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--- lindblad-equation [June 09, 2026 at 13:37] – Ivan Janevski
+++ lindblad-equation [June 13, 2026 at 03:13] (current) – external edit 127.0.0.1
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 # Lindblad equation
-**Lindblad equation** (also known as the **master equation** or the **GKSL equation**) is an equation in quantum mechanics that generalizes the Schrodinger equation for open quantum systems.
+**Lindblad equation** (also called the **GKSL equation**, after Gorini, Kossakowski, Sudarshan, and Lindblad) is the most general Markovian master equation for the [[density-matrix|density matrix]] $\rho$ of an open quantum system. It extends the [[von-neumann-equation|von Neumann equation]] by adding a dissipator term that models incoherent processes — energy loss, dephasing, and other forms of coupling to an environment.
+$$\frac{\mathrm{d}\rho}{\mathrm{d}t} = -\frac{i}{\hbar}[H,\,\rho] + \sum_k\!\left(L_k\rho L_k^\dagger - \frac{1}{2}\{L_k^\dagger L_k,\,\rho\}\right)$$
+## Coherent term
+The first term $-\frac{i}{\hbar}[H, \rho]$ is the von Neumann term. It describes the unitary, reversible part of the evolution driven by the system Hamiltonian $H$ — the same dynamics that the [[schrodinger-equation|Schrödinger equation]] captures for pure states. When the dissipator is zero (no environment coupling), the Lindblad equation reduces exactly to the von Neumann equation.
+## Dissipator
+The sum $\sum_k\!\left(L_k\rho L_k^\dagger - \frac{1}{2}\{L_k^\dagger L_k, \rho\}\right)$ is the dissipator. Each term corresponds to one incoherent channel indexed by $k$. The $L_k\rho L_k^\dagger$ part is the quantum jump: the environment has "observed" the system through channel $k$, collapsing it toward a new state. The $-\frac{1}{2}\{L_k^\dagger L_k, \rho\}$ part — with $\{A, B\} = AB + BA$ the anticommutator — is the no-jump correction that keeps $\text{tr}(\rho) = 1$ by accounting for the probability that no jump has occurred. Together, the two parts form a completely positive trace-preserving (CPTP) map on $\rho$.
+## Jump operators
+The operators $L_k$ are called **jump operators** (or **Lindblad operators**). Their choice encodes the physics of the environment — different environments couple to the system in different ways, and each coupling mechanism gets its own $L_k$. The six examples below cover the most common cases for qubits and bosonic modes.
+### 1. Relaxation (T1 / amplitude damping)
+Relaxation is spontaneous decay from the excited state $\lvert 1\rangle$ down to the ground state $\lvert 0\rangle$. Physically this is what happens when a two-level atom emits a photon, or when a superconducting qubit dissipates energy into its microwave environment. The jump operator is the **lowering operator** $\sigma^-$, which maps $\lvert 1\rangle \to \lvert 0\rangle$ and annihilates $\lvert 0\rangle$:
+$$L = \sqrt{\gamma}\,\sigma^-, \qquad
+\sigma^- = \lvert 0\rangle\langle 1\rvert = \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$$
+Its Hermitian adjoint is the **raising operator** $\sigma^+ = (\sigma^-)^\dagger = \lvert 1\rangle\langle 0\rvert$:
+$$\sigma^+ = \begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}$$
+Note that $\sigma^+\sigma^- = \lvert 1\rangle\langle 1\rvert$ is the projector onto the excited state. With $L^\dagger L = \gamma\,\sigma^+\sigma^-$, the special-case Lindblad equation (single channel, $H = 0$) is:
+$$\frac{\mathrm{d}\rho}{\mathrm{d}t} = \gamma\!\left(\sigma^-\rho\,\sigma^+ - \frac{1}{2}\{\sigma^+\sigma^-,\,\rho\}\right)$$
+In components this drives $\dot\rho_{11} = -\gamma\rho_{11}$ (excited population decays at rate $\gamma = 1/T_1$) and $\dot\rho_{01} = -\frac{\gamma}{2}\rho_{01}$ (coherences also decay, at half the rate). The ground-state population grows accordingly: $\dot\rho_{00} = +\gamma\rho_{11}$.
