only up to some transformation which is a symmetry of the action. Thus if $U$ is a symmetry of the action and $U^\dagger U = 1$, we can choose a twisted boundary condition for the generic field $\phi$ of the form $$\notag
\phi(x_i + L) = U \phi(x_i)
,$$ while maintaining the single valuedness of the action. Twisted boundary conditions are by no means new [@Gross:1982at; @Roberge:1986mm; @Wiese:1991ku; @Luscher:1996sc; @Bucarelli:1998mu; @Guagnelli:2003hw; @Kiskis:2002gr; @Kiskis:2003rd; @Kim:2002np; @Kim:2003xt], and there has been renewed interest in their utility at producing continuous hadron momentum [@Bedaque:2004kc; @deDivitiis:2004kq; @Sachrajda:2004mi; @Bedaque:2004ax; @Tiburzi:2005hg; @Flynn:2005in; @Guadagnoli:2005be; @Aarts:2006wt].

Obtaining hadronic states at continuous values of momentum does not solve the problem of coarse grained sampling of form factors, as the momentum transfer will still be quantized. In [@Tiburzi:2005hg] it was shown that matrix elements of flavor changing operators, however, can be accessed at continuous values of the momentum transfer $\bm{q}$, of the form $\bm{q} = 2 \pi \bm{n} / L + \delta \bm{\theta}/L$, where $\delta \bm{\theta}$ is a continuous parameter and is equal to the difference of twist angles of the flavors changed. As such flavor-changing operators have no self-contractions, the difference of twists is implemented only in the valence quark 