Finite square-root measurement packing bound for cq states

For a realized codebook, set ${\mathrm{\Gamma }}_m=\mathrm{\Pi }{\mathrm{\Pi }}_{X_m}\mathrm{\Pi }$ and $S={\sum }_m{\mathrm{\Gamma }}_m$. Since each factor is a projection, $0⪯{\mathrm{\Gamma }}_m⪯I$. Define ${\mathrm{\Lambda }}_m=S^{-1\mathrm{/}2}{\mathrm{\Gamma }}_m{S}^{-1\mathrm{/}2}$ on $\mathrm{supp}S$ and zero on its kernel; add ${\mathrm{\Lambda }}_0=I-{\sum }_m{\mathrm{\Lambda }}_m$. By [F4], all ${\mathrm{\Lambda }}_m$ are nonnegative and ${\sum }_m{\mathrm{\Lambda }}_m$ is the support projection of $S$, so $({\mathrm{\Lambda }}_0,{\mathrm{\Lambda }}_1,\dots ,{\mathrm{\Lambda }}_M)$ is a POVM.

For $j\mathrm{\ne }m$, independence of $X_j$ and $X_m$ gives $\mathbb{E}\mathrm{tr}({\mathrm{\Gamma }}_j{\sigma }_{X_m})=\mathrm{tr}(\mathrm{\Pi }\mathbb{E}[{\mathrm{\Pi }}_{X_j}]\mathrm{\Pi }\bar{\sigma })=\mathrm{tr}((\mathrm{\Pi }\mathbb{E}[{\mathrm{\Pi }}_{X_j}]\mathrm{\Pi })(\mathrm{\Pi }\bar{\sigma }\mathrm{\Pi }))\le D^{-1}\mathrm{tr}(\mathrm{\Pi }\mathbb{E}[{\mathrm{\Pi }}_{X_j}]\mathrm{\Pi })\le d\mathrm{/}D\mathrm{.}$ The first equality factors the finite independent expectation into the product of its two means and distributes the finite trace sum using [F8]; the second uses [F7] and the support of the first factor. For the first inequality, $D^{-1}\mathrm{\Pi }-\mathrm{\Pi }\bar{\sigma }\mathrm{\Pi }$ is nonnegative, so [F6] applied to its product with the nonnegative first factor gives the trace bound. For the last inequality, $I-\mathrm{\Pi }$ is nonnegative and [F6]–[F7] give $\mathrm{tr}(\mathrm{\Pi }{\mathrm{\Pi }}_x\mathrm{\Pi })=\mathrm{tr}({\mathrm{\Pi }}_x\mathrm{\Pi })\le \mathrm{tr}({\mathrm{\Pi }}_x)\le d$.

For the transmitted codeword $X_m$, [F3] applied with $S_0={\mathrm{\Gamma }}_m$ and $T_0={\sum }_{j\mathrm{\ne }m}{\mathrm{\Gamma }}_j$ gives the operator inequality $I-{\mathrm{\Lambda }}_m⪯2(I-{\mathrm{\Gamma }}_m)+4{\sum }_{j\mathrm{\ne }m}{\mathrm{\Gamma }}_j$. Since ${\sigma }_{X_m}$ is nonnegative by [F1], [F6] applied to its product with the nonnegative difference of the right and left sides preserves the order after taking trace, and trace linearity [F8] expands the resulting right side to $\mathrm{tr}((I-{\mathrm{\Lambda }}_m){\sigma }_{X_m})\le 2\mathrm{tr}((I-{\mathrm{\Gamma }}_m){\sigma }_{X_m})+4{\sum }_{j\mathrm{\ne }m}\mathrm{tr}({\mathrm{\Gamma }}_j{\sigma }_{X_m})\mathrm{.}$ The gentle bound [F2] for the global projection gives $\mathrm{\parallel }\mathrm{\Pi }{\sigma }_{X_m}\mathrm{\Pi }-{\sigma }_{X_m}{\mathrm{\parallel }}_1\le 2\sqrt{u}$. Since $\mathrm{\parallel }{\mathrm{\Pi }}_{X_m}\mathrm{\parallel }\le 1$, trace-norm duality [F5], cyclicity [F7], and ${\mathrm{\Gamma }}_m=\mathrm{\Pi }{\mathrm{\Pi }}_{X_m}\mathrm{\Pi }$ then give $\mathrm{tr}({\mathrm{\Gamma }}_m{\sigma }_{X_m})=\mathrm{tr}({\mathrm{\Pi }}_{X_m}\mathrm{\Pi }{\sigma }_{X_m}\mathrm{\Pi })\ge \mathrm{tr}({\mathrm{\Pi }}_{X_m}{\sigma }_{X_m})-2\sqrt{u}\ge 1-u-2\sqrt{u}\mathrm{.}$

Average step 2.1 over all messages and codebooks. Its first term is at most $2(u+2\sqrt{u})$ by the acceptance estimate there, and each of the $M-1$ competing terms is at most $4d\mathrm{/}D$ by step 1.2. Thus the expected average error has the claimed bound. Because the codebook sample space is finite, at least one realization has error no greater than its expectation.