Appendix A The reduction formula ‣ MATH425

For the full field $\phi(x)$ , the EoM is not just $(\partial^{2}+m^{2})\phi(x)$ . However, we can still calculate this as

	$\displaystyle\int{\rm d}^{3}x\ {\rm e}^{ik\cdot x}(\partial^{2}+m^{2})\phi(x)$	$\displaystyle=\int{\rm d}^{3}x\ {\rm e}^{ik\cdot x}(\partial_{t}^{2}-\vec{% \nabla}^{2}+m^{2})\phi(x)\stackrel{{\scriptstyle*}}{{=}}\int{\rm d}^{3}x\ (% \vec{\nabla}^{2}{\rm e}^{ik\cdot x})(\partial_{t}^{2}+m^{2})\phi(x)$		(462)
		$\displaystyle=\int{\rm d}^{3}x\ {\rm e}^{ik\cdot x}(\partial_{t}^{2}+\vec{k}^{% 2}+m^{2})\phi(x)=\int{\rm d}^{3}x\ {\rm e}^{ik\cdot x}(\partial_{t}^{2}+E_{% \vec{k}}^{2})\phi(x)\,,$		(463)

where we have used integration-by-parts at $*$ . This makes no assumption on the structure of the field except that it falls of quickly enough so that the boundary conditions do not contribute. Further, we have used that $\vec{k}^{2}+m^{2}=E_{\vec{k}}^{2}$ . Consider now

$\displaystyle i{\rm e}^{ik\cdot x}(\partial_{t}^{2}+E_{\vec{k}}^{2})$	$\displaystyle={\rm e}^{ik\cdot x}(E_{\vec{k}}\partial_{t}+i\partial_{t}^{2}+iE% _{\vec{k}}^{2}-E_{\vec{k}}\partial_{t})$
	$\displaystyle={\rm e}^{ik\cdot x}(E_{\vec{k}}+i\partial_{t})\partial_{t}+{\rm e% }^{ik\cdot x}iE_{\vec{k}}(E_{\vec{k}}+i\partial_{t})$
	$\displaystyle={\rm e}^{ik\cdot x}(E_{\vec{k}}+i\partial_{t})\partial_{t}+(% \partial_{t}{\rm e}^{ik\cdot x})(E_{\vec{k}}+i\partial_{t})=\partial_{t}{\rm e% }^{ik\cdot x}(E_{\vec{k}}+i\partial_{t})\,,$	(464)

where the derivative always acts to its right. This means our original expression becomes

\displaystyle\int{\rm d}^{3}x\ {\rm e}^{ik\cdot x}(\partial^{2}+m^{2})\phi(x)

\displaystyle=-i\int{\rm d}^{3}x\ \partial_{t}\Big{[}{\rm e}^{ik\cdot x}(E_{% \vec{k}}+i\partial_{t})\phi(x)\Big{]}\,.

(465)

The field $\phi(x)$ here is still the full interacting field which we know little about. However, if we integrate $t=-\infty$ to $t=+\infty$ , the derivative turns the expression into its boundary terms

\displaystyle\int_{-\infty}^{+\infty}{\rm d}t\int{\rm d}^{3}x\ {\rm e}^{ik% \cdot x}(\partial^{2}+m^{2})\phi(x)

\displaystyle=-i\int{\rm d}^{3}x\ {\rm e}^{ik\cdot x}(E_{\vec{k}}+i\partial_{t% })\phi(x)\Big{|}_{t=-\infty}^{t=+\infty}\,.

(466)

In these limits, we actually do understand the field $\phi$ as the in and out fields that fulfil the free KG equation and can be written in terms of $a$ and $a^{\dagger}$ operators. However, we need to keep renormalisation in mind. The free field $\phi_{0}$ is related to the interacting field through (426), modifying (134) and (137).

