分推广为四维源微分，得到
⟨ 0 ∣ T φ ( x 1 ) ⋯ φ ( x n ) ∣ 0 ⟩ = ∏ a = 1 n 1 i δ δ J ( x a ) Z 0 [ J ] ∣ J = 0 . (8.30) \langle0|T\varphi(x_1)\cdots\varphi(x_n)|0\rangle
=\left.
\prod_{a=1}^{n}\frac1i\frac{\delta}{\delta J(x_a)}
Z_0[J]\right|_{J=0}.
\tag{8.30} ⟨ 0∣ Tφ ( x 1 ) ⋯ φ ( x n ) ∣0 ⟩ = a = 1 ∏ n i 1 δ J ( x a ) δ Z 0 [ J ] J = 0 . ( 8.30 ) 泛函导数的归一化也与先前的胞元体积相配。沿式（8.5） ，四维源作用量写成 ∑ r v 4 J r φ r \sum_rv_4J_r\varphi_r ∑ r v 4 J r φ r ∂ / ∂ J r \partial/\partial J_r ∂ / ∂ J r i v 4 φ r iv_4\varphi_r i v 4 φ r v 4 − 1 ∂ / ∂ J r v_4^{-1}\partial/\partial J_r v 4 − 1 ∂ / ∂ J r δ 4 ( x r − x s ) \delta^4(x_r-x_s) δ 4 ( x r − x s ) δ r s / v 4 \delta_{rs}/v_4 δ rs / v 4
记 F x [ J ] = ∫ d 4 y Δ ( x − y ) J ( y ) F_x[J]=\int d^4y\,\Delta(x-y)J(y) F x [ J ] = ∫ d 4 y Δ ( x − y ) J ( y ) D x = i − 1 δ / δ J ( x ) \mathscr D_x=i^{-1}\delta/\delta J(x) D x = i − 1 δ / δ J ( x ) Δ ( z ) = Δ ( − z ) \Delta(z)=\Delta(-z) Δ ( z ) = Δ ( − z )
D x Z 0 = F x Z 0 , D y D x Z 0 = [ Δ ( x − y ) i + F x F y ] Z 0 . (8.31) \mathscr D_xZ_0=F_xZ_0,\qquad
\mathscr D_y\mathscr D_xZ_0
=\left[\frac{\Delta(x-y)}i+F_xF_y\right]Z_0 .
\tag{8.31} D x Z 0 = F x Z 0 , D y D x Z 0 = [ i Δ ( x − y ) + F x F y ] Z 0 . ( 8.31 ) 第一项已不含源，第二项则仍带两份源；因而在 J = 0 J=0 J = 0
C ( x , y ) ≡ ⟨ 0 ∣ T φ ( x ) φ ( y ) ∣ 0 ⟩ = Δ ( x − y ) i , Δ = i C . (8.32) C(x,y)\equiv\langle0|T\varphi(x)\varphi(y)|0\rangle
=\frac{\Delta(x-y)}i,\qquad \Delta=iC .
\tag{8.32} C ( x , y ) ≡ ⟨ 0∣ Tφ ( x ) φ ( y ) ∣0 ⟩ = i Δ ( x − y ) , Δ = i C . ( 8.32 ) 传播子与时间序矩阵元由此相差一个 i i i D Δ = δ 4 D\Delta=\delta^4 D Δ = δ 4 D C = − i δ 4 DC=-i\delta^4 D C = − i δ 4
模式展开也给出同一结果。由第3节的场展开和式（3.20） ，有 φ ( x ) = ∫ d k ~ [ a e i k x + a † e − i k x ] \varphi(x)=\int d\widetilde k\,[ae^{ikx}+a^\dagger e^{-ikx}] φ ( x ) = ∫ d k [ a e ik x + a † e − ik x ]
⟨ 0 ∣ φ ( x ) φ ( y ) ∣ 0 ⟩ = ∫ d k ~ d p ~ e i k x − i p y ( 2 π ) 3 2 ω k δ 3 ( k − p ) = ∫ d k ~ e i k ( x − y ) . (8.33) \begin{aligned}
\langle0|\varphi(x)\varphi(y)|0\rangle
&=\int d\widetilde k\,d\widetilde p\,
e^{ikx-ipy}
(2\pi)^3\,2\omega_{\mathbf k}\delta^3(\mathbf k-\mathbf p)\\
&=\int d\widetilde k\,e^{ik(x-y)}.
