Maybe this is more appropriate for Math stackexchange, but this question regards the solution we use in order to find representation for massive / massless spin-1 particle.

When $$(\Box + m^2)A_\mu = 0$$ for $\mu = 0,1,2,3$, general solution is of the form

$$A_\mu(x) = \Sigma_i \int \frac{d^3\vec{p}}{(2\pi)^3} a_i(\vec{p})\epsilon_\mu^i(p)e^{ipx}$$ Why? I have not studied partial differential equations enough to know why this polarization vector is necessary. Even a reference would be helpful, since I just could not find a place to read about this.

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    $\begingroup$ I've not studied QFT, but that looks a lot like Green's function. $\endgroup$ – Kyle Kanos Mar 6 '15 at 3:16
  • $\begingroup$ The polarisation vector has nothing to do with the equation, it's just the collection of the four a priori independent expansion coefficients into a single vector. Also, your integration measure is missing a $\frac{1}{\sqrt{2E_p}}$ to be Lorentz invariant. $\endgroup$ – ACuriousMind Mar 6 '15 at 11:00
  • $\begingroup$ @KyleKanos: Greens functions don't concern a differential equation which is homogenous everywhere. Regarding thread pointing out that $(ip)^2=-m^2$ might help. $\endgroup$ – Nikolaj-K Mar 6 '15 at 11:07
  • $\begingroup$ You have a linear equation that is solved by plane waves. The general solution is then just a superposition of such plane waves. It's basically just a Fourier transform of the k-space solution. $\endgroup$ – FenderLesPaul Mar 6 '15 at 14:35
  • $\begingroup$ @NikolajK: This is true, but not having studied QFT, I'm not familiar with the homogeneity of this equation versus the Helmholtz equation that it looks like. Thank you for the correction. $\endgroup$ – Kyle Kanos Mar 6 '15 at 18:09

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