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Optional unit · institution-dependent · Interactive lesson

Amplitude damping, phase damping and depolarisation

Nearly every qubit error budget is written in three channels. This lesson builds each one from an environment you can state in a line, draws what it does to the Bloch ball, and lets successive time intervals turn channel strengths into T₁ and T₂, including the reason no qubit, however well isolated, keeps its coherence past 2T₁.

UAQM-14.02 · Lesson 2 of 9 in this unit. Position is not a completion score.
Starting model: Amplitude damping, phase damping and depolarisationLeft: the x–z section of the Bloch ball and its image (ellipse) after t = 0.70 T₁ of amplitude damping and pure dephasing with T₂ = 2.0 T₁; dashed, the path of a pure state started at θ₀ = 90°; arrow, its Bloch vector now. Right: its coherence |ρ_eg| (solid) and ceiling √(ρₑₑ ρ_gg) (dashed) for t from 0 to 3T₁, scale 0 to ½. T₂ above 2T₁ is drawn only to show why it is impossible.Bloch section at t = 0.70 T₁, start θ₀ = 90°T₂ = 2.0 T₁ needs T₁/Tφ = 0.00|e⟩|g⟩solid |ρeg|, dashed √(ρₑₑ ρgg)t from 0 to 3 T₁
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

Populations fall as probabilities and coherences as amplitudes, so T₂ stops at 2T₁.

The key idea

Amplitude damping, phase damping and depolarisation are the qubit maps ρ → Σₖ KₖρKₖ† left by an environment that respectively absorbs the excitation, keeps a record of σz, or replaces the state by I/2 with probability p.

Example. Hydrogen in a 2p–1s superposition, left for 1.00 ns with T₁ = 1.60 ns, suffers amplitude damping with p = 1 − e(−1.00/1.596) = 0.466: its excited population keeps 0.534 of its value and its coherence √0.534 = 0.731.

The complete physical model and assumptions

A qubit that starts uncorrelated with its environment and evolves with it is left, once the environment is traced out, with a Kraus map, and three couplings cover most of what a lone qubit suffers. If the coupling can hand the excitation over, U|e⟩|0E⟩ = √(1 − p)|e⟩|0E⟩ + √p|g⟩|1E⟩, the result is amplitude damping: the excited population is multiplied by 1 − p but the coherence only by √(1 − p), because a coherence is a product of amplitudes and survives only in the branch where nothing was emitted. If the coupling commutes with σz, the environment can only keep a record of which state the qubit is in, and the coherence is multiplied by the overlap of the two records: phase damping. If, with probability p, one of I, σₓ, σy and σz chosen at random acts, the state is replaced by I/2 with probability p: depolarisation. On the Bloch ball the three squeeze towards the ground-state pole, towards the z axis and towards the centre. When each interval meets a fresh environment, channel strengths must compose by multiplication, and the only continuous solutions are exponentials: populations relax in T₁, coherences in T₂, with 1/T₂ = 1/(2T₁) + 1/Tφ, and positivity alone forbids T₂ above 2T₁. All of that is derived; the rate 1/T₁ = Γ is quoted from Unit 12. The model stops where the environment remembers: slow noise gives a Gaussian decay that an echo reverses, and a strongly coupled cavity hands the photon back.

1Amplitude damping: the excitation is handed over

Units are SI and nothing here is relativistic, so the course metric (+, −, −, −) never enters. Write ρeg = ⟨e|ρ|g⟩ and take the Bloch vector with z = ρₑₑ − ρgg and |ρeg| = ½√(x² + y²), so the ground state is the south pole. Let the environment start empty, in |0E⟩. The ground state has nothing to give, U|g⟩|0E⟩ = |g⟩|0E⟩, while U|e⟩|0E⟩ = √(1 − p)|e⟩|0E⟩ + √p|g⟩|1E⟩. Topic 14.01's rule Kₖ = ⟨kE|U|0E⟩ gives K₀ = |g⟩⟨g| + √(1 − p)|e⟩⟨e| and K₁ = √p|g⟩⟨e|, with K₀†K₀ + K₁†K₁ = I, derived. So ρₑₑ → (1 − p)ρₑₑ, ρgg → ρgg + pρₑₑ and ρeg → √(1 − p)ρeg. The square root is the point: coherence survives only in the K₀ branch, where the excited amplitude shrank by √(1 − p) and the ground amplitude did not, while in the K₁ branch the environment holds a photon, an orthogonal record. At p = 0.36 the excited population keeps 0.64 and the coherence 0.80. The ball becomes an ellipsoid with semi-axes √(1 − p), √(1 − p) and 1 − p, centred at z = −p and touching the ground pole; p = 0 is the identity, p = 1 sends every state to |g⟩.

