Particle current operator in general vs Particle current operator for tight binding Hamiltonian

Question

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}$

Answer 1

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.

Answer 2

The fermions in the tight-binding model are not the same ones as in the "general" model, and therefore there should not be an equality between the two current operators. The tight-binding model is an approximation that allows one to take into account much of the effect of the lattice ions. In general there are overlaps between the atomic wavefunctions from which the tight binding fermions are constructed.

The current operator given by Mahan is consistent with the tight-binding Hamiltonian, and should be used whenever this approximation is used. The "bare" current operator will not satisfy particle conservation within this model.

This kind of thing often repeats itself in physics: you pick the relevant, low-energy effective model that describes your system (in this case, the tight-binding Hamiltonian and the operators' commutation relations) and derive other quantities from there.