スクイーズド状態
⟨ a ^ 2 ⟩ = I m [ α cosh  2 r − α ∗ e i φ sinh  2 r ] {\displaystyle \langle {\hat {a}}_{2}\rangle =Im[\alpha \cosh 2r-\alpha ^{*}e^{i\phi }\sinh 2r]}
直交位相振幅の分散については、 φ = 0 {\displaystyle \phi =0} のとき、
⟨ ( Δ a ^ 1 ) 2 ⟩ = e − 2 r 4 {\displaystyle \langle (\Delta {\hat {a}}_{1})^{2}\rangle ={\frac {e^{-2r}}{4}}}
⟨ ( Δ a ^ 2 ) 2 ⟩ = e 2 r 4 {\displaystyle \langle (\Delta {\hat {a}}_{2})^{2}\rangle ={\frac {e^{2r}}{4}}}
また φ = π / 2 {\displaystyle \phi =\pi /2} のときは、
⟨ ( Δ a ^ 1 ) 2 ⟩ = e 2 r 4 {\displaystyle \langle (\Delta {\hat {a}}_{1})^{2}\rangle ={\frac {e^{2r}}{4}}}