	$\displaystyle Z^{1/2}_{\phi}\phi_{\rm in}$	$\displaystyle=\lim_{t\to-\infty}\phi(t)=\int\frac{{\rm d}^{3}k}{(2\pi)^{3}}% \frac{1}{\sqrt{2E_{\vec{k}}}}\Big{[}a_{\rm in}(\vec{k}){\rm e}^{-ik\cdot x}+a_% {\rm in}^{\dagger}(\vec{k}){\rm e}^{ik\cdot x}\Big{]}\,,$		(467)
	$\displaystyle Z^{1/2}_{\phi}\phi_{\rm out}$	$\displaystyle=\lim_{t\to+\infty}\phi(t)=\int\frac{{\rm d}^{3}k}{(2\pi)^{3}}% \frac{1}{\sqrt{2E_{\vec{k}}}}\Big{[}a_{\rm out}(\vec{k}){\rm e}^{-ik\cdot x}+a% _{\rm out}^{\dagger}(\vec{k}){\rm e}^{ik\cdot x}\Big{]}\,.$		(468)

Let us therefore calculate for a free field $\phi_{0}$ which we will either identify with $\phi_{\rm in}$ or $\phi_{\rm out}$

\displaystyle\begin{split}{\rm e}^{ik\cdot x}(E_{\vec{k}}+i\partial_{t})\phi_{% 0}(x)&=\int\frac{{\rm d}^{3}p}{(2\pi)^{3}}\frac{1}{\sqrt{2E_{\vec{p}}}}{\rm e}% ^{ik\cdot x}(E_{\vec{k}}+i\partial_{t})\Big{[}a(\vec{p}){\rm e}^{-ip\cdot x}+a% ^{\dagger}(\vec{p}){\rm e}^{ip\cdot x}\Big{]}\\ &=\int\frac{{\rm d}^{3}p}{(2\pi)^{3}}\frac{1}{\sqrt{2E_{\vec{p}}}}\Big{[}a(% \vec{p})(E_{\vec{k}}+E_{\vec{p}}){\rm e}^{-i(p-k)\cdot x}+a^{\dagger}(\vec{p})% (E_{\vec{k}}-E_{\vec{p}}){\rm e}^{-i(-p-k)\cdot x}\Big{]}\,.\end{split}

(469)

Integrating over ${\rm d}^{3}x$ and using (88)

$\displaystyle\int{\rm d}^{3}x\ {\rm e}^{ik\cdot x}(E_{\vec{k}}+i\partial_{t})% \phi_{0}(x)$	$\displaystyle=\int\frac{{\rm d}^{3}p}{(2\pi)^{3}}\frac{1}{\sqrt{2E_{\vec{p}}}}% \Big{[}a(\vec{p})(E_{\vec{k}}+E_{\vec{p}})(2\pi)^{3}{\rm e}^{-i(E_{\vec{p}}-E_% {\vec{k}})t}\delta^{(3)}(\vec{p}+\vec{k})$
	$\displaystyle\qquad\qquad+a^{\dagger}(\vec{p})(E_{\vec{k}}-E_{\vec{p}})(2\pi)^% {3}{\rm e}^{-i(-E_{\vec{p}}-E_{\vec{k}})t}\delta^{(3)}(-\vec{p}+\vec{k})\Big{]}$
	$\displaystyle=\sqrt{2E_{\vec{k}}}\ a(\vec{k})\,.$	(470)

This means we have just found another way of expressing the destruction operator $a$ . We can substitute this into our expression and use $a_{\rm in}$ for $t\to-\infty$ and $a_{\rm out}$ for $t\to+\infty$ (using ${\rm d}^{4}x$ instead of ${\rm d}t{\rm d}^{3}x$ )

\displaystyle\begin{split}\int{\rm d}^{4}x\ {\rm e}^{ik\cdot x}(\partial^{2}+m% ^{2})\phi(x)&=-iZ_{\phi}^{-1/2}\sqrt{2E_{\vec{k}}}\Big{(}a_{\rm out}(\vec{k})-% a_{\rm in}(\vec{k})\Big{)}\,,\\ \int{\rm d}^{4}x\ {\rm e}^{-ik\cdot x}(\partial^{2}+m^{2})\phi(x)&=+iZ_{\phi}^% {-1/2}\sqrt{2E_{\vec{k}}}\Big{(}a^{\dagger}_{\rm out}(\vec{k})-a^{\dagger}_{% \rm in}(\vec{k})\Big{)}\,.\end{split}

(471)

In the limit of the free field, the two operators are identical so that this vanishes – as expected by the EoM for $\phi_{0}(x)$ .

In (142), we have seen how to calculate the probability of a scattering from the amplitude

\displaystyle P\sim|{}_{o}\langle{f}|{i}\rangle_{i}|^{2}\,.