2Phase damping: the environment keeps a record of σz

A coupling that commutes with σz cannot move population; it can only let the environment learn which state the qubit is in. Then U = |g⟩⟨g| ⊗ Ug + |e⟩⟨e| ⊗ Uₑ, and from |E⟩ the environment ends in |Eg⟩ = Ug|E⟩ or |Eₑ⟩ = Uₑ|E⟩. Tracing it out leaves ρₑₑ and ρgg alone and multiplies ρeg by ⟨Eg|Eₑ⟩, derived: the better the two records can be told apart, the less coherence survives. Make the environment a qubit tilted by ±θ, |Eg⟩ = cos θ|0E⟩ + sin θ|1E⟩ and |Eₑ⟩ = cos θ|0E⟩ − sin θ|1E⟩. Then K₀ = cos θ I and K₁ = −sin θ σz, a phase flip with probability q = sin²θ, and the coherence factor 1 − 2q = cos 2θ is the overlap: θ = 30° gives q = 0.25 and keeps half the coherence. An environment that flips only when the qubit is excited, with probability λ, gives the different-looking pair |g⟩⟨g| + √(1 − λ)|e⟩⟨e| and √λ|e⟩⟨e|, the same channel whenever √(1 − λ) = 1 − 2q: topic 14.01's unitary freedom. On the ball x and y shrink by 1 − 2q and z stays, so states slide towards the z axis; q = ½ is a σz measurement nobody read.

3Depolarisation: a random Pauli erases the state

Replace the qubit by I/2 with probability p: ρ → (1 − p)ρ + p I/2. Kraus operators need I/2 written as an average. With σ₀ = I, the rules σᵢσⱼσᵢ = −σⱼ for i ≠ j and σⱼσⱼσⱼ = σⱼ give Σμ σμσⱼσμ = (1 + 1 − 1 − 1)σⱼ = 0, while Σμ σμIσμ = 4I. So for any ρ = ½(I + r⋅σ), Σμ σμρσμ = 2I, derived: applying I, σₓ, σy or σz at random, each with probability ¼, erases every Bloch vector. Hence ρ → (1 − 3p/4)ρ + (p/4)(σₓρσₓ + σyρσy + σzρσz), with K₀ = √(1 − 3p/4) I and Kᵢ = √(p/4) σᵢ, and r → (1 − p)r: the only one of the three channels with no preferred axis, and the only one that drives every state to the centre. Every pure state keeps fidelity 1 − p/2, so p = 0.06 leaves 0.97. The weight 1 − 3p/4 stays non-negative up to p = 4/3, where r → −r/3, so the map stays physical past the point where p reads as a probability. At p = 2 it would give r → −r, which maps the ball onto itself, yet on half of a singlet it leaves an eigenvalue of −½: positive, not completely positive, as topic 14.01 warned.

4Successive intervals compose into exponentials

A channel has a strength, not yet a clock. Let amplitude damping over a time t have strength p(t), and suppose each interval meets a fresh environment, the photon emitted earlier having left: the Markov condition, which topic 14.03 earns. Running t and then s must equal running t + s, and the factors on the excited population multiply: 1 − p(t + s) = [1 − p(t)][1 − p(s)], with p(0) = 0. Logarithms make ln[1 − p(t)] additive, and an additive continuous function is linear, so 1 − p(t) = e(−t/T₁), derived. Then ρₑₑ(t) = ρₑₑ(0)e(−t/T₁) and the coherence factor is √(1 − p) = e(−t/(2T₁)): amplitude damping alone gives T₂ = 2T₁. For t ≪ T₁, p ≈ t/T₁ is the golden-rule probability Γt, so T₁ = 1/Γ, Unit 12's Weisskopf-Wigner rate, quoted. Hydrogen 2p has Γ = 6.26 × 10⁸ s⁻¹, so T₁ = 1.60 ns, as measured, and T₂ = 3.19 ns. A coherence decaying as e(−t/T₂) gives a Lorentzian line of full width 1/(πT₂); for rubidium-87's D2 line, measured lifetime 26.24 ns, that is 6.07 MHz, the natural width of the line. Phase damping composes the same way, 1 − 2q(t) = e(−t/Tφ).

