kraus-operator
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| kraus-operator [May 22, 2026 at 23:47] – Ivan Janevski | kraus-operator [May 25, 2026 at 13:55] (current) – external edit 127.0.0.1 | ||
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| # Kraus operator | # Kraus operator | ||
| - | **Kraus | + | **Kraus |
| $$\mathcal{E}(\rho) = \sum_k K_k\rho K_k^\dagger$$ | $$\mathcal{E}(\rho) = \sum_k K_k\rho K_k^\dagger$$ | ||
| + | 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 |
| - | $$K_0 = I - \frac{i}{\hbar}H dt - \frac{1}{2}$$ | + | |
| + | ## 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 | ||
| + | |||
| + | $$K_0 = I - \frac{i}{\hbar}H\,dt - \frac{1}{2}L^\dagger L\,dt, \qquad K_1 = \sqrt{dt}\, | ||
| + | |||
| + | Substituting into $\rho(t+dt) = K_0\rho K_0^\dagger + K_1\rho K_1^\dagger$ and expanding to first order in $dt$ recovers the Lindblad equation. | ||
| - | You then use: | ||
| - | $$\rho(t + dt) = K_0\rho K_0^\dagger + K_1\rrho K_1^\dagger$$ | ||
kraus-operator.1779493638.txt.gz · Last modified: by Ivan Janevski
