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I'm learning about the mean-field approximation and I want to rewrite the following Hamiltonian, which is the simplified part of the interaction term for Graphene:

$$\hat{H}^1_I = g_1 c_2 \sum_{\langle i,j \rangle} \hat{D}_{ij}^{\dagger}\hat{D}_{ij}.$$ Where $\langle i,j \rangle$ means the sum over nearest neighbors and $\hat{D}_{ij}=\hat{a}_{i \downarrow}\hat{b}_{j \uparrow} - \hat{a}_{i \uparrow} \hat{b}_{j \downarrow}$. The Hamiltonian is described in two species of electrons, which live at different sub-lattices, A and B. The operator $\hat{a}_{i \sigma}$ creates an electron on lattice site $i$ with spin $\sigma$.

So I define the following mean-field parameter: $$\Delta_{ij}=\langle\hat{D}_{ij} \rangle$$

and thus I also have the following for $\hat{D}_{ij}^{\dagger}$, $\Delta_{ij}^{\dagger}=\langle \hat{D}_{ij} ^{\dagger}\rangle$.

Now as far as I understand mean-field approximations I have to do the following to rewrite the Hamiltonian:

$$ \hat{H}^1_I = g_1 c_2 \sum_{\langle i,j \rangle} \left( \Delta_{ij}+\delta D_{ij} \right) (\Delta_{ij}^{\dagger} + \delta D_{ij}^{\dagger})= \Delta_{ij} \ \Delta_{ij}^{\dagger} + \Delta_{ij} \ \delta D_{ij}^{\dagger} + \Delta_{ij}^{\dagger} \ \delta D_{ij}. $$

But now I don't understand what I should do, because I took at the higher orders of the small deviations of the mean-field. Should I put the $\hat{D}_{ij}$ and $\hat{D}_{ij}^{\dagger}$ back in? Because I want my hamiltonian to be an operator. Why would this be allowed and what should I do with the higher order terms of $\delta$?

I don't really understand what I do when I do a mean-field approximation. As far as I understand you somehow say that all interactions can be seen as one mean contribution. But why would I then get the operators back in again.

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  • $\begingroup$ Is $\hat{D}_{ij}=c_{i}^{\dagger} c_{j}^{}$ ? Further if you are working at zero temperature, and want to find approximate ground state wave function and energy, after doing mean field approximation, your hamiltonian is quadratic, which can be diagonalized in principle to get $|\text{ground state}\rangle$ and corresponding energy in terms of $\Delta_{ij}$, which needs to be fixed self consistently $\Delta_{ij}=\langle\text{ground state}|\hat{D}_{ij}|\text{ground state}\rangle$. $\endgroup$ – Sunyam Oct 28 '17 at 18:26
  • $\begingroup$ I have edited my question to explain more about $\hat{D}_{ij}$ $\endgroup$ – DeanTheMachine Oct 28 '17 at 18:36

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