5Dephasing adds rates, and positivity caps T₂ at 2T₁

Phase damping leaves populations alone and both channels multiply ρeg by a number, so they commute and their coherence factors multiply: e(−t/(2T₁))e(−t/Tφ) = e(−t/T₂), with 1/T₂ = 1/(2T₁) + 1/Tφ, derived. Rates add, times do not, and since 1/Tφ ≥ 0, T₂ ≤ 2T₁. The bound outlives these models. A 2 × 2 density matrix has det ρ ≥ 0, so |ρeg|² ≤ ρₑₑρgg. Take a pure state with a small excited population a, so |ρeg(0)|² = a(1 − a). If populations relax to the ground state in T₁ and coherences decay in T₂, positivity at time t needs a(1 − a)e(−2t/T₂) ≤ ae(−t/T₁)[1 − ae(−t/T₁)], and as a → 0 that is e(−2t/T₂) ≤ e(−t/T₁): T₂ ≤ 2T₁ for every environment, derived. A coherence is bounded by the square root of the excited population, so it can fall at most half as fast. A transmon measured at T₁ = 62 μs and T₂ = 85 μs therefore has 1/Tφ = 1/85 − 1/124 μs⁻¹, Tφ = 270 μs. A report of T₂ = 120 μs beside T₁ = 50 μs cannot be right: for a = 0.01, after 100 μs ρ would have the eigenvalue −5.2 × 10⁻⁴.

6Where the exponential laws stop

Everything above rests on the fresh environment, which needs the environment's correlation time τc to be short against every time of interest. For an atom in vacuum τc is about 1/ω₀, 6.5 × 10⁻¹⁷ s for Lyman-α, seven orders of magnitude below T₁ = 1.60 ns. It fails in three ways. At times comparable with τc, survival falls quadratically in t, as the short-time expansion of e(−iHt/ħ) requires, not linearly; the deviation is real but far too early to resolve for an atom. In a cavity whose coupling g exceeds both its loss and Γ, the excitation comes back, the population oscillates at Unit 11's vacuum Rabi frequency and no composition law holds, so topic 14.03's master equation fails there too. And slow noise remembers: if a qubit's frequency is fixed during each run but scatters between runs with r.m.s. spread σ, the coherence factor averages to e(−σ²t²/2), a Gaussian; for σ/2π = 10 kHz it reaches 1/e at 22.5 μs. That decay, written T₂*, reverses under a spin echo, which no composing channel allows. Finite temperature, which adds absorption in proportion to n̄, is topic 14.04's.

2

Key equations

Read the units, assumptions and limits before substituting.

Amplitude damping

K₀ = |g⟩⟨g| + √(1 − p) |e⟩⟨e|, K₁ = √p |g⟩⟨e|

Derived from U|e⟩|0E⟩ = √(1 − p)|e⟩|0E⟩ + √p|g⟩|1E⟩. On the Bloch ball x and y shrink by √(1 − p) and z + 1 by 1 − p: towards the ground pole.

p, dimensionless, is the probability that an excitation has passed to the environment; K₀†K₀ + K₁†K₁ = I

Phase damping

K₀ = √(1 − q) I, K₁ = √q σz, so ρeg → (1 − 2q) ρeg

Derived from U|g⟩|E⟩ = |g⟩|Eg⟩ and U|e⟩|E⟩ = |e⟩|Eₑ⟩. Populations are untouched; x and y shrink by 1 − 2q, towards the z axis.

q from 0 to ½, dimensionless; 1 − 2q = ⟨Eg|Eₑ⟩, the real overlap of the environment's two records of σz

Depolarisation

(1 − p)ρ + p I/2 = (1 − 3p/4)ρ + (p/4)(σₓρσₓ + σyρσy + σzρσz)

Derived from Σμ σμρσμ = 2I for any unit-trace ρ, with σ₀ = I. Every Bloch vector shrinks by 1 − p, uniformly towards the centre.

p dimensionless; the Kraus weights need 1 − 3p/4 ≥ 0, so the map is completely positive for 0 ≤ p ≤ 4/3

Composition forces an exponential

1 − p(t + s) = [1 − p(t)][1 − p(s)] ⇒ 1 − p(t) = e(−t/T₁)

Derived: a continuous f with f(t + s) = f(t)f(s) and f(0) = 1 is an exponential. With Unit 12's Γ, quoted, T₁ = 1/Γ = 1.60 ns for hydrogen 2p.

t, s and T₁ in s; holds when each interval meets a fresh environment; for t ≪ T₁, p ≈ t/T₁ = Γt

Coherence time with pure dephasing

1/T₂ = 1/(2T₁) + 1/Tφ

Derived: the two channels commute and their coherence factors multiply, so rates add. Amplitude damping alone gives T₂ = 2T₁, 3.19 ns for hydrogen 2p.

T₂, T₁, Tφ in s; |ρeg(t)| = |ρeg(0)| e(−t/T₂), and phase damping alone gives 1 − 2q(t) = e(−t/Tφ)

Positivity caps the coherence time

|ρeg|² ≤ ρₑₑ ρgg ⇒ T₂ ≤ 2T₁

Derived for any environment relaxing the qubit to |g⟩: a pure state with ρₑₑ = a ≪ 1 has |ρeg|² ≈ a, and ae(−2t/T₂) may never exceed ae(−t/T₁).

