Claude Itzykson and Jean-Berard Zuber,2006,Quantum Field Theory,6-1-4

6-1-4 スカラー電磁力学

6-1-4 Scalar Electrodynamics

ファインマン則の導出の最後の例として、スカラー電磁力学、すなわち電荷を持つスピンゼロ粒子の電気力学のケースを紹介します。この理論に基づく基本的な過程の予測を、フェルミオンの量子電磁力学（QED）の予測と比較することで、後者におけるスピンの役割を理解することができます。理論的な観点からも、スカラー電磁力学は興味深いケースです。なぜなら、この理論の相互作用ラグランジアンには微分項が含まれており、そのため相互作用ハミルトニアンの負の値とは異なる形をしているからです。

As a last illustration of the derivation of Feynman rules, we will now present the case of scalar electrodynamics, i.e., electrodynamics of charged spin-zero particles. It allows us to compare the predictions of this theory for basic processes with those of fermionic quantum electrodynamics, and therefore to understand the role of spin in the latter. On theoretical grounds, scalar electrodynamics is an interesting case, since its interaction lagrangian contains derivatives and differs from the negative of the interaction hamiltonian.

電荷を持つ自由なスカラー場（複素スカラー場）のラグランジアンから始めます。

We start from the lagrangian of a free charged scalar field

$$\begin{array}{}\text{(6-50)}& {\mathcal{L}}_0(\phi )={\partial }_{\mu }{\phi }^{†}{\partial }^{\mu }\phi -m^2{\phi }^{†}\phi \end{array}$$

（注意：$\frac{1}{2}$の因子はありません！）最小限の置換（共変微分）${\partial }_{\mu }\phi \to ({\partial }_{\mu }+ie{A}_{\mu })\phi$を行います。さらに、式 (6-25) で示された電磁場のラグランジアン${\mathcal{L}}_0^{\gamma }(A)$を加えると、次のようになります。

(caution, no factor $\frac{1}{2}$!) and perform the minimal substitution ${\partial }_{\mu }\phi \to ({\partial }_{\mu }+ie{A}_{\mu })\phi$. Adding the electromagnetic lagrangian ${\mathcal{L}}_0^{\gamma }(A)$ of Eq. (6-25) yields

$$\begin{array}{rcl}\text{(6-51a)}& \mathcal{L}& =& {\mathcal{L}}_0(\phi )+{\mathcal{L}}_0^{\gamma }(A)+{\mathcal{L}}_{\text{int}}\text{(6-51b)}& {\mathcal{L}}_{\text{int}}& =& -ie{A}^{\mu }({\phi }^{†}{\overleftrightarrow{\partial }}_{\mu }\phi )+e^2{A}^2{\phi }^{†}\phi & =& -ie{A}^{\mu }({\phi }^{†}{\partial }_{\mu }\phi -({\partial }_{\mu }{\phi }^{†})\phi )+e^2{A}^2{\phi }^{†}\phi \end{array}$$

相互作用ラグランジアンには微分項が含まれているため、正準運動量が次のように修正されます。

Since the interaction lagrangian contains derivatives, the canonical momenta are modified:

$$\begin{array}{rcl}{\pi }^{†}& =& \frac{\partial \mathcal{L}}{\partial ({\partial }_0{\phi }^{†})}={\partial }_0\phi +ie{A}^0\phi \\ \text{(6-52)}& \pi & =& \frac{\partial \mathcal{L}}{\partial ({\partial }_0\phi )}={\partial }_0{\phi }^{†}-ie{A}^0{\phi }^{†}\end{array}$$

ハミルトニアンは次のようになります。

The hamiltonian reads

$$\begin{array}{rcl}H& =& H_0^{\gamma }+H_0(\phi ,\pi )+H_{\text{int}}\\ \text{(6-53)}& H_0(\phi ,\pi )+H_{\text{int}}& =& \int d^{3}x(\pi \dot{\phi }+{\pi }^{†}{\dot{\phi }}^{†}-{\mathcal{L}}_0-{\mathcal{L}}_{\text{int}})\end{array}$$

ここで、$H_0$は式 (3-76) で示された自由ハミルトニアンです。

Here H_0 is the free hamiltonian of Eq. (3-76):

$$\begin{array}{}\text{(6-54)}& H_0(\phi ,\pi )=\int d^{3}x({\pi }^{†}\pi +{\partial }_k{\phi }^{†}{\partial }_k\phi +m^2{\phi }^{†}\phi )\end{array}$$

