Differences

This shows you the differences between two versions of the page.

--- kraus-operator [May 25, 2026 at 13:55] – external edit 127.0.0.1
+++ kraus-operator [June 13, 2026 at 03:13] (current) – external edit 127.0.0.1
@@ Line 4: / Line 4: @@
 $$\mathcal{E}(\rho) = \sum_k K_k\rho K_k^\dagger$$
-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.
+The Kraus operators must satisfy the **completeness relation** $\sum_k K_k^\dagger K_k = I$, which ensures the map is trace-preserving ($\text{tr}(\mathcal{E}(\rho)) = \text{tr}(\rho) = 1$). Any set of matrices satisfying this condition defines a valid quantum channel. The decomposition into Kraus operators is not unique — the same channel can be expressed with different sets of $K_k$.
 ## Examples
-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: $K_0 = \sqrt{1-p}\,I$, $K_1 = \sqrt{p/3}\,X$, $K_2 = \sqrt{p/3}\,Y$, $K_3 = \sqrt{p/3}\,Z$.
+The six channels below all act on a single qubit. For each one the Kraus operators are stated, the completeness relation is verified, and the channel action $\mathcal{E}(\rho)$ is worked out in matrix form.
+### 1. Bit flip channel
+With probability $1-p$ nothing happens; with probability $p$ the $X$ gate is applied, swapping $\lvert 0\rangle \leftrightarrow \lvert 1\rangle$. Two Kraus operators:
+$$K_0 = \sqrt{1-p}\,I = \sqrt{1-p}\begin{pmatrix}1&0\\0&1\end{pmatrix}, \qquad
+K_1 = \sqrt{p}\,X = \sqrt{p}\begin{pmatrix}0&1\\1&0\end{pmatrix}$$
+Completeness (using $X^\dagger X = I$):
+$$K_0^\dagger K_0 + K_1^\dagger K_1 = (1-p)I + pX^\dagger X = (1-p)I + pI = I \quad\checkmark$$
+Channel action. First compute $X\rho X$:
+$$X\rho X = \begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}\rho_{00}&\rho_{01}\\\rho_{10}&\rho_{11}\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix} = \begin{pmatrix}\rho_{11}&\rho_{10}\\\rho_{01}&\rho_{00}\end{pmatrix}$$
+Then:
+$$\mathcal{E}(\rho) = (1-p)\rho + pX\rho X = \begin{pmatrix}(1-p)\rho_{00}+p\rho_{11} & (1-p)\rho_{01}+p\rho_{10} \\ (1-p)\rho_{10}+p\rho_{01} & (1-p)\rho_{11}+p\rho_{00}\end{pmatrix}$$
+The populations $\rho_{00}$ and $\rho_{11}$ mix toward each other. At $p=1/2$ both reach $1/2$ — the state is maximally mixed in the computational basis.
+### 2. Phase flip channel
+With probability $p$ the $Z$ gate is applied, which maps $\lvert 0\rangle \to \lvert 0\rangle$ and $\lvert 1\rangle \to -\lvert 1\rangle$. This adds a random $\pi$ phase to the $\lvert 1\rangle$ state without changing populations:
+$$K_0 = \sqrt{1-p}\,I, \qquad
+K_1 = \sqrt{p}\,Z = \sqrt{p}\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$
+Completeness (using $Z^2 = I$):
+$$K_0^\dagger K_0 + K_1^\dagger K_1 = (1-p)I + pZ^2 = I \quad\checkmark$$
+Channel action. First compute $Z\rho Z$:
+$$Z\rho Z = \begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}\rho_{00}&\rho_{01}\\\rho_{10}&\rho_{11}\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix} = \begin{pmatrix}\rho_{00}&-\rho_{01}\\-\rho_{10}&\rho_{11}\end{pmatrix}$$
+Then:
+$$\mathcal{E}(\rho) = (1-p)\rho + pZ\rho Z = \begin{pmatrix}\rho_{00} & (1-2p)\rho_{01} \\ (1-2p)\rho_{10} & \rho_{11}\end{pmatrix}$$
+Populations are completely unchanged. Coherences are suppressed by the factor $(1-2p)$: at $p=0$ nothing changes, at $p=1/2$ coherences are fully destroyed, at $p=1$ they are restored with opposite sign. This is the discrete-time model of $T_2$ dephasing.
