# Scalar product in Fock space and Coulomb interaction in second quantization

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?

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!

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 $$\int d^3{\bf r}\frac{e^{i\bf k\cdot r}}{r}\propto\frac{1}{k^2}$$

• Yes, of course, that's is what I had done, and I don't reach the correct expression. I am guessing there is somo conceptual mistake in the derivation of the equation I obtained, which I haven't seen anywhere... But I don't see what I am doing wrong. Thanks! – Qwertuy Apr 4 '17 at 15:38