Passive beam-splitter photon statistics

The matrix $C$ is unitary because $C^{\ast }C=I_2$ follows from $\mathrm{\mid }t{\mathrm{\mid }}^2+\mathrm{\mid }r{\mathrm{\mid }}^2=1$. Set $S=C$, so $U=S,V=0$ satisfy $U{U}^{\ast }-V{V}^{\ast }=I_2$ and $U{V}^T=V{U}^T$. By [F2] the finite-mode Bogoliubov statement supplies an implementer $M$ with $M^{\ast }aM=Sa$. Multiplying by $M$ and $M^{\ast }$ gives $Ma{M}^{\ast }=S^{-1}a$; taking adjoints gives $M{a}^{\ast }{M}^{\ast }={\bar{S}}^{-1}{a}^{\ast }=S^T{a}^{\ast }=C^T{a}^{\ast }$ because $S=C$ is unitary. Thus the first creator transforms to $t{a}_1^{\ast }+r{a}_2^{\ast }$ and the second to $-\bar{r}{a}_1^{\ast }+\bar{t}{a}_2^{\ast }$. These identities hold on the invariant transported Schwartz space supplied by the implementer.

The transformed vacuum has ${\sum }_{j=1}^2\mathrm{\parallel }{a}_{j}M\mathrm{\Omega }{\mathrm{\parallel }}^2=0$ by [F2], so $a_{1}M\mathrm{\Omega }=a_{2}M\mathrm{\Omega }=0$. Expanding $M\mathrm{\Omega }$ in the complete occupation basis [F1], the equation $a_1\psi =0$ forces every coefficient with first occupation $n_1>0$ to vanish by the ladder formula; $a_2\psi =0$ then forces every coefficient with $n_2>0$ to vanish. Hence $M\mathrm{\Omega }=e^{i\varphi }\mathrm{\Omega }$ for some real $\varphi$. Replacing $M$ by $U_C=e^{-i\varphi }M$ preserves the conjugation identities and gives $U_C\mathrm{\Omega }=\mathrm{\Omega }$. Every occupation vector is a polynomial in the creators applied to $\mathrm{\Omega }$, so these identities show that $U_C$ maps the finite-particle span into itself. The inverse creator identities give the same inclusion for $U_C^{\ast }$; hence this span is invariant in both directions and the conjugation identities hold there.

For $\mathrm{\mid }n,0⟩=(a_1^{\ast }{)}^n\mathrm{\Omega }\mathrm{/}\sqrt{n!}$, repeated use of the creation identity in [F1], the conjugation identity from step 1.1, and $U_C\mathrm{\Omega }=\mathrm{\Omega }$ from step 2.1 give $U_C\mathrm{\mid }n,0⟩=(t{a}_1^{\ast }+r{a}_2^{\ast }{)}^n\mathrm{\Omega }\mathrm{/}\sqrt{n!}={\sum }_{k=0}^n\sqrt{\left(\genfrac{}{}{0px}{}{n}{k}\right)}{t}^k{r}^{n-k}\mathrm{\mid }k,n-k⟩$. The coefficient follows by choosing which $k$ of the $n$ commuting creation factors contribute $a_1^{\ast }$ and using $\mathrm{\mid }k,n-k⟩=(a_1^{\ast }{)}^k(a_2^{\ast }{)}^{n-k}\mathrm{\Omega }\mathrm{/}\sqrt{k!(n-k)!}$.

Orthogonality of the output occupation basis makes the joint ideal number-count probability the squared modulus of the coefficient in step 3.1. Since $\mathrm{\mid }t{\mathrm{\mid }}^2=\eta$ and $\mathrm{\mid }r{\mathrm{\mid }}^2=1-\eta$, for $0<\eta <1$ this is the displayed binomial law; at the endpoints the expansion has a single surviving basis vector. The declared number PVM and its Born rule [A1, F4] assign these probabilities.

By [F1, F3, F5], the families $\mathrm{\mid }m,n⟩$ and $\mathrm{\mid }m⟩\otimes \mathrm{\mid }n⟩$ are complete orthonormal families in ${\mathcal{F}}^{(2)}$ and $\mathcal{F}\widehat{\otimes }\mathcal{F}$, respectively. On finite sums define ${\mathcal{J}}_0\sum c_{mn}\mathrm{\mid }m,n⟩=\sum c_{mn}\mathrm{\mid }m⟩\otimes \mathrm{\mid }n⟩$; orthonormality makes its squared norm $\sum \mathrm{\mid }{c}_{mn}{\mathrm{\mid }}^2$, so it is well defined and inner-product preserving. For $\psi \in {\mathcal{F}}^{(2)}$, approximate by finite sums ${\psi }_j$ and define $\mathcal{J}\psi ={\lim }_j{\mathcal{J}}_0{\psi }_j$. The norm identity makes the images Cauchy in the complete product and the limit independent of the approximating sequence; it preserves linearity, norm and inner product. Its range is closed: if $\mathcal{J}{\psi }_j$ converges, the same norm identity makes ${\psi }_j$ Cauchy in the complete Fock space, so its limit maps to that range limit. The range contains the complete product family and hence is dense, so it is the whole product. This proves that $\mathcal{J}$ is unitary, uniquely fixed by the occupation map, with $\mathcal{J}\mathrm{\Omega }=\mathrm{\mid }0⟩\otimes \mathrm{\mid }0⟩$. The coherent input series converges by [F3] and this unitary identification. Apply the bounded unitary $U_C$ to its norm-convergent partial sums. Each output occupation coefficient has only one contributing input total $n=k+\mathrm{\ell }$; step 3.1 gives it as $e^{-\mathrm{\mid }\alpha {\mathrm{\mid }}^2\mathrm{/}2}{\alpha }^{k+\mathrm{\ell }}{t}^k{r}^{\mathrm{\ell }}\mathrm{/}\sqrt{k!\mathrm{\ell }!}$. The product of the two norm-convergent coherent series exists in the completed product by [F5]; approximating each factor by finite partial sums is legitimate since $\mathrm{\parallel }u\otimes v-u^{\mathrm{\prime }}\otimes v^{\mathrm{\prime }}\mathrm{\parallel }\le \mathrm{\parallel }u-u^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }+\mathrm{\parallel }{u}^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{\parallel }v-v^{\mathrm{\prime }}\mathrm{\parallel }$. Its product-basis coefficients are the same, because $\mathrm{\mid }t\alpha {\mathrm{\mid }}^2+\mathrm{\mid }r\alpha {\mathrm{\mid }}^2=\mathrm{\mid }\alpha {\mathrm{\mid }}^2$. Completeness of that basis therefore proves $\mathcal{J}{U}_C\mathrm{\mid }\alpha ,0⟩=\mathrm{\mid }t\alpha ⟩\otimes \mathrm{\mid }r\alpha ⟩$, retaining the exact vacuum phase and port convention.