The Lindblad master equation

The quantum channel of the last section answered “what a single evolution can do to $\hat\rho$”. But it is a snapshot: a state goes in, a state comes out, and time is flattened into the map.

The questions in the lab, however, are continuous.

The phenomenon first: two time constants

To interrogate a superconducting qubit, the standard procedure is two experiments.

Experiment one: excite it to $\ket{1}$, wait a time $t$, and measure whether it is still in $\ket{1}$. The result is a clean exponential decay, $P_1(t)=\ee^{-t/T_1}$. This $T_1$ is called the energy relaxation time.

Experiment two: prepare it in the superposition $\frac{1}{\sqrt2}(\ket{0}+\ket{1})$, wait a time $t$, and then measure its interference fringes (a Ramsey experiment). The fringe contrast also decays exponentially, with a time constant written $T_2$, called the coherence time.

Two experiments, two constants — and experiment finds $T_2\le2T_1$ every single time, with no exception on record. The von Neumann equation of chapter 3, $\ii\hbar\,\dd\hat\rho/\dd t=[\hat H,\hat\rho]$, explains neither: it is unitary, purity is conserved, nothing decays. Last section’s channels can describe decay, but only as a discrete one-shot deal — they cannot say “what happens each microsecond”.

What we need is: a differential equation, every small step of which is a legitimate quantum channel.

From channel to equation: one extra assumption required

Slice the evolution into slabs of time: $\hat\rho(t+\dd t)=\mathcal{E}_{\dd t}\big(\hat\rho(t)\big)$. For this sentence to become a differential equation, a non-trivial assumption is hiding inside — $\mathcal{E}_{\dd t}$ must depend only on the current $\hat\rho(t)$, not on the history.

This is the Markov approximation. In plain words: the environment forgets. The information the system leaks into the environment (that escaped photon, that carried-away phonon) is gone for good; the environment does not hand it back, and does not hold a grudge to settle later. The condition for this to hold is an environment that is “big and fast”: its own correlation time is far shorter than the system’s relaxation time, so at every instant the system faces a freshly “laundered” environment. Add the requirement that the system-environment coupling is weak (treatable as a perturbation — the Born approximation), and together they are the Born–Markov approximation — the only approximation in this section, and the only place things could go wrong.

Under these two assumptions, ask “what is the most general Markovian equation preserving the three properties of a density matrix”, and the answer is a theorem (Gorini–Kossakowski–Sudarshan–Lindblad, 1976; we state it without a rigorous derivation):

$$\boxed{\; \frac{\dd\hat\rho}{\dd t} =-\frac{\ii}{\hbar}[\hat H,\hat\rho] +\sum_k\gamma_k\Big(\hat L_k\hat\rho\hat L_k^\dagger -\tfrac12\big\{\hat L_k^\dagger\hat L_k,\hat\rho\big\}\Big) \;}\tag{10.4.1}$$

A full worked example: the damped two-level atom

A two-level atom ($\ket{0}$ ground, $\ket{1}$ excited, gap $\hbar\omega_0$) sits in vacuum and emits spontaneously. The environment does exactly one thing: pull the atom from $\ket{1}$ down to $\ket{0}$. The jump operator is just the lowering operator

$$\hat L=\hat\sigma_-=\ket{0}\bra{1},\qquad \text{rate}\ \gamma\tag{10.4.3}$$

Solving the damped atom: from equation to T₁ and T₂basic~9 minexpand

Step 1: write the matrix-element equations. Take $\hat H=\hbar\omega_0\ket{1}\bra{1}$ and compute term by term.

The jump term: $\hat\sigma_-\hat\rho\,\hat\sigma_+=\ket{0}\bra{1}\hat\rho\ket{1}\bra{0}=\rho_{11}\ket{0}\bra{0}$ — it moves the excited-state population into the ground state.

The anticommutator term: $\hat\sigma_+\hat\sigma_-=\ket{1}\bra{1}$, so

$$\tfrac12\{\ket{1}\bra{1},\hat\rho\} =\begin{pmatrix}0&\tfrac12\rho_{01}\\ \tfrac12\rho_{10}&\rho_{11}\end{pmatrix}\tag{10.4.4}$$

The Hamiltonian term merely gives the off-diagonal a rotating phase $+\ii\omega_0\rho_{01}$. Assembling, the equations for the four matrix elements:

$$\dot\rho_{11}=-\gamma\rho_{11},\qquad \dot\rho_{00}=+\gamma\rho_{11},\qquad \dot\rho_{01}=\Big(\ii\omega_0-\frac{\gamma}{2}\Big)\rho_{01}\tag{10.4.5}$$

($\rho_{10}=\rho_{01}^*$ needs no separate solution.) Notice the equations have already decoupled — integrate directly.

Step 2: read the populations.

$$\rho_{11}(t)=\rho_{11}(0)\,\ee^{-\gamma t} \qquad\Longrightarrow\qquad T_1=\frac{1}{\gamma}\tag{10.4.6}$$

The excited state decays exponentially — exactly the curve of “experiment one”. $\rho_{00}$ grows to match; the total trace is conserved.

