系列文章：Srednicki QFT 共 97 篇
上一节要求物理质量和散射截面不随μ \mu μ m m m α \alpha α
同一个拉格朗日量的两种写法
要把改变尺度与保持理论不变这两件事联系起来，先比较拉格朗日量的两种参数化。在d = 6 − ϵ d=6-\epsilon d = 6 − ϵ
L = − 1 2 Z φ ( ∂ φ ) 2 − 1 2 Z m m 2 φ 2 + 1 6 Z g g μ ~ ϵ / 2 φ 3 + Y φ , L = − 1 2 ( ∂ φ 0 ) 2 − 1 2 m 0 2 φ 0 2 + 1 6 g 0 φ 0 3 + Y 0 φ 0 . (28.1) \begin{aligned}
\mathcal L
&=-\frac12Z_\varphi(\partial\varphi)^2
-\frac12Z_m m^2\varphi^2
+\frac16Z_g g\widetilde\mu^{\epsilon/2}\varphi^3+Y\varphi,\\
\mathcal L
&=-\frac12(\partial\varphi_0)^2
-\frac12m_0^2\varphi_0^2+\frac16g_0\varphi_0^3+Y_0\varphi_0.
\end{aligned}
\tag{28.1} L L = − 2 1 Z φ ( ∂ φ ) 2 − 2 1 Z m m 2 φ 2 + 6 1 Z g g μ ϵ /2 φ 3 + Y φ , = − 2 1 ( ∂ φ 0 ) 2 − 2 1 m 0 2 φ 0 2 + 6 1 g 0 φ 0 3 + Y 0 φ 0 . ( 28.1 ) 两种写法表示同一个拉格朗日量，因此各个场单项式的系数必须对应。先从动能项确定φ 0 = Z φ φ \varphi_0=\sqrt{Z_\varphi}\varphi φ 0 = Z φ φ φ 2 , φ 3 , φ \varphi^2,\varphi^3,\varphi φ 2 , φ 3 , φ
m 0 2 Z φ = Z m m 2 , g 0 Z φ 3 / 2 = Z g g μ ~ ϵ / 2 , Y 0 Z φ = Y , m 0 = Z φ − 1 / 2 Z m 1 / 2 m , g 0 = Z φ − 3 / 2 Z g g μ ~ ϵ / 2 , Y 0 = Z φ − 1 / 2 Y . (28.2) \begin{gathered}
m_0^2Z_\varphi=Z_m m^2,\qquad
g_0Z_\varphi^{3/2}=Z_g g\widetilde\mu^{\epsilon/2},\qquad
Y_0\sqrt{Z_\varphi}=Y,\\
m_0=Z_\varphi^{-1/2}Z_m^{1/2}m,\qquad
g_0=Z_\varphi^{-3/2}Z_g g\widetilde\mu^{\epsilon/2},\qquad
Y_0=Z_\varphi^{-1/2}Y.
\end{gathered}
\tag{28.2} m 0 2 Z φ = Z m m 2 , g 0 Z φ 3/2 = Z g g μ ϵ /2 , Y 0 Z φ = Y , m 0 = Z φ − 1/2 Z m 1/2 m , g 0 = Z φ − 3/2 Z g g μ ϵ /2 , Y 0 = Z φ − 1/2 Y . ( 28.2 ) 这里的质量平方根取正质量分支。φ 3 \varphi^3 φ 3 Z φ \sqrt{Z_\varphi} Z φ 3 / 2 3/2 3/2 Y Y Y
重整化群导数所比较的是同一个理论：改变减除尺度时，固定裸场、全部裸参数和外部动量，允许重整化场及参数随之变化。于是式（28.1） 的第二行保持不变，第一行则通过系数与场归一化的共同变化表示同一内容。这也解释了上一节为什么要求物理量的显式尺度依赖与参数跑动相抵消。这里g g g [ g 0 ] = ϵ / 2 [g_0]=\epsilon/2 [ g 0 ] = ϵ /2 μ \mu μ μ ~ \widetilde\mu μ
反项的极点怎样决定耦合变化
在M S ‾ \overline{\mathrm{MS}} MS
Z φ = 1 + ∑ n ≥ 1 a n ( α ) ϵ n , Z m = 1 + ∑ n ≥ 1 b n ( α ) ϵ n , Z g = 1 + ∑ n ≥ 1 c n ( α ) ϵ n . (28.3) Z_\varphi=1+\sum_{n\ge1}\frac{a_n(\alpha)}{\epsilon^n},\qquad
Z_m=1+\sum_{n\ge1}\frac{b_n(\alpha)}{\epsilon^n},\qquad
Z_g=1+\sum_{n\ge1}\frac{c_n(\alpha)}{\epsilon^n}.
