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{iehbar}{m} sum limits_i vec{A} nabla_i $, used the fourier transform of the vector potential $vec{A}$, i.e. $vec{A(vec{r})}=sum limits_{vec{q}} vec{aleft(vec{q}right)} e^{ivec{q}vec{r}}$ and he ended up with Eq. 3.97:
$$H=frac{-ehbar}{m} sum_{vec{k},vec{q}} vec{k} , vec{aleft(vec{q}right)} c_{vec{k}+vec{q}}^{dagger} c_{vec{k}}$$
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

Motionx · Accepted Answer

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

Hamiltonian in second quantization

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