We want to write the Hamiltonian in the second quantized notation for a system of fermions interacting under the Yukawa potential. First we deal with the kinetic energy term which is a one body operator. In the formalism of second quantization this is written as

$${\mathcal H}_0= \sum_{ij} c^{\dagger}_i c_j \langle i \vert \hat{T} \vert j \rangle ,$$ (1)

where $\hat{T} = \hat{p}^2/2m$, the states $\vert j \rangle$ define an arbitrary basis of the single particle Hilbert space, and $c^{\dagger}_i$ and $c_j$ are fermionic operators obeying anticommutation relation $\{ c_j , c^{\dagger}_i \} = \delta _{ji}$. To represent ${\mathcal H}_0$ in the position basis we use the completeness condition for the single particle Hilbert space, namely

$$\int d^3 {\bf r}\vert {\bf r}\rangle \langle {\bf r}\vert = 1.$$

We get,

$${\mathcal H}_0= \sum_{ij} c^{\dagger}_i c_j \int d^3 {\bf r}... ...^{\prime } \rangle \langle {\bf r}^{\prime } \vert j \rangle .$$

Now, $\langle i \vert {\bf r}\rangle = \phi _i ({\bf r})$ is the representation of the state $\vert i \rangle$ in the position basis (wavefunction). We also note that

$$\langle {\bf r}\vert \frac{\hat{p}^2}{2m} \vert {\bf r}^{\pr... ...ac{\hbar^2}{2m} \nabla^2 \delta ({\bf r}- {\bf r}^{\prime }).$$

Then,

$${\mathcal H}_0= \int d^3 {\bf r}\left( \sum_i c^{\dagger}_i ... ...r^2}{2m} \nabla^2 \left( \sum_j c_j \phi _j ({\bf r}) \right).$$

We define the creation field operator as

$$\psi^{\dagger}({\bf r}) = \sum_i c^{\dagger}_i \phi _i^{\ast}({\bf r}),$$

and similarly the annihilation operator $\psi ({\bf r})$. Then we can write

$${\mathcal H}_0= - \frac{\hbar^2}{2m} \int d^3 {\bf r}\psi^{\dagger}({\bf r}) \nabla^2 \psi ({\bf r}).$$ (2)

The two body interaction term is written as

$$\h1 = \frac{1}{2} \sum_{ijkl} c^{\dagger}_i c^{\dagger}_j c_k c_l \langle i j \vert \hat{V} \vert l k \rangle ,$$ (3)

where $\vert i j \rangle = \vert i \rangle \otimes \vert j \rangle$ is a state of the two particle Hilbert space. Using the completeness condition

$$\int d^3 {\bf r}d^3 {\bf r}^{\prime } \vert {\bf r}{\bf r}^{\prime } \rangle \langle {\bf r}{\bf r}^{\prime } \vert = 1$$

for the two particle Hilbert space, we get

$$\h1 = \frac{1}{2} \sum_{ijkl} \int d^3 {\bf r}d^3 {\bf r}^{\prime } c^{... ...({\bf r}- {\bf r}^{\prime }) \langle {\bf r}{\bf r}^{\prime } \vert l k \rangle$$

$$= \frac{1}{2} \sum_{ijkl} \int d^3 {\bf r}d^3 {\bf r}^{\prime } c^{... ... }) V({\bf r}- {\bf r}^{\prime }) \phi _l ({\bf r}^{\prime }) \phi _k ({\bf r})$$

$$= \frac{1}{2} \int d^3 {\bf r}d^3 {\bf r}^{\prime } \psi^{\dagger}(... ...prime }) V({\bf r}- {\bf r}^{\prime }) \psi ({\bf r}^{\prime }) \psi ({\bf r}).$$ (4)

We will now rewrite the Hamiltonian in the momentum basis. For this we define the Fourier transform of the field operators as

$$\psi ({\bf r}) = \frac{1}{V^{1/2}} \sum_{{\bf k}} c_{{\bf k}} e^{i {\bf k}\cdot {\bf r}},$$ (5)

and similarly for $\psi^{\dagger}({\bf r})$. Here $V$ is the volume of the system. Using this definition in equation (2) we get,

$${\mathcal H}_0 = - \left( \frac{\hbar^2}{2m} \right) \frac{1}{V} \sum_{{\bf k}_1 {... ...^3 {\bf r}e^{-i {\bf k}_1 \cdot {\bf r}} \nabla^2 e^{i {\bf k}_2 \cdot {\bf r}}$$

$$= \sum_{{\bf k}} \left( \frac{\hbar^2 k^2}{2m} \right) c^{\dagger}_{{\bf k}} c_{{\bf k}}.$$ (6)

In the above equation we have used the definition of Kronecker delta

$$\int d^3 {\bf r}e^{i ({\bf k}_2 - {\bf k}_1) \cdot {\bf r}} = V \delta _{{\bf k}_2, {\bf k}_1}.$$

To write the interaction term in the momentum basis we first introduce a new variable ${\bf\rho}= {\bf r}- {\bf r}^{\prime }$. From equation (4) we get

$$\h1 = \frac{1}{2} \int d^3 {\bf r}^{\prime } d^3 {\bf\rho}\psi^{\dagger... ...{\prime }) V(\r ) \psi ({\bf r}^{\prime }) \psi ({\bf r}^{\prime } + {\bf\rho})$$

$$= \frac{1}{2V^2} \sum_{{\bf k}_1 \cdots {\bf k}_4} c^{\dagger}_{{\b... ...f k}_4)} \int d^3 {\bf\rho}V(\r ) e^{-i {\bf\rho}\cdot ({\bf k}_1- {\bf k}_4)}.$$

Now,

$$\int d^3 {\bf r}^{\prime } e^{-i {\bf r}^{\prime } \cdot ({\... ...)} = V \delta _{{\bf k}_1 + {\bf k}_2, {\bf k}_3 + {\bf k}_4},$$

which gives overall momentum conservation for the two particle scattering process. We write ${\bf k}_4= {\bf k}, {\bf k}_3 = {\bf k}^{\prime }, {\bf k}_2 = {\bf k}^{\prime } - {\bf q}, {\bf k}_1 = {\bf k}+ {\bf q}$. Then we get,

$$\h1 = \frac{1}{2V}\sum_{{\bf k},{\bf k}^{\prime },{\bf q}} c... ...bf k}} \int d^3 {\bf\rho}V(\r ) e^{-i {\bf q}\cdot {\bf\rho}}.$$ (7)

The Yukawa potential is given by

$$V(\r ) = \frac{A}{4 \pi \r } e^{- \lambda \r }.$$

Its Fourier transform is given by

$$\int d^3 {\bf\rho}V(\r ) e^{-i {\bf q}\cdot {\bf\rho}} = 2 \pi \int d \r\r ^2 \int_{-1}^1 dz \frac{A}{4 \pi \r } e^{- \lambda \r } e^{-i q \r z}$$

$$= \frac{A}{q} \int d \r e^{- \lambda \r } \sin(q \r )$$

$$= \frac{A}{q^2 + \lambda ^2}.$$ (8)