また、$H_{\text{int}}$は次の形をとります（空間インデックス$k$は1,2,3の値を取ります）。

and $H_{\text{int}}$ takes the form (the spatial index $k$ takes the values 1, 2, 3)

$$\begin{array}{}\text{(6-55)}& H_{\text{int}}=\int d^{3}x{\mathcal{H}}_{\text{int}}=\int d^{3}x[ie{A}^0({\pi }^{†}{\phi }^{†}-\pi \phi )+ie{A}^k({\phi }^{†}{\overleftrightarrow{\partial }}_k\phi )\cancel{-}+e^2{A}^k{A}_k{\phi }^{†}\phi ]\end{array}$$ $$\begin{array}{rcl}{\mathcal{H}}_{\text{int}}& =& {\pi }^{†}{\dot{\phi }}^{†}+\pi \dot{\phi }-{\mathcal{L}}_0-{\mathcal{L}}_{\text{int}}-{\mathcal{H}}_0\\ & =& {\pi }^{†}(\pi +ie{A}^0{\phi }^{†})+\pi ({\pi }^{†}-ie{A}^0\phi )-({\partial }_{\mu }{\phi }^{†}{\partial }^{\mu }\phi -\cancel{m^2{\phi }^{†}\phi })-\{-ie{A}^{\mu }({\phi }^{†}{\partial }_{\mu }\phi -({\partial }_{\mu }{\phi }^{†})\phi )+e^2{A}^2{\phi }^{†}\phi \}-({\pi }^{†}\pi +{\partial }_k{\phi }^{†}{\partial }_k\phi +\cancel{m^2{\phi }^{†}\phi })\\ & =& \cancel{{\pi }^{†}\pi }+ie{A}^0{\pi }^{†}{\phi }^{†}+\pi {\pi }^{†}-ie{A}^0\pi \phi -{\partial }_0{\phi }^{†}{\partial }^0\phi +\cancel{{\partial }_k{\phi }^{†}{\partial }^k\phi }+ie{A}^0{\phi }^{†}{\partial }_0\phi -ie{A}^0({\partial }_0^{†}\phi )\phi +ie{A}^k{\phi }^{†}{\partial }_k\phi -ie{A}^k({\partial }_k{\phi }^{†})\phi -e^2{A}^0{A}_0{\phi }^{†}\phi +e^2{A}^k{A}_k{\phi }^{†}\phi +\cancel{{\pi }^{†}\pi }-\cancel{{\partial }_k{\phi }^{†}{\partial }_k\phi }\\ & =& ie{A}^0({\pi }^{†}{\phi }^{†}-\pi \phi )+ie{A}^k({\phi }^{†}{\partial }_k\phi -({\partial }_k{\phi }^{†})\phi )+e^2{A}^k{A}_k{\phi }^{†}\phi +\pi {\pi }^{†}-{\partial }_0{\phi }^{†}{\partial }^0\phi -({\partial }_0{\phi }^{†})ie{A}^0\phi +ie{A}^0{\phi }^{†}{\partial }_0\phi -e^2{A}^0{A}_0{\phi }^{†}\phi \\ & =& ie{A}^0({\pi }^{†}{\phi }^{†}-\pi \phi )+ie{A}^k({\phi }^{†}{\partial }_k\phi -({\partial }_k{\phi }^{†})\phi )+e^2{A}^k{A}_k{\phi }^{†}\phi +{\cancel{\pi \pi }}^{†}-\cancel{({\partial }_0{\phi }^{†}-ie{A}^0{\phi }^{†})({\partial }_0\phi +ie{A}^0\phi )}\end{array}$$

