Hamiltonian in second quantization

Question

I'm reading Tinhkam's Superconductivity book and I'm not able to understand how he ended up with Eq. 3.97.

He started with a Hamiltonian $H=\frac{ie\hbar}{m} \sum \limits_i \vec{A} \nabla_i $, used the fourier transform of the vector potential $\vec{A}$, i.e. $\vec{A(\vec{r})}=\sum \limits_{\vec{q}} \vec{a\left(\vec{q}\right)} e^{i\vec{q}\vec{r}}$ and he ended up with Eq. 3.97:

$$H=\frac{-e\hbar}{m} \sum_{\vec{k},\vec{q}} \vec{k} \, \vec{a\left(\vec{q}\right)} c_{\vec{k}+\vec{q}}^{\dagger} c_{\vec{k}}$$

I know that a single particle operator $ \Omega_i$ in second quantization can be written in the following form: $$\sum_i \Omega_i= \sum_{k,k^{'}} \langle k^{'}| \Omega |k \rangle \, c_{k^{'}}^{\dagger}c_{k} = \sum_{k,q} \langle k+q| \Omega |k \rangle c_{k+q}^{\dagger}c_{k}$$ where the summation with respect to i runs over all the particles. This is probably used here, but I don't see from where the $\vec{k}$ is coming from, how he got rid of the exponential factor and the bra and ket ...

Greetings

Answer 1

The sum over the single particle operator $\vec{A}\vec{\nabla_i}$ can be written as follows

$$\sum_{k,q} \langle k+q| \vec{A} \vec{\nabla} |k \rangle c_{k+q}^{\dagger}c_{k}.$$

Using the above equation and inserting the completeness relation into the hamilonian yields

$$H=\frac{ie\hbar}{m} \sum_{\vec{k},\vec{q}} \sum_{r} \langle k+q| r\rangle \langle r |\sum_{k'} \vec{a} \,(\vec{k'}) \, e^{i\vec{k'}\vec{r}} \, \vec{\nabla}|k \rangle c_{\vec{k}+\vec{q}}^{\dagger} c_{\vec{k}}$$

Knowing that $\langle r| k\rangle \sim e^{i\vec{k}\vec{r}} $, $\sum_{r}e^{i(\vec{k}-\vec{k'})\vec{r}} = \delta(\vec{k}-\vec{k'}) $ and $-i\hbar \vec{\nabla} = \hbar \vec{k}$ one obtains $$H=\frac{-e\hbar}{m} \sum_{\vec{k},\vec{q}} \vec{k} \vec{a(\vec{q})}c_{\vec{k}+\vec{q}}^{\dagger} c_{\vec{k}}$$