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Optional unit · institution-dependent · Interactive lesson

Pointer states: what decoherence explains and leaves open

Quantum Mechanics II said a measurement entangles a system with a pointer and left the pointer abstract. Here the pointer is a bath of oscillators: we compute exactly how much interference it removes and how fast, find which states it leaves alone, and then say plainly what the calculation does not deliver, which is an outcome.

UAQM-14.09 · Lesson 9 of 9 in this unit. Position is not a completion score.
Starting model: Pointer states: what decoherence explains and leaves openExact coherence |ρ↑↓(t)|/|ρ↑↓(0)| = exp[−Γ(t)] of a qubit coupled through σ_z to vacuum oscillators, time in periods 2π/Ω. Solid: modes spaced by δ = Ω/8 in a Gaussian band of r.m.s. width Ω/4 about Ω, record strength S = 2.0; it falls to e^(−S) = 0.135 and returns to 1 at 2π/δ = 8 periods. Dashed: one mode with the same S, dipping to 0.018 every period.coherence |ρ↑↓(t)| / |ρ↑↓(0)|solid: band of modesdashed: one modee(−S)106122π/δt in periods 2π/Ω
Start with a picture. Change the model in the lab below.

Optional unit · institution-dependent. Units 14 and 15 are the course’s optional extension into open systems and quantum gases. Unit 14 builds on radiation and decay; Unit 15 builds on second quantisation. Keep the fifteen-unit numbering and follow your institution’s required syllabus. Optional work is not a required completion gate.

1

Build the model

The bath copies σz into its oscillators: interference goes, and no outcome arrives.

The key idea

Pointer states are the states of a system that its coupling to an environment records without disturbing: they stay pure and unentangled, while a superposition of two of them loses its interference as the environment's records of them become distinguishable.

Example. A qubit whose σz is recorded by phonons with S = Σₖ4|gₖ|²/ωₖ² = 3.5 keeps |↑⟩ and |↓⟩ intact, while (|↑⟩ + |↓⟩)/√2 keeps only e(−3.5) = 0.030 of its coherence at long times; the NV centre in diamond, S ≈ 3.5, emits about 3% of its light in its zero-phonon line.

The complete physical model and assumptions

Quantum Mechanics II modelled a measurement as entanglement with a pointer, c↑|↑⟩|P↑⟩ + c↓|↓⟩|P↓⟩, and noted that the system's coherence is then multiplied by ⟨P↓|P↑⟩. This lesson builds the pointer out of an environment and computes that overlap. Couple a qubit through σz to a bath of oscillators in their vacuum. σz commutes with the whole Hamiltonian, so no energy leaves the qubit and |↑⟩ and |↓⟩ are never disturbed; but each oscillator feels a force whose sign is set by σz and is displaced to +αₖ or −αₖ accordingly. The two branches drag apart two coherent states in every mode, and the coherence picks up the product of their overlaps, exp[−Σₖ4|gₖ|²(1 − cos ωₖt)/ωₖ²]. One mode hands the coherence back every period; a continuum does not, and whether the loss is total is decided by the slowest modes. The states the coupling leaves unentangled, the eigenstates of the system operator it contains, are the pointer states, provided the coupling outweighs the system's own Hamiltonian; for a weakly damped oscillator, whose Hamiltonian only carries one coherent state into another, they are coherent states. The model stops in three places. It is exact only because σz is conserved. It says nothing about an outcome, since qubit and bath together remain in a pure superposition of both branches. And the weights |c↑|² and |c↓|² come from the trace rule it assumes, not from anything it derives.