\tag{28.3} Z φ = 1 + n ≥ 1 ∑ ϵ n a n ( α ) , Z m = 1 + n ≥ 1 ∑ ϵ n b n ( α ) , Z g = 1 + n ≥ 1 ∑ ϵ n c n ( α ) . ( 28.3 ) 每个极点系数又可以按α \alpha α
a 1 = − α 6 + O ( α 2 ) , b 1 = − α + O ( α 2 ) , c 1 = − α + O ( α 2 ) . (28.4) a_1=-\frac{\alpha}{6}+O(\alpha^2),\qquad
b_1=-\alpha+O(\alpha^2),\qquad
c_1=-\alpha+O(\alpha^2).
\tag{28.4} a 1 = − 6 α + O ( α 2 ) , b 1 = − α + O ( α 2 ) , c 1 = − α + O ( α 2 ) . ( 28.4 ) 由于一圈只有简单极点，n ≥ 2 n\ge2 n ≥ 2 O ( α 2 ) O(\alpha^2) O ( α 2 ) n n n
耦合的裸量关系包含场因子和顶角因子的乘积。先将裸耦合平方，再用对数把这一乘积化为和，定义
α 0 = g 0 2 ( 4 π ) 3 = μ ~ ϵ α e G ( α , ϵ ) , G = ln ( Z g 2 Z φ − 3 ) = ∑ n ≥ 1 G n ( α ) ϵ n . (28.5) \alpha_0=\frac{g_0^2}{(4\pi)^3}
=\widetilde\mu^\epsilon\alpha\,e^{G(\alpha,\epsilon)},\qquad
G=\ln(Z_g^2Z_\varphi^{-3})
=\sum_{n\ge1}\frac{G_n(\alpha)}{\epsilon^n}.
\tag{28.5} α 0 = ( 4 π ) 3 g 0 2 = μ ϵ α e G ( α , ϵ ) , G = ln ( Z g 2 Z φ − 3 ) = n ≥ 1 ∑ ϵ n G n ( α ) . ( 28.5 ) 其中α 0 \alpha_0 α 0 ϵ \epsilon ϵ α \alpha α G G G ln ( 1 + u ) = u − u 2 / 2 + ⋯ \ln(1+u)=u-u^2/2+\cdots ln ( 1 + u ) = u − u 2 /2 + ⋯ u u u ϵ − 1 \epsilon^{-1} ϵ − 1 u 2 u^2 u 2
ln Z φ = a 1 ϵ + a 2 − a 1 2 / 2 ϵ 2 + ⋯ , G 1 = 2 c 1 − 3 a 1 = − 3 2 α + O ( α 2 ) . (28.6) \ln Z_\varphi=\frac{a_1}{\epsilon}
+\frac{a_2-a_1^2/2}{\epsilon^2}+\cdots,\qquad
G_1=2c_1-3a_1=-\frac32\alpha+O(\alpha^2).
\tag{28.6} ln Z φ = ϵ a 1 + ϵ 2 a 2 − a 1 2 /2 + ⋯ , G 1 = 2 c 1 − 3 a 1 = − 2 3 α + O ( α 2 ) . ( 28.6 ) 二重极点中的1 / 2 1/2 1/2
记B ϵ = d α / d ln μ B_\epsilon=d\alpha/d\ln\mu B ϵ = d α / d ln μ ln α 0 = G + ln α + ϵ ln μ ~ \ln\alpha_0=G+\ln\alpha+\epsilon\ln\widetilde\mu ln α 0 = G + ln α + ϵ ln μ
0 = ( ∂ α G + 1 α ) B ϵ + ϵ , 0 = ( 1 + α ∑ n ≥ 1 G n ′ ϵ n ) B ϵ + ϵ α . (28.7) 0=\left(\partial_\alpha G+\frac1\alpha\right)B_\epsilon+\epsilon,
\qquad
0=\left(1+\alpha\sum_{n\ge1}\frac{G_n'}{\epsilon^n}\right)B_\epsilon
+\epsilon\alpha.
\tag{28.7} 0 = ( ∂ α G + α 1 ) B ϵ + ϵ , 0 = ( 1 + α n ≥ 1 ∑ ϵ n G n ′ ) B ϵ + ϵ α . ( 28.7 ) 为解出B ϵ B_\epsilon B ϵ − ϵ α ( 1 + α ∂ α G ) − 1 -\epsilon\alpha(1+\alpha\partial_\alpha G)^{-1} − ϵ α ( 1 + α ∂ α G ) − 1 ϵ \epsilon ϵ − ϵ α -\epsilon\alpha − ϵ α ϵ 0 \epsilon^0 ϵ 0 ϵ → 0 \epsilon\to0 ϵ → 0
B ϵ = − ϵ α + β ( α ) . (28.8) B_\epsilon=-\epsilon\alpha+\beta(\alpha).