Step 3: read the coherences.

$$\rho_{01}(t)=\rho_{01}(0)\,\ee^{\ii\omega_0t}\,\ee^{-\gamma t/2} \qquad\Longrightarrow\qquad T_2=\frac{2}{\gamma}=2T_1\tag{10.4.7}$$

The coherence rotates at $\omega_0$ (the Ramsey fringes themselves) while decaying at $\gamma/2$ (the falling fringe contrast).

Why exactly half the decay rate? $\rho_{11}\propto|c_1|^2$ is the square of an amplitude, while $\rho_{01}\propto c_0c_1^*$ contains the amplitude only to the first power — the amplitude decays at $\gamma/2$, so its square naturally decays at $\gamma$. The relation between $1-\gamma$ and $\sqrt{1-\gamma}$ in last section’s amplitude damping reappears here in differential form.

Step 4: add pure dephasing, and get the general relation. A real environment does more than absorb energy — it also kicks the phase alone: low-frequency noise jitters $\omega_0$ at random, no energy is lost, and the phase cannot be kept. This corresponds to a second jump operator $\hat L'=\hat\sigma_z$ with rate $\gamma_\varphi/2$. Since $\hat\sigma_z^2=\hat 1$, this dissipator simplifies to $\tfrac{\gamma_\varphi}{2}(\hat\sigma_z\hat\rho\hat\sigma_z-\hat\rho)$: the diagonal does not move at all, while the off-diagonal decays by an extra $\gamma_\varphi$. Hence

$$\frac{1}{T_2}=\frac{\gamma}{2}+\gamma_\varphi=\frac{1}{2T_1}+\frac{1}{T_\varphi}\tag{10.4.8}$$

$T_\varphi\equiv1/\gamma_\varphi$ is called the pure dephasing time. From $\gamma_\varphi\ge0$ it follows at once that

$$T_2\le 2T_1\tag{10.4.9}$$

Losing energy necessarily costs coherence (even if that is the only thing the environment does, $T_2$ is at most $2T_1$), while losing coherence need not cost any energy at all. Phase is the more delicate commodity — that sentence is the trailer for the entire decoherence story.

∎Key formulas

• Lindblad equation

  $$\frac{\dd\hat\rho}{\dd t}=-\frac{\ii}{\hbar}[\hat H,\hat\rho]+\sum_k\gamma_k\Big(\hat L_k\hat\rho\hat L_k^\dagger-\tfrac12\{\hat L_k^\dagger\hat L_k,\hat\rho\}\Big)$$

  The most general Markovian equation preserving the three properties of a density matrix

• Solution for the damped atom

  $$\rho_{11}(t)=\rho_{11}(0)\ee^{-\gamma t},\quad \rho_{01}(t)=\rho_{01}(0)\ee^{\ii\omega_0 t-\gamma t/2}$$

  Populations decay at γ, coherences at γ/2 — an amplitude versus its square

• The two time constants

  $$T_1=\frac1\gamma,\qquad \frac{1}{T_2}=\frac{1}{2T_1}+\frac{1}{T_\varphi}$$

  T₁ energy relaxation; T₂ coherence time; T_φ pure dephasing

• The famous inequality

  $$T_2\le 2T_1$$

  Losing energy costs coherence; losing coherence need not cost energy; equality = no pure dephasing

?Self-check4 questions

1. 1.

   The physical role of the anticommutator term −½{L†L, ρ} in the Lindblad equation is:

   • To describe the correction from the "no jump occurred" branch, and, together with the jump term, keep the total trace intact
   • To describe the environment giving energy back to the system
   • Pure mathematical bookkeeping with no physical meaning
   • To describe measurement collapse

   Submit
2. 2.

   For a damped atom with spontaneous emission only (jump operator σ₋, rate γ), why do the coherences decay at half the rate of the populations?

   • Populations ∝ amplitude squared, coherences ∝ amplitude to the first power; the amplitude decays at γ/2
   • Because there are two coherences (ρ₀₁ and ρ₁₀), which split the decay between them
   • It is an artefact of the Born-Markov approximation
   • Because the Hamiltonian term cancels half of the decay

   Submit
3. 3.

   A qubit has T₁ = 30 μs and pure dephasing time T_φ = 60 μs. Find T₂ (in μs).

   μs100% relative tolerance

   Submit
4. 4.

   Which statements about T₂ ≤ 2T₁ are correct? (Select all that apply.)

   Select all that apply

   • Equality holds if and only if there is no pure dephasing (T_φ → ∞)
   • A measured T₂ = 2.5 T₁ could only be a measurement error (within the Markovian framework)
   • Pure dephasing loses coherence but no energy, so it lowers T₂ without touching T₁
   • If T₁ is infinite, then T₂ must be infinite too

   Submit

What comes next

We can solve the equation now, but the most important physical question is still hanging: what entitles the environment to go straight for the coherences? Why is it precisely the phase between $\ket{0}+\ket{1}$ that gets wiped, and not something else? Who decreed the “preferred” basis?

The next section returns to the original question — why Schrödinger’s cat is never seen — gives decoherence a physical picture, and works out a few astonishing numbers.