free real KG fieldにて。

D_{F}\left(x-y\right) =\theta\left(x^{0}-y^{0}\right)\left\langle 0\right|\phi\left(x\right)\phi\left(y\right)\left|0\right\rangle +\theta\left(y^{0}-x^{0}\right)\left\langle 0\right|\phi\left(y\right)\phi\left(x\right)\left|0\right\rangle \equiv\left\langle 0\right|T\phi\left(x\right)\phi\left(y\right)\left|0\right\rangle

T\phi\left(x\right)\phi\left(y\right) \equiv\underbrace{\theta\left(x^{0}-y^{0}\right)\phi\left(x\right)\phi\left(y\right)}_{\mathrm{future}\ \leftarrow\ \mathrm{past}}+\theta\left(y^{0}-x^{0}\right)\phi\left(y\right)\phi\left(x\right)

ここで、Heaviside step functionの定義を思い出そう。

\theta\left(x^{0}-y^{0}\right)=\lim_{\varepsilon\rightarrow+0}\frac{-1}{2\pi i}\overset{+\infty}{\underset{-\infty}{\int}}dz\frac{1}{z+i\varepsilon}e^{-iz\left(x^{0}-y^{0}\right)}
\theta\left(y^{0}-x^{0}\right)=\lim_{\varepsilon\rightarrow+0}\frac{+1}{2\pi i}\overset{+\infty}{\underset{-\infty}{\int}}dz\frac{1}{z-i\varepsilon}e^{+iz\left(y^{0}-x^{0}\right)}

これを用いて、本格的に計算をはじめよう。

\left\langle 0\right|T\phi\left(x\right)\phi\left(y\right)\left|0\right\rangle

=\int\frac{d^{3}pd^{3}q}{\left(2\pi\right)^{3}\left(2\pi\right)^{3}}\frac{1}{\sqrt{2E_{\mathbf{p}}E_{\mathbf{q}}}}\theta\left(x^{0}-y^{0}\right)e^{-ip\cdot x+iq\cdot y}\underbrace{\left\langle 0\right|a_{\mathbf{p}}a_{\mathbf{q}}^{\dagger}\left|0\right\rangle }_{\left(2\pi\right)^{3}\delta^{\left(3\right)}\left(\mathbf{p-q}\right)}

\ \ \ \ \ +\int\frac{d^{3}pd^{3}q}{\left(2\pi\right)^{3}\left(2\pi\right)^{3}}\frac{1}{\sqrt{2E_{\mathbf{p}}E_{\mathbf{q}}}}\theta\left(y^{0}-x^{0}\right)e^{ip\cdot x-iq\cdot y}\underbrace{\left\langle 0\right|a_{\mathbf{p}}a_{\mathbf{k}}^{\dagger}\left|0\right\rangle }_{\left(2\pi\right)^{3}\delta^{\left(3\right)}\left(\mathbf{q-p}\right)}

=\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\frac{1}{2E_{\mathbf{p}}}\left\{ \theta\left(x^{0}-y^{0}\right)e^{-ip\cdot(x-y)}+\theta\left(y^{0}-x^{0}\right)e^{ip\cdot(x-y)}\right\}

=\overset{+\infty}{\underset{-\infty}{\int}}\frac{idz}{2\pi}\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\frac{1}{2E_{\mathbf{p}}}\left\{ \frac{e^{-iz\left(x^{0}-y^{0}\right)}e^{-ip\cdot(x-y)}}{z+i\varepsilon}-\frac{e^{+iz\left(y^{0}-x^{0}\right)}e^{ip\cdot(x-y)}}{z-i\varepsilon}\right\}

=\overset{+\infty}{\underset{-\infty}{\int}}\frac{idz}{2\pi}\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\frac{1}{2E_{\mathbf{p}}}\left\{ \frac{e^{-i\left(z+E_{\mathbf{p}}\right)\left(x^{0}-y^{0}\right)}e^{+i\mathbf{p}\cdot\left(\mathbf{x-y}\right)}}{z+i\varepsilon}-\underset{\mathbf{p}\rightarrow-\mathbf{p}}{\underbrace{\frac{e^{i\left(-z+E_{\mathbf{p}}\right)\left(x^{0}-y^{0}\right)}e^{-i\mathbf{p}\cdot\left(\mathbf{x-y}\right)}}{z-i\varepsilon}}}\right\}

=\overset{+\infty}{\underset{-\infty}{\int}}\frac{idz}{2\pi}\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\frac{1}{2E_{\mathbf{p}}}\left\{ \underset{z+E_{\mathbf{p}}\rightarrow p^{0}}{\underbrace{\frac{e^{-i\left(z+E_{\mathbf{p}}\right)\left(x^{0}-y^{0}\right)}e^{+i\mathbf{p}\cdot\left(\mathbf{x-y}\right)}}{z+i\varepsilon}}}-\underset{z-E_{\mathbf{p}}\rightarrow p^{0}}{\underbrace{\frac{e^{i\left(-z+E_{\mathbf{p}}\right)\left(x^{0}-y^{0}\right)}e^{+i\mathbf{p}\cdot\left(\mathbf{x-y}\right)}}{z-i\varepsilon}}}\right\}

=\overset{+\infty}{\underset{-\infty}{\int}}\frac{idp^{0}}{2\pi}\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\frac{1}{2E_{\mathbf{p}}}\left\{ \frac{e^{-ip^{0}\left(x^{0}-y^{0}\right)}e^{+i\mathbf{p}\cdot\left(\mathbf{x-y}\right)}}{p^{0}-E_{\mathbf{p}}+i\varepsilon}-\frac{e^{-ip^{0}\left(x^{0}-y^{0}\right)}e^{+i\mathbf{p}\cdot\left(\mathbf{x-y}\right)}}{p^{0}+E_{\mathbf{p}}-i\varepsilon}\right\}

=\int\frac{d^{4}p}{\left(2\pi\right)^{4}}\frac{i}{2E_{\mathbf{p}}}\left\{ \frac{1}{p^{0}-E_{\mathbf{p}}+i\varepsilon}-\frac{1}{p^{0}+E_{\mathbf{p}}-i\varepsilon}\right\} e^{-ip\cdot\left(x-y\right)}

=\lim_{\varepsilon\rightarrow+0}\int\frac{d^{4}p}{\left(2\pi\right)^{4}}\frac{i}{2E_{\mathbf{p}}}\left\{ \frac{\left(p^{0}+E_{\mathbf{p}}-i\varepsilon\right)-\left(p^{0}-E_{\mathbf{p}}+i\varepsilon\right)}{\left(p^{0}-E_{\mathbf{p}}+i\varepsilon\right)\left(p^{0}+E_{\mathbf{p}}-i\varepsilon\right)}\right\} e^{-ip\cdot\left(x-y\right)}

=\lim_{\varepsilon\rightarrow+0}\int\frac{d^{4}p}{\left(2\pi\right)^{4}}\frac{i}{2E_{\mathbf{p}}}\left\{ \frac{2E_{\mathbf{p}}-i\varepsilon}{\left(p^{0}-E_{\mathbf{p}}+i\varepsilon\right)\left(p^{0}+E_{\mathbf{p}}-i\varepsilon\right)}\right\} e^{-ip\cdot\left(x-y\right)}

それゆえ、

\left\langle 0\right|T\phi\left(x\right)\phi\left(y\right)\left|0\right\rangle =\lim_{\varepsilon\rightarrow+0}\int\frac{d^{4}p}{\left(2\pi\right)^{4}}\frac{i}{p^{2}-m^{2}+i\varepsilon}e^{-ip\cdot\left(x-y\right)}

なのである。