# Becchi-Rouet-Stora-Tyutin symmetry/Landau’s gauge free solutions

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Landau’s gauge free solutions

Decomposing the vector potential in its physical and unphysical parts, $$A_\mu(x)=A^{(ph)}_\mu(x)+A^{(u)}_\mu(x)$$, the general solution of electrodynamic equations in Landau’s gauge reads as follows $A^{(ph)}_\mu(x)=\int {d^4 k\over(2\pi)^{3/2}}e^{-ik\cdot x}\theta(k_0)\left[\delta(k^2)\sum_{h=\pm 1}\epsilon_\mu(\vec k, h) a(\vec k,h)\right]+ c.-c.\ ,$ $A^{(u)}_\mu(x)=i\int {d^4 k\over(2\pi)^{3/2}}e^{-ik\cdot x}\theta(k_0)\left[\delta(k^2)\left(k_\mu\alpha(\vec k) -\bar k_\mu {\beta(\vec k)\over k\cdot\bar k}\right)- k_\mu\delta'(k^2)\beta(\vec k)\right]+ c.-c.\ ,$ $b(x)=\int {d^4 k\over(2\pi)^{3/2}}e^{-ik\cdot x}\theta(k_0)\delta(k^2)\beta(\vec k)+ c.-c.\$ where:

• $$c.-c.$$ means complex conjugate;
• $$\delta(k^2)$$ and $$\delta'(k^2)$$ are Dirac's delta measure and its derivative;
• $$\epsilon_\mu(\vec k, h)$$ for $$h=\pm 1$$ are space-like circular polarization vectors such that:
• $$\epsilon\cdot k=\epsilon\cdot \bar k=0$$,
• $$\epsilon^*_\mu(\vec k, h)=\epsilon_\mu(-\vec k, h)$$;
• $$\bar k$$ is the parity reflected image of $$k$$.

The polarization vectors define the unpolarized photon density matrix $\sum_{h=\pm}\epsilon _\mu(\vec k, h)\epsilon^*_\nu(\vec k, h)=-g_{\mu\nu}+{k_\mu\bar k_\nu+\bar k_\mu k_\nu\over k\cdot\bar k}\ .$ It is easy to verify, using the identity $$x\delta'(x)=-\delta(x)$$ and $$x\delta(x)=0$$, that for a generic choice of the functions $$a\ ,\alpha\ ,\ \beta$$ the above equations give the general solution to the Landau's gauge free field equations.