# Some question on the definition of flux in the projective construction?

+ 1 like - 0 dislike
1228 views

Here I have some confusing points about the definition of flux in the projective construction. For example, consider the same mean-field Hamiltonian in my previous question, and assume the $2\times 2$ complex matrix $\chi_{ij}$ has the form $\begin{pmatrix} t_{ij}& \Delta_{ij}\\ \Delta_{ij}^* & -t_{ij}^* \end{pmatrix}$. Consider a loop with $n$ links on the 2D lattice, the flux through this loop can be defined as the phase of $tr(\chi_1 \cdots\chi_n)$, where $\chi_i=\begin{pmatrix} t_i& \Delta_i\\ \Delta_i^* & -t_i^* \end{pmatrix},i=1,2,...,n$ representing the $i$ th link. And due to the identity $\chi_i^*=-\sigma_y\chi_i\sigma_y$, it's easy to show that $[tr(\chi_1 \cdots\chi_n)]^*=(-1)^ntr(\chi_1 \cdots\chi_n)$, which means that for an even loop, the flux is always $0$ or $\pi$; while for an odd loop, the flux is always $\pm\frac{\pi}{2}$. My questions are as follows:

(1)When $\chi_{ij}=\begin{pmatrix} t_{ij}& 0\\ 0 & -t_{ij}^* \end{pmatrix}$, the mean-field Hamiltonian can be rewritten as $H_{MF}=\sum(t_{ij}f_{i\sigma}^\dagger f_{j\sigma}+H.c.)$, if we define the flux through a loop $i\rightarrow j\rightarrow k\rightarrow \cdots\rightarrow l\rightarrow i$ as the phase of $t_{ij}t_{jk}\cdots t_{li}$, then the flux may take any real value in addition to the above only allowed values $0,\pi,\pm\frac{\pi}{2}$. So which definition of flux is correct?

(2)If $tr(\chi_1 \cdots\chi_n)=0$, how we define the flux(now the phase is highly uncertain)?

Thank you very much.

This post imported from StackExchange Physics at 2014-03-09 08:40 (UCT), posted by SE-user K-boy
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverf$\varnothing$owThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.