# Hamiltonian in second quantization

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

• Well, do you know what $\langle r |k\rangle$ corresponds to? Mar 21, 2021 at 14:51
• you mean the kronecker delta, i.e. $\langle r | k \rangle = \delta_{r,k}$? Mar 21, 2021 at 14:58
• No. Think about how the momentum and position-space representations are related in QM. Then try to insert the completeness relation in the matrix element of the operator, i.e. try to calculate $\langle k^\prime|\mathbb{I}\, \Omega|k\rangle$. You should then see how the $k$ appears etc. Mar 21, 2021 at 15:03
• thanks, I think I see now :) Mar 21, 2021 at 17:06
• You could then consider to answer your own question, because otherwise it would remain in the category unanswered. Mar 22, 2021 at 12:22

The sum over the single particle operator $$\vec{A}\vec{\nabla_i}$$ can be written as follows
$$\sum_{k,q} \langle k+q| \vec{A} \vec{\nabla} |k \rangle c_{k+q}^{\dagger}c_{k}.$$
$$H=\frac{ie\hbar}{m} \sum_{\vec{k},\vec{q}} \sum_{r} \langle k+q| r\rangle \langle r |\sum_{k'} \vec{a} \,(\vec{k'}) \, e^{i\vec{k'}\vec{r}} \, \vec{\nabla}|k \rangle c_{\vec{k}+\vec{q}}^{\dagger} c_{\vec{k}}$$
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 $$H=\frac{-e\hbar}{m} \sum_{\vec{k},\vec{q}} \vec{k} \vec{a(\vec{q})}c_{\vec{k}+\vec{q}}^{\dagger} c_{\vec{k}}$$