Kraus operators in an identity-channel representation are scalar

For a unit vector $\psi$, define $P_{\psi }z=⟨z,\psi ⟩\psi$. The adjoint identity [F6] rewrites $K_j{P}_{\psi }{K}_j^{\ast }z=⟨z,K_j\psi ⟩K_j\psi$, so the assumed channel identity gives ${\sum }_j{K}_j{P}_{\psi }{K}_j^{\ast }=P_{\psi }$. If $u\perp \psi$, then $⟨P_{\psi }u,u⟩=0$ and $⟨K_j{P}_{\psi }{K}_j^{\ast }u,u⟩=\mathrm{\mid }⟨u,K_j\psi ⟩{\mathrm{\mid }}^2$. Thus ${\sum }_j\mathrm{\mid }⟨u,K_j\psi ⟩{\mathrm{\mid }}^2=0$, so each inner product vanishes. For a fixed $j$, write $K_j\psi =c\psi +v$ with $v\perp \psi$ using [F5]. The vanishing inner product with $u=v$ gives $0=⟨v,K_j\psi ⟩=\mathrm{\parallel }v{\mathrm{\parallel }}^2$, so $v=0$ and $K_j\psi \in \mathrm{span}\{\psi \}$.

If $\dim H=1$, each linear endomorphism is scalar, so the conclusion holds. If $\dim H\ge 2$, choose an orthonormal basis $(e_1,\dots ,e_n)$ by [F4]. Step 1.1 gives $K_j{e}_r=c_{jr}{e}_r$. For $r\mathrm{\ne }s$, applying step 1.1 to $(e_r+e_s)\mathrm{/}\sqrt{2}$ shows $c_{jr}{e}_r+c_{js}{e}_s\in \mathrm{span}\{e_r+e_s\}$, hence $c_{jr}=c_{js}$. Thus $c_{jr}$ is independent of $r$ for each $j$.

In either dimension case, each $K_j$ acts by one scalar on a basis of $H$, and therefore $K_j=c_j{I}_H$ on all of $H$.