0
$\begingroup$

If I have a lattice Hamiltonian, say for example the Hubbard model $$H = \sum_{j,k, \sigma} t_{j,k} \hat{c}^\dagger_{j \sigma}\hat{c}_{k \sigma} + U\sum_{j} \hat{n}_{j \uparrow}\hat{n}_{j \downarrow},$$ and I want to make it interact with an electromagnetic field, in the dipolar approximation I would write it $$H_I = -\vec{\hat{D}} \cdot \vec{\hat{E}}, $$ where the electric field is given by boson operators: $$\vec{\hat{E}}=\sum_{\vec{k},\lambda}w_k^{-1/2} \vec{e}_\lambda(\vec{k}) \left(\hat{b}_\lambda(\vec{k}) - \hat{b}^\dagger_\lambda(\vec{k}) \right)$$ (up to a prefactor).

But how can I write in the lattice model the dipolar operator $\vec{\hat{D}}$? That is, I would like to write the dipolar operator, usually written in space representation as $\sum_i q_i \hat{r}_i,$ in terms of the lattice operators $\hat{c}_{j\sigma}$ \ $\hat{c}^\dagger_{j\sigma}$

$\endgroup$

1 Answer 1

0
$\begingroup$

If each lattice site is considered a state, then the usual second quantization procedure means expanding the wave function as $$ \psi_\sigma(\mathbf{r})=\sum_{j,\sigma}\phi\left(\mathbf{r}-\mathbf{R}_j\right)\chi_\sigma c_{j,\sigma}, $$ where $\phi\left(\mathbf{r}-\mathbf{R}_j\right)$ is the orbital or a Wannier function centered on the $j$-th site. (Wannier functions are orthogonal, and this form a real basis, but one usually uses non-orthogonal functions, whose oevrlap is related to tight-binding approximation hopping integral.)

Given any operator in space representation, e.g., $$ H_I=-e\mathbf{r}\mathbf{E},$$ we can now convert it to the second quantized form.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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