Scalar product in Fock space and Coulomb interaction in second quantization

Question

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?

Answer 1

Ok. I have realized what was wrong. Actually, it is a quite subtle detail which, in my opinion, is not well-explained in most text-books. For those interested: It was in the appendix about Second Quantization of the book "Density Functional Theory: An Advanced Course" by Engel and Dreizler, where I finally clarified my doubts.

Following the above mentioned book's appendix's notation, let me call $\vert a b )=\vert a \rangle \otimes \vert b \rangle $ and $\vert a b \rangle=\frac{1}{\sqrt{2}}\left( \vert a b )-\vert \vert b a ) \right).$

The KEY is then that it is NOT true, as I had written in my question (I will omit spin now for clarity), that $$V=\sum_{k_1 k_2 k_3 k_4} \langle k_1 k_2 \vert V \vert k_3 k_4 \rangle c^\dagger_{k_1} c^\dagger_{k_2} c_{k_4}c_{k_3}.$$ The correct identity is: $$V=\frac{1}{2}\sum_{k_1 k_2 k_3 k_4} \left( k_1 k_2 \vert V \vert k_3 k_4 \right) c^\dagger_{k_1} c^\dagger_{k_2} c_{k_4}c_{k_3},$$ and then all the computations follow easily (NOTE: The factor $1/2$ is not that important; that was just a minor mistake. The crucial point is the change from $\langle ... \rangle$ to $( ... ).$

The proof of the above identity is easy. It just uses the form of the resolution of the identity in Fock space, $\frac{1}{2}\sum_{a b}\vert a b \rangle \langle a b \vert=$ Id (the 2 comes from the overcompleteness of the basis set) and the symmetry of the interaction, $(a b \vert V \vert c d)=(b a \vert V \vert d c).$

The thing is that until yesterday, I had never stopped to derive that formula (the form of the 2-body interaction in second quantization) by my own, and since a lot of textbooks are not too specific in this matter, I believed the wrong formula to be the correct one (actually, I've seen the other formula written down -with the $\frac{1}{2}$ factor- in many places, which, although understandable, I think is a notational disaster and potentially very confusing for begginers like myself).

For students: I guess the bottom line is: Do the calculations on your own without consulting anything. Only when you are able to do that can you be sure of having learnt something!

Answer 2

You can try to convert the argument from $r_1, r_2$ into $r_1, r_2-r_1$. With this, you then integrate out $r_2-r_1$, which is something like $$\int d^3{\bf r}\frac{e^{i\bf k\cdot r}}{r}\propto\frac{1}{k^2} $$