+### 3. Bit-phase flip channel
+With probability $p$ the $Y$ gate is applied. Since $Y = iXZ$, this simultaneously flips the bit and applies a phase — hence the name. Two Kraus operators:
+$$K_0 = \sqrt{1-p}\,I, \qquad
+K_1 = \sqrt{p}\,Y = \sqrt{p}\begin{pmatrix}0&-i\\i&0\end{pmatrix}$$
+Completeness (using $Y^\dagger Y = I$): $(1-p)I + pI = I$ ✓
+Channel action. First compute $Y\rho Y$:
+$$Y\rho Y = \begin{pmatrix}0&-i\\i&0\end{pmatrix}\begin{pmatrix}\rho_{00}&\rho_{01}\\\rho_{10}&\rho_{11}\end{pmatrix}\begin{pmatrix}0&i\\-i&0\end{pmatrix} = \begin{pmatrix}\rho_{11}&-\rho_{10}\\-\rho_{01}&\rho_{00}\end{pmatrix}$$
+Then:
+$$\mathcal{E}(\rho) = (1-p)\rho + pY\rho Y = \begin{pmatrix}(1-p)\rho_{00}+p\rho_{11} & (1-p)\rho_{01}-p\rho_{10} \\ (1-p)\rho_{10}-p\rho_{01} & (1-p)\rho_{11}+p\rho_{00}\end{pmatrix}$$
+Populations mix as in the bit-flip channel. Coherences pick up an extra sign flip from the $Y$ phase: the $(1,2)$ element mixes $\rho_{01}$ and $-\rho_{10}$ rather than $+\rho_{10}$. At $p=1/2$ coherences are destroyed regardless.
+### 4. Depolarizing channel
+With probability $1-p$ nothing happens; with probability $p/3$ each, one of $X$, $Y$, $Z$ is applied. The channel is symmetric under all single-qubit unitaries. Four Kraus operators:
+$$K_0 = \sqrt{1-p}\,I, \qquad
+K_1 = \sqrt{\frac{p}{3}}\,X, \qquad
+K_2 = \sqrt{\frac{p}{3}}\,Y, \qquad
+K_3 = \sqrt{\frac{p}{3}}\,Z$$
+Completeness (using $X^2 = Y^2 = Z^2 = I$):
+$$\sum_k K_k^\dagger K_k = (1-p)I + \frac{p}{3}(X^2 + Y^2 + Z^2) = (1-p)I + pI = I \quad\checkmark$$
+Channel action, using the Pauli identity $X\rho X + Y\rho Y + Z\rho Z = 2\,\text{tr}(\rho)\,I - \rho = 2I - \rho$ (holds for any $2\times 2$ density matrix):
+$$\mathcal{E}(\rho) = (1-p)\rho + \frac{p}{3}(X\rho X + Y\rho Y + Z\rho Z) = (1-p)\rho + \frac{p}{3}(2I - \rho)$$
+$$\mathcal{E}(\rho) = \left(1 - \frac{4p}{3}\right)\rho + \frac{2p}{3}I$$
+In components:
+$$\mathcal{E}(\rho) = \begin{pmatrix}\left(1-\frac{2p}{3}\right)\rho_{00}+\frac{2p}{3}\rho_{11} & \left(1-\frac{4p}{3}\right)\rho_{01} \\ \left(1-\frac{4p}{3}\right)\rho_{10} & \left(1-\frac{2p}{3}\right)\rho_{11}+\frac{2p}{3}\rho_{00}\end{pmatrix}$$
+Both populations and coherences shrink toward the maximally mixed state $I/2$. The steady state $\rho = I/2$ is reached when $p = 3/4$.
+### 5. Amplitude damping (T1 / relaxation)
+Amplitude damping models spontaneous emission: with probability $\gamma$ the qubit emits a photon and drops from $\lvert 1\rangle$ to $\lvert 0\rangle$. For a continuous-time process, $\gamma = 1 - e^{-t/T_1}$. Two Kraus operators:
+$$K_0 = \begin{pmatrix}1&0\\0&\sqrt{1-\gamma}\end{pmatrix}, \qquad
+K_1 = \begin{pmatrix}0&\sqrt{\gamma}\\0&0\end{pmatrix}$$
+$K_0$ is the no-emission evolution: the ground state amplitude stays fixed, the excited state amplitude shrinks by $\sqrt{1-\gamma}$. $K_1$ is the emission event: excitation $\lvert 1\rangle$ collapses to ground $\lvert 0\rangle$, and the $\sqrt{\gamma}$ prefactor ensures the correct probability.