(472)

The in-state $|{i}\rangle_{i}$ (out-state $|{f}\rangle_{o}$ ) is created using $a^{\dagger}_{\rm in}$ ( $a^{\dagger}_{\rm out}$ ) from the vacuum $|{\Omega}\rangle$ . For a process of $n$ particles to $m$ particles, we have

	$\displaystyle\|{i}\rangle_{i}$	$\displaystyle=\sqrt{2E_{\vec{p}_{1}}}a^{\dagger}_{\rm in}(p_{1})\ \cdots\ % \sqrt{2E_{\vec{p}_{n}}}a^{\dagger}_{\rm in}(p_{n})\|{\Omega}\rangle=\bigg{(}% \prod_{j=1}^{n}\sqrt{2E_{\vec{p}_{j}}}a^{\dagger}_{\rm in}(p_{j})\bigg{)}\|{% \Omega}\rangle\,,$		(473)
	$\displaystyle\langle{f}\|_{o}$	$\displaystyle=\langle{\Omega}\|\sqrt{2E_{\vec{q}_{1}}}a_{\rm out}(q_{1})\ % \cdots\ \sqrt{2E_{\vec{q}_{m}}}a_{\rm out}(q_{m})=\langle{\Omega}\|\bigg{(}% \prod_{k=1}^{m}\sqrt{2E_{\vec{q}_{k}}}a_{\rm out}(q_{k})\bigg{)}\,.$		(474)

The amplitude therefore becomes

\displaystyle{}_{o}\langle{f}|{i}\rangle_{i}=\langle{\Omega}|\bigg{(}\prod_{j=% 1}^{n}\sqrt{2E_{\vec{p}_{j}}}a_{\rm out}(p_{j})\bigg{)}\bigg{(}\prod_{k=1}^{m}% \sqrt{2E_{\vec{q}_{k}}}a^{\dagger}_{\rm in}(q_{k})\bigg{)}|{\Omega}\rangle\,,

(475)

The operator product is naturally time-ordered so let us enforce this henceforth. We can now replace the $a^{\dagger}_{\rm in}$ and $a_{\rm out}$ using (471)

	$\displaystyle{}_{o}\langle{f}\|{i}\rangle_{i}=\langle{\Omega}\|T\Bigg{\{}\prod_{% j=1}^{n}\bigg{[}i$	$\displaystyle Z_{\phi}^{-1/2}\int{\rm d}^{4}x_{j}\ {\rm e}^{ip_{j}\cdot x_{j}}% (\partial_{j}^{2}+m^{2})\phi(x_{j})+\sqrt{2E_{\vec{p}_{j}}}a_{\rm in}(\vec{p}_% {j})\bigg{]}$
		$\displaystyle\times\prod_{k=1}^{m}\bigg{[}iZ_{\phi}^{-1/2}\int{\rm d}^{4}y_{k}% \ {\rm e}^{-iq_{k}\cdot y_{k}}(\partial_{k}^{2}+m^{2})\phi(y_{k})+\sqrt{2E_{% \vec{q}_{k}}}a^{\dagger}_{\rm out}(\vec{q}_{k})\bigg{]}\Bigg{\}}\|{\Omega}% \rangle\,.$		(476)

Note how the $a^{\dagger}_{\rm out}(\vec{q}_{k})$ is currently all the way to the right of the expression, even though it is taken at very early time. This means that time-ordering pushes it all the way to the left. Similarly, the $a_{\rm in}(\vec{p}_{j})$ will pushed to the right where it acts on the vacuum $|{\Omega}\rangle$ . Dropping these disconnected terms, we have and identifying the left-hand side with the $\mathcal{S}$ matrix element $\langle{f}|\mathcal{S}|{i}\rangle$

	$\displaystyle{}_{o}\langle{f}\|{i}\rangle_{i}=\langle{f}\|\mathcal{S}\|{i}\rangle% =\int\bigg{[}\prod_{j=1}^{n}{\rm d}^{4}x_{j}\$	$\displaystyle\frac{i}{\sqrt{Z_{\phi}}}{\rm e}^{ip_{j}\cdot x_{j}}(\partial_{j}% ^{2}+m^{2})\bigg{]}\bigg{[}\prod_{k=1}^{m}{\rm d}^{4}y_{k}\ \frac{i}{\sqrt{Z_{% \phi}}}{\rm e}^{-iq_{k}\cdot y_{k}}(\partial_{k}^{2}+m^{2})\bigg{]}$
		$\displaystyle\times\langle{\Omega}\|T\Big{\{}\phi(x_{1})\cdots\phi(x_{n})\cdot% \phi(y_{1})\cdots\phi(y_{m})\Big{\}}\|{\Omega}\rangle\,.$		(477)

