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In the Landauer formula, $G=(e^2/h)\mathrm{tr}(t^{†}t)$ when the flux-normalized transmission matrix $t$ includes every incident spin channel. The trace already sums both spin states if both are represented; multiplying this expression by another factor of two would count them twice.

For a Hamiltonian independent of spin with two identical, uncoupled spin sectors, one may instead calculate a single sector $t_0$ and write $G=(2e^2/h)\mathrm{tr}(t_0^{†}{t}_0)$. For example, one perfectly transmitted spatial mode gives $2e^2/h$ with both sectors present, and $e^2/h$ when only one sector propagates. With spin-orbit coupling, keep the full matrix even when its eigenchannels have no definite spin projection.