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

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.