= Chiral routing in an unequal-channel ballistic cross
{title2=$R_+=(h/e^2)(1/M-1/N),\quad R_-=0$}
For left-deflected injection in the left 1,top 2,right 3,bottom 4 geometry, use 1 to 2 with $M$ <modes>, 2 to 3 with $M$, 3 to 4 with $N$, and 4 to 1 with $M$ plus 4 to 3 with $N-M$. The zero-current probes give $V_2=V_1$ and $V_3=MV_1/N$, hence $R_{14,23}=(h/e^2)(1/M-1/N)$. Reversing the transmission matrix gives equal probe potentials and zero <bend resistance>. Fixed-lead resistance need not be even in field; reciprocity also exchanges current and <voltage> pairs.
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