advanced34 min studyLesson 415 in Physics

Amplitude damping and energy relaxation

Compute decay channels, distinguish population and coherence rates, and impose thermal detailed balance.

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The question for today

What you’ll understand

Apply the amplitude-damping channel to a general qubit.
Relate decay probability to a constant energy-relaxation rate.
Compute a finite-temperature stationary population from upward and downward rates.

Energy relaxation transfers excitation between a system and its surroundings. For a qubit in a vacuum-like bath, the simplest channel takes excited population toward the ground state while attenuating coherence at a different rate. Its non-unital character is physically important: unlike a random rotation, it can purify the system by cooling it. A thermal bath adds upward transitions and changes both the final population and the relevant relaxation time.

The decay channel keeps the no-emission branch

Choose $|0\rangle$ as the ground state and $|1\rangle$ as the excited state. Over a specified interval, let the decay probability of an initially excited state be $\lambda$ . The jump operator transfers excitation downward; a second operator preserves the ground amplitude and reduces the excited amplitude. Both are needed for normalization. Treating the jump alone as a deterministic operation would confuse a detected-emission outcome with an experiment in which the environment record is ignored.

Applying the two Kraus terms shows that the lost excited population enters the ground population, while off-diagonal terms acquire the square root of the survival probability. This square root follows from amplitudes, not a second population-decay process. The map leaves the ground state fixed, but it does not leave the identity fixed. At complete decay every initial state becomes ground, demonstrating that dissipative reduced evolution need not monotonically decrease purity.

K_0=\begin{pmatrix}1&0\\0&\sqrt{1-\lambda}\end{pmatrix},\quad K_1=\begin{pmatrix}0&\sqrt\lambda\\0&0\end{pmatrix},\quad \rho' =\begin{pmatrix}\rho_{00}+\lambda\rho_{11}&\sqrt{1-\lambda}\rho_{01}\\\sqrt{1-\lambda}\rho_{10}&(1-\lambda)\rho_{11}\end{pmatrix}

All matrix entries and λ are dimensionless, with 0 ≤ λ ≤ 1. The map is exact for this specified channel; interpreting it as vacuum emission assumes no upward excitation and no additional dephasing during the interval.↔ Scroll wide equations horizontally.

Continuous relaxation fixes the factor of two

For a time-independent downward rate, the Lindblad direction is the lowering operator and its dissipator gives exponential excited-population survival. Composing decay channels multiplies the survival factors, yielding the semigroup relation between probability and elapsed time. Coherence decays at half the downward population rate in a frame removing the coherent energy splitting. Thus the absence of pure dephasing does not mean coherence is preserved during energy loss.

A physical atom can have short-time nonexponential decay, driving, thermal excitation, or spectral structure, none of which is captured by the constant-rate vacuum model. Its use requires a suitable weak-coupling and Markov regime. Even in that regime, the energy of the reduced qubit is not conserved: the mean emitted energy equals the decrease of the excitation energy when no drive does work. This is compatible with unitary conservation of total energy for an appropriately closed system–bath description.

\rho_{11}(t)=e^{-\gamma_\downarrow t}\rho_{11}(0),\quad \rho_{01}(t)=e^{-\gamma_\downarrow t/2}\rho_{01}(0),\quad\lambda(t)=1-e^{-\gamma_\downarrow t},\quad T_1=\gamma_\downarrow^{-1}

The rate γ↓ is in s⁻¹, t and T1 in seconds, and the coherence formula is in the interaction frame. These expressions assume constant vacuum damping, no drive, and no extra dephasing.↔ Scroll wide equations horizontally.

A thermal reservoir must also excite

At nonzero temperature, let the upward rate be $\gamma_\uparrow$ . The excited population receives upward influx proportional to the ground population and downward outflux proportional to the excited population. Their sum sets the decay rate of deviations from stationarity, while their ratio sets the stationary population. For a single equilibrium reservoir at positive temperature, detailed balance makes the upward rate smaller than the downward rate by the Boltzmann factor of the positive level gap.