+### 2. Pure dephasing (T2)
+Pure dephasing destroys off-diagonal coherences with no energy exchange — the populations $\rho_{00}$ and $\rho_{11}$ stay fixed while the off-diagonals $\rho_{01}$ and $\rho_{10}$ decay. Physically this comes from random fluctuations in the qubit's transition frequency (e.g. charge noise in superconducting qubits, or collisions in atomic systems) that scramble the relative phase between $\lvert 0\rangle$ and $\lvert 1\rangle$.
+$$L = \sqrt{\frac{\gamma_\phi}{2}}\,\sigma_z, \qquad
+\sigma_z = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}$$
+Because $\sigma_z^2 = I$, we have $L^\dagger L = (\gamma_\phi/2)\,I$, so the anticommutator term simplifies: $\frac{1}{2}\{L^\dagger L, \rho\} = \frac{\gamma_\phi}{2}\rho$. The special-case Lindblad equation becomes:
+$$\frac{\mathrm{d}\rho}{\mathrm{d}t} = \frac{\gamma_\phi}{2}\!\left(\sigma_z\rho\,\sigma_z - \rho\right)$$
+In components: $\dot\rho_{00} = 0$, $\dot\rho_{11} = 0$, $\dot\rho_{01} = -\gamma_\phi\,\rho_{01}$, $\dot\rho_{10} = -\gamma_\phi\,\rho_{10}$. The coherences decay exponentially at rate $\gamma_\phi = 1/T_\phi$. Combined with relaxation, the total dephasing rate is $1/T_2 = 1/(2T_1) + 1/T_\phi$.
+### 3. Thermal excitation
+At nonzero temperature the environment contains thermal photons, which can excite the qubit from $\lvert 0\rangle$ to $\lvert 1\rangle$. This is the reverse process from relaxation. The jump operator is the **raising operator** $\sigma^+$:
+$$L = \sqrt{\gamma_\uparrow}\,\sigma^+, \qquad
+\sigma^+ = \lvert 1\rangle\langle 0\rvert = \begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}$$
+The thermal excitation rate is $\gamma_\uparrow = \gamma\,\bar{n}$, where $\bar{n} = (e^{\hbar\omega/k_BT} - 1)^{-1}$ is the mean photon number of the bath at the qubit frequency $\omega$ and temperature $T$. Note that $\sigma^-\sigma^+ = \lvert 0\rangle\langle 0\rvert$ is the projector onto the ground state. The special-case Lindblad equation is:
+$$\frac{\mathrm{d}\rho}{\mathrm{d}t} = \gamma_\uparrow\!\left(\sigma^+\rho\,\sigma^- - \frac{1}{2}\{\sigma^-\sigma^+,\,\rho\}\right)$$
+In components: $\dot\rho_{00} = -\gamma_\uparrow\rho_{00}$ (ground population depletes) and $\dot\rho_{11} = +\gamma_\uparrow\rho_{00}$. Together with the relaxation channel ($L = \sqrt{\gamma}\,\sigma^-$, rate $\gamma = \gamma_\downarrow$), the steady state is the thermal equilibrium $\rho_\infty = \text{diag}(1/(1+\bar{n}),\; \bar{n}/(1+\bar{n}))$, i.e. the qubit thermalises to the bath temperature.
+### 4. Depolarizing / isotropic noise
+Depolarizing noise applies random Pauli errors in all three directions with equal probability, driving $\rho$ toward the maximally mixed state $I/2$. It requires three separate jump operators, one per Pauli:
+$$L_x = \sqrt{\frac{p}{3}}\,\sigma_x, \qquad
+L_y = \sqrt{\frac{p}{3}}\,\sigma_y, \qquad
+L_z = \sqrt{\frac{p}{3}}\,\sigma_z$$
+The three Pauli matrices in full:
+$$\sigma_x = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}, \qquad
+\sigma_y = \begin{pmatrix}0 & -i \\ i & 0\end{pmatrix}, \qquad
+\sigma_z = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}$$
+All three satisfy $\sigma_j^2 = I$, so $\frac{1}{2}\{L_j^\dagger L_j, \rho\} = \frac{p}{3}\rho$ for each $j$. The special-case Lindblad equation (H = 0, three channels):
+$$\frac{\mathrm{d}\rho}{\mathrm{d}t} = \frac{p}{3}\sum_{j\in\{x,y,z\}}\!\left(\sigma_j\rho\,\sigma_j - \rho\right)$$
+Using the Pauli identity $\sum_j \sigma_j\rho\,\sigma_j = 2\,\text{tr}(\rho)\,I - \rho = 2I - \rho$ (which holds for any $2\times 2$ matrix $\rho$ with $\text{tr}(\rho) = 1$), this simplifies to:
+$$\frac{\mathrm{d}\rho}{\mathrm{d}t} = \frac{p}{3}(2I - \rho - 3\rho) = \frac{4p}{3}\!\left(\frac{I}{2} - \rho\right)$$
+The fixed point is $\rho = I/2$: any initial state is driven toward the completely mixed state at rate $4p/3$.
