Jump Operators

What a jump operator is

In the Lindblad equation, each operator $L_k$ in the dissipator is called a jump operator (or Lindblad operator). The name comes from the quantum-trajectory picture: in a single run of an experiment, the system evolves smoothly until, at random times, it "jumps" — $|\psi\rangle \to L_k|\psi\rangle/\lVert L_k|\psi\rangle\rVert$ — corresponding to a detection event in the bath (a photon emitted, a phonon scattered). Averaging over all such trajectories reproduces the smooth density-matrix evolution. We make this precise in the trajectories lesson; here we read off what different $L_k$ do.

The dissipator, channel by channel

For a single channel,

\mathcal{D}[\rho] = L\,\rho\,L^\dagger - \tfrac12\{L^\dagger L,\rho\}.

The first term places the system into the range of $L$ ; the second removes weight from the support of $L^\dagger L$ . Three canonical single-qubit jump operators cover most of practice.

Amplitude damping (energy relaxation)

L = \sqrt{\gamma}\,\sigma_- = \sqrt{\gamma}\,|0\rangle\langle1| .

This sends the excited state to the ground state, modelling spontaneous emission. Population flows $|1\rangle \to |0\rangle$ at rate $\gamma$ , and coherence decays at rate $\gamma/2$ . The fixed point is the ground state $|0\rangle$ .

Pure dephasing (phase damping)

L = \sqrt{\tfrac{\gamma}{2}}\,\sigma_z .

Here $L^\dagger L = \tfrac{\gamma}{2}\mathbb{1}$ commutes with everything, so the populations $\rho_{00},\rho_{11}$ never change. Because $\sigma_z$ negates the off-diagonal elements, the dissipator drives the coherences to zero:

\dot\rho_{01} = -\gamma\,\rho_{01}\quad\Rightarrow\quad \rho_{01}(t)=\rho_{01}(0)\,e^{-\gamma t}.

This is decoherence without dissipation — energy is conserved, only phase information is lost, exactly the $T_\phi$ process introduced earlier.

Incoherent excitation and thermal baths

A bath at finite temperature both de-excites and excites the qubit, modelled by two jump operators

L_- = \sqrt{\gamma_\downarrow}\,\sigma_-,\qquad L_+ = \sqrt{\gamma_\uparrow}\,\sigma_+ ,

with rates fixed by detailed balance, $\gamma_\uparrow/\gamma_\downarrow = e^{-\hbar\omega/k_BT}$ . The steady state is then the thermal (Gibbs) state at temperature $T$ , not the ground state — a point we return to when discussing steady states.

Try it

This is a numerical exercise — your code should return a number. For pure dephasing with $L=\sqrt{\gamma/2}\,\sigma_z$ , the coherence obeys $\rho_{01}(t)=\rho_{01}(0)e^{-\gamma t}$ . With $\rho_{01}(0)=0.5$ , $\gamma=0.8$ , $t=1.25$ , compute and return $\rho_{01}(t)$ .

Run your code to see the quantum state.