I am referring Mahan Many-Particle Physics. There are 2 particle current operators -one in general and one for the tight binding Hamiltonian. How do we go from the general current operator (1.195 in Mahan)
$ j_{i}(\mathbf{r})=\frac{1}{2mi}\left(\psi^{\dagger}(\mathbf{r})\nabla\psi(\mathbf{r})-\psi(\mathbf{r})\nabla\psi^{\dagger}(\mathbf{r})\right)
$
[whose Fourier transform is
$\int d^{3}r\,\,\exp(-i\mathbf{q}.\mathbf{r})\,\, j_{i}(\mathbf{r}) = \frac{1}{2mi}\sum_{\sigma\beta}c_{\sigma}^{\dagger}c_{\beta}\int d^{3}r\,\,\exp(-i\mathbf{q}.\mathbf{r})\,\left(\phi_{\alpha}^{\star}(\mathbf{r})\nabla\phi_{\beta}(\mathbf{r})-\phi_{\beta}(\mathbf{r})\nabla\phi_{\alpha}^{\star}(\mathbf{r})\right)$
giving $j_{l}(q) = \frac{1}{m}\sum_{k\alpha}\left(k+\frac{q}{2}\right)c_{k+q,\sigma}^{\dagger}c_{k,\sigma}
$]
to (1.204) current operator $
j=-iw\sum_{j\delta\sigma}\delta c_{j+\delta,\sigma}^{\dagger}c_{j\sigma}$
for the Tight Binding Hamiltonian $H=w\sum_{j\delta\sigma}c_{j+\delta,\sigma}^{\dagger}c_{j,\sigma}+\frac{1}{2}\sum_{ij,ss^{\prime}}n_{is}n_{js^{\prime}}V_{ij}$? (here $\phi(r)$
are plane waves.) I mean we use an alternative definition of current in terms of polarization to get to the form of current operator in real space for the tight binding Hamiltonian but on fourier transforming this operator, it should return us a form similar to the current operator generally defined in Fourier space– I don't see how we will get the mass m of $j_{l}(q)$ on taking the fourier transform of $ j=-iw\sum_{j\delta\sigma}\delta c_{j+\delta,\sigma}^{\dagger}c_{j\sigma}$
Best Answer
You appear to be neglecting the Bloch functions.
$$\psi^\dagger(r) = \sum_j\int_{BZ}\!\!\!d\vec{q}\,\,e^{iq\cdot r} u_{q,j}(r)c^\dagger_j(q)$$
where $u_{q,j}(r)$ is the Bloch function and $j$ is some band/spin/whatever index. If you plug that into the formula you do not obtain what you wrote because there are terms with $\nabla u$. that you did not include. You need to treat these terms to connect the two formulae.