\tag{28.8} B ϵ = − ϵ α + β ( α ) . ( 28.8 ) 这一形式同时利用了纯极点减除和逐阶有限性：前者限制可能出现的调节器幂次，后者使负幂系数为零。
将式（28.8） 代回裸量的导数关系，按有限项和各重极点重新排列，便有
0 = β − α 2 G 1 ′ + ∑ n ≥ 1 α β G n ′ − α 2 G n + 1 ′ ϵ n . (28.9) 0=\beta-\alpha^2G_1'
+\sum_{n\ge1}
\frac{\alpha\beta G_n'-\alpha^2G_{n+1}'}{\epsilon^n}.
\tag{28.9} 0 = β − α 2 G 1 ′ + n ≥ 1 ∑ ϵ n α β G n ′ − α 2 G n + 1 ′ . ( 28.9 ) 例如，− ϵ α -\epsilon\alpha − ϵ α G n + 1 ′ / ϵ n + 1 G_{n+1}'/\epsilon^{n+1} G n + 1 ′ / ϵ n + 1 n n n
β = α 2 G 1 ′ , G n + 1 ′ = β α G n ′ = α G 1 ′ G n ′ , G n ( 0 ) = 0. (28.10) \beta=\alpha^2G_1',\qquad
G_{n+1}'=\frac{\beta}{\alpha}G_n'
=\alpha G_1'G_n',\qquad G_n(0)=0.
\tag{28.10} β = α 2 G 1 ′ , G n + 1 ′ = α β G n ′ = α G 1 ′ G n ′ , G n ( 0 ) = 0. ( 28.10 ) 最后的边界值来自零耦合时Z = 1 Z=1 Z = 1 G 2 ′ = α ( G 1 ′ ) 2 G_2'=\alpha(G_1')^2 G 2 ′ = α ( G 1 ′ ) 2
先在一圈精度内代入G 1 = − 3 α / 2 G_1=-3\alpha/2 G 1 = − 3 α /2
β ( α ) = − 3 2 α 2 + O ( α 3 ) , d g d ln μ = − ϵ 2 g − 3 g 3 4 ( 4 π ) 3 + O ( g 5 ) . (28.11) \beta(\alpha)=-\frac32\alpha^2+O(\alpha^3),\qquad
\frac{dg}{d\ln\mu}
=-\frac{\epsilon}{2}g-\frac{3g^3}{4(4\pi)^3}+O(g^5).
\tag{28.11} β ( α ) = − 2 3 α 2 + O ( α 3 ) , d ln μ d g = − 2 ϵ g − 4 ( 4 π ) 3 3 g 3 + O ( g 5 ) . ( 28.11 ) 第二式利用了α ∝ g 2 \alpha\propto g^2 α ∝ g 2 β g / g = B ϵ / ( 2 α ) \beta_g/g=B_\epsilon/(2\alpha) β g / g = B ϵ / ( 2 α )
递推还告诉我们，一圈信息能确定高阶结果中的哪些部分。若在G 1 G_1 G 1
G 2 = 9 8 α 2 , G n = ( − 3 / 2 ) n n α n , G = − ln ( 1 + 3 α 2 ϵ ) . (28.12) G_2=\frac98\alpha^2,\qquad
G_n=\frac{(-3/2)^n}{n}\alpha^n,\qquad
G=-\ln\left(1+\frac{3\alpha}{2\epsilon}\right).
\tag{28.12} G 2 = 8 9 α 2 , G n = n ( − 3/2 ) n α n , G = − ln ( 1 + 2 ϵ 3 α ) . ( 28.12 ) 把一般G n G_n G n （28.10） ，两边的耦合幂次与系数相同；n = 1 n=1 n = 1 G 1 G_1 G 1 α 2 , α 3 , … \alpha^2,\alpha^3,\ldots α 2 , α 3 , …
质量反常量纲
质量可以按同样的思路处理。对m 0 m_0 m 0 M \mathcal M M
M = 1 2 ln Z m Z φ = ∑ n ≥ 1 M n ϵ n , M 1 = 1 2 ( b 1 − a 1 ) = − 5 12 α + O ( α 2 ) . (28.13) \mathcal M=\frac12\ln\frac{Z_m}{Z_\varphi}
=\sum_{n\ge1}\frac{\mathcal M_n}{\epsilon^n},\qquad
\mathcal M_1=\frac12(b_1-a_1)=-\frac5{12}\alpha+O(\alpha^2).
