Identical histograms can hide orthogonal states
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Erasing the signs leaves sample probabilities unchanged and changes the state. Let \(\chi_n=|c_{12}(n)|/\sqrt{48}\). Both states are normalized and
\[
|\psi_n|^2=|\chi_n|^2\quad\text{for every }n,
\qquad\langle\psi,\chi\rangle=0.
\]
The orthogonality follows from equal positive-support and negative-support energies: each contributes \(24\) before division by \(48\). A sample-basis probability histogram therefore cannot reconstruct the coherent state.