on both the $H$ and $V$ polarizations of the signal beam. Hence, if the signal-beam input is in the arbitrarily-polarized single-photon state, $$|\psi_S\rangle = c_H |1\rangle_{H}|0\rangle_{V} + c_V |0\rangle_{H}|1\rangle_{V},
\vspace*{.075in}$$ where $|c_H|^2 + |c_V|^2 = 1$, then, except for a physically unimportant absolute phase, the signal-beam output will also be in the state $|\psi_S\rangle$.*

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