General form of electromagnetic vertex function in QFT

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How did the authors write the general form of the electromagnetic vertex function out of nowhere?
I am studying a beginner's book on QFT.

In a chapter on electromagnetic form factors, the authors have written, using normalized states,
$$\begin{eqnarray}
\langle \vec{p'}, s'| j_\mu (x) |\vec{p}, s \rangle \ = \ \exp(-i \ q \cdot x) \langle \vec{p'}, s'| j_\mu (0) |\vec{p}, s \rangle \nonumber \\
\Rightarrow \ \langle \vec{p'}, s'| j_\mu (x) |\vec{p}, s \rangle \ = \ \dfrac{\exp(-i \ q \cdot x)}{\sqrt{2 E_p V} \sqrt{2 E_{p'} V}} \bar{u}_{s'} (\vec{p'}) e \Gamma_\mu(p, p') u_s(\vec{p}) \nonumber
\end{eqnarray}$$
where ##q = p - p'##, ##E_p = p^0##, ##E_{p'} = p'^0##, ##\Gamma## is the vertex function, ##u_s## is the plane wave solution of the Dirac Equation, ##\bar{u}_s## is the Dirac conjugate of ##u_s##, and other symbols have their usual meanings.

After this, the authors have said that the most general form of ##\Gamma## is $$ \Gamma_\mu \ = \ \gamma_\mu(F_1 + \tilde{F}_1 \gamma_5) \ + (\ i F_2 + \tilde{F}_2 \gamma_5) \sigma_{\mu \nu} q^\nu \ + \tilde{F}_3 q_\mu \not\!q\gamma_5 \ + \ q_\mu (F_4 + \tilde{F}_4\gamma_5),$$ where ##\sigma_{\mu \nu} \ = \ \frac{i}{2} \left[\gamma_\mu, \ \gamma_\nu\right]## and ##\gamma_5 \ = \ \frac{i}{4!} \epsilon_{\mu\nu\lambda\rho} \gamma^\mu \gamma^\nu \gamma^\lambda \gamma^\rho##, all the ##F##'s are the electromagnetic form factors and other symbols have their usual meanings.

I understand that the book is for beginners, but how did the authors, out of nowhere, write down the general form for ##\Gamma##?
 
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The products of one or more ##\gamma## and ##\gamma_5##, forms a basis for all matrices of dimension 4. ##\Gamma## can thus be written in a basis of them. You need then also to take into account e.g. Lorentz invariance in order to arrive at the expression for ##\Gamma##.
 
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