This result is known as the Lehmann-Symanzik-Zimmermann (LSZ) reduction formula and it is the last missing piece of our discussion. We can transform this into momentum space as well where we replace $\partial_{j}^{2}\to-p_{j}^{2}$

	$\displaystyle\langle{f}\|\mathcal{S}\|{i}\rangle=\int\bigg{[}\prod_{j=1}^{n}{\rm d% }^{4}x_{j}\$	$\displaystyle\frac{-i}{\sqrt{Z_{\phi}}}{\rm e}^{ip_{j}\cdot x_{j}}(p_{j}^{2}-m% ^{2})\bigg{]}\bigg{[}\prod_{k=1}^{m}{\rm d}^{4}y_{k}\ \frac{-i}{\sqrt{Z_{\phi}% }}{\rm e}^{-iq_{k}\cdot y_{k}}(q_{k}^{2}-m^{2})\bigg{]}$
		$\displaystyle\times\langle{\Omega}\|T\Big{\{}\phi(x_{1})\cdots\phi(x_{n})\cdot% \phi(y_{1})\cdots\phi(y_{m})\Big{\}}\|{\Omega}\rangle\,.$		(478)

We can re-interpret this by moving the factors of $p_{j}^{2}-m^{2}$ and $\sqrt{Z_{\phi}}$ to the other side

\displaystyle\begin{split}\bigg{[}\prod_{j=1}^{n}\frac{i\sqrt{Z_{\phi}}}{p_{j}% ^{2}-m^{2}}\bigg{]}\bigg{[}\prod_{k=1}^{m}\frac{i\sqrt{Z_{\phi}}}{q_{k}^{2}-m^% {2}}\bigg{]}\langle{f}|\mathcal{S}|{i}\rangle=\int\prod_{j=1}^{n}&{\rm d}^{4}x% _{j}\ {\rm e}^{ip_{j}\cdot x_{j}}\int\prod_{k=1}^{m}{\rm d}^{4}y_{k}\ {\rm e}^% {-iq_{k}\cdot y_{k}}\\ &\times\langle{\Omega}|T\Big{\{}\phi(x_{1})\cdots\phi(x_{n})\cdot\phi(y_{1})% \cdots\phi(y_{m})\Big{\}}|{\Omega}\rangle\,.\end{split}

(479)

This is the relation we have been implicitly using in Section 3 when we related the $\mathcal{S}$ matrix element to the (Fourier-transformed) correlation function.

The correlation function $\langle{\Omega}|T\{\phi\cdots\phi\}|{\Omega}\rangle$ still contains the non-amputated pieces that we were trying to figure out in (228). In general, these terms exist and do contribute to the off-shell correlation function

\displaystyle\langle{\Omega}|T\big{\{}\phi\cdots\phi\big{\}}|{\Omega}\rangle=% \Big{(}\text{amputated}\Big{)}\times\Big{(}\text{non-amputated}\Big{)}\,.

(480)

The non-amputated pieces are just the bare propagator $\langle{\Omega}|T\{\phi_{0}\phi_{0}\}|{\Omega}\rangle$ which we calculated in (409)

\displaystyle\begin{split}\langle{\Omega}|T\big{\{}\phi_{0}\phi_{0}\big{\}}|{% \Omega}\rangle&=\includegraphics{bmlimages/notes-178.svg}\bml@image@depth{178}% \bmlDescription{A lone propagator}+\includegraphics{bmlimages/notes-179.svg}% \bml@image@depth{179}\bmlDescription{A propagator with a tadpole}+% \includegraphics{bmlimages/notes-180.svg}\bml@image@depth{180}\bmlDescription{% A propagator with two tadpoles side by side}+\includegraphics{bmlimages/notes-% 181.svg}\bml@image@depth{181}\bmlDescription{A propagator with a sunset}\\ &\qquad\qquad+\includegraphics{bmlimages/notes-182.svg}\bml@image@depth{182}% \bmlDescription{A propagator with a sunset next to a tadpole}+\includegraphics% {bmlimages/notes-183.svg}\bml@image@depth{183}\bmlDescription{A propagator % with three tadpoles side by side}+\includegraphics{bmlimages/notes-184.svg}% \bml@image@depth{184}\bmlDescription{A propagator with a sunset which itself % has a tadpole}+\cdots\,.\end{split}

