= Unequal-channel zero-field ballistic cross
{title2=$R_{14,23}(0)=-h/[e^2(N-M)]$}
Number the leads left 1,top 2,right 3,bottom 4. Let narrow leads 1,2 have $M$ spin-resolved <modes> and wide leads 3,4 have $N>M$. Straight-through injection sends $M$ <modes> between 1 and 3 and between 2 and 4; each excess $N-M$ <mode> connects 3 and 4. With current 1 to 4, $V_4=0$ and zero current at 2,3, $V_2=0$, $V_3=MV_1/N$ and $I=(e^2/h)M(N-M)V_1/N$. Thus $R_{14,23}=-h/[e^2(N-M)]$. At $N=M$ this ideal network disconnects the current source from its drain.
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