Skip to content
中

Module 12

Open systems and decoherence

Answer one concrete question: why does the macroscopic world look classical?

What you will see

  • The Bloch vector shrinking from the surface towards the centre — a pure state becoming mixed
  • Phase damping flattens the equator; energy relaxation drags everything towards the north pole
  • More environmental degrees of freedom means faster loss of coherence — far too fast to measure

Assumed background

  • Density matrices, pure versus mixed states (module 07)
  • The Bloch sphere (module 05)

Everything so far has dealt with closed systems: unitary evolution, pure states staying pure, no information lost.

Real systems are never closed. Air molecules hit them, thermal radiation shines on them, their own vibrations join in. This module asks what happens once the environment is included.

The answer is short — coherences disappear extremely fast, and classical probabilities survive.

Decoherence: how a point on the sphere falls inside it

Pure states live on the sphere, mixed states inside it. Switch on the coupling to the environment and watch the arrow shrink — and the whole unit sphere squash into an ellipsoid.

Loading 3D scene…

Initial state
90°
120°
Coupling to the environment
0.50
0.15
1.20

Phase damping eats only the coherences (the equatorial directions); energy relaxation also drags the populations towards the ground state (the north pole).

Evolution and display
t = 0.00
Speed
Time constants
T₁ (energy relaxation)6.67
T₂ (decoherence)1.74
T₂ ≤ 2T₁0.26 ≤ 2 ✓
ρ=12(I^+r⃗⋅σ⃗),Tr(ρ2)=1+∣r⃗∣22\rho = \tfrac12\left(\hat I + \vec r\cdot\vec\sigma\right),\qquad \mathrm{Tr}(\rho^2)=\tfrac{1+|\vec r|^2}{2}
|r⃗| = 1.0000Purity Tr(ρ²) = 1.0000Coherence |ρ₀₁| = 0.5000Population ρ₁₁ = 0.5000
  • Bloch vector r⃗
  • Channel ellipsoid (image of the unit sphere)
  • Trajectory

What to look for

  • Set both couplings to zero first: the arrow precesses steadily on the sphere and never shortens. That is unitary evolution of a closed system.
  • Switch on phase damping alone: the horizontal component collapses fast while the vertical one does not move. The ellipsoid is squashed into a vertical needle — coherence is gone, yet the probability of "up or down" is untouched.
  • Switch on energy relaxation alone: the whole ellipsoid shrinks and drifts towards the north pole. The system is dumping energy into the environment.
  • Set θ = 0 (exactly at the north pole) with phase damping only: nothing happens. The ground state has no coherence to lose — decoherence only means something for superpositions.
  • Push γφ to its maximum: the coherence is gone within a precession cycle or two. For a real macroscopic object that time is of order 10⁻³¹ seconds.

✕ "Decoherence is wavefunction collapse"

It is not. Decoherence turns a superposition into something that looks like a classical mixture (the density matrix goes diagonal), but it does not explain why a single outcome occurs each time. The measurement problem is compressed, not removed.

✕ "The environment drains the energy, so coherence disappears"

Not necessarily. Pure phase damping exchanges no energy at all — the populations do not move — and still destroys coherence completely. What is lost is phase information, not energy.

✕ "A mixed state just means we do not know which pure state it is"

The same density matrix arises from infinitely many different ensembles of pure states, and they are physically indistinguishable. "It really is some pure state, we just do not know which" has no operational content.

Think it through

  1. Taken together, system plus environment evolve unitarily and a pure state stays pure. Where, then, does the entropy of the system come from?
  2. What does the inequality T₂ ≤ 2T₁ mean geometrically? (Hint: compare how fast the ellipsoid shrinks equatorially and vertically.)
  3. A quantum computer must finish enough gates within T₂. With 20 ns per gate and T₂ = 100 μs, roughly how many operations fit? Enough for an algorithm needing 10⁶ gates?

From pure to mixed

A qubit’s density matrix can be written

ρ=12(I^+r⃗⋅σ⃗),Tr(ρ2)=1+∣r⃗∣22.\rho = \frac12\left(\hat I + \vec r\cdot\vec\sigma\right),\qquad \mathrm{Tr}(\rho^2) = \frac{1+|\vec r|^2}{2}.
  • ∣r⃗∣=1|\vec r| = 1: pure, on the surface;
  • ∣r⃗∣<1|\vec r| < 1: mixed, inside;
  • ∣r⃗∣=0|\vec r| = 0: maximally mixed, at the centre, carrying no information at all.

Decoherence is the process of ∣r⃗∣|\vec r| getting shorter. The arrow contracting in the scene is exactly that.

Two decays, two time scales

ProcessWhat it doesTime constantGeometric picture
Phase damping (dephasing)Destroys ρ01\rho_{01}, the relative phaseT2T_2The ball is squashed into a vertical needle
Amplitude damping (relaxation)System dumps energy and falls to the ground stateT1T_1The ball is dragged towards the north pole

In general T2≤2T1T_2 \le 2T_1: coherence goes at least as fast as energy. In quantum computing it is usually T2T_2 that limits you.

Switch the two channels on separately in the scene and watch whether the arrow is squashed or dragged — the two geometries are entirely different.

The Lindblad equation

Demand only that the evolution keep the density matrix positive and trace-preserving (a completely positive trace-preserving map), and assume the environment has no memory (the Markov approximation). The most general form is then

dρdt=−iℏ[H^,ρ]+∑kγk(L^kρL^k†−12{L^k†L^k,ρ}).\frac{\dd\rho}{\dd t} = -\frac{\ii}{\hbar}[\hat H,\rho] + \sum_k \gamma_k\left(\hat L_k\rho \hat L_k^\dagger - \tfrac12\{\hat L_k^\dagger \hat L_k,\rho\}\right).

The first term is the familiar unitary evolution; the second (the dissipator) is the environment’s contribution. Taking L^=σ^z\hat L = \hat\sigma_z gives pure dephasing; L^=σ^−\hat L = \hat\sigma_- gives amplitude damping. The two sliders in the scene are these two γk\gamma_k.

Key formulas

  • Bloch form

    ρ=12(I^+r⃗⋅σ⃗)\rho = \tfrac12(\hat I + \vec r\cdot\vec\sigma)

    Purity = (1+|r|²)/2

  • Coherence decay

    ρ01(t)=ρ01(0)e−t/T2\rho_{01}(t) = \rho_{01}(0)\ee^{-t/T_2}

    Dephasing

  • Population relaxation

    ρ11(t)=ρ11(0)e−t/T1\rho_{11}(t)=\rho_{11}(0)\ee^{-t/T_1}

    Energy flowing to the environment

  • Lindblad equation

    ρ˙=−iℏ[H^,ρ]+∑kγkD[L^k]ρ\dot\rho = -\tfrac{\ii}{\hbar}[\hat H,\rho]+\sum_k\gamma_k\mathcal{D}[\hat L_k]\rho

    The most general Markovian evolution

Think it through

  1. Include the environment and the whole thing evolves unitarily as a pure state. Where, then, does the system’s entropy come from? (Keywords: entanglement entropy, reduced density matrix.)
  2. What does T2≤2T1T_2 \le 2T_1 mean geometrically? (Hint: compare the shrinking rates of the ellipsoid along the equator and along the vertical.)
  3. Error correction has to finish before decoherence destroys the information. If a single gate takes 20 ns and T2=100T_2 = 100 μs, roughly how many operations fit?

Go deeper · matching textbook sections

The 3D scenes build the picture; the full derivations and exercises live in the textbook.

Having finished this module