+Completeness:
+$$K_0^\dagger K_0 + K_1^\dagger K_1 = \begin{pmatrix}1&0\\0&1-\gamma\end{pmatrix} + \begin{pmatrix}0&0\\0&\gamma\end{pmatrix} = I \quad\checkmark$$
+Channel action. No-emission term:
+$$K_0\rho K_0^\dagger = \begin{pmatrix}1&0\\0&\sqrt{1-\gamma}\end{pmatrix}\begin{pmatrix}\rho_{00}&\rho_{01}\\\rho_{10}&\rho_{11}\end{pmatrix}\begin{pmatrix}1&0\\0&\sqrt{1-\gamma}\end{pmatrix} = \begin{pmatrix}\rho_{00}&\sqrt{1-\gamma}\,\rho_{01}\\\sqrt{1-\gamma}\,\rho_{10}&(1-\gamma)\rho_{11}\end{pmatrix}$$
+Emission term:
+$$K_1\rho K_1^\dagger = \begin{pmatrix}0&\sqrt{\gamma}\\0&0\end{pmatrix}\begin{pmatrix}\rho_{00}&\rho_{01}\\\rho_{10}&\rho_{11}\end{pmatrix}\begin{pmatrix}0&0\\\sqrt{\gamma}&0\end{pmatrix} = \begin{pmatrix}\gamma\rho_{11}&0\\0&0\end{pmatrix}$$
+Sum:
+$$\mathcal{E}(\rho) = \begin{pmatrix}\rho_{00}+\gamma\rho_{11} & \sqrt{1-\gamma}\,\rho_{01} \\ \sqrt{1-\gamma}\,\rho_{10} & (1-\gamma)\rho_{11}\end{pmatrix}$$
+The excited population $\rho_{11}$ decays by factor $(1-\gamma)$; that probability flows into the ground state $\rho_{00}$. Coherences shrink by $\sqrt{1-\gamma}$ — half the rate (in exponent) of the population decay, matching the $T_2 = 2T_1$ limit.
+### 6. Phase damping (T2 / dephasing)
+Phase damping destroys coherences without changing populations. It models random scattering events that shift the qubit frequency but do not cause energy transitions. Two Kraus operators:
+$$K_0 = \begin{pmatrix}1&0\\0&\sqrt{1-\lambda}\end{pmatrix}, \qquad
+K_1 = \begin{pmatrix}0&0\\0&\sqrt{\lambda}\end{pmatrix}$$
+Completeness:
+$$K_0^\dagger K_0 + K_1^\dagger K_1 = \begin{pmatrix}1&0\\0&1-\lambda\end{pmatrix} + \begin{pmatrix}0&0\\0&\lambda\end{pmatrix} = I \quad\checkmark$$
+Channel action:
+$$K_0\rho K_0^\dagger = \begin{pmatrix}\rho_{00}&\sqrt{1-\lambda}\,\rho_{01}\\\sqrt{1-\lambda}\,\rho_{10}&(1-\lambda)\rho_{11}\end{pmatrix}$$
+$$K_1\rho K_1^\dagger = \begin{pmatrix}0&0\\0&\lambda\rho_{11}\end{pmatrix}$$
+$$\mathcal{E}(\rho) = \begin{pmatrix}\rho_{00} & \sqrt{1-\lambda}\,\rho_{01} \\ \sqrt{1-\lambda}\,\rho_{10} & \rho_{11}\end{pmatrix}$$
+Populations are exactly preserved. Coherences shrink by $\sqrt{1-\lambda}$. Setting $\lambda = 1 - e^{-t/T_2}$ gives the continuous-time $T_2$ decay: $\rho_{01}(t) = \rho_{01}(0)\,e^{-t/(2T_2)}$.
