Beam splitter scattering matrix ($\mathbf{b}=S\mathbf{a}$)

For a linear, fixed-frequency beam splitter, choose power-normalized complex amplitudes $a_1,a_2$ at its input optical ports and $b_1,b_2$ at its output optical ports: $|a_j{|}^2$ and $|b_j{|}^2$ are powers. The scalar description treats one specified electromagnetic polarization and matching transverse profiles; other fields require additional components. With the geometrical ordering in the diagram,Here $t,t^{\prime }$ are transmission complex amplitudes per unit input and $r,r^{\prime }$ are reflection complex amplitudes per unit input. In this power normalization, a single input gives $T=|t{|}^2$ and $R=|r{|}^2$. Raw electric field amplitudes in different media generally need additional power-normalization factors.

If these two output optical ports account for all outgoing power and the device has no loss or gain, $S$ is a unitary matrix:The dagger is the Hermitian transpose and a star is complex conjugation. To prove the condition, substitute $\mathbf{b}=S\mathbf{a}$ into conservation of total power for every input vector: ${\mathbf{a}}^{†}(S^{†}S-I)\mathbf{a}=0$. Inputs $(1,0)$ and $(0,1)$ fix the diagonal entries; $(1,1)$ and $(1,i)$ fix the real and imaginary parts of the off-diagonal entry. A lossy two-output description instead has $S^{†}S\le I$ for a passive device.^[ref]

One consistent ideal symmetric 50:50 choice isEach reflection carries $+\pi /2$ relative wave phase in this convention. The columns have unit vector norm and zero inner product, directly verifying that $S$ is a unitary matrix. Choosing a different phase convention for complex amplitudes at the optical ports replaces $S$ by $D_{\mathrm{out}}S{D}_{\mathrm{in}}^{-1}$, where the $D$ are diagonal matrices whose entries have modulus one; the input complex amplitudes must be transformed consistently. Observable powers are unchanged. The displayed $i$ is a convention for this ideal model, not a universal reflection rule for every coating or incidence geometry.

Since $HZH=X$, unitary conjugation gives $H{e}^{-i\beta Z/2}H=e^{-i\beta X/2}$. Multiplying by the two outer $R_z(\alpha )$ rotations givesThe case $\alpha =\beta =\pi /2$ equals $-iH$. Equal output probabilities for one input do not by themselves identify an optical component with the Hadamard gate; relative wave phases also matter.

For two identical bosonic input populations of $N$ in a balanced two-mode mixer, only even occupancies occur at either output. Expanding $[(b_1^{†}{)}^2-(b_2^{†}{)}^2{]}^N$ and retaining normalized number-state factorials gives the displayed law. Its sum is one by the generating function $(1-4z{)}^{-1/2}(1-4z{)}^{-1/2}=(1-4z{)}^{-1}$.