\tag{28.13} M = 2 1 ln Z φ Z m = n ≥ 1 ∑ ϵ n M n , M 1 = 2 1 ( b 1 − a 1 ) = − 12 5 α + O ( α 2 ) . ( 28.13 ) 这里的b 1 b_1 b 1
现在m 0 = m e M m_0=me^{\mathcal M} m 0 = m e M m 0 m_0 m 0
d ln m d ln μ = − B ϵ ∂ α M = α M 1 ′ + ∑ n ≥ 1 α M n + 1 ′ − β M n ′ ϵ n . (28.14) \begin{aligned}
\frac{d\ln m}{d\ln\mu}
&=-B_\epsilon\partial_\alpha\mathcal M\\
&=\alpha\mathcal M_1'
+\sum_{n\ge1}
\frac{\alpha\mathcal M_{n+1}'-\beta\mathcal M_n'}{\epsilon^n}.
\end{aligned}
\tag{28.14} d ln μ d ln m = − B ϵ ∂ α M = α M 1 ′ + n ≥ 1 ∑ ϵ n α M n + 1 ′ − β M n ′ . ( 28.14 ) 参数质量的尺度导数必须逐阶有限，因此负幂系数给出与耦合相同的递推结构，留下的有限项则确定质量反常量纲：
M n + 1 ′ = β α M n ′ , γ m = d ln m d ln μ = α M 1 ′ = − 5 12 α + O ( α 2 ) . (28.15) \mathcal M_{n+1}'=\frac{\beta}{\alpha}\mathcal M_n',\qquad
\gamma_m=\frac{d\ln m}{d\ln\mu}
=\alpha\mathcal M_1'
=-\frac5{12}\alpha+O(\alpha^2).
\tag{28.15} M n + 1 ′ = α β M n ′ , γ m = d ln μ d ln m = α M 1 ′ = − 12 5 α + O ( α 2 ) . ( 28.15 ) 这里γ m \gamma_m γ m m m m d m 2 / d ln μ = 2 γ m m 2 dm^2/d\ln\mu=2\gamma_m m^2 d m 2 / d ln μ = 2 γ m m 2 m / μ m/\mu m / μ γ m − 1 \gamma_m-1 γ m − 1
场的归一化和Callan–Symanzik方程
场归一化的改变还会传入关联函数。即使所描述的物理过程相同，用不同归一化场写出的传播子也会随尺度变化。先在正则化维数定义重整化传播子与裸传播子：
Δ ( k 2 ) = i ∫ d d x e i k x ⟨ 0 ∣ T φ ( x ) φ ( 0 ) ∣ 0 ⟩ , Δ 0 ( k 2 ) = i ∫ d d x e i k x ⟨ 0 ∣ T φ 0 ( x ) φ 0 ( 0 ) ∣ 0 ⟩ = Z φ Δ ( k 2 ) . (28.16) \begin{aligned}
\boldsymbol\Delta(k^2)
&=i\int d^dx\,e^{ikx}\langle0|T\varphi(x)\varphi(0)|0\rangle,\\
\boldsymbol\Delta_0(k^2)
&=i\int d^dx\,e^{ikx}\langle0|T\varphi_0(x)\varphi_0(0)|0\rangle
=Z_\varphi\boldsymbol\Delta(k^2).
\end{aligned}
\tag{28.16} Δ ( k 2 ) Δ 0 ( k 2 ) = i ∫ d d x e ik x ⟨ 0∣ Tφ ( x ) φ ( 0 ) ∣0 ⟩ , = i ∫ d d x e ik x ⟨ 0∣ T φ 0 ( x ) φ 0 ( 0 ) ∣0 ⟩ = Z φ Δ ( k 2 ) . ( 28.16 ) 对于实标量的时间序二点函数，平移不变性与两个场的交换使它在x ↦ − x x\mapsto-x x ↦ − x e + i k x e^{+ikx} e + ik x k 2 k^2 k 2 i i i Δ = 1 / ( k 2 + m 2 − i 0 ) \boldsymbol\Delta=1/(k^2+m^2-i0) Δ = 1/ ( k 2 + m 2 − i 0 ) d d d
为了求场归一化的尺度变化，定义A = ln Z φ = ∑ n ≥ 1 A n / ϵ n \mathcal A=\ln Z_\varphi=\sum_{n\ge1}\mathcal A_n/\epsilon^n A = ln Z φ = ∑ n ≥ 1 A n / ϵ n A 1 = a 1 \mathcal A_1=a_1 A 1 = a 1
d ln Z φ d ln μ = − α a 1 ′ + ∑ n ≥ 1 β A n ′ − α A n + 1 ′ ϵ n . (28.17) \frac{d\ln Z_\varphi}{d\ln\mu}
=-\alpha a_1'
+\sum_{n\ge1}
\frac{\beta\mathcal A_n'-\alpha\mathcal A_{n+1}'}{\epsilon^n}.
