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A coherent pump has a Poisson distribution of photon numbers. Second-harmonic conversion couples a pump number state $|n⟩$ to a state with two fewer pump photons with a matrix element proportional to $\sqrt{n(n-1)}$. Different number components therefore evolve at different rates, and the pump becomes correlated with the harmonic mode. Its reduced state is generally neither pure nor exactly Poisson coherent. An undepleted classical-pump approximation suppresses precisely this backaction; it cannot prove that the exact finite-depletion quantum state stays coherent. Vacuum input or zero interaction are trivial exceptions.

Let $A=a^{†2}$, $B=a^2$ and $N=a^{†}a$. The oscillator algebra gives $[A,B]=-4N-2$, $[N,A]=2A$ and $[N,B]=-2B$. Substituting into the definitions proves closure:The factors depend on the stated normalization: $K_3$ has an extra factor one-half relative to $N+1/2$. The Robertson uncertainty relation consequently givesFor a coherent state with $n=|\alpha {|}^2$, normal ordering givesThus $⟨K_1⟩=\mathrm{Re}{\alpha }^2$ and $⟨K_2⟩=-\mathrm{Im}{\alpha }^2$. In either variance, the fourth-power and $n^2$ terms cancel against the square of the mean, leavingThe variance product is exactly $4⟨K_3{⟩}^2$, so the coherent state saturates the stated bound.

Amplitude-squared squeezing in either of these axes meansThe other variance must then rise sufficiently to respect the product inequality. A rotated field-squared quadrature can reveal squeezing in another direction; the criterion concerns the declared quadrature and reference, not merely a reduction in mean optical intensity.