Editing LSZ reduction formula (section)

==The reduction formula for scalars==
The asymptotic relations are all that is needed to obtain the LSZ reduction formula. For future convenience we start with the matrix element:

:<math>\mathcal M=\langle \beta\ \mathrm{out}|\mathrm{T}\varphi(y_1)\ldots\varphi(y_n)|\alpha\ \mathrm{in}\rangle </math>

which is slightly more general than an ''S''-matrix element. Indeed, <math>\mathcal M</math> is the expectation value of the [[Path-ordering|time-ordered product]] of a number of fields <math>\varphi(y_1)\cdots\varphi(y_n)</math> between an ''out'' state and an ''in'' state. The ''out'' state can contain anything from the vacuum to an undefined number of particles, whose momenta are summarized by the index {{mvar|β}}. The ''in'' state contains at least a particle of momentum {{mvar|p}}, and possibly many others, whose momenta are summarized by the index {{mvar|α}}. If there are no fields in the time-ordered product, then <math>\mathcal M</math> is obviously an ''S''-matrix element. The particle with momentum {{mvar|p}} can be 'extracted' from the ''in'' state by use of a creation operator:

:<math> \mathcal M=\sqrt{2\omega_p}\ \left \langle \beta\ \mathrm{out} \bigg| \mathrm T\left[\varphi(y_1)\ldots\varphi(y_n)\right] a_{\mathrm{in}}^\dagger(\mathbf p) \bigg|\alpha'\ \mathrm{in} \right \rangle </math>

where the prime on <math>\alpha</math> denotes that one particle has been taken out. With the assumption that no particle with momentum {{mvar|p}} is present in the ''out'' state, that is, we are ignoring forward scattering, we can write:

:<math>\mathcal M=\sqrt{2\omega_p}\ \left \langle \beta\ \mathrm{out} \bigg| 
\left\{ 
\mathrm T\left[\varphi(y_1)\ldots\varphi(y_n)\right] a_{\mathrm{in}}^\dagger (\mathbf p)- a_{\mathrm{out}}^\dagger(\mathbf p) \mathrm T\left[\varphi(y_1)\ldots\varphi(y_n)\right] 
\right\}
\bigg|\alpha'\ \mathrm{in} \right \rangle</math>

because <math>a_{\mathrm{out}}^\dagger</math> acting on the left gives zero. Expressing the construction operators in terms of ''in'' and ''out'' fields, we have:

:<math>\mathcal M=-i\sqrt{2\omega_p}\ \int \mathrm{d}^3x f_p(x)\overleftrightarrow{\partial_0} \left\langle \beta\ \mathrm{out} \bigg| 
\left\{
\mathrm T\left[\varphi(y_1)\ldots\varphi(y_n)\right] \varphi_{\mathrm{in}}(x)- \varphi_{\mathrm{out}}(x) \mathrm T\left[\varphi(y_1)\ldots\varphi(y_n)\right] 
\right\}
\bigg|\alpha'\ \mathrm{in}\right \rangle</math>

Now we can use the asymptotic condition to write:

:<math>\mathcal M= -i\sqrt{\frac{2\omega_p}{Z}} \left\{ \lim_{x^0\to-\infty} \int \mathrm{d}^3x f_p(x)\overleftrightarrow{\partial_0} \langle \beta\ \mathrm{out}| \mathrm T\left[\varphi(y_1)\ldots\varphi(y_n)\right] \varphi(x) |\alpha'\ \mathrm{in}\rangle-\lim_{x^0\to\infty} \int \mathrm{d}^3x f_p(x)\overleftrightarrow{\partial_0} \langle \beta\ \mathrm{out}| \varphi(x) \mathrm T\left[\varphi(y_1)\ldots\varphi(y_n)\right] |\alpha'\ \mathrm{in}\rangle \right\}</math>

Then we notice that the field {{math|''φ''(''x'')}} can be brought inside the time-ordered product, since it appears on the right when {{math|''x''<sup>0</sup> → −∞}} and on the left when {{math|''x''<sup>0</sup> → ∞}}:

:<math>\mathcal M=-i\sqrt{\frac{2\omega_p}{Z}} \left(\lim_{x^0\to-\infty}-\lim_{x^0\to \infty}\right) \int \mathrm{d}^3x f_p(x) \overleftrightarrow{\partial_0} \langle \beta\ \mathrm{out}| \mathrm T\left[\varphi(x)\varphi(y_1)\ldots\varphi(y_n)\right] |\alpha'\ \mathrm{in} \rangle </math>

In the following, {{mvar|x}} dependence in the time-ordered product is what matters, so we set:

:<math>\langle \beta\ \mathrm{out}| \mathrm T\left[\varphi(x)\varphi(y_1)\ldots\varphi(y_n)\right] |\alpha'\ \mathrm{in}\rangle= \eta(x) </math>

It can be shown by explicitly carrying out the time integration that:{{refn|group=note|Pulling the operators from time-ordering is not entirely trivial since neither <math>\mathcal \Box+m^2</math> nor <math>\mathcal \int dx^0</math> commutes with time-ordering <math>\mathrm T</math>. When we apply both the differential and the integral operators, however, the problems cancel out and the combined operator commutes with time-ordering.<ref name="Ticciati" />}}

:<math> \mathcal M=i\sqrt{\frac{2\omega_p}{Z}} \int \mathrm{d}(x^0)\partial_0 \int \mathrm{d}^3x f_p(x)\overleftrightarrow{\partial_0}\eta(x)</math>

so that, by explicit time derivation, we have:

:<math>\mathcal M=i\sqrt{\frac{2\omega_p}{Z}} \int \mathrm{d}^4 x\left\{f_p(x)\partial_0^2\eta(x)-\eta(x)\partial_0^2 f_p(x)\right\}</math>

By its definition we see that {{math|&nbsp;''f<sub>p</sub>''&nbsp;(''x'')}} is a solution of the Klein–Gordon equation, which can be written as:

:<math>\partial_0^2f_p(x)=\left(\Delta-m^2\right) f_p(x)</math>

Substituting into the expression for <math>\mathcal M</math> and integrating by parts, we arrive at:

:<math>\mathcal M=i\sqrt{\frac{2\omega_p}{Z}} \int \mathrm{d}^4 x f_p(x)\left(\partial_0^2-\Delta+m^2\right)\eta(x)</math>

That is:

:<math> \mathcal M=\frac{i}{(2\pi)^{\frac{3}{2}} Z^{\frac{1}{2}}} \int \mathrm{d}^4 x e^{-ip\cdot x} \left(\Box+m^2\right)\langle \beta\ \mathrm{out}| \mathrm T\left[\varphi(x)\varphi(y_1)\ldots\varphi(y_n)\right] |\alpha'\ \mathrm{in}\rangle</math>

Starting from this result, and following the same path another particle can be extracted from the ''in'' state, leading to the insertion of another field in the time-ordered product. A very similar routine can extract particles from the ''out'' state, and the two can be iterated to get vacuum both on right and on left of the time-ordered product, leading to the general formula:

:<math>\langle p_1,\ldots,p_n\ \mathrm{out}|q_1,\ldots,q_m\ \mathrm{in}\rangle=\int \prod_{i=1}^{m} \left\{\mathrm{d}^4x_i \frac{i e^{-iq_i\cdot x_i} \left(\Box_{x_i}+m^2\right)}{(2\pi)^{\frac{3}{2}} Z^{\frac{1}{2}}}  \right\} \prod_{j=1}^{n} \left\{ \mathrm{d}^4y_j \frac{i e^{ip_j\cdot y_j}\left(\Box_{y_j}+m^2\right)}{(2\pi)^{\frac{3}{2}} Z^{\frac{1}{2}}}  \right\} \langle \Omega|\mathrm{T} \varphi(x_1)\ldots\varphi(x_m)\varphi(y_1)\ldots\varphi(y_n)|\Omega\rangle</math>

Which is the LSZ reduction formula for Klein–Gordon scalars. It gains a much better looking aspect if it is written using the Fourier transform of the correlation function:

:<math> \Gamma \left (p_1,\ldots,p_n \right )=\int \prod_{i=1}^{n} \left\{\mathrm{d}^4x_i e^{i p_i\cdot x_i} \right\} \langle \Omega|\mathrm{T}\ \varphi(x_1)\ldots\varphi(x_n)|\Omega\rangle</math>

Using the inverse transform to substitute in the LSZ reduction formula, with some effort, the following result can be obtained:

:<math>\langle p_1,\ldots,p_n\ \mathrm{out}|q_1,\ldots,q_m\ \mathrm{in}\rangle= \prod_{i=1}^{m} \left\{-\frac{i\left(p_i^2-m^2\right)}{(2\pi)^{\frac{3}{2}} Z^{\frac{1}{2}}} \right\} \prod_{j=1}^{n} \left\{ -\frac{i\left(q_j^2-m^2\right)}{(2\pi)^{\frac{3}{2}} Z^{\frac{1}{2}}}  \right\} \Gamma \left (p_1,\ldots,p_n;-q_1,\ldots,-q_m \right )</math>

Leaving aside normalization factors, this formula asserts that ''S''-matrix elements are the residues of the poles that arise in the Fourier transform of the correlation functions as four-momenta are put on-shell.