\tag{28.17} d ln μ d ln Z φ = − α a 1 ′ + n ≥ 1 ∑ ϵ n β A n ′ − α A n + 1 ′ . ( 28.17 ) 重整化场及其传播子的尺度变化应当有限，因此负幂部分同样相消。场的反常量纲由剩下的有限部分确定，定义为
γ φ = 1 2 d ln Z φ d ln μ = − 1 2 α a 1 ′ = α 12 + O ( α 2 ) , d φ d ln μ ∣ φ 0 = − γ φ φ . (28.18) \gamma_\varphi=\frac12\frac{d\ln Z_\varphi}{d\ln\mu}
=-\frac12\alpha a_1'
=\frac{\alpha}{12}+O(\alpha^2),\qquad
\left.\frac{d\varphi}{d\ln\mu}\right|_{\varphi_0}
=-\gamma_\varphi\varphi.
\tag{28.18} γ φ = 2 1 d ln μ d ln Z φ = − 2 1 α a 1 ′ = 12 α + O ( α 2 ) , d ln μ d φ φ 0 = − γ φ φ . ( 28.18 ) 定义中的1 / 2 1/2 1/2 Y 0 Y_0 Y 0 （28.2） 给出d Y / d ln μ = γ φ Y dY/d\ln\mu=\gamma_\varphi Y d Y / d ln μ = γ φ Y Y Y Y
在外部k k k Δ 0 = Z φ Δ \boldsymbol\Delta_0=Z_\varphi\boldsymbol\Delta Δ 0 = Z φ Δ
0 = Z φ [ ( ∂ ∂ ln μ + B ϵ ∂ ∂ α + γ m m ∂ ∂ m ) Δ + 2 γ φ Δ ] . (28.19) 0=Z_\varphi\left[
\left(\frac{\partial}{\partial\ln\mu}
+B_\epsilon\frac{\partial}{\partial\alpha}
+\gamma_m m\frac{\partial}{\partial m}\right)\boldsymbol\Delta
+2\gamma_\varphi\boldsymbol\Delta\right].
\tag{28.19} 0 = Z φ [ ( ∂ ln μ ∂ + B ϵ ∂ α ∂ + γ m m ∂ m ∂ ) Δ + 2 γ φ Δ ] . ( 28.19 ) 显式μ \mu μ m , α m,\alpha m , α Y Y Y ϵ → 0 \epsilon\to0 ϵ → 0
[ ∂ ∂ ln μ + β ( α ) ∂ ∂ α + γ m ( α ) m ∂ ∂ m + 2 γ φ ( α ) ] Δ ( k 2 ; μ , α , m ) = 0. (28.20) \left[
\frac{\partial}{\partial\ln\mu}
+\beta(\alpha)\frac{\partial}{\partial\alpha}
+\gamma_m(\alpha)m\frac{\partial}{\partial m}
+2\gamma_\varphi(\alpha)
\right]\boldsymbol\Delta(k^2;\mu,\alpha,m)=0.
\tag{28.20} [ ∂ ln μ ∂ + β ( α ) ∂ α ∂ + γ m ( α ) m ∂ m ∂ + 2 γ φ ( α ) ] Δ ( k 2 ; μ , α , m ) = 0. ( 28.20 ) 上一节的一圈自能可以具体展示这些尺度变化怎样相互补偿。令z = k 2 z=k^2 z = k 2 m , α m,\alpha m , α （27.6） 求显式尺度导数，有
∂ Π ∂ ln μ = − α ∫ 0 1 d x [ m 2 + x ( 1 − x ) z ] = − α ( m 2 + z 6 ) . (28.21) \frac{\partial\Pi}{\partial\ln\mu}
=-\alpha\int_0^1dx\,[m^2+x(1-x)z]
=-\alpha\left(m^2+\frac z6\right).
\tag{28.21} ∂ ln μ ∂ Π = − α ∫ 0 1 d x [ m 2 + x ( 1 − x ) z ] = − α ( m 2 + 6 z ) . ( 28.21 ) 记D = ∂ ln μ + β ∂ α + γ m m ∂ m \mathcal D=\partial_{\ln\mu}+\beta\partial_\alpha+\gamma_m m\partial_m D = ∂ l n μ + β ∂ α + γ m m ∂ m β \beta β
D Δ − 1 = α ( m 2 + z 6 ) + 2 γ m m 2 + O ( α 2 ) = α 6 ( z + m 2 ) + O ( α 2 ) . (28.22) \begin{aligned}
\mathcal D\boldsymbol\Delta^{-1}
&=\alpha\left(m^2+\frac z6\right)+2\gamma_m m^2+O(\alpha^2)\\
&=\frac{\alpha}{6}(z+m^2)+O(\alpha^2).
