Editing LSZ reduction formula (section)

==Field strength normalization==
The reason of the normalization factor {{mvar|Z}} in the definition of ''in'' and ''out'' fields can be understood by taking that relation between the vacuum and a single particle state <math>|p\rangle</math> with four-moment on-shell:

:<math>\langle 0|\varphi(x)|p\rangle= \sqrt Z \langle 0|\varphi_{\mathrm{in}}(x)|p\rangle + \int \mathrm{d}^4y \Delta_{\mathrm{ret}}(x-y) \langle 0|j(y)|p\rangle</math>

Remembering that both {{mvar|φ}} and {{math|''φ''<sub>in</sub>}} are scalar fields with their Lorentz transform according to:

:<math>\varphi(x)=e^{iP\cdot x}\varphi(0)e^{-iP\cdot x}</math>

where {{mvar|P<sup>μ</sup>}} is the four-momentum operator, we can write:

:<math> e^{-ip\cdot x}\langle 0|\varphi(0)|p\rangle= \sqrt Z e^{-ip\cdot x} \langle 0|\varphi_{\mathrm{in}}(0)|p\rangle + \int \mathrm{d}^4y \Delta_{\mathrm{ret}}(x-y)\langle 0|j(y)|p\rangle</math>

Applying the Klein–Gordon operator {{math|∂<sup>2</sup> + ''m''<sup>2</sup>}} on both sides, remembering that the four-moment {{mvar|p}} is on-shell and that {{math|Δ<sub>ret</sub>}} is the Green's function of the operator, we obtain:

:<math> 0=0 + \int \mathrm{d}^4y \delta^4(x-y) \langle 0|j(y)|p\rangle; \quad\Leftrightarrow\quad \langle 0|j(x)|p\rangle=0</math>

So we arrive to the relation:

:<math>\langle 0|\varphi(x)|p\rangle= \sqrt Z \langle 0|\varphi_{\mathrm{in}}(x)|p\rangle </math>

which accounts for the need of the factor {{mvar|Z}}. The ''in'' field is a free field, so it can only connect one-particle states with the vacuum. That is, its expectation value between the vacuum and a many-particle state is null. On the other hand, the interacting field can also connect  many-particle states to the vacuum, thanks to interaction, so the expectation values on the two sides of the last equation are different, and need a normalization factor in between. The right hand side can be computed explicitly, by expanding the ''in'' field in creation and annihilation operators:

:<math>\langle 0|\varphi_{\mathrm{in}}(x)|p\rangle= \int \frac{\mathrm{d}^3q}{(2\pi)^{\frac{3}{2}}(2\omega_q)^{\frac{1}{2}}} e^{-iq\cdot x} \langle 0|a_{\mathrm{in}}(\mathbf q)|p\rangle= \int \frac{\mathrm{d}^3q}{(2\pi)^{\frac{3}{2}}} e^{-iq\cdot x} \langle 0|a_{\mathrm{in}}(\mathbf q)a^\dagger_{\mathrm{in}}(\mathbf p)|0\rangle </math>

Using the commutation relation between {{math|''a''<sub>in</sub>}} and <math>a^\dagger_{\mathrm{in}}</math> we obtain:

:<math> \langle 0|\varphi_{\mathrm{in}}(x)|p\rangle= \frac{e^{-ip\cdot x}}{(2\pi)^{\frac{3}{2}}} </math>

leading to the relation:

:<math>\langle 0|\varphi(0)|p\rangle= \sqrt \frac{Z}{(2\pi)^3}</math>

by which the value of {{mvar|Z}} may be computed, provided that one knows how to compute <math>\langle 0|\varphi(0)|p\rangle</math>.