+

+⋯.\displaystyle\begin{split}\langle{\Omega}|T\big{\{}\phi_{0}\phi_{0}\big{\}}|{% \Omega}\rangle&=\includegraphics{bmlimages/notes-178.svg}\bml@image@depth{178}% \bmlDescription{A lone propagator}+\includegraphics{bmlimages/notes-179.svg}% \bml@image@depth{179}\bmlDescription{A propagator with a tadpole}+% \includegraphics{bmlimages/notes-180.svg}\bml@image@depth{180}\bmlDescription{% A propagator with two tadpoles side by side}+\includegraphics{bmlimages/notes-% 181.svg}\bml@image@depth{181}\bmlDescription{A propagator with a sunset}\\ &\qquad\qquad+\includegraphics{bmlimages/notes-182.svg}\bml@image@depth{182}% \bmlDescription{A propagator with a sunset next to a tadpole}+\includegraphics% {bmlimages/notes-183.svg}\bml@image@depth{183}\bmlDescription{A propagator % with three tadpoles side by side}+\includegraphics{bmlimages/notes-184.svg}% \bml@image@depth{184}\bmlDescription{A propagator with a sunset which itself % has a tadpole}+\cdots\,.\end{split}

(481)

Some of these terms such as the third, fifth, and sixth are different from the others in that they can be cut in half and just expressed through two or more copies. The terms for which this is not possible are called 1PI. If we bundle all the 1PI corrections into a blob, we can write

$\displaystyle-i\acs{1PI}$ 1PI\displaystyle-i\acs{1PI}	$\displaystyle=\includegraphics{bmlimages/notes-185.svg}\bml@image@depth{185}% \bmlDescription{A propagator with a 1PI blob}=\includegraphics{bmlimages/notes% -186.svg}\bml@image@depth{186}\bmlDescription{A propagator with a tadpole}+% \includegraphics{bmlimages/notes-187.svg}\bml@image@depth{187}\bmlDescription{% A propagator with a sunset}+\includegraphics{bmlimages/notes-188.svg}% \bml@image@depth{188}\bmlDescription{A propagator with a sunset which itself % has a tadpole}+\cdots\,,$ = + + +⋯,\displaystyle=\includegraphics{bmlimages/notes-185.svg}\bml@image@depth{185}% \bmlDescription{A propagator with a 1PI blob}=\includegraphics{bmlimages/notes% -186.svg}\bml@image@depth{186}\bmlDescription{A propagator with a tadpole}+% \includegraphics{bmlimages/notes-187.svg}\bml@image@depth{187}\bmlDescription{% A propagator with a sunset}+\includegraphics{bmlimages/notes-188.svg}% \bml@image@depth{188}\bmlDescription{A propagator with a sunset which itself % has a tadpole}+\cdots\,,	(482)
$\displaystyle\langle{\Omega}\|T\big{\{}\phi_{0}\phi_{0}\big{\}}\|{\Omega}\rangle$	$\displaystyle=\includegraphics{bmlimages/notes-189.svg}\bml@image@depth{189}% \bmlDescription{A lone propagator}+\includegraphics{bmlimages/notes-190.svg}% \bml@image@depth{190}\bmlDescription{A propagator with a 1PI blob}+% \includegraphics{bmlimages/notes-191.svg}\bml@image@depth{191}\bmlDescription{% A propagator with two 1PI blobs}+\cdots$ + + +⋯\displaystyle=\includegraphics{bmlimages/notes-189.svg}\bml@image@depth{189}% \bmlDescription{A lone propagator}+\includegraphics{bmlimages/notes-190.svg}% \bml@image@depth{190}\bmlDescription{A propagator with a 1PI blob}+% \includegraphics{bmlimages/notes-191.svg}\bml@image@depth{191}\bmlDescription{% A propagator with two 1PI blobs}+\cdots
	$\displaystyle=\frac{i}{p^{2}-m_{0}^{2}}+\frac{i}{p^{2}-m_{0}^{2}}(-i\acs{1PI})% \frac{i}{p^{2}-m_{0}^{2}}+\frac{i}{p^{2}-m_{0}^{2}}(-i\acs{1PI})\frac{i}{p^{2}% -m_{0}^{2}}(-i\acs{1PI})\frac{i}{p^{2}-m_{0}^{2}}\,.$ 1PI)⁢ip2−m02+ip2−m02⁢(−i⁢1PI)⁢ip2−m02⁢(−i⁢1PI)⁢ip2−m02.\displaystyle=\frac{i}{p^{2}-m_{0}^{2}}+\frac{i}{p^{2}-m_{0}^{2}}(-i\acs{1PI})% \frac{i}{p^{2}-m_{0}^{2}}+\frac{i}{p^{2}-m_{0}^{2}}(-i\acs{1PI})\frac{i}{p^{2}% -m_{0}^{2}}(-i\acs{1PI})\frac{i}{p^{2}-m_{0}^{2}}\,.	(483)

