240 Appendix F: Heavy Quark Production Radiation Derivation 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 ∫ d $^{3}$Q$^{2}$( x ) = 1 4 ∫ d $^{3}$x d $^{3}$k (2 π ) 3 d $^{3}$k ′ (2 π ) 3 ( Q ( k ) e − ik · x + Q $^{∗}$( k ) e ik · x ) ×$^{(}$Q ( k $^{′}$) e − ik $^{′}$· x + Q $^{∗}$( k $^{′}$) e ik $^{′}$· x ) = 1 4 ∫ d $^{3}$k (2 π ) $_{3}$Q ( k ) Q ( − k ) e − 2 ik $^{0}$t + Q $^{∗}$( k ) Q $^{∗}$( − k ) e 2 ik $^{0}$t + 2 Q ( k ) Q $^{∗}$( k ) = ∫ 1 2 d $^{3}$k (2 π ) $_{3}$Q ( k ) Q ( − k ) cos(2 k $^{0}$t ) + Q ( k ) Q $^{∗}$( k ) , (F.69) 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 ⃗ E$_{k}$ · ⃗ E$_{−}$$_{k}$ = $^{[}$i ( k 0 ⃗ A$_{k}$ − ⃗ k A 0 $_{k}$) $^{][}$i ( k 0 ⃗ A$_{−}$$_{k}$ + ⃗ k A 0 − $_{k}$) ] = k $^{0}$A0 $_{k}$⃗ k · ⃗ A$_{−}$$_{k}$ + ⃗ k $^{2}$A0 $_{k}$A 0 − k − k 0 2 ⃗ A$_{k}$ · ⃗ A$_{−}$$_{k}$ − k $^{0}$A0 − $_{k}$⃗ k · ⃗ A$_{k}$ = ⃗ k $^{2}$A0 $_{k}$A 0 − k − k 0 2 ⃗ A$_{k}$ · ⃗ A$_{−}$$_{k}$ (F.70) = ⃗ E ∗ k · ⃗ E ∗ − $_{k}$, (F.71)