# Question about the derivation from 2d Bose Hubbard to Quantum Rotor model

It is well-known that the Bose-Hubbard model (BH) and Quantum Rotor model (QR) can be mapped to each other under certain constraints (here I'm focusing on the 2d case). I was trying to get this mapping using a path integral representation of the partition function of the BH hamiltonian: $$H_{b.h}= -t \sum_{\langle i, j \rangle} \left( b^{\dagger}_i b_j+b^{\dagger}_j b_i\right) + \frac{U}{2}\sum_i \hat{n}_i \left( \hat{n}_i - 1 \right)-\mu \sum_i \hat{n}_i$$ it is clear that one can rewrite the interaction and the chemical potential part into a form like: $$\frac{U}{2}\sum_i \left( \hat{n}_i - \frac{\mu}{U} - \frac{1}{2} \right)^2-\frac{U}{2}\left( \frac{\mu}{U} + \frac{1}{2} \right)^2$$ and we can see that the average density (in the case without hopping term) would just be $$n_{ave}=\frac{\mu}{U} + \frac{1}{2}$$ however, when we try to write the partition function $Z=$Tr$\left[ e^{-\beta H_{b.h}} \right]$ in terms of path integral, we need first rewrite the interaction part as \begin{align} \frac{U}{2}\sum_i \hat{n}_i \left( \hat{n}_i - 1 \right) &= \frac{U}{2}\sum_i b^{\dagger}_i b_i b^{\dagger}_i b_i-b^{\dagger}_i b_i \\ &= \frac{U}{2}\sum_i b^{\dagger}_i b^{\dagger}_i b_i b_i \end{align} and in the path integral formalism, we would get: \begin{align} Z &= \int D\psi^* D\psi\ e^{-S} \\ S &= \int_{0}^{\tau} d\tau\ \left( \sum_{i}\psi^*_i \partial_{\tau}\psi\ +\frac{U}{2}\psi^*_i \psi^*_i \psi_i \psi_i-\mu \psi^*_i\psi_i-t\sum_{\langle i, j\rangle}\psi^*_i\psi^*_j+\psi^*_j\psi_i \right) \\ &= \int_{0}^{\tau} d\tau\ \left( \sum_{i}\psi^*_i \partial_{\tau}\psi\ +\frac{U}{2}( \psi^*_i \psi_i-\frac{\mu}{U})^2-\frac{\mu^2}{2U}-t\sum_{\langle i, j\rangle}\psi^*_i\psi_j+\psi^*_j\psi_i \right) \end{align} now if I use the method mentioned in the paper by M. P. A Fisher and G. Grinstein, which is to expand the bosonic fielld operator around it's most probable amplitude $n_0 = \mu/U$, i.e. $$\psi_i = \left( \sqrt{n_0}+\delta \psi_i\right)e^{i\theta_i}$$ and integrate out the amplitude fluctuation field $\delta\psi_i$, under certain approximation, I would get \begin{align} Z &= \int D\theta e^{-S[\theta]} \\ S[\theta] &= \int_{0}^{\beta}d\tau \sum_i i\ n_0 \partial_\tau \theta_i+\frac{1}{2U}\left( \partial_\tau \theta_i \right)^2-J\sum_{\langle i, j\rangle} \cos\left( \theta_i-\theta_j \right) \end{align} so I can say if the $n_0=\mu/U$ is an integer, the first imaginary berry phase term would diappear due to the periodic boundary condition of $\theta$, and we essentially arive at the partition function of a QR model.

My question is, it is obvious that the average density of the original BH hamiltonian $n_{ave}$ is different from the expanding amplitude of the bosonic filed $n_0$ by a factor of $1/2$, and from my derivation, when $n_0 \in \mathbb{Z}$ ($n_{ave}$ not an integer) the mapping is "exact", however, in the literature, people tends to say that when the average dentsity of BH model ($n_{ave}$ I assume?) is an integer, there is the mapping between two models. So I wish someone who is familiar in this area can tell me what is the true result here? and is the appraoch I'm using here is correct? Any other approach to show the mapping between the two models is also welcome.