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For the Bose-Hubbard model: $$H=-t\sum_{\langle i,j \rangle}(b_i^{\dagger}b_j+b_j^{\dagger}b_i)+K\sum_{i}(\hat{n}-n_0)^2$$ In the large-filling limit, we can replace $\hat{n}$ by $\hat{n}+n_0$, and also $\hat{n}_j$ by $-i\frac{\partial}{\partial \theta_j}$, after the which the on-site repulsion term in the Hamiltonian becomes the kinetic term for the phase angle.

However, for the hopping term, why can we simply replace $b_j^{\dagger}$ with $e^{i\theta_j}$? Basically, we are supposed to get the final phase-only quantum rotor model given as the following: $$H=-2t\sum_{\langle i,j \rangle}cos(\theta_i-\theta_j)+K\sum_j\left(-i\frac{\partial}{\partial \theta_j}\right)^2$$

Source: Eq.(14.3) from "advanced solid state physics 2e" by P. Philips.

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You can simply replace $b^{\dagger}_{j}$ with $e^{i\theta_{j}}$ because the Hamiltonian (real) remains the same under this change. It can seen as follows (assuming $\theta_{i}$ and $\theta_{j}$ commute) --

$b^{\dagger}_{i}b_{j} + b^{\dagger}_{j}b_{i} = e^{i\theta_{i}}e^{-i\theta_{j}} + e^{i\theta_{j}}e^{-i\theta_{i}}$ = 2$\mathbb{R}~( e^{i(\theta_{i} - \theta_{j})})$ = 2 $\cos(\theta_{i} - \theta_{j})$

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  • $\begingroup$ but how can be sure this $\theta\equiv -i\mathrm{log}b^{\dagger}$ is the same phase variable conjugate to particle number? $\endgroup$ – M. Zeng Aug 8 '18 at 21:43
  • $\begingroup$ I think it implies by restricting to the first branch sheet (so $\theta$ takes value only in 0 to $2\pi$). I don't see any other explanation. $\endgroup$ – R.G.J Aug 8 '18 at 21:56
  • $\begingroup$ I mean, even if it is restricted to $[0,2\pi)$, it still doesn't necessarily imply it's the same phase variable. $\endgroup$ – M. Zeng Aug 8 '18 at 22:00
  • $\begingroup$ also, another problem is it seems to assume the modulus of $b^{\dagger}$ is 1. $\endgroup$ – M. Zeng Aug 8 '18 at 22:03

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