Trace Preservation

Complete positivity makes a map physically allowed; trace preservation makes it deterministic — it guarantees that a normalized state stays normalized, so no probability leaks away. Together they form the CPTP condition that defines a quantum channel. This lesson shows exactly where trace preservation lives in each representation and how to check it.

The condition on the map

A channel $\mathcal{E}$ is trace preserving if

\operatorname{tr}\,\mathcal{E}(\rho) = \operatorname{tr}\rho \qquad \text{for all } \rho.

By linearity it is enough to demand $\operatorname{tr}\,\mathcal{E}(X) = \operatorname{tr}X$ on a basis of operators $X$ , since every operator is a linear combination of density operators.

In the Kraus picture: the completeness relation

Write $\mathcal{E}(\rho) = \sum_k K_k\rho K_k^\dagger$ . Using cyclicity of the trace,

\operatorname{tr}\,\mathcal{E}(\rho) = \sum_k \operatorname{tr}\big(K_k \rho K_k^\dagger\big) = \sum_k \operatorname{tr}\big(K_k^\dagger K_k\,\rho\big) = \operatorname{tr}\!\Big[\Big(\textstyle\sum_k K_k^\dagger K_k\Big)\rho\Big].

This equals $\operatorname{tr}\rho = \operatorname{tr}(\mathbb{1}\,\rho)$ for all $\rho$ precisely when the bracket is the identity. Hence

\boxed{\;\sum_k K_k^\dagger K_k = \mathbb{1}\;}

is necessary and sufficient for trace preservation. This is the completeness relation, the single equation you check to certify that a CP map is a genuine channel. Note that $\sum_k K_k K_k^\dagger$ (the other order) need not be the identity — when it is, the channel is additionally unital ( $\mathcal{E}(\mathbb{1}) = \mathbb{1}$ ), a strictly stronger property.

In the Choi picture: a partial trace

The same condition reads, on the Choi operator $J(\mathcal{E}) = \sum_{ij}\mathcal{E}(|i\rangle\langle j|)\otimes|i\rangle\langle j|$ ,

\operatorname{tr}_B\, J(\mathcal{E}) = \mathbb{1}_A,

i.e. tracing out the output system of the Choi operator leaves the identity on the input system. So "CPTP" is the pair of conditions $J(\mathcal{E}) \succeq 0$ (complete positivity) and $\operatorname{tr}_B J = \mathbb{1}_A$ (trace preservation) — two clean tests on one operator.

Trace non-increasing maps and instruments

If instead $\sum_k K_k^\dagger K_k \preceq \mathbb{1}$ , the map is trace non-increasing: $\operatorname{tr}\,\mathcal{E}(\rho) \le \operatorname{tr}\rho$ . The "missing" trace is the probability that some measurement outcome did not occur. A collection of such maps $\{\mathcal{E}_m\}$ with $\sum_m \mathcal{E}_m$ trace preserving is a quantum instrument: each $\mathcal{E}_m$ is the (unnormalized) post-measurement update for outcome $m$ , and the normalization $\operatorname{tr}\,\mathcal{E}_m(\rho)$ is the probability of $m$ .

Try it

The amplitude-damping channel (next lesson) has Kraus operators

K_0 = \begin{pmatrix} 1 & 0 \\ 0 & \sqrt{1-\gamma} \end{pmatrix}, \qquad K_1 = \begin{pmatrix} 0 & \sqrt{\gamma} \\ 0 & 0 \end{pmatrix},

with $\gamma = 0.3$ . Verify trace preservation by computing $M = K_0^\dagger K_0 + K_1^\dagger K_1$ and returning the Frobenius distance $\big\lVert M - \mathbb{1}\big\rVert_F = \sqrt{\sum_{ij} |M_{ij} - \delta_{ij}|^2}$ . A genuine channel gives exactly $0$ .

Run your code to see the quantum state.