式 (6-52) を式 (6-55) に代入すると、${\mathcal{H}}_{\text{int}}(\phi )=-{\mathcal{L}}_{\text{int}}(A,\phi )-e^2{A}^{02}{\phi }^{†}\phi$の関係式が得られます。

$$\begin{array}{rcl}{\mathcal{H}}_{\text{int}}& =& ie{A}^0\{({\partial }_0\phi +ie{A}^0\phi ){\phi }^{†}-({\partial }_0{\phi }^{†}-ie{A}^0{\phi }^{†})\phi \}+ie{A}^k({\phi }^{†}{\partial }_k\phi -({\partial }_k{\phi }^{†})\phi )+e^2{A}^k{A}_k{\phi }^{†}\phi \\ & =& ie{A}^0({\partial }_0\phi ){\phi }^{†}-e^2{A}^{02}\phi {\phi }^{†}-ie{A}^0({\partial }_0{\phi }^{†})\phi -e^2{A}^{02}{\phi }^{†}\phi +ie{A}^k({\phi }^{†}{\partial }_k\phi -({\partial }_k{\phi }^{†})\phi )+e^2{A}^k{A}_k{\phi }^{†}\phi \\ & =& ie(A^0{\phi }^{†}{\partial }_0\phi +A^k{\phi }^{†}{\partial }_k\phi )-ie(A^0({\partial }_0{\phi }^{†})\phi +A^k({\partial }_k{\phi }^{†})\phi )-e^2(A^{02}\phi {\phi }^{†}-A^k{A}_k{\phi }^{†}\phi )-e^2{A}^{02}{\phi }^{†}\phi \\ & =& -\{-ie{A}^{\mu }({\phi }^{†}{\partial }_{\mu }\phi -({\partial }_{\mu }{\phi }^{†})\phi )+e^2{A}^2{\phi }^{†}\phi \}-e^2{A}^{02}{\phi }^{†}\phi \\ & =& -{\mathcal{L}}_{\text{int}}-e^2{A}^{02}{\phi }^{†}\phi \end{array}$$

一方で、相互作用表示への正準変換を行うこともできます。このとき、$\phi$、$\pi$、$A$はそれぞれ${\phi }_{\text{int}}$、${\pi }_{\text{int}}$、$A_{\text{int}}$に変換され、また${\pi }_{\text{in}}={\partial }_0{\phi }_{\text{in}}^{†}$となります。したがって、この表示では次のようになります。

If we substitute (6-52) into (6-55), we observe that ${\mathcal{H}}_{\text{int}}(\phi )=-{\mathcal{L}}_{\text{int}}(A,\phi )-e^2{A}^{02}{\phi }^{†}\phi$. On the other hand, we may perform a canonical transformation to the interaction representation; $\phi$, $\pi$, $A$ are changed into ${\phi }_{\text{int}}$, ${\pi }_{\text{int}}$, $A_{\text{int}}$ and ${\pi }_{\text{in}}={\partial }_0{\phi }_{\text{in}}^{†}$. Thus, in this representation

$$\begin{array}{rcl}{\mathcal{H}}_{\text{int}}& =& ie{A}_{\text{in}}^0({\pi }_{\text{in}}^{†}{\phi }_{\text{in}}^{†}-{\pi }_{\text{in}}{\phi }_{\text{in}})+ie{A}_{\text{in}}^k{\phi }_{\text{in}}^{†}{\overleftrightarrow{\partial }}_k{\phi }_{\text{in}}\cancel{-}+e^2{A}_{\text{in},k}{A}_{\text{in}}^k{\phi }_{\text{in}}^{†}{\phi }_{\text{in}}\\ & =& ie{A}_{\text{in}}^0(({\partial }_0{\phi }_{\text{in}}){\phi }_{\text{in}}^{†}-({\partial }_0{\phi }_{\text{in}}^{†}){\phi }_{\text{in}})-ie{A}_{\text{in}}^k(-{\phi }_{\text{in}}^{†}{\partial }_k{\phi }_{\text{in}}+({\partial }_k{\phi }_{\text{in}}^{†}){\phi }_{\text{in}})+e^2{A}_{\text{in},k}{A}_{\text{in}}^k{\phi }_{\text{in}}^{†}{\phi }_{\text{in}}\\ & =& ie{A}_{\text{in}}^{\mu }({\phi }_{\text{in}}^{†}{\partial }_{\mu }{\phi }_{\text{in}}-({\partial }_{\mu }{\phi }_{\text{in}}^{†}){\phi }_{\text{in}})+e^2{A}_{\text{in}}^{02}{\phi }_{\text{in}}^{†}{\phi }_{\text{in}}-(e^2{A}_{\text{in}}^{02}{\phi }_{\text{in}}^{†}{\phi }_{\text{in}}-e^2{A}_{\text{in},k}{A}_{\text{in}}^k{\phi }_{\text{in}}^{†}{\phi }_{\text{in}})\\ & =& ie{A}_{\text{in}}^{\mu }{\phi }_{\text{in}}^{†}{\overleftrightarrow{\partial }}_{\mu }{\phi }_{\text{in}}-e^2{A}_{\text{in}}^2{\phi }_{\text{in}}^{†}{\phi }_{\text{in}}+e^2{A}_{\text{in}}^{02}{\phi }_{\text{in}}^{†}{\phi }_{\text{in}}\\ \text{(6-56)}& & =& -{\mathcal{L}}_{\text{int}}(A_{\text{in}},{\phi }_{\text{in}})+e^2{A}_{\text{in}}^{02}{\phi }_{\text{in}}^{†}{\phi }_{\text{in}}\end{array}$$