+### 5. Bosonic loss (cavity / photon loss)
+For a bosonic mode (cavity field, mechanical oscillator, LC circuit), photon loss into the environment is described by the bosonic annihilation operator $a$. In the Fock (photon-number) basis $\{\lvert 0\rangle, \lvert 1\rangle, \lvert 2\rangle, \ldots\}$, $a$ acts as $a\lvert n\rangle = \sqrt{n}\,\lvert n-1\rangle$ — it removes one photon and lowers the mode by one excitation level.
+$$L = \sqrt{\kappa}\,a, \qquad
+a = \begin{pmatrix}0 & \sqrt{1} & 0 & 0 & \cdots \\ 0 & 0 & \sqrt{2} & 0 & \cdots \\ 0 & 0 & 0 & \sqrt{3} & \cdots \\ \vdots & & & & \ddots\end{pmatrix}$$
+The rate $\kappa = 1/T_{\text{cavity}}$ is the photon loss rate (inverse cavity lifetime). The Hermitian adjoint $a^\dagger$ is the creation operator; $a^\dagger a = \hat{n}$ is the photon number operator. The special-case Lindblad equation (H = 0, single channel):
+$$\frac{\mathrm{d}\rho}{\mathrm{d}t} = \kappa\!\left(a\rho\, a^\dagger - \frac{1}{2}\{a^\dagger a,\,\rho\}\right)$$
+In the Fock basis, the matrix element $\rho_{mn} = \langle m\rvert\rho\lvert n\rangle$ evolves as:
+$$\dot\rho_{mn} = \kappa\!\left(\sqrt{(m+1)(n+1)}\;\rho_{m+1,\,n+1} - \frac{m+n}{2}\,\rho_{mn}\right)$$
+The mean photon number decays exponentially: $d\langle\hat{n}\rangle/dt = -\kappa\langle\hat{n}\rangle$. The steady state is the vacuum $\lvert 0\rangle\langle 0\rvert$.
+### 6. Bosonic dephasing (photon number dephasing)
+Bosonic dephasing preserves photon number populations $\rho_{nn}$ but destroys coherences between Fock states with different photon numbers. The jump operator is the number operator $\hat{n} = a^\dagger a$:
+$$L = \sqrt{\kappa_\phi}\,\hat{n} = \sqrt{\kappa_\phi}\,a^\dagger a, \qquad
+a^\dagger a = \begin{pmatrix}0 & 0 & 0 & \cdots \\ 0 & 1 & 0 & \cdots \\ 0 & 0 & 2 & \cdots \\ \vdots & & & \ddots\end{pmatrix}$$
+Physically this arises when the mode frequency fluctuates randomly (e.g. due to a dispersive coupling to a fluctuating qubit, or dielectric loss that is proportional to photon number). The special-case Lindblad equation (H = 0, single channel):
+$$\frac{\mathrm{d}\rho}{\mathrm{d}t} = \kappa_\phi\!\left(\hat{n}\,\rho\,\hat{n} - \frac{1}{2}\{\hat{n}^2,\,\rho\}\right)$$
+In the Fock basis the result is particularly clean. Since $\hat{n}\lvert n\rangle = n\lvert n\rangle$ is diagonal, every matrix element picks up a simple factor:
+$$\dot\rho_{mn} = -\frac{\kappa_\phi}{2}(m - n)^2\,\rho_{mn}$$
+Diagonal elements $(m = n)$ are exactly preserved. Off-diagonal elements decay at a rate proportional to the squared difference in photon number, so states separated by many quanta dephase much faster than adjacent Fock states.