1Build the pointer out of oscillators

SI units with ħ kept explicit; no four-vector appears, and where the course needs one its metric is (+, −, −, −). Quoted from Quantum Mechanics II: a measurement entangles, c↑|↑⟩|P↑⟩ + c↓|↓⟩|P↓⟩, and leaves the system's coherence multiplied by ⟨P↓|P↑⟩. Now make the pointer a bath: H = ½ħω₀σz + Σₖħωₖbₖ†bₖ + ħσzΣₖ(gₖbₖ† + gₖ*bₖ), every mode in its vacuum at t = 0, so each mode is topic 11.07's driven oscillator with a constant drive whose sign is set by σz. In the interaction picture the coupling is ħσzΣₖ(gₖbₖ†e(iωₖt) + gₖ*bₖe(−iωₖt)). Its commutator at two times is a c-number times σz² = 1, so the Magnus series ends at second order and U(t) = e(iφ) exp[σzΣₖ(αₖbₖ† − αₖ*bₖ)], with φ(t) a c-number phase and αₖ = gₖ(1 − e(iωₖt))/ωₖ: derived. That is a displacement by +αₖ when σz = +1 and by −αₖ when σz = −1, and the phase φ is the same for both. So |↑⟩|0⟩ becomes |↑⟩|+α⟩, still a product state, and |↓⟩|0⟩ becomes |↓⟩|−α⟩. Only superpositions become entangled.

2Compute the overlap the bath leaves behind

Start from (c↑|↑⟩ + c↓|↓⟩)|0⟩. At time t the state is c↑|↑⟩|+α⟩ + c↓|↓⟩|−α⟩, and tracing out the bath gives ρ↑↓ = c↑c↓* ∏ₖ⟨−αₖ|αₖ⟩. Quoted from topic 11.07: ⟨β|α⟩ = exp(−|α|²/2 − |β|²/2 + β*α), so ⟨−α|α⟩ = e(−2|α|²), real and positive. With |1 − e(iωt)|² = 2(1 − cos ωt), each mode contributes 2|αₖ|² = 4|gₖ|²(1 − cos ωₖt)/ωₖ², and restoring the qubit's own precession, ρ↑↓(t) = ρ↑↓(0) e(−iω₀t) exp[−Γ(t)] with Γ(t) = Σₖ4|gₖ|²(1 − cos ωₖt)/ωₖ²: derived. Checks. Dimensions: gₖ/ωₖ is a pure number. No coupling: Γ = 0. Short times: Γ ≈ 2t²Σₖ|gₖ|², quadratic as second-order perturbation theory requires, so there is no rate until the bath has had time to respond. One mode: Γ swings between 0 and 8|g|²/ω² and is zero again at t = 2π/ω, when the displaced oscillator has come back to its vacuum and the record is wiped. Monroe and co-workers (1996) drove a ⁹Be⁺ ion's motion 83 nm apart in its two spin branches, against a 7 nm wavepacket: an overlap of e(−17.6) ≈ 2 × 10⁻⁸.

3Many modes: when the loss stops being undone

For many modes define J(ω) = 4Σₖ|gₖ|²δ(ω − ωₖ), in s⁻¹, so that Γ(t) = ∫J(ω)(1 − cos ωt)/ω² dω. Two questions decide the outcome. Do the cosines dephase? For a continuum, ∫J cos(ωt)/ω² dω dies away once t exceeds the inverse bandwidth, by the Riemann-Lebesgue lemma, and Γ settles at S = ∫J/ω² dω. Is S finite? For an ohmic bath, J = λω exp(−ω/ωc), it is not, and the exact integral is Γ = (λ/2)ln(1 + ωc²t²): derived. The coherence falls to zero, but only as (ωc t)(−λ), carried there by ever slower modes. For acoustic phonons coupled through a deformation potential, J ∝ ω³ at low ω, S is finite and the coherence stops at e(−S): the record is incomplete. At zero temperature e(−S) is also the weight of the zero-phonon line, quoted, and the NV centre in diamond, which puts about 3% of its light there, has S ≈ 3.5. A thermal bath multiplies each term by coth(ħωₖ/2kB T), quoted, which turns the ohmic power law into exponential decay. A discrete bath with spacing δ returns after 2π/δ; for any macroscopic body that is longer than anyone will wait.

