Discrete gauge cancellation in two-path interference ($W_C={\prod }_{(i,j)\in C}{U}_{ij}$)

A four-vertex model proves the essential gauge invariance without solving an Electron scattering problem. Use two directed paths $S\to U\to D$ and $S\to L\to D$. Assign an edge phase $a_{ij}$ and a complex unit link $U_{ij}=e^{i{a}_{ij}}$ to each directed edge, with $U_{ji}=U_{ij}^{-1}$. These links represent discretized magnetic line integrals multiplied by $q/\hslash$. Under a gauge transformation with arbitrary vertex phases ${\chi }_i$,Both path products acquire the same factor $e^{i({\chi }_D-{\chi }_S)}$. Their gauge-invariant ratio is the closed-loop holonomyEvery intermediate vertex phase cancels. For any fixed nonnegative real amplitudes $a,b$, the interference quantity $|a{P}_1+b{P}_2{|}^2$ is unchanged too. A gauge-dependent open-path phase consequently cannot be interpreted as an observable by itself.

For an exact example, measure phases in units of $\pi$ and order edges as $(SU,UD,SL,LD)$. Choose edge phases $(0,0,0,1/2)$ and vertex phases $({\chi }_S,{\chi }_U,{\chi }_L,{\chi }_D)=(1/3,-1/4,2/5,1/7)\pi$. The transformed edge phases areThe upper and lower sums become $-4\pi /21$ and $13\pi /42$; their difference remains $-\pi /2$. With amplitudes $a=1$, $b=1/2$, the intensity remains $5/4$. Sweeping one physical loop phase through $2\pi$ instead produces extrema $9/4$ and $1/4$, agreeing with the interferometric visibility example above.