+## Qubit
+A qubit is the simplest open quantum system: a two-level system with states $\lvert 0\rangle$ (ground) and $\lvert 1\rangle$ (excited). Its density matrix is a $2\times 2$ matrix:
+$$\rho = \begin{pmatrix}\rho_{00} & \rho_{01} \\ \rho_{10} & \rho_{11}\end{pmatrix}$$
+The diagonal entries $\rho_{00}$ and $\rho_{11}$ are real, non-negative **populations** summing to one ($\rho_{00} + \rho_{11} = 1$). The off-diagonal entries $\rho_{01} = \rho_{10}^*$ are complex **coherences**. A pure state $\lvert\psi\rangle = \alpha\lvert 0\rangle + \beta\lvert 1\rangle$ corresponds to $\rho_{00} = |\alpha|^2$, $\rho_{11} = |\beta|^2$, $\rho_{01} = \alpha\beta^*$.
+The standard qubit Hamiltonian describes a two-level system with energy splitting $\hbar\omega$ between ground and excited state:
+$$H = \frac{\hbar\omega}{2}\sigma_z = \frac{\hbar\omega}{2}\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$
+To write the full Lindblad equation for a realistic qubit, use two jump operators — one for T1 relaxation and one for pure dephasing:
+$$L_1 = \sqrt{\gamma}\,\sigma^- = \sqrt{\gamma}\begin{pmatrix}0&1\\0&0\end{pmatrix}, \qquad
+L_2 = \sqrt{\frac{\gamma_\phi}{2}}\,\sigma_z = \sqrt{\frac{\gamma_\phi}{2}}\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$
+Now work out each term of $\dot\rho = -\frac{i}{\hbar}[H,\rho] + \sum_k\!\left(L_k\rho L_k^\dagger - \frac{1}{2}\{L_k^\dagger L_k,\rho\}\right)$ in matrix form.
+**Coherent term.** The commutator $[H, \rho]$:
+$$[\sigma_z, \rho] = \begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}\rho_{00}&\rho_{01}\\\rho_{10}&\rho_{11}\end{pmatrix} - \begin{pmatrix}\rho_{00}&\rho_{01}\\\rho_{10}&\rho_{11}\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix} = \begin{pmatrix}0&2\rho_{01}\\-2\rho_{10}&0\end{pmatrix}$$
+$$-\frac{i}{\hbar}[H,\rho] = -\frac{i\omega}{2}\begin{pmatrix}0&2\rho_{01}\\-2\rho_{10}&0\end{pmatrix} = \begin{pmatrix}0&-i\omega\rho_{01}\\i\omega\rho_{10}&0\end{pmatrix}$$
+The coherent term drives oscillation of the coherences at frequency $\omega$ (Larmor precession). It leaves populations untouched.
+**T1 dissipator.** With $L_1^\dagger L_1 = \gamma\,\sigma^+\sigma^- = \gamma\begin{pmatrix}0&0\\0&1\end{pmatrix}$:
+$$\sigma^-\rho\,\sigma^+ = \begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}\rho_{00}&\rho_{01}\\\rho_{10}&\rho_{11}\end{pmatrix}\begin{pmatrix}0&0\\1&0\end{pmatrix} = \begin{pmatrix}\rho_{11}&0\\0&0\end{pmatrix}$$
+$$\frac{1}{2}\{L_1^\dagger L_1,\rho\} = \frac{\gamma}{2}\left\{\begin{pmatrix}0&0\\0&1\end{pmatrix},\rho\right\} = \frac{\gamma}{2}\begin{pmatrix}0&\rho_{01}\\\rho_{10}&2\rho_{11}\end{pmatrix}$$
+$$L_1\rho L_1^\dagger - \frac{1}{2}\{L_1^\dagger L_1,\rho\} = \gamma\begin{pmatrix}\rho_{11}&0\\0&0\end{pmatrix} - \frac{\gamma}{2}\begin{pmatrix}0&\rho_{01}\\\rho_{10}&2\rho_{11}\end{pmatrix} = \gamma\begin{pmatrix}\rho_{11}&-\tfrac{1}{2}\rho_{01}\\-\tfrac{1}{2}\rho_{10}&-\rho_{11}\end{pmatrix}$$
+**Dephasing dissipator.** With $L_2^\dagger L_2 = (\gamma_\phi/2)\,\sigma_z^2 = (\gamma_\phi/2)\,I$ (since $\sigma_z^2 = I$):