4Which states survive: the pointer basis

|↑⟩ and |↓⟩ are special because the coupling contains σz and nothing else, so the bath copies them undisturbed. The general test is how fast a pure state loses purity: for a dissipator with jump operator L and rate γ, −d Tr ρ²/dt = 2γ(⟨L†L⟩ − |⟨L⟩|²) at ρ = |ψ⟩⟨ψ|, derived because Tr(ρ[H, ρ]) = 0. It vanishes only when ψ is an eigenstate of L. When the coupling outweighs the system's own Hamiltonian HS, those eigenstates are the pointer states and their superpositions are what the environment destroys; Zurek called the test the predictability sieve. An HS that does not commute with L rotates them into superpositions, so selection is a competition. A weakly damped oscillator settles it. By topic 14.05, L = a, whose eigenstates are coherent states, and ħωa†a turns |α⟩ into |αe(−iωt)⟩, still one. So |α⟩ stays pure, while the cat |α⟩ + |−α⟩, with ⟨a⟩ = 0, loses purity at 2κ|α|²tanh(|α|²), topic 14.05's rate 2κ|α|² once |α|² ≳ 2. If HS dominates and the bath cannot exchange energy at the system's frequencies, energy eigenstates are selected instead (Paz and Zurek, 1999, quoted).

5What decoherence explains

It explains two things. First, lost interference: once Γ ≫ 1, every measurement on the qubit alone returns the statistics of ρS = |c↑|²|↑⟩⟨↑| + |c↓|²|↓⟩⟨↓|, so no Ramsey fringe and no ⟨σₓ⟩ reveals the superposition. The loss is exactly the information now in the bath: for equal weights the fringe visibility is V = |⟨E↓|E↑⟩|, two pure records differ by the trace distance D = √(1 − V²), quoted, and so V² + D² = 1: what the fringes lose, the records gain. Second, a preferred basis. Quantum Mechanics II's pointer could not say why the record is of |↑⟩ and |↓⟩ rather than of (|↑⟩ ± |↓⟩)/√2, since an entangled state with equal weights can be expanded either way. Here the Hamiltonian decides: only eigenstates of the coupled operator stay unentangled, so the basis is fixed by physics, not bookkeeping. Both are seen. Brune and co-workers (1996) watched the interference between two coherent field states in a cavity fade faster the further apart they were, and Hornberger and co-workers (2003) saw fullerene fringes fade exponentially with background-gas pressure, each collision adding to the record.

6What it leaves open

Decoherence never removes a branch. The qubit-plus-bath state is c↑|↑⟩|E↑⟩ + c↓|↓⟩|E↓⟩, as pure as it began, and ρS is diagonal only because the trace discards the correlations. Read ρS as the qubit being up or down, unknown which, and an improper mixture, obtained by ignoring a partner, has been mistaken for a proper one, an ensemble of definite states. The two agree on every measurement of the qubit and disagree on joint ones, where an observable sensitive to both branches still shows the interference. One mode does hand the coherence back at t = 2π/ω; with many, only the impracticality of reversing the bath prevents it. So decoherence explains why interference becomes unobservable and which basis is preferred. It does not explain why one outcome occurs, and the weights |c↑|² and |c↓|² enter through the trace rule, which it assumes. That is the measurement problem, and it stands. Copenhagen, Everett's relative states, de Broglie-Bohm pilot waves and objective-collapse models such as GRW answer it differently, the last with testable predictions; this course names them only.

2

Key equations

Read the units, assumptions and limits before substituting.

Pure-dephasing model of qubit and bath

H = ½ħω₀σz + Σₖ ħωₖbₖ†bₖ + ħσz Σₖ (gₖbₖ† + gₖ*bₖ)

The model, assumed. Since [σz, H] = 0 the populations never move and no energy leaves the qubit, T₁ = ∞: whatever coherence is lost is lost to a record.