\end{aligned}
\tag{28.22} D Δ − 1 = α ( m 2 + 6 z ) + 2 γ m m 2 + O ( α 2 ) = 6 α ( z + m 2 ) + O ( α 2 ) . ( 28.22 ) 再对倒数求导，就得到D Δ = − ( α / 6 ) Δ + O ( α 2 ) \mathcal D\boldsymbol\Delta=-(\alpha/6)\boldsymbol\Delta+O(\alpha^2) D Δ = − ( α /6 ) Δ + O ( α 2 ) + 2 γ φ Δ +2\gamma_\varphi\boldsymbol\Delta + 2 γ φ Δ
外腿归一化怎样消去尺度变化
对连通的 n n n Z φ 1 / 2 Z_\varphi^{1/2} Z φ 1/2
G c , 0 ( n ) = Z φ n / 2 G c ( n ) , D G c ( n ) = − n γ φ G c ( n ) . G_{c,0}^{(n)}=Z_\varphi^{n/2}G_c^{(n)},\qquad
\mathcal D G_c^{(n)}=-n\gamma_\varphi G_c^{(n)}. G c , 0 ( n ) = Z φ n /2 G c ( n ) , D G c ( n ) = − n γ φ G c ( n ) . 这里 D \mathcal D D A n = ∏ j = 1 n Δ ( k j ) − 1 G c ( n ) \mathcal A_n=\prod_{j=1}^n\boldsymbol\Delta(k_j)^{-1}G_c^{(n)} A n = ∏ j = 1 n Δ ( k j ) − 1 G c ( n ) D Δ − 1 = 2 γ φ Δ − 1 \mathcal D\boldsymbol\Delta^{-1}=2\gamma_\varphi\boldsymbol\Delta^{-1} D Δ − 1 = 2 γ φ Δ − 1
D A n = ( 2 n γ φ − n γ φ ) A n = n γ φ A n . \mathcal D\mathcal A_n
=(2n\gamma_\varphi-n\gamma_\varphi)\mathcal A_n
=n\gamma_\varphi\mathcal A_n. D A n = ( 2 n γ φ − n γ φ ) A n = n γ φ A n . 对于存在孤立稳定单粒子极点的外腿，写 Δ ( k 2 ) ∼ R / ( k 2 + M 2 − i 0 ) \boldsymbol\Delta(k^2)\sim R/(k^2+M^2-i0) Δ ( k 2 ) ∼ R / ( k 2 + M 2 − i 0 ) D M 2 = 0 \mathcal D M^2=0 D M 2 = 0 D R = − 2 γ φ R \mathcal D R=-2\gamma_\varphi R D R = − 2 γ φ R LSZ 约化 在壳化，并把固定的 i i i
T n = R n / 2 A n ∣ o n s h e l l , D T n = ( − n γ φ + n γ φ ) R n / 2 A n ∣ o n s h e l l = 0. \begin{aligned}
\mathcal T_n&=R^{n/2}\mathcal A_n\big|_{\rm on\ shell},\\
\mathcal D\mathcal T_n
&=(-n\gamma_\varphi+n\gamma_\varphi)
R^{n/2}\mathcal A_n\big|_{\rm on\ shell}=0.
\end{aligned} T n D T n = R n /2 A n on shell , = ( − n γ φ + n γ φ ) R n /2 A n on shell = 0. 关联函数与截腿核都随场的归一化变化，而每条外腿的 R \sqrt R R
沿跑动参数求解
传播子的尺度方程同时包含质量和耦合的导数，可以沿它们的跑动轨迹化为普通微分方程。取t = ln ( μ / μ r ) t=\ln(\mu/\mu_r) t = ln ( μ / μ r ) d α / d t = β d\alpha/dt=\beta d α / d t = β d m / d t = γ m m dm/dt=\gamma_m m d m / d t = γ m m （28.20） 化为
d d t Δ ( k ; μ r e t , α ( t ) , m ( t ) ) = − 2 γ φ ( α ( t ) ) Δ , (28.23) \frac{d}{dt}\boldsymbol\Delta(k;\mu_re^t,\alpha(t),m(t))
=-2\gamma_\varphi(\alpha(t))\boldsymbol\Delta,
\tag{28.23} d t d Δ ( k ; μ r e t , α ( t ) , m ( t )) = − 2 γ φ ( α ( t )) Δ , ( 28.23 ) 用积分因子解这一线性方程，从参照尺度积分到所求尺度，得到
Δ ( t ) = exp [ − 2 ∫ 0 t d u γ φ ( α ( u ) ) ] Δ ( 0 ) . (28.24) \boldsymbol\Delta(t)
=\exp\left[-2\int_0^tdu\,\gamma_\varphi(\alpha(u))\right]
\boldsymbol\Delta(0).