ρₑₑ and ρgg are populations and ρeg = ⟨e|ρ|g⟩, all dimensionless; the inequality is det ρ ≥ 0
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

0.70 T₁
0 to 3 T₁; steps of 0.05.
2.0 T₁
0.2 to 4 T₁; steps of 0.1.
90 °
10 to 170 °; steps of 10.

Try this. At the opening T₂ = 2T₁ the ellipse touches the ground pole and an equator state falls along z = x² − 1. Lower T₂ and the ball collapses towards the z axis first. Then push T₂ to 4T₁ with θ₀ = 130°: the coherence climbs over its ceiling and the ellipse bulges out beside the ground pole.

Live results

EXCITED POPULATION ρₑₑ
0.248
COHERENCE |ρeg|
0.352
CEILING √(ρₑₑ ρgg)
0.432
NEEDED T₁/Tφ
0.00

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

Interactive Advanced Quantum Mechanics modelLeft: the x–z section of the Bloch ball and its image (ellipse) after t = 0.70 T₁ of amplitude damping and pure dephasing with T₂ = 2.0 T₁; dashed, the path of a pure state started at θ₀ = 90°; arrow, its Bloch vector now. Right: its coherence |ρ_eg| (solid) and ceiling √(ρₑₑ ρ_gg) (dashed) for t from 0 to 3T₁, scale 0 to ½. T₂ above 2T₁ is drawn only to show why it is impossible.Bloch section at t = 0.70 T₁, start θ₀ = 90°T₂ = 2.0 T₁ needs T₁/Tφ = 0.00|e⟩|g⟩solid |ρeg|, dashed √(ρₑₑ ρgg)t from 0 to 3 T₁
Left: the x–z section of the Bloch ball and its image (ellipse) after t = 0.70 T₁ of amplitude damping and pure dephasing with T₂ = 2.0 T₁; dashed, the path of a pure state started at θ₀ = 90°; arrow, its Bloch vector now. Right: its coherence |ρeg| (solid) and ceiling √(ρₑₑ ρgg) (dashed) for t from 0 to 3T₁, scale 0 to ½. T₂ above 2T₁ is drawn only to show why it is impossible.

Predict → change one variable → check

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

Saved baseline. Elapsed time t: 0.70 T₁; Coherence time T₂: 2.0 T₁; Starting polar angle θ₀: 90 °

Baseline 0.70; choose 0 to 3, in steps of 0.05.
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

The commonest error is to treat T₂ as the life of a superposition that cannot outlast the excited state it involves, so T₂ ≤ T₁, or to add rates as 1/T₂ = 1/T₁ + 1/Tφ. Both treat a coherence like a population. A population is a probability, the square of an amplitude; ρeg is a product of two amplitudes, cₑ cg*, and amplitude damping shrinks only cₑ, by √(1 − p). So the coherence falls at half the population's rate, and T₂ = 2T₁ is the radiative limit, not an anomaly: hydrogen's 2p–1s coherence lasts 3.19 ns against a 1.60 ns lifetime, and a radiatively broadened line has full width 1/(πT₂) = Γ/2π, not the Γ/π that T₂ = T₁ would give. Dephasing can only pull T₂ below 2T₁; nothing pushes it above.

5

Practice & worked examples

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

EasyExample 1

A hydrogen atom is prepared in (√3|e⟩ + |g⟩)/2, with |e⟩ = 2p and |g⟩ = 1s, and left in the vacuum for 0.712 ns, which with T₁ = 1.596 ns gives amplitude damping with p = 0.360. Work in the frame rotating at the transition frequency. Find the density matrix and Bloch vector afterwards, and the probability that a photon has been emitted.

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 transmon is measured to have T₁ = 62 μs and an echo coherence time T₂ = 85 μs. Find its pure-dephasing time, then split a 40 μs idle into an amplitude-damping channel and a phase-damping channel, giving p and q. Starting from |+⟩ = (|e⟩ + |g⟩)/√2, find the Bloch vector and the fidelity with |+⟩ after the idle.

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.

HardExample 3

Each gate on qubit A acts as a depolarising channel with p₁ = 2.0 × 10⁻³. Qubit A starts as half of the Bell pair |Φ⁺⟩ = (|gg⟩ + |ee⟩)/√2, and partner B is never touched. Find the channel left after N = 300 gates, the eigenvalues of the pair's state, and, using the partial-transpose test, the largest number of gates after which A and B are still entangled.

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 superconducting qubit relaxes to its ground state with T₁ = 50 μs, and an echo experiment on the same device reports a coherence time T₂ = 80 μs. What does this imply?
7

Review & continue

Amplitude damping, phase damping and depolarisation are the qubit maps ρ → Σₖ KₖρKₖ† left by an environment that respectively absorbs the excitation, keeps a record of σz, or replaces the state by I/2 with probability p.

  • 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.