ハミルトニアンと負のラグランジアンは、非共変な項によって異なっている（この項の前の符号の変化に注意）。これは理論の共変性を損なうように思われる。

The hamiltonian and minus the lagrangian differ by a noncovariant term (notice the change of sign in front of this term) which seems to jeopardize the covariance of the theory.

しかしながら、理論の典型的なグリーン関数を考えてみよう。

Consider, however, a typical Green function of the theory

$$\begin{array}{rcl}& & G(x_1,\cdots ,x_n;x_{n+1},\cdots ,x_{2n};y_1,\cdots ,y_p)\\ & =& ⟨0|T\phi (x_1)\cdots \phi (x_n){\phi }^{†}(x_{n+1})\cdots {\phi }^{†}(x_{2n})A(y_1)\cdots A(y_p)|0⟩\\ \text{(6-57)}& & =& \frac{⟨0|T{\phi }_{\text{in}}(x_1)\cdots A_{\text{in}}(y_p)\exp [-i\int d^{4}z{\mathcal{H}}_{\text{int}}(z)]|0⟩}{⟨0|T\exp [-i\int d^{4}z{\mathcal{H}}_{\text{int}}(z)]|0⟩}\end{array}$$

導出項の時系列積には、非共変項の第二の原因が存在します。共変な結果を得るためには、これら二種類の非共変項が完全に打ち消し合うことが期待されます。実際に、そのようになります。

There is a second source of noncovariant terms in the chronological products of derivative terms. To end up with covariant results, we expect these two types of noncovariant terms to cancel exactly. This is indeed what happens.

摂動展開における紫外発散に関連する問題を回避するため、この性質について正式な証明を示します。

We give a formal proof of this property, since we sidestep the problems connected with the ultraviolet divergences of the perturbation expansion.

導出項によって生じる非共変な縮約について、より詳しく見ていきます。以下では、「in」指数を省略します。関連する伝播関数を列挙します。

Let us look more closely at the noncovariant contractions induced by the derivative terms. In the following, the “in” indices will be dropped. We list the relevant propagators

$$\begin{array}{}\text{(6-58a)}& ⟨0|T\phi (x){\phi }^{†}(y)|0⟩& =& ⟨0|T\phi (y){\phi }^{†}(x)|0⟩=i\int \frac{d^{4}k}{(2\pi {)}^4}\frac{e^{-ik\cdot (x-y)}}{k^2-m^2+i\epsilon }\text{(6-58b)}& ⟨0|T{\partial }_{\mu }\phi (x){\phi }^{†}(y)|0⟩& =& {\partial }_{\mu }^x⟨0|T\phi (x){\phi }^{†}(y)|0⟩\text{(6-58c)}& ⟨0|T{\partial }_{\mu }\phi (x){\partial }_{\nu }{\phi }^{†}(y)|0⟩& =& {\partial }_{\mu }^x{\partial }_{\nu }^y⟨0|T\phi (x){\phi }^{†}(y)|0⟩-i{g}_{\mu 0}{g}_{\nu 0}{\delta }^4(x-y)\end{array}$$

$\phi$の伝播関数だけが非共変な項を含むため、以下の議論では場$A$を古典的なものとみなし、$\phi$および${\phi }^{†}$のすべてのグリーン関数を生成汎関数を導入することで扱うことにします。

