This is a doubt into which I've run trying to compute the form of the Coulombian interaction in second quantization in a basis of plane waves.
Let $k$ denote the momentum and $r$ the position, and let me use $a$ and $b$ for spin variables, so that for example $\vert r a \rangle$ denotes a state of a particle with position $r$ and spin $a$ and $\vert k b \rangle$ a state of a particle with momentum $k$ and spin $b.$ Knowing that it holds (asumming the normalization constant to be one for ease of notation) that $\langle ra \vert kb \rangle = \exp(i kr) \delta_{a,b},$ I would like to compute the following scalar product in Fock space: $$\langle k a, k^\prime a^\prime \vert r b, r^\prime b^\prime \rangle.$$
My argument: By definition, $\vert r b, r^\prime b^\prime \rangle=\frac{1}{\sqrt{2}} \left(\vert r b \rangle \otimes \vert r^\prime b^\prime \rangle-\vert r^\prime b^\prime \rangle \otimes \vert r b \rangle \right)$ and analogously for $\langle k a, k^\prime a^\prime \vert.$ Therefore, one quickly finds that the scalar product is just the determinant of the matrix consisting of all possible pairings, that is: $$\langle k a, k^\prime a^\prime \vert r b, r^\prime b^\prime \rangle=\langle ka \vert rb \rangle \langle k^\prime a^\prime \vert r^\prime b^\prime \rangle- \langle ka \vert r^\prime b^\prime \rangle \langle k^\prime a^\prime \vert rb\rangle,$$ which gives the result: $$\exp(-ikr-ik^\prime r^\prime)\delta_{a,b}\delta_{a^\prime,b^\prime}-\exp(ikr^\prime+ik^\prime r)\delta_{a,b^\prime} \delta_{a^\prime, b}.$$
Once I have this, I want to use this result to compute the matrix element $\langle k_1 a_1,k_2 a_2 \vert V \vert k_3 a_3,k_4 a_4 \rangle$ of the Coulomb interaction, wich by definition satisfies (in adequate units): $$\langle r_1 b_1,r_2 b_2 \vert V \vert r_3 b_3,r_4 b_4 \rangle=\frac{1}{\vert r_1-r_2 \vert}\delta(r_3-r_1)\delta(r_4-r_2)\delta_{b_3,b_1}\delta_{b4,b2}.$$ By employing twice the resolution of the identity, we have: $$\langle k_1 a_1,k_2 a_2 \vert V \vert k_3 a_3,k_4 a_4 \rangle=$$ $$\sum_{ b_1,b_2,b_3,b_4 } \int dr_1 dr_2 dr_3 dr_4 \langle k_1 a_1,k_2a_2 \vert r_1 b_1,r_2b_2 \rangle \langle r_1 b_1,r_2 b_2 \vert V \vert r_3 b_3,r_4 b_4 \rangle \langle r_3b_3,r_4b_4 \vert k_3a_3,k_4a_4 \rangle.$$ Using the scalar product above, I get: $$\sum_{b_1,b_2}\int dr_1 dr_2 \left( e^{-ik_1r_1-ik_2 r_2}\delta_{a_1,b_1}\delta_{a_2,b_2} -e^{ik_1r_2+ik_2 r_1}\delta_{a_1,b_2} \delta_{a_2, b_1}\right)$$ $$\times \frac{1}{\vert r_1-r_2 \vert} \left( e^{-ik_3r_1-ik_4 r_2}\delta_{a_3,b_1}\delta_{a_4,b_2}-e^{ik_3r_2+ik_4 r_1}\delta_{a_3,b_2} \delta_{a_4, b_1}\right).$$
Next I plug this result into the expansion $$\sum_{k_1 s_1,k_2 s_2,k_3 s_3, k_4 s_4} \langle k_1 a_1,k_2 a_2 \vert V \vert k_3 a_3,k_4 a_4 \rangle c^{\dagger}_{k_1 s_1}c^{\dagger}_{k_2 s_2}c_{k_4 s_4}c_{k_3 s_3}$$
BUT (this is the problem) I am unable to derive from here the second-quantized form of the Coulombian interaction, which is $\frac{1}{2}\sum_{k_1,k_2,q,s_1,s_2} q^{-2} c^{\dagger}_{k_1,s_1}c^{\dagger}_{k_2, s_2}c_{k_4-q,s_2}c_{k_3+q,s_1}.$ Things just don't add up. There are some exponentials which seem to be wrong-placed, and I don't see where can I possible have made a mistake. What am I doing wrong?
