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--- kraus-operator [May 22, 2026 at 23:49] – Ivan Janevski
+++ kraus-operator [May 25, 2026 at 13:55] (current) – external edit 127.0.0.1
@@ Line 1: / Line 1: @@
 # Kraus operator
-**Kraus operators** are used to decompose into operator form a quantum channel $\mathcal{E}: \rho \rightarrow \mathcal{E}(\rho)$ which is a CPTP map i.e. it sends density matrices into density matrices.
+**Kraus operators** $\{K_k\}$ are a set of matrices used to represent a quantum channel — a completely positive trace-preserving (CPTP) map $\mathcal{E}$ that sends density matrices to density matrices. Any physically valid quantum channel can be expressed in Kraus form.
 $$\mathcal{E}(\rho) = \sum_k K_k\rho K_k^\dagger$$
+The Kraus operators must satisfy the completeness relation $\sum_k K_k^\dagger K_k = I$, which ensures the map is trace-preserving (i.e., $\text{tr}(\mathcal{E}(\rho)) = \text{tr}(\rho) = 1$). Any set of matrices satisfying this condition defines a valid quantum channel.
-To derive the Lindbald master equation you need the following
+## Examples
-$$K_1 = \sqrt{dt}L$$
+For a single qubit undergoing bit-flip noise with probability $p$, the Kraus operators are $K_0 = \sqrt{1-p}\,I$ (no error) and $K_1 = \sqrt{p}\,X$ (bit flip). For depolarizing noise with probability $p$, there are four Kraus operators: $K_0 = \sqrt{1-p}\,I$, $K_1 = \sqrt{p/3}\,X$, $K_2 = \sqrt{p/3}\,Y$, $K_3 = \sqrt{p/3}\,Z$.
-$$K_0 = I - \frac{i}{\hbar}H dt - \frac{1}{2}$$
+## Deriving the Lindblad equation
+Kraus operators are the link between the [[lindbald-equation|Lindblad master equation]] and the operator-sum representation. For an infinitesimal time step $dt$ with Lindblad jump operator $L$ and Hamiltonian $H$, the Kraus operators are:
+$$K_0 = I - \frac{i}{\hbar}H\,dt - \frac{1}{2}L^\dagger L\,dt, \qquad K_1 = \sqrt{dt}\,L$$
+Substituting into $\rho(t+dt) = K_0\rho K_0^\dagger + K_1\rho K_1^\dagger$ and expanding to first order in $dt$ recovers the Lindblad equation.
-You then use:
-$$\rho(t + dt) = K_0\rho K_0^\dagger + K_1\rrho K_1^\dagger$$