= Solution
Denote the normalized initial field by $|F\rangle$, and its shifted packets by
$$
|F_-\rangle=\frac1{\sqrt k}\sum_{n=1}^k|n\rangle,\qquad
|F_+\rangle=\frac1{\sqrt k}\sum_{n=3}^{k+2}|n\rangle.
$$
Here the photon-number kets are <orthogonal> <Fock states>. Counting shared number labels gives
$$
\langle F|F_-\rangle=\langle F|F_+\rangle=r_k=\frac{k-1}{k},\qquad
\langle F_-|F_+\rangle=s_k=\frac{\max(k-2,0)}k.
$$
After the interaction, linearity gives the joint state
$$
|\Omega\rangle=\frac1{\sqrt2}\left[
|0\rangle(\alpha|F\rangle+\beta|F_+\rangle)
+|1\rangle(\alpha|F_-\rangle-\beta|F\rangle)\right].
$$
Take the <partial trace> over the field. Put $d=|\alpha|^2-|\beta|^2$ and $c=\alpha\beta^*+\alpha^*\beta$. In the ordered <qubit> basis $(|0\rangle,|1\rangle)$, the reduced <density matrix> is
$$
\boxed{\rho=\frac12\begin{pmatrix}
1+r_kc&r_kd+s_k\alpha^*\beta-\alpha\beta^*\\
r_kd+s_k\alpha\beta^*-\alpha^*\beta&1-r_kc
\end{pmatrix}.}
$$
For example, its upper off-diagonal entry is half the overlap of the field accompanying $|1\rangle$ with the field accompanying $|0\rangle$, fixing the conjugation order. The <trace> is one, and positivity follows because this is the <partial trace> of a normalized <pure state>. For $k\ge2$, $s_k=1-2/k$; at $k=1$ it is zero, not the continuation $-1$. This is the <photon-number reference for a Hadamard gate> channel, not an actual coherent-state packet.
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