\end{aligned}
\tag{8.33} ⟨ 0∣ φ ( x ) φ ( y ) ∣0 ⟩ = ∫ d k d p e ik x − i p y ( 2 π ) 3 2 ω k δ 3 ( k − p ) = ∫ d k e ik ( x − y ) . ( 8.33 ) 对易子系数抵消 p \mathbf p p ( 2 π ) 3 2 ω p (2\pi)^3 2\omega_{\mathbf p} ( 2 π ) 3 2 ω p x 0 > y 0 x^0>y^0 x 0 > y 0 x , y x,y x , y （8.20） 除以 i i i
高阶关联的组合计数也可以直接沿用第7节。现在插入标号由时间改成时空点，每个配对的核由 G G G Δ \Delta Δ
⟨ 0 ∣ T φ 1 φ 2 φ 3 φ 4 ∣ 0 ⟩ = 1 i 2 [ Δ 12 Δ 34 + Δ 13 Δ 24 + Δ 14 Δ 23 ] , φ a = φ ( x a ) , Δ a b = Δ ( x a − x b ) . (8.34) \begin{aligned}
\langle0|T\varphi_1\varphi_2\varphi_3\varphi_4|0\rangle
=\frac1{i^2}\big[
&\Delta_{12}\Delta_{34}
+\Delta_{13}\Delta_{24}
+\Delta_{14}\Delta_{23}\big],\\
&\varphi_a=\varphi(x_a),\qquad
\Delta_{ab}=\Delta(x_a-x_b).
\end{aligned}
\tag{8.34} ⟨ 0∣ T φ 1 φ 2 φ 3 φ 4 ∣0 ⟩ = i 2 1 [ Δ 12 Δ 34 + Δ 13 Δ 24 + Δ 14 Δ 23 ] , φ a = φ ( x a ) , Δ ab = Δ ( x a − x b ) . ( 8.34 ) 对于一般 2 n 2n 2 n n n n 2 n n ! 2^n n! 2 n n ! 2 n n ! 2^n n! 2 n n ! 1 / i 1/i 1/ i i − n i^{-n} i − n
⟨ 0 ∣ T φ ( x 1 ) ⋯ φ ( x 2 n ) ∣ 0 ⟩ = 1 i n ∑ P ∏ ( a , b ) ∈ P Δ ( x a − x b ) . (8.35) \langle0|T\varphi(x_1)\cdots\varphi(x_{2n})|0\rangle
=\frac1{i^n}\sum_{\mathcal P}
\prod_{(a,b)\in\mathcal P}\Delta(x_a-x_b).
\tag{8.35} ⟨ 0∣ Tφ ( x 1 ) ⋯ φ ( x 2 n ) ∣0 ⟩ = i n 1 P ∑ ( a , b ) ∈ P ∏ Δ ( x a − x b ) . ( 8.35 ) 奇数点函数为零，而上式不重复的配对数为 ( 2 n − 1 ) ! ! (2n-1)!! ( 2 n − 1 )!!
复标量场的外源与配对
把两个独立、质量相同的实场组合为 ϕ = ( φ 1 + i φ 2 ) / 2 \phi=(\varphi_1+i\varphi_2)/\sqrt2 ϕ = ( φ 1 + i φ 2 ) / 2
L 0 + L J = − ∂ μ ϕ † ∂ μ ϕ − m 2 ϕ † ϕ + J † ϕ + ϕ † J . \mathcal L_0+\mathcal L_J
=-\partial^\mu\phi^\dagger\partial_\mu\phi-m^2\phi^\dagger\phi
+J^\dagger\phi+\phi^\dagger J. L 0 + L J = − ∂ μ ϕ † ∂ μ ϕ − m 2 ϕ † ϕ + J † ϕ + ϕ † J . 令 J = ( J 1 + i J 2 ) / 2 J=(J_1+iJ_2)/\sqrt2 J = ( J 1 + i J 2 ) / 2 J 1 φ 1 + J 2 φ 2 J_1\varphi_1+J_2\varphi_2 J 1 φ 1 + J 2 φ 2
Z 0 [ J † , J ] = exp [ i 2 ∫ d 4 x d 4 y ( J 1 ( x ) Δ ( x − y ) J 1 ( y ) + J 2 ( x ) Δ ( x − y ) J 2 ( y ) ) ] = exp [ i ∫ d 4 x d 4 y J † ( x ) Δ ( x − y ) J ( y ) ] . \begin{aligned}
Z_0[J^\dagger,J]
&=\exp\!\left[\frac i2\int d^4x\,d^4y\,
\bigl(J_1(x)\Delta(x-y)J_1(y)+J_2(x)\Delta(x-y)J_2(y)\bigr)\right]\\
&=\exp\!\left[i\int d^4x\,d^4y\,
J^\dagger(x)\Delta(x-y)J(y)\right].