+$$\sigma_z\rho\,\sigma_z = \begin{pmatrix}\rho_{00}&-\rho_{01}\\-\rho_{10}&\rho_{11}\end{pmatrix}, \qquad \frac{1}{2}\{L_2^\dagger L_2,\rho\} = \frac{\gamma_\phi}{2}\rho$$
+$$L_2\rho L_2^\dagger - \frac{1}{2}\{L_2^\dagger L_2,\rho\} = \frac{\gamma_\phi}{2}(\sigma_z\rho\,\sigma_z - \rho) = \begin{pmatrix}0&-\gamma_\phi\rho_{01}\\-\gamma_\phi\rho_{10}&0\end{pmatrix}$$
+**Full equation.** Sum all three contributions:
+$$\dot\rho = \begin{pmatrix}0&-i\omega\rho_{01}\\i\omega\rho_{10}&0\end{pmatrix} + \gamma\begin{pmatrix}\rho_{11}&-\tfrac{1}{2}\rho_{01}\\-\tfrac{1}{2}\rho_{10}&-\rho_{11}\end{pmatrix} + \begin{pmatrix}0&-\gamma_\phi\rho_{01}\\-\gamma_\phi\rho_{10}&0\end{pmatrix}$$
+$$\dot\rho = \begin{pmatrix}\gamma\rho_{11} & \left(-i\omega - \tfrac{\gamma}{2} - \gamma_\phi\right)\rho_{01} \\ \left(i\omega - \tfrac{\gamma}{2} - \gamma_\phi\right)\rho_{10} & -\gamma\rho_{11}\end{pmatrix}$$
+Reading off the four equations for the matrix entries:
+$$\dot\rho_{00} = +\gamma\,\rho_{11}$$
+$$\dot\rho_{11} = -\gamma\,\rho_{11}$$
+$$\dot\rho_{01} = \left(-i\omega - \frac{1}{T_2}\right)\rho_{01}, \qquad \frac{1}{T_2} = \frac{\gamma}{2} + \gamma_\phi$$
+$$\dot\rho_{10} = \left(+i\omega - \frac{1}{T_2}\right)\rho_{10}$$
+These are the **optical Bloch equations**. The solutions are direct:
+$$\rho_{11}(t) = \rho_{11}(0)\,e^{-\gamma t} = \rho_{11}(0)\,e^{-t/T_1}$$
+$$\rho_{01}(t) = \rho_{01}(0)\,e^{(-i\omega - 1/T_2)\,t}$$
+The excited population decays exponentially at rate $\gamma = 1/T_1$. The coherences oscillate at the qubit frequency $\omega$ while their magnitude decays at rate $1/T_2 = \gamma/2 + \gamma_\phi$. Since $T_2 \leq 2T_1$, dephasing is always at least as fast as the T1-limited floor.
+## Markov approximation
+The Lindblad equation is valid under the **Markov approximation**: the environment is assumed to be memoryless, forgetting the interaction faster than the system evolves. This is analogous to an interrupt-driven peripheral that completes its transaction well within one main-loop cycle — the system never needs to track the peripheral's internal state. When the bath correlation time $\tau_{\text{bath}}$ is comparable to the system timescale, memory effects matter and the full non-Markovian dynamics must be used instead.
+## Derivation
+The Lindblad equation can be derived from the quantum channel picture. Over a short time step $dt$, a quantum channel is described by a set of **Kraus operators** $\{K_k\}$ satisfying the completeness relation $\sum_k K_k^\dagger K_k = I$. The density matrix evolves as:
+$$\rho(t + dt) = \sum_k K_k\,\rho(t)\,K_k^\dagger$$
+This is the most general evolution that is linear, completely positive, and trace-preserving. For a single dissipation channel described by a jump operator $L$, use two Kraus operators:
+$$K_0 = I - \frac{i}{\hbar}H\,dt - \frac{1}{2}L^\dagger L\,dt, \qquad K_1 = \sqrt{dt}\;L$$
+$K_1$ is the **jump** Kraus operator: it fires with probability proportional to $dt$, representing one quantum jump. $K_0$ is the **no-jump** Kraus operator: it captures the coherent Hamiltonian evolution plus the accumulated "penalty" from not jumping.
