Proposition: A residual separation with identical records

Equal calibration records need not determine residual quantum states or residual source expectations. There is a qubit example with residual trace-distance gap $1/2$.

Proof. Compare the identity channel and computational dephasing, each with a single fixed reported outcome. Both record effects are $I$. On $|+⟩⟨+|$, their residuals are $|+⟩⟨+|$ and $I/2$. Their difference is $X/2$, with eigenvalues $1/2,-1/2$. The trace distance is $1/2$, and the residual observable $X$ has expectations one and zero. Tensor the example with any unchanged calibration table to obtain identical calibration records as well.