Kraus Representation Limits

A normal channel between separable Hilbert spaces has a countable Kraus representation, but that statement supplies neither uniform convergence of finite truncations nor differentiable Kraus operators along a smooth time evolution. The canonical Kraus Representation article explains operator-sum physics. The countable representation theorem establishes the convergence and dilation result; here those results are used to diagnose approximation and dynamical limits.

Required background. Kraus Representations distinguishes strong operator completeness from trace-norm output convergence.

Kraus truncation removes a measurable amount of probability

Suppose

$$\Phi(X)=\sum_{j=1}^\infty K_jXK_j^\dagger,\qquad \sum_{j=1}^\infty K_j^\dagger K_j=I$$

with the completeness sum converging strongly. Set $R_N=\sum_{j\le N}K_j^\dagger K_j$ and $F_N(X)=\sum_{j\le N}K_jXK_j^\dagger$. For a state $\rho$, the omitted output is positive and has trace

$$\varepsilon_N(\rho)=\operatorname{Tr}[\rho(I-R_N)].$$

Consequently, $\|\Phi(\rho)-F_N(\rho)\|_1=\varepsilon_N(\rho)\to0$. The truncated operation is CP and trace nonincreasing; normalizing its output makes it a conditioned branch and generally destroys linearity.

A linear trace-preserving repair is

$$\widehat\Phi_N(X) =F_N(X)+\operatorname{Tr}[(I-R_N)X]\tau,$$

where $\tau$ is any fixed output state. The missing operation is CP because it is a positive functional followed by preparation of $\tau$. The repaired map is therefore a channel, and

$$\|\widehat\Phi_N(\rho)-\Phi(\rho)\|_1 \le2\varepsilon_N(\rho).$$

This bound has an experimental interpretation: two devices that agree on all retained outcomes can disagree only on the omitted branch. The factor two is the maximal trace-norm difference between two subnormalized states of equal trace.

Uniform accuracy follows if $\|I-R_N\|\to0$, since $\varepsilon_N(\rho)\le\|I-R_N\|$. Strong completeness alone gives only state-dependent convergence. A countably resolved projective measurement is the immediate counterexample: for each finite cutoff an input can be supported entirely on an omitted projector.

Smooth channels can need square-root Kraus amplitudes

Consider amplitude damping of a qubit, with decay rate $\gamma>0$:

$$K_0(t)=|0\rangle\langle0|+e^{-\gamma t/2}|1\rangle\langle1|, \qquad K_1(t)=\sqrt{1-e^{-\gamma t}}\,|0\rangle\langle1|.$$

The channel is analytic in $t$ as a linear superoperator: populations involve $e^{-\gamma t}$ and coherences involve $e^{-\gamma t/2}$. Yet $K_1(t)\sim\sqrt{\gamma t}\,|0\rangle\langle1|$ has no finite right derivative at zero.

This square root is natural. A jump with probability of order $t$ has an amplitude of order $\sqrt t$. It also prevents a misleading derivation of the generator. In a finite Kraus family with $K_0(0)=I$, $K_j(0)=0$ for $j>0$, and all $K_j$ differentiable in operator norm at zero, the latter terms contribute only $O(t^2)$ to $K_j(t)\rho K_j(t)^\dagger$. They cannot produce a nonzero first-order jump term.

The conclusion concerns such a smooth Kraus choice, not smoothness of the channel. Different Kraus representations also mix through environment isometries, so differentiating a particular list introduces representation choices absent from the superoperator.

There is no normalized maximally entangled vector for infinitely many levels

The finite-dimensional Choi construction uses $d^{-1/2}\sum_{j=1}^d|j,j\rangle$. On a countably infinite basis, a vector with equal nonzero Schmidt coefficients is not normalizable. The formal sum $\sum_j|j,j\rangle$ is not an input state for $\mathrm{id}\otimes\Phi$.

Choose instead a faithful reference density operator $\sigma=\sum_jp_j|j\rangle\langle j|$, where every $p_j>0$ and $\sum_jp_j=1$. Its normalized purification defines

$$|\Omega_\sigma\rangle=\sum_j\sqrt{p_j}|j,j\rangle,\qquad J_\sigma(\Phi) =(\mathrm{id}\otimes\Phi)(|\Omega_\sigma\rangle\langle\Omega_\sigma|).$$

This is a legitimate state with reference marginal $\sigma$. It determines a normal channel uniquely because its reference blocks satisfy

$$(\langle i|\otimes I)J_\sigma(\Phi)(|j\rangle\otimes I) =\sqrt{p_ip_j}\,\Phi(|i\rangle\langle j|).$$

All $p_j$ are nonzero, so every matrix-unit output can be recovered; density and channel boundedness then determine all trace-class outputs.

Recovery is poorly conditioned on high levels: dividing by $\sqrt{p_ip_j}$ has no uniform bound as $p_j\to0$. A small trace-norm error in one faithful-reference state therefore does not imply small unrestricted channel discrimination error. The input-cutoff example supplies a concrete family: its reference-state outputs converge, while its worst-case channel error stays maximal.

Exercises

Ancilla-assisted truncation. Prove the repaired Kraus error bound for a joint input $\rho_{AR}$ with an untouched reference.

Solution

The omitted true output is positive with trace $\varepsilon_N(\rho_A)$. The repaired branch is $\tau_B$ tensored with the positive reference operator $\operatorname{Tr}_A[((I-R_N)^{1/2}\otimes I)\rho_{AR}((I-R_N)^{1/2}\otimes I)]$. It has the same trace. Their difference has norm at most $2\varepsilon_N(\rho_A)$. The error is thus controlled by the input marginal’s omitted probability even with entangled probes.

Nonfaithful reference. What information is lost if one reference weight $p_k$ vanishes?

Solution

Every block involving index $k$ vanishes independently of the channel. In particular, the channel’s response to $|k\rangle\langle k|$ is invisible. The reference state then determines the channel only on the supported input subspace, not on the whole input algebra.

References

• Holevo, Alexander S. Quantum Systems, Channels, Information: A Mathematical Introduction, 2nd ed. De Gruyter, 2019. doi:10.1515/9783110642490. Infinite-dimensional channels, operator sums, and normal-state descriptions.
• Watrous, John. The Theory of Quantum Information. Cambridge University Press, 2018. Author’s text. Kraus freedom, Choi representations, and channel norms; the faithful-reference construction above specifies the infinite-dimensional replacement.