Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 53 / 4 / ii / Solution

Reality of $E_{ij}^T$ and the given polarization relation imply ${\psi }_{(p)}(-\mathbf{k})={\psi }_{(p)}(\mathbf{k}{)}^{\ast }$. Insert the Fourier expansion twice into each quadratic term. The spatial integral givesThis cancels the two factors of $(2\pi {)}^{-3/2}$. Set ${\mathbf{k}}^{\prime }=-\mathbf{k}$ and use $M_{ij}^{(q)}(-\mathbf{k})=M_{ij}^{(q)}(\mathbf{k}{)}^{\ast }$. The unit-normalization condition eliminates cross-polarization terms, givingFor the gradient term, $(i{k}_l)(i{k}_l^{\prime })$ becomes $k^2$ after the delta-function substitution. ThusThe full momentum-space integral and its reality relation are those of real fields; there is no additional independent copy of the negative-wavevector modes.