\tag{28.24} Δ ( t ) = exp [ − 2 ∫ 0 t d u γ φ ( α ( u )) ] Δ ( 0 ) . ( 28.24 ) 积分过程中动量k k k
一圈时定义W = 1 + 3 2 α r t W=1+\tfrac32\alpha_rt W = 1 + 2 3 α r t 上一节 已经给出联立跑动解α = α r / W \alpha=\alpha_r/W α = α r / W m = m r W − 5 / 18 m=m_rW^{-5/18} m = m r W − 5/18 d ln W / d t = 3 α / 2 d\ln W/dt=3\alpha/2 d ln W / d t = 3 α /2 ∫ 0 t γ φ d u = ln W / 18 \int_0^t\gamma_\varphi\,du=\ln W/18 ∫ 0 t γ φ d u = ln W /18
φ ( t ) = W − 1 / 18 φ ( 0 ) , Δ ( t ) = W − 1 / 9 Δ ( 0 ) . (28.25) \varphi(t)=W^{-1/18}\varphi(0),\qquad
\boldsymbol\Delta(t)=W^{-1/9}\boldsymbol\Delta(0).
\tag{28.25} φ ( t ) = W − 1/18 φ ( 0 ) , Δ ( t ) = W − 1/9 Δ ( 0 ) . ( 28.25 ) 这两式表示同一裸场在不同重整化尺度下的归一化关系，其中的演化参数是对数能标。使用这些表达式时，须有W > 0 W>0 W > 0
它们也给出了高能计算的一个方便步骤。若原参数在低尺度μ r \mu_r μ r k 2 k^2 k 2 μ h = k 2 \mu_h=\sqrt{k^2} μ h = k 2
Δ ( k ; μ r ) = W h 1 / 9 Δ ( k ; μ h ) , W h = 1 + 3 4 α r ln k 2 μ r 2 . (28.26) \boldsymbol\Delta(k;\mu_r)
=W_h^{1/9}\boldsymbol\Delta(k;\mu_h),\qquad
W_h=1+\frac34\alpha_r\ln\frac{k^2}{\mu_r^2}.
\tag{28.26} Δ ( k ; μ r ) = W h 1/9 Δ ( k ; μ h ) , W h = 1 + 4 3 α r ln μ r 2 k 2 . ( 28.26 ) 当m 2 ≪ k 2 m^2\ll k^2 m 2 ≪ k 2 1 / k 2 1/k^2 1/ k 2 W h 1 / 9 W_h^{1/9} W h 1/9
Δ ( k ; μ r ) = 1 k 2 [ 1 + α r 12 ln k 2 μ r 2 + 有限一圈项 + ⋯ ] . (28.27) \boldsymbol\Delta(k;\mu_r)
=\frac1{k^2}\left[1+\frac{\alpha_r}{12}
\ln\frac{k^2}{\mu_r^2}+\text{有限一圈项}+\cdots\right].
\tag{28.27} Δ ( k ; μ r ) = k 2 1 [ 1 + 12 α r ln μ r 2 k 2 + 有限一圈项 + ⋯ ] . ( 28.27 ) 共同对数的系数与式（27.7） 中直接计算的自能相同，因而跑动方程将这一对数包含进了尺度换算因子。对于上一节的包容率，耦合跑动所给的领先因子同样可以写为
α ( s ) 2 = α r 2 [ 1 + 3 4 α r ln ( s / μ r 2 ) ] 2 = α r 2 [ 1 − 3 2 α r ln s μ r 2 + ⋯ ] . (28.28) \alpha(\sqrt s)^2
=\frac{\alpha_r^2}
{[1+\tfrac34\alpha_r\ln(s/\mu_r^2)]^2}
=\alpha_r^2\left[1-\frac32\alpha_r\ln\frac s{\mu_r^2}+\cdots\right].
\tag{28.28} α ( s ) 2 = [ 1 + 4 3 α r ln ( s / μ r 2 ) ] 2 α r 2 = α r 2 [ 1 − 2 3 α r ln μ r 2 s + ⋯ ] . ( 28.28 ) 这将一圈β \beta β
还可以保留有限的ϵ \epsilon ϵ ϵ ≠ 0 \epsilon\ne0 ϵ = 0 d α / d t = − ϵ α − 3 α 2 / 2 d\alpha/dt=-\epsilon\alpha-3\alpha^2/2 d α / d t = − ϵ α − 3 α 2 /2
D ϵ ( t ) = 1 + 3 α r 2 ϵ ( 1 − e − ϵ t ) , α ( t ) = α r e − ϵ t D ϵ ( t ) . (28.29) D_\epsilon(t)=1+\frac{3\alpha_r}{2\epsilon}(1-e^{-\epsilon t}),
\qquad
\alpha(t)=\frac{\alpha_re^{-\epsilon t}}{D_\epsilon(t)}.
