Calculation of Radiation Energy in Peskin&Schroeder

Question

In Peskin&Schroeder QFT section 6.1, they try to compute the radiation energy of soft bremsstrahlung classically where the momentum-space amplitude of vector potential is $$\mathcal{A}^{\mu}(\mathbf{k}) = \frac{-e}{|\mathbf{k}|}\left(\frac{p'^{\mu}}{k\cdot p'}-\frac{p^{\mu}}{k\cdot p}\right) \tag{6.7}$$ The momentum-space amplitude of the radiation field are $$\mathcal{E}(\mathbf{k}) = -i\mathbf{k}\mathcal{A}^0(\mathbf{k})+ik^{0}\mathcal{A}(\mathbf{k})\tag{6.8}$$ $$\mathcal{B}(\mathbf{k}) = i\mathbf{k}\times\mathcal{A}(\mathbf{k})\tag{6.8}$$ and the energy after several calculation $$\text{Energy} = \frac{1}{2}\int{\frac{d^3k}{(2\pi)^3}\, \mathcal{E}(\mathbf{k})\cdot\mathcal{E}^*(\mathbf{k})}\tag{6.11}$$ They introduce two transverse unit polarization vectors and the part parallel to $\mathbf{k}$ in $\mathcal{E}$ vanishes, $$\mathcal{E}(\mathbf{k})\cdot\mathcal{E}^*(\mathbf{k}) = \sum_{\lambda = 1, 2}|\mathbf{\epsilon}_{\lambda}(\mathbf{k})\cdot\mathcal{E}(\mathbf{k})|^2 = |\mathbf{k}|^2\sum_{\lambda = 1, 2}|\mathbf{\epsilon}_{\lambda}(\mathbf{k})\cdot\mathcal{A}(\mathbf{k})|^2$$ In the following calculation, they claim we can freely change $\mathbf{\epsilon}, \mathbf{p}, \text{and }\mathbf{p'}$ into 4-vectors in this expression, which confused me.

Can anyone explain why we can extand 3-vectors into 4-vectors here?

Answer 1

It is just the gauge choice that the time component of the vector field doesn’t have corresponding canonical momentum, and thus we have the time component of polarization is always $0$ (or equivalently $A^0=0$), so we can upgrade 3D to 3D+1D without worrying this would affect the equation.