ω₀, ωₖ and gₖ in s⁻¹; bₖ and bₖ† dimensionless; σz = |↑⟩⟨↑| − |↓⟩⟨↓|; each term of H in J.

Each branch displaces every mode

|↑⟩|0⟩ → |↑⟩|+α(t)⟩ and |↓⟩|0⟩ → |↓⟩|−α(t)⟩, with αₖ(t) = gₖ(1 − e(iωₖt))/ωₖ

Derived: the Magnus series stops at second order, and that term is a c-number times σz² = 1, a phase common to both branches. The σz eigenstates stay product states.

Interaction picture; |±α⟩ is the coherent state ±αₖ in every mode k; αₖ is dimensionless, with |αₖ|² quanta in mode k.

Vacuum decoherence factor

ρ↑↓(t) = ρ↑↓(0) e(−iω₀t) exp[−Σₖ 4|gₖ|²(1 − cos ωₖt)/ωₖ²]

Derived: Quantum Mechanics II's pointer overlap, computed. Short times give 1 − 2t²Σₖ|gₖ|², the second-order result; a single mode gives the coherence back every 2π/ω.

ρ↑↓ = ⟨↑|ρS|↓⟩, dimensionless; ρ↑↑ and ρ↓↓ stay constant. Each term is 2|αₖ|², from ⟨−α|α⟩ = e(−2|α|²).

Ohmic bath at zero temperature

|ρ↑↓(t)/ρ↑↓(0)| = (1 + ωc²t²)(−λ/2)

Derived. The loss is complete, but only as the power law (ωc t)(−λ), and only because modes of ever lower frequency keep adding to the record.

For 4Σₖ|gₖ|²δ(ω − ωₖ) = λω exp(−ω/ωc): λ dimensionless, cutoff ωc in s⁻¹, t in s.

Record strength and the long-time plateau

|ρ↑↓(∞)/ρ↑↓(0)| = e(−S), with S = Σₖ 4|gₖ|²/ωₖ²

Derived, as the cosines dephase. At zero temperature e(−S) is also the zero-phonon-line weight, quoted: about 3% for the NV centre in diamond (measured), so S ≈ 3.5 there.

S is dimensionless, the Huang-Rhys factor; finite when the sum converges at low ω, as for acoustic phonons in 3D.

Purity loss and the predictability sieve

−d Tr ρ²/dt = 2γ(⟨L†L⟩ − |⟨L⟩|²) at a pure state ρ = |ψ⟩⟨ψ|

Derived. Zero only for eigenstates of L: σz eigenstates for L = σz, coherent states for L = a, which ħωa†a keeps coherent. Those are the pointer states.

For a dissipator γ(LρL† − ½L†Lρ − ½ρL†L), γ in s⁻¹, L dimensionless; the Hamiltonian drops out of d Tr ρ²/dt.
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

2.0
0.2 to 4; steps of 0.1.
8
4 to 12; steps of 1.

Try this. Set S to 3.5, the NV centre's value, and the plateau sits near 3%; then raise Ω/δ from 4 to 12 and watch the recurrence move out with the number of modes. Evenly spaced modes all rephase at 2π/δ, so a finite bath always hands the coherence back: the record was spread over the modes, never erased.