\tag{28.29} D ϵ ( t ) = 1 + 2 ϵ 3 α r ( 1 − e − ϵ t ) , α ( t ) = D ϵ ( t ) α r e − ϵ t . ( 28.29 ) 求这个解时，先令q = 1 / α q=1/\alpha q = 1/ α q ′ − ϵ q = 3 / 2 q'-\epsilon q=3/2 q ′ − ϵ q = 3/2 e − ϵ t e^{-\epsilon t} e − ϵ t m / m r = D ϵ − 5 / 18 m/m_r=D_\epsilon^{-5/18} m / m r = D ϵ − 5/18 φ / φ r = D ϵ − 1 / 18 \varphi/\varphi_r=D_\epsilon^{-1/18} φ / φ r = D ϵ − 1/18 ϵ → 0 \epsilon\to0 ϵ → 0 ( 1 − e − ϵ t ) / ϵ → t (1-e^{-\epsilon t})/\epsilon\to t ( 1 − e − ϵ t ) / ϵ → t
固定点上的异常幂律
最后考察一种特殊情形。设某理论的β \beta β α ∗ \alpha_* α ∗ m = 0 m=0 m = 0 α \alpha α γ ∗ = γ φ ( α ∗ ) \gamma_*=\gamma_\varphi(\alpha_*) γ ∗ = γ φ ( α ∗ )
( ∂ ∂ ln μ + 2 γ ∗ ) Δ ( k 2 ) = 0 , Δ ( k 2 ) = μ − 2 γ ∗ A ( k 2 ) . (28.30) \left(\frac{\partial}{\partial\ln\mu}+2\gamma_*\right)
\boldsymbol\Delta(k^2)=0,
\qquad
\boldsymbol\Delta(k^2)=\mu^{-2\gamma_*}A(k^2).
\tag{28.30} ( ∂ ln μ ∂ + 2 γ ∗ ) Δ ( k 2 ) = 0 , Δ ( k 2 ) = μ − 2 γ ∗ A ( k 2 ) . ( 28.30 ) 这个微分方程先确定了尺度依赖。若所考察的无质量真空还不含另一有量纲尺度，洛伦兹不变性使A A A k 2 k^2 k 2 [ Δ ] = − 2 [\boldsymbol\Delta]=-2 [ Δ ] = − 2 [ A ] = − 2 + 2 γ ∗ [A]=-2+2\gamma_* [ A ] = − 2 + 2 γ ∗ k 2 > 0 k^2>0 k 2 > 0
Δ ( k 2 ) = C ( α ∗ ) k 2 ( μ 2 k 2 ) − γ ∗ . (28.31) \boldsymbol\Delta(k^2)
=\frac{C(\alpha_*)}{k^2}
\left(\frac{\mu^2}{k^2}\right)^{-\gamma_*}.
\tag{28.31} Δ ( k 2 ) = k 2 C ( α ∗ ) ( k 2 μ 2 ) − γ ∗ . ( 28.31 ) 对ln μ \ln\mu ln μ − 2 γ ∗ -2\gamma_* − 2 γ ∗ μ − 2 γ ∗ ( k 2 ) − 1 + γ ∗ \mu^{-2\gamma_*}(k^2)^{-1+\gamma_*} μ − 2 γ ∗ ( k 2 ) − 1 + γ ∗ − 2 -2 − 2 k 2 − i 0 k^2-i0 k 2 − i 0
固定μ \mu μ k − 2 + 2 γ ∗ k^{-2+2\gamma_*} k − 2 + 2 γ ∗ Δ φ \Delta_\varphi Δ φ k 2 Δ φ − d k^{2\Delta_\varphi-d} k 2 Δ φ − d x ↦ x / b x\mapsto x/b x ↦ x / b b − d b^{-d} b − d b 2 Δ φ b^{2\Delta_\varphi} b 2 Δ φ
Δ φ = d − 2 2 + γ ∗ . (28.32) \Delta_\varphi=\frac{d-2}{2}+\gamma_*.
\tag{28.32} Δ φ = 2 d − 2 + γ ∗ . ( 28.32 ) 场的反常量纲由此直接进入了标度维数，临界现象中的异常幂律正体现这种变化。对于此处的实φ 3 \varphi^3 φ 3 β = − 3 α 2 / 2 \beta=-3\alpha^2/2 β = − 3 α 2 /2 β ( α ∗ ) = 0 \beta(\alpha_*)=0 β ( α ∗ ) = 0
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