The geometric series can be summed to result in the propagator

\displaystyle\langle{\Omega}|T\big{\{}\phi_{0}\phi_{0}\big{\}}|{\Omega}\rangle

\displaystyle=\frac{i}{p^{2}-m_{0}^{2}-\acs{1PI}}\,.

1PI.\displaystyle=\frac{i}{p^{2}-m_{0}^{2}-\acs{1PI}}\,.

(484)

This is to be compared to the equivalent renormalised expression $\langle{\Omega}|T\{\phi\phi\}|{\Omega}\rangle=iZ_{\phi}/(p^{2}-m^{2})$

\displaystyle\langle{\Omega}|T\big{\{}\phi_{0}\phi_{0}\big{\}}|{\Omega}\rangle

\displaystyle=\frac{i}{p^{2}-m_{0}^{2}-\acs{1PI}}\sim\frac{iZ_{\phi}}{p^{2}-m^% {2}}+\text{regular}\,.

1PI∼i⁢Zϕp2−m2+regular.\displaystyle=\frac{i}{p^{2}-m_{0}^{2}-\acs{1PI}}\sim\frac{iZ_{\phi}}{p^{2}-m^% {2}}+\text{regular}\,.

(485)

Since the LSZ formula requires us to pick out only the singular terms of the correlation function when calculating $\mathcal{S}$ matrix elements, we have

\displaystyle\big{(}\sqrt{Z_{\phi}}\big{)}^{n+m}\langle{f}|\mathcal{S}|{i}% \rangle=\langle{\Omega}|T\big{\{}\phi\cdots\phi\big{\}}|{\Omega}\rangle\Big{|}% _{\text{singular}}=Z_{\phi}^{n+m}\ \langle{\Omega}|T\big{\{}\phi\cdots\phi\big% {\}}|{\Omega}\rangle\Big{|}_{\text{amputated}}\,.

(486)

This is exactly what we stated in (229) without proving it

\displaystyle\langle{f}|\mathcal{S}|{i}\rangle=\big{(}\sqrt{Z_{\phi}}\big{)}^{% n+m}\langle{\Omega}|T\big{\{}\phi\cdots\phi\big{\}}|{\Omega}\rangle\Big{|}_{% \text{amputated}}\,.

(487)

Confusingly this result is also sometimes referred to as the LSZ formula and it is our main recipe for calculating $\mathcal{S}$ matrix elements: calculate the connected and amputated Feynman diagrams using the correlation function, take the external legs on-shell and multiply with $\sqrt{Z_{\phi}}$ for each particle.

	$\displaystyle\|{i}\rangle_{i}$	$\displaystyle=\sqrt{2E_{\vec{p}_{1}}}a^{\dagger}_{\rm in}(p_{1})\ \cdots\ % \sqrt{2E_{\vec{p}_{n}}}a^{\dagger}_{\rm in}(p_{n})\|{\Omega}\rangle=\bigg{(}% \prod_{j=1}^{n}\sqrt{2E_{\vec{p}_{j}}}a^{\dagger}_{\rm in}(p_{j})\bigg{)}\|{% \Omega}\rangle\,,$		(473)
	$\displaystyle\langle{f}\|_{o}$	$\displaystyle=\langle{\Omega}\|\sqrt{2E_{\vec{q}_{1}}}a_{\rm out}(q_{1})\ % \cdots\ \sqrt{2E_{\vec{q}_{m}}}a_{\rm out}(q_{m})=\langle{\Omega}\|\bigg{(}% \prod_{k=1}^{m}\sqrt{2E_{\vec{q}_{k}}}a_{\rm out}(q_{k})\bigg{)}\,.$		(474)

Appendix A The Lehmann-Symanzik-Zimmermann reduction formula