content_image/312141.jpg content_image/312144.jpg content_image/312145.jpg
polarization vectors (whose sum in fact obeys ∑ 3 λ =1 ϵ µ $_{λ}$ϵ ∗ ν λ = − g$_{µν}$ + k $^{µ}$k$^{ν}$/µ $^{2}$). As a result we won’t use any of the tricks exploited in the E&M calculation; rather we will plug our solution of A$_{Proca}$ into the formula for E$_{Proca}$ , Eq. (F.21), and brute force manipulate it. 

We are interested in finding quantities of the form ∫ d $^{3}$xQ$^{2}$( x ) where Q ( x ) = Re ∫ d $^{3}$k Q ( k ) exp( − ik · x ) / (2 π ) $^{3}$. Then 

<img src='content_image/312141.jpg'>

where we are taking k = ( k $^{0}$, ⃗ k ) and − k = ( k $^{0}$, − ⃗ k ). We get the last line because, as we will see below, A ∗ µ = A µ ⇒ Q ( k ) Q ( − k ) = Q $^{∗}$( k ) Q $^{∗}$( − k ) for all the quantities of interest. 

From Eq. (F.45) we see that 

<img src='content_image/312144.jpg'>

<img src='content_image/312145.jpg'>