Since the φ propagators are the only ones with noncovariant terms, we may consider the field A as classical in the following argument and handle all Green functions of φ and φ† by introducing a generating functional

$$\begin{array}{rcl}Z(j)& \equiv & ⟨0|T\exp -i\int d^{4}z[{\mathcal{H}}_{\text{int}}(y)+j^{\ast }(y)\phi (y)+j(y){\phi }^{†}(y)]|0⟩\\ & =& ⟨0|T\sum _r\frac{e^r}{r!}\int d^4{z}_1\cdots d^4{z}_r{A}^{{\mu }_1}(z_1){\phi }^{†}(z_1){\overleftrightarrow{\partial }}_{{\mu }_1}\phi (z_1)\cdots A^{{\mu }_r}(z_r){\phi }^{†}(z_r){\overleftrightarrow{\partial }}_{{\mu }_r}\phi (z_r)\\ \text{(6-59)}& & & \times \exp -i\int d^{4}z(-e^2{A}^2{\phi }^{†}\phi +e^2{A}^{02}{\phi }^{†}\phi +j^{\ast }\phi +j{\phi }^{†})|0⟩\end{array}$$

相互作用ハミルトニアンのうち、導関数を含む部分のみを e のべき展開しています。これは、非共変な縮約 (6-58c) を生じるためです。ここで、新たな記法を導入します。共変な時系列積は、(6-58) 式の T 積と一致しますが、(6-58c) の最後の項を除きます。

Only the derivative part of the interaction hamiltonian has been expanded in powers of e, since it gives rise to the noncovariant contraction (6-58c). We introduce a further notation: the covariant chronological product coincides with the T product of Eqs. (6-58), but for the last term of (6-58c):

$$\begin{array}{}\text{(6-60)}& ⟨0|\hat{T}{\partial }_{\mu }\phi (x){\partial }_{\nu }{\phi }^{†}(y)|0⟩\equiv {\partial }_{\mu }^x{\partial }_{\nu }^y⟨0|T\phi (x){\phi }^{†}(y)|0⟩\end{array}$$

非共変な縮約$-i{g}_{\mu 0}{g}_{\nu 0}{\delta }^4(x-y)$の寄与を調べましょう。r個の頂点の中から(2s)個を選ぶ方法は$\left(\begin{array}{c}r\\ 2s\end{array}\right)=\frac{r!}{(2s)!(r-2s)!}$通りあり、それらをs組に縮約する方法は$(2s!)!/2^{s}s!$通りあります。各非共変な縮約は、2つの頂点に対して次のようになります。

Let us examine the contribution of s noncovariant contractions $-i{g}_{\mu 0}{g}_{\nu 0}{\delta }^4(x-y)$. There are $\left(\begin{array}{c}r\\ 2s\end{array}\right)=r!/(2s)!(r-2s)!$ choices of (2s) vertices among r, and $(2s!)/2^{s}s!$ ways of contracting them into s pairs. Each noncovariant contraction of two vertices gives

$$\begin{array}{r}e{A}^{\mu }(x)\phi (x)[⟨0|T{\partial }_{\mu }{\phi }^{†}(x){\partial }_{\nu }\phi (y)|0⟩-⟨0|\hat{T}{\partial }_{\mu }{\phi }^{†}(x){\partial }_{\nu }\phi (y)|0⟩](-e)A^{\nu }(y){\phi }^{†}(y)+(\phi \leftrightarrow {\phi }^{†})\\ =2i{e}^2{A}_0^2{\phi }^{†}(x)\phi (x){\delta }^4(x-y)\end{array}$$

ここで、ウィック順序を暗黙的に仮定しています。したがって、次の結果を得ます。

where we have implicitly assumed Wick ordering. Therefore, we obtain

$$\begin{array}{rcl}Z(j)& =& ⟨0|\hat{T}\sum _{r=0}^{\mathrm{\infty }}\frac{e^r}{r!}{[\int d^4{z}^{\prime }{A}^{\mu }(z^{\prime }){\phi }^{†}(z^{\prime }){\overleftrightarrow{\partial }}_{\mu }\phi (z^{\prime })]}^r\sum _{s=0}^{\mathrm{\infty }}\frac{(i{e}^2{)}^s}{s!}{[\int d^{4}z”A_0^2(z”){\phi }^{†}(z”)\phi (z”)]}^s\\ \text{(6-61)}& & & \times \exp -i\int d^{4}z(-e^2{A}^2{\phi }^{†}\phi +e^2{A}_0^2{\phi }^{†}\phi +j{\phi }^{†}+j^{\ast }\phi |0⟩\end{array}$$