Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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}$

share|improve this question

2 Answers 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.

share|improve this 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.

share|improve this answer
    
still where will I get the mass m from (last line in my question) ? And I think you should see Mahan as I have just written the resulting current operator of the general and tight-binding Hamiltonian directly from there. –  cleanplay Jul 10 '13 at 18:43
    
@user25957: If, in the first formula, by $c_k$ you mean the operator that creates a free electron of momentum $k$, then what you wrote is correct. But that $c_k$ is not the same $c_k$ as in the tight-binding formula, which creates a complicated superposition of free waves. –  BebopButUnsteady Jul 10 '13 at 20:49
1  
@user25957: You get the right mass because the correct formula is approximately $v(k)c^\dagger_{k+Q/2}c_{k-Q/2}$, where $v$ is the group velocity. Near a band minimum $v\sim k/m_{eff}$. The two formula have a complicated connection - The tight binding model has absorbed all the information about the mass and the lattice potential and repackaged it into the dispersion relation. –  BebopButUnsteady Jul 10 '13 at 20:53

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.