+## Qubit
+A qubit's state space is two-dimensional, so $\rho$ is a $2\times 2$ matrix. A general qubit density matrix has the form:
+$$\rho = \begin{pmatrix}\rho_{00} & \rho_{01} \\ \rho_{10} & \rho_{11}\end{pmatrix}$$
+where the diagonal entries $\rho_{00}$ and $\rho_{11}$ are real, non-negative **populations** (probabilities of finding the qubit in $\lvert 0\rangle$ and $\lvert 1\rangle$ respectively), and the off-diagonal entries $\rho_{01} = \rho_{10}^*$ are complex **coherences** that capture quantum superposition. The trace condition $\rho_{00} + \rho_{11} = 1$ reduces the free parameters to three real numbers (e.g. $\rho_{00}$, $\text{Re}\,\rho_{01}$, $\text{Im}\,\rho_{01}$).
+Because both $\rho$ and every $K_k$ are $2\times 2$ matrices, the channel $\mathcal{E}(\rho) = \sum_k K_k\rho K_k^\dagger$ is just a sum of $2\times 2$ matrix products. To make this fully concrete, here is the amplitude damping channel worked out entry by entry.
+Start from $\rho = \begin{pmatrix}\rho_{00}&\rho_{01}\\\rho_{10}&\rho_{11}\end{pmatrix}$ and the Kraus operators:
+$$K_0 = \begin{pmatrix}1&0\\0&\sqrt{1-\gamma}\end{pmatrix}, \qquad K_1 = \begin{pmatrix}0&\sqrt{\gamma}\\0&0\end{pmatrix}$$
+Multiply out $K_0\rho$:
+$$K_0\rho = \begin{pmatrix}1&0\\0&\sqrt{1-\gamma}\end{pmatrix}\begin{pmatrix}\rho_{00}&\rho_{01}\\\rho_{10}&\rho_{11}\end{pmatrix} = \begin{pmatrix}\rho_{00}&\rho_{01}\\\sqrt{1-\gamma}\,\rho_{10}&\sqrt{1-\gamma}\,\rho_{11}\end{pmatrix}$$
+Then multiply by $K_0^\dagger = K_0$ (it is real and diagonal):
+$$K_0\rho K_0^\dagger = \begin{pmatrix}\rho_{00}&\rho_{01}\\\sqrt{1-\gamma}\,\rho_{10}&\sqrt{1-\gamma}\,\rho_{11}\end{pmatrix}\begin{pmatrix}1&0\\0&\sqrt{1-\gamma}\end{pmatrix} = \begin{pmatrix}\rho_{00}&\sqrt{1-\gamma}\,\rho_{01}\\\sqrt{1-\gamma}\,\rho_{10}&(1-\gamma)\rho_{11}\end{pmatrix}$$
+Multiply out $K_1\rho$:
+$$K_1\rho = \begin{pmatrix}0&\sqrt{\gamma}\\0&0\end{pmatrix}\begin{pmatrix}\rho_{00}&\rho_{01}\\\rho_{10}&\rho_{11}\end{pmatrix} = \begin{pmatrix}\sqrt{\gamma}\,\rho_{10}&\sqrt{\gamma}\,\rho_{11}\\0&0\end{pmatrix}$$
+Then multiply by $K_1^\dagger = \begin{pmatrix}0&0\\\sqrt{\gamma}&0\end{pmatrix}$:
+$$K_1\rho K_1^\dagger = \begin{pmatrix}\sqrt{\gamma}\,\rho_{10}&\sqrt{\gamma}\,\rho_{11}\\0&0\end{pmatrix}\begin{pmatrix}0&0\\\sqrt{\gamma}&0\end{pmatrix} = \begin{pmatrix}\gamma\rho_{11}&0\\0&0\end{pmatrix}$$
+Add the two terms:
+$$\mathcal{E}(\rho) = K_0\rho K_0^\dagger + K_1\rho K_1^\dagger = \begin{pmatrix}\rho_{00}+\gamma\rho_{11} & \sqrt{1-\gamma}\,\rho_{01} \\ \sqrt{1-\gamma}\,\rho_{10} & (1-\gamma)\rho_{11}\end{pmatrix}$$
+Reading off the entries: the excited population $\rho_{11}$ decays by $\gamma$, that probability flows into the ground population $\rho_{00}$, and coherences shrink by $\sqrt{1-\gamma}$. No approximation was made — this is an exact result for one time step.
 ## 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:
+Kraus operators are the link between the [[lindblad-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.
+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. The full derivation is in the [[lindblad-equation#Derivation|Lindblad equation article]].