\end{aligned} Z 0 [ J † , J ] = exp [ 2 i ∫ d 4 x d 4 y ( J 1 ( x ) Δ ( x − y ) J 1 ( y ) + J 2 ( x ) Δ ( x − y ) J 2 ( y ) ) ] = exp [ i ∫ d 4 x d 4 y J † ( x ) Δ ( x − y ) J ( y ) ] . 第二行利用 Δ ( x − y ) = Δ ( y − x ) \Delta(x-y)=\Delta(y-x) Δ ( x − y ) = Δ ( y − x ) J , J † J,J^\dagger J , J † i − 1 δ / δ J † i^{-1}\delta/\delta J^\dagger i − 1 δ / δ J † ϕ \phi ϕ i − 1 δ / δ J i^{-1}\delta/\delta J i − 1 δ / δ J ϕ † \phi^\dagger ϕ † J J J J † J^\dagger J †
⟨ 0 ∣ T ϕ ( x ) ϕ † ( y ) ∣ 0 ⟩ = Δ ( x − y ) i , ⟨ T ϕ ( x ) ϕ ( y ) ⟩ = ⟨ T ϕ † ( x ) ϕ † ( y ) ⟩ = 0. \langle0|T\phi(x)\phi^\dagger(y)|0\rangle=\frac{\Delta(x-y)}i,
\qquad
\langle T\phi(x)\phi(y)\rangle
=\langle T\phi^\dagger(x)\phi^\dagger(y)\rangle=0. ⟨ 0∣ Tϕ ( x ) ϕ † ( y ) ∣0 ⟩ = i Δ ( x − y ) , ⟨ Tϕ ( x ) ϕ ( y )⟩ = ⟨ T ϕ † ( x ) ϕ † ( y )⟩ = 0. 一般关联函数在两种场的数目相等时才能非零。含 n n n ϕ \phi ϕ n n n ϕ † \phi^\dagger ϕ † n ! n! n !
⟨ 0 ∣ T ϕ ( x 1 ) ⋯ ϕ ( x n ) ϕ † ( y 1 ) ⋯ ϕ † ( y n ) ∣ 0 ⟩ = 1 i n ∑ σ ∈ S n ∏ a = 1 n Δ ( x a − y σ ( a ) ) . \langle0|T\phi(x_1)\cdots\phi(x_n)
\phi^\dagger(y_1)\cdots\phi^\dagger(y_n)|0\rangle
=\frac1{i^n}\sum_{\sigma\in S_n}
\prod_{a=1}^n\Delta(x_a-y_{\sigma(a)}). ⟨ 0∣ Tϕ ( x 1 ) ⋯ ϕ ( x n ) ϕ † ( y 1 ) ⋯ ϕ † ( y n ) ∣0 ⟩ = i n 1 σ ∈ S n ∑ a = 1 ∏ n Δ ( x a − y σ ( a ) ) . 这里 S n S_n S n n n n i − 2 [ Δ ( x 1 − y 1 ) Δ ( x 2 − y 2 ) + Δ ( x 1 − y 2 ) Δ ( x 2 − y 1 ) ] i^{-2}[\Delta(x_1-y_1)\Delta(x_2-y_2)+\Delta(x_1-y_2)\Delta(x_2-y_1)] i − 2 [ Δ ( x 1 − y 1 ) Δ ( x 2 − y 2 ) + Δ ( x 1 − y 2 ) Δ ( x 2 − y 1 )]
下一节 将相互作用写成作用于 Z 0 [ J ] Z_0[J] Z 0 [ J ]
← 第 7 节 · 章节地图 · 第 9 节 →