# Absence of fermion sign problem at half filling

It is said that in the hubbard model, at $$\mu = 0$$ , there is no sign problem. I do not see why $$\mu = 0$$ is necessary in the above argument? For the hubbard hamiltonian, \begin{aligned} H=-t & \sum_{\langle i j\rangle, \sigma}\left(c_{i \sigma}^{\dagger} c_{j \sigma}+c_{i \sigma}^{\dagger} c_{j \sigma}\right)-\mu \sum_{i}\left(n_{i+}+n_{i-}\right) \\ &+U \sum_{i}\left(n_{i+}-\frac{1}{2}\right)\left(n_{i-}-\frac{1}{2}\right) \end{aligned} $$\mu = 0$$ corresponds to $$n = n_+ + n_{-}=1$$, where $$n$$ is the average occupancy. Under particle-hole transformation, $$c_{\mathrm{i} \sigma}^{\dagger} \rightarrow d_{\mathrm{i} \sigma}=(-1)^{\mathrm{i}} c_{\mathrm{i} \sigma}^{\dagger}, c_{\mathrm{i} \sigma} \rightarrow d_{\mathrm{i} \sigma}^{\dagger}=(-1)^{\mathrm{i}} c_{\mathrm{i} \sigma}$$

\begin{aligned} n_{\mathrm{i} \sigma} \equiv c_{\mathrm{i} \sigma}^{\dagger} c_{\mathrm{i} \sigma}=1-d_{\mathrm{i} \sigma}^{\dagger} d_{\mathrm{i} \sigma} \equiv 1-\tilde{n}_{\mathrm{i} \sigma}\end{aligned} In the determinant monte-carlo procedure, \begin{aligned} \operatorname{det} \mathrm{O}^{+} & =\underset{\{n\}}{\mathcal{T}} r \prod_{\ell} e^{-\Delta \tau \mathbf{K}_{+}} e^{-\lambda \sum_{i} s_{i}(\ell) n_{i+}} \\ & =\underset{\{\tilde{n}\}}{\mathcal{T}} r \prod_{\ell} e^{-\Delta \tau \tilde{\mathbf{K}}_{+}} e^{\lambda \sum_{i} s_{i}(\ell) \tilde{n}_{\mathrm{i-}}} e^{-\lambda \sum_{i} s_{i}(\ell)} \\ & =e^{-\lambda \sum_{i \ell} s_{i}(\ell)} \operatorname{det} O^{-}\end{aligned} so that the product $$\operatorname{det} \mathrm{O}^{+}\times\operatorname{det} \mathrm{O}^{-}$$ is positive. The above relation concerning the determinant is true for any value of $$n(\mu \neq 0)$$, to my understanding.

I do not understand why when $$\mu \neq 0$$, the above argument is said to be false.