Arbitrary positive upward and downward rates still define a valid GKSL model, but calling them thermal requires that additional ratio and a consistent gap. Pure downward damping is the zero-temperature limit for a positive gap. Equal rates give equal stationary populations, corresponding to the infinite-temperature limit of this two-level model. Negative or inverted effective temperatures require a separate population-inversion interpretation rather than being silently identified with an ordinary positive-temperature bath.

\dot p_1=\gamma_\uparrow(1-p_1)-\gamma_\downarrow p_1,\qquad p_{1,*}=\frac{\gamma_\uparrow}{\gamma_\uparrow+\gamma_\downarrow},\qquad\frac{\gamma_\uparrow}{\gamma_\downarrow}=e^{-\beta\varepsilon}

The population p1 is dimensionless, both rates are in s⁻¹, ε > 0 is the energy gap in joules, and β = 1/(kBT) has units J⁻¹. The stationary formula requires a nonzero sum of rates; the final relation additionally assumes a single equilibrium thermal reservoir and the appropriate weak-coupling detailed-balance regime.↔ Scroll wide equations horizontally.

Work it through

Decay of a coherent equal superposition

Apply the vacuum amplitude-damping channel with $\lambda=3/4$ to $|+\rangle$ . Find the full output, its purity, and mean excitation energy for a gap $\varepsilon$ , taking ground energy zero.

The input has every matrix entry equal to one half. Survival is $1-\lambda=1/4$ , so the excited population becomes $1/8$ and the ground population becomes $1/2+(3/4)(1/2)=7/8$ .
The coherence factor is $\sqrt{1/4}=1/2$ , leaving each off-diagonal entry $1/4$ . The output determinant is $(7/8)(1/8)-(1/4)^2=3/64$ , so it is positive and normalized.
For a normalized two-level matrix, purity is one minus twice the determinant: $1-6/64=29/32$ . Mean excitation energy is $\varepsilon\rho_{11}=\varepsilon/8$ , down from $\varepsilon/2$ .

The result

The output is [[7/8, 1/4], [1/4, 1/8]], with purity 29/32 and energy ε/8. The emitted mean energy is 3ε/8; the final state is approaching the pure ground state.

The living notebook

Explore the model

Interactive lab

Interactive lab · markov

Downward population transfer in a discrete analogue

The chain always starts in state 0; identify that state with the excited population for this analogy. Reset and set both controls to p = 0.50 and q = 0. Predict 50% in state 1 after one Step and 75% after two Steps; state 1 is the absorbing ground analogue. This is exact classical population transfer, with no coherence or emitted-photon state and no continuous decay-rate control.

P(state 0)P(state 1)Stationary distribution

P(state 1) at step 0: 0%
Stationary P(state 1): 60%
Chain behavior: Monotone convergence

Transition p · 0 → 10.3Transition q · 1 → 00.2

t = 0 s

Exact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA one-way population process can approach a pure basis population, but the classical instrument cannot show the coherence square-root law.

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Read the checkpoint explanation

The excited amplitude carries the square root of the survival probability. The Lindblad equation therefore gives coherence decay γ/2 and excited-population decay γ in the interaction frame.

Your turn

Try it yourself

Let $\gamma_\downarrow=3\,\mathrm{s}^{-1}$ and $\gamma_\uparrow=1\,\mathrm{s}^{-1}$ . Start fully excited. Find the stationary excited population and the population at $t=(\ln3)/4\,\mathrm{s}$ . What value of $\beta\varepsilon$ would make these rates thermal?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Evolve the deviation from the stationary population at the sum of the two rates.

Compare with the solution

The rate equation is $\dot p=1-4p$ , so $p_*=1/4$ and $p(t)=1/4+(3/4)e^{-4t}$ . The exponential is $1/3$ at the given time, giving $p=1/2$ . Detailed balance requires $\gamma_\uparrow/\gamma_\downarrow=e^{-\beta\varepsilon}=1/3$ , hence $\beta\varepsilon=\ln3$ , where $\beta=1/(k_BT)$ . This positive value is consistent with an ordinary thermal bath for a positive gap.

Keep the idea

Take it with you

Vacuum damping multiplies coherence by the square root of population survival.
Energy relaxation can purify a qubit as it approaches its ground state.
Thermal populations require a detailed-balance ratio as well as positive rates.

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