Live results

LONG-TIME COHERENCE e(−S)
0.135
RECORD DISTINGUISHABILITY
0.991
ONE-MODE DIP e(−2S)
0.018
MODES WITHIN Ω ± Ω/2
9 modes

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Advanced Quantum Mechanics modelExact coherence |ρ↑↓(t)|/|ρ↑↓(0)| = exp[−Γ(t)] of a qubit coupled through σ_z to vacuum oscillators, time in periods 2π/Ω. Solid: modes spaced by δ = Ω/8 in a Gaussian band of r.m.s. width Ω/4 about Ω, record strength S = 2.0; it falls to e^(−S) = 0.135 and returns to 1 at 2π/δ = 8 periods. Dashed: one mode with the same S, dipping to 0.018 every period.coherence |ρ↑↓(t)| / |ρ↑↓(0)|solid: band of modesdashed: one modee(−S)106122π/δt in periods 2π/Ω
Exact coherence |ρ↑↓(t)|/|ρ↑↓(0)| = exp[−Γ(t)] of a qubit coupled through σz to vacuum oscillators, time in periods 2π/Ω. Solid: modes spaced by δ = Ω/8 in a Gaussian band of r.m.s. width Ω/4 about Ω, record strength S = 2.0; it falls to e(−S) = 0.135 and returns to 1 at 2π/δ = 8 periods. Dashed: one mode with the same S, dipping to 0.018 every period.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Record strength S: 2.0; Band centre over spacing, Ω/δ: 8

Baseline 2.0; choose 0.2 to 4, in steps of 0.1.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

Decoherence is often read as collapse: once the bath has recorded σz, the qubit really is in |↑⟩ or in |↓⟩, and ρS merely expresses which one we do not know. Every step of this lesson's calculation is unitary, so that cannot be right. The joint state is c↑|↑⟩|E↑⟩ + c↓|↓⟩|E↓⟩ at every time, both branches present, and ρS becomes diagonal because the partial trace throws the correlations away. That improper mixture agrees with a genuine ensemble of up and down qubits on every measurement of the qubit alone, but not on measurements of qubit and bath together, and with a single bath mode the coherence returns after one period, which no story of hidden definite values allows. Decoherence makes the alternatives look classical; it does not choose one.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

A ⁴⁰Ca⁺ ion's motional mode, ω/2π = 2.00 MHz, starts in its ground state, and a spin-dependent force couples it to the ion's spin as ħgσz(b + b†) with g/2π = 0.50 MHz. The spin starts in (|↑⟩ + |↓⟩)/√2. Find the largest |α| and when it occurs, the smallest |ρ↑↓| and the wavepacket separation then, when |ρ↑↓| is back to ½, and the separation needed for an overlap of 1%.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

MediumExample 2

A qubit's σz couples to an ohmic bath at zero temperature, J(ω) = 4Σₖ|gₖ|²δ(ω − ωₖ) = λω exp(−ω/ωc), with λ = 0.20 and ωc = 1.0 × 10¹² s⁻¹. Find the coherence factor at 1.0 ps, 1.0 ns and 1.0 µs, and the times to reach 1/e and 1%. Then remove every mode below ωₘᵢₙ = 1.0 × 10⁶ s⁻¹, as a finite sample would, and find where the coherence stops.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 6 steps revealed.

HardExample 3

A molecule's mirror-image forms |L⟩ and |R⟩ (σz = ±1) tunnel through HS = −½ħΔσₓ, Δ = 2.0 × 10⁵ s⁻¹. Gas collisions record which form it is in: in the Markov limit, the term (γφ/2)(σzρσz − ρ) with γφ = 5.0 × 10⁹ s⁻¹. Find the purity-loss rates of |L⟩ and of (|L⟩ + |R⟩)/√2, the isolated |L⟩-to-|R⟩ time, and the time for |L⟩ to relax halfway to an even mixture. Name the pointer basis, and test against ammonia's 23.9 GHz inversion line.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 6 steps revealed.

6

Quick check

Select an answer and explain it before checking.

A qubit prepared in (|↑⟩ + |↓⟩)/√2 is coupled through σz to a large bath of oscillators in their vacuum. Later the coherence factor exp[−Γ(t)] has fallen to 10⁻⁶. Which statement about that moment is correct?
7

Review & continue

Pointer states are the states of a system that its coupling to an environment records without disturbing: they stay pure and unentangled, while a superposition of two of them loses its interference as the environment's records of them become distinguishable.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.