+First, verify that these satisfy the completeness relation to first order in $dt$ (anything proportional to $dt^2$ is dropped). Take the dagger of $K_0$:
+$$K_0^\dagger = I + \frac{i}{\hbar}H\,dt - \frac{1}{2}L^\dagger L\,dt$$
+Then:
+$$K_0^\dagger K_0 = \left(I + \frac{i}{\hbar}H\,dt - \frac{1}{2}L^\dagger L\,dt\right)\!\left(I - \frac{i}{\hbar}H\,dt - \frac{1}{2}L^\dagger L\,dt\right)$$
+Expand and drop $dt^2$ terms (e.g. $H^2 dt^2$ and $H L^\dagger L\,dt^2$ all vanish). The surviving terms are:
+$$K_0^\dagger K_0 = I - L^\dagger L\,dt$$
+The jump term gives:
+$$K_1^\dagger K_1 = \left(\sqrt{dt}\;L\right)^\dagger\!\left(\sqrt{dt}\;L\right) = dt\;L^\dagger L$$
+Sum them:
+$$K_0^\dagger K_0 + K_1^\dagger K_1 = I - L^\dagger L\,dt + L^\dagger L\,dt = I \quad\checkmark$$
+The completeness relation holds to first order. Now expand the full evolution $\rho(t+dt) = K_0\rho K_0^\dagger + K_1\rho K_1^\dagger$.
+Start with the no-jump term $K_0\rho K_0^\dagger$. Multiply out, keeping only first-order terms in $dt$:
+$$K_0\rho K_0^\dagger
+= \left(I - \frac{i}{\hbar}H\,dt - \frac{1}{2}L^\dagger L\,dt\right)\rho\left(I + \frac{i}{\hbar}H\,dt - \frac{1}{2}L^\dagger L\,dt\right)$$
+$$= \rho + \frac{i}{\hbar}\rho H\,dt - \frac{1}{2}\rho L^\dagger L\,dt
+     - \frac{i}{\hbar}H\rho\,dt
+     - \frac{1}{2}L^\dagger L\rho\,dt + \mathcal{O}(dt^2)$$
+Recognise the commutator $[H, \rho] = H\rho - \rho H$ and the anticommutator $\{L^\dagger L, \rho\} = L^\dagger L\rho + \rho L^\dagger L$:
+$$K_0\rho K_0^\dagger = \rho - \frac{i}{\hbar}[H,\,\rho]\,dt - \frac{1}{2}\{L^\dagger L,\,\rho\}\,dt$$
+Now the jump term $K_1\rho K_1^\dagger$:
+$$K_1\rho K_1^\dagger = \sqrt{dt}\;L\;\rho\;L^\dagger\sqrt{dt} = L\rho L^\dagger\,dt$$
+Add both terms together:
+$$\rho(t+dt) = \rho - \frac{i}{\hbar}[H,\,\rho]\,dt - \frac{1}{2}\{L^\dagger L,\,\rho\}\,dt + L\rho L^\dagger\,dt$$
+Subtract $\rho(t)$ from both sides and divide by $dt$:
+$$\frac{\mathrm{d}\rho}{\mathrm{d}t} = -\frac{i}{\hbar}[H,\,\rho] + L\rho L^\dagger - \frac{1}{2}\{L^\dagger L,\,\rho\}$$
+That is the Lindblad equation for a single jump channel. For multiple independent channels, each with its own Kraus pair $(K_0^{(k)}, K_1^{(k)})$ and jump operator $L_k$, the no-jump Kraus operator picks up one $-\frac{1}{2}L_k^\dagger L_k\,dt$ term per channel:
+$$K_0 = I - \frac{i}{\hbar}H\,dt - \frac{1}{2}\sum_k L_k^\dagger L_k\,dt$$
+Repeating the same expansion yields the general form:
+$$\frac{\mathrm{d}\rho}{\mathrm{d}t} = -\frac{i}{\hbar}[H,\,\rho] + \sum_k\!\left(L_k\rho L_k^\dagger - \frac{1}{2}\{L_k^\dagger L_k,\,\rho\}\right)$$
+## List of code implementations
+ - [[lindblad-equation-qutip|Lindblad equation (QuTiP)]]
-$$\frac{\mathrm d \rho}{\mathrm dt} = - \frac{i}{\hbar}[H, \rho]$$