University Advanced Quantum Mechanics · unit UAQM-14
Open Quantum Systems and the Master Equation
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Reduced dynamics and the Kraus representation: infer an input with uncertainty
The Kraus representation writes the evolution of a system that shares a unitary with an environment as ρ → Σₖ Kₖ ρ Kₖ†, a sum of operators acting on the system alone whose only constraint is Σₖ Kₖ†Kₖ = 1. Use the supplied topic model Y_out = 0.5*(1+(cosd(th)^2*cosd(a)+sind(th)^2*(2*p-1))^2+cosd(th)^2*sind(a)^2), where Y_out is PURITY Tr ρ_S² (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. a is Initial Bloch angle α of S, measured in °. Keep p (Excitation probability p of E) = 0; th (Coupling angle gt) = 45 ° fixed. For this exercise use only 19.125 ° ≤ a ≤ 160.875 ° on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.93316718, with a bounded reading uncertainty ±0.0099678. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.014175 °, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
PURITY Tr ρ_S² against a
Predictions from the supplied Reduced dynamics and the Kraus representation model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): PURITY Tr ρ_S²
Horizontal axis (x): a: Initial Bloch angle α of S / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Initial Bloch angle α of S / ° | PURITY Tr ρ_S² |
|---|---|
| 19.125 | 0.5272161802032339 |
| 54.5625 | 0.6880232491909773 |
| 90 | 0.875 |
| 125.4375 | 0.9779305232542341 |
| 160.875 | 0.9996192034366729 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(a) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(19.125) = 0.527216 and Y_out(160.875) = 0.999619. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives a_line = 19.125 + (0.933167 − 0.527216)(160.875 − 19.125)/(0.999619 − 0.527216) = 140.935 °. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives a = 105.593 °; substituting back gives Y_out = 0.93316718.
- The interpolation error in the input is |a_line − a| = 35.3428 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.92319938 and 0.94313498. Sort the roots even if the response decreases: 102.484 ° ≤ a ≤ 108.997 °.
- The lower and upper input deviations are 3.1089 ° and 3.40472 °. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer a ≈ 105.593 °; compatible interval [102.484, 108.997] °. Endpoint interpolation gives 140.935 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Amplitude damping, phase damping and depolarisation: infer an input with uncertainty
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 σ_z, or replaces the state by I/2 with probability p. Use the supplied topic model Y_out = 1/k-0.5, where Y_out is NEEDED T₁/T_φ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. k is Coherence time T₂, measured in T₁. Keep t (Elapsed time t) = 0.7 T₁; th (Starting polar angle θ₀) = 90 ° fixed. For this exercise use only 0.60375 T₁ ≤ k ≤ 3.59625 T₁ on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = -0.088337604, with a bounded reading uncertainty ±0.020039. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown k by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00029925 T₁, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
NEEDED T₁/T_φ against k
Predictions from the supplied Amplitude damping, phase damping and depolarisation model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): NEEDED T₁/T_φ
Horizontal axis (x): k: Coherence time T₂ / T₁
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: Coherence time T₂ / T₁ | NEEDED T₁/T_φ |
|---|---|
| 0.60375 | 1.1563146997929605 |
| 1.351875 | 0.23971336107258445 |
| 2.0999999999999996 | -0.023809523809523725 |
| 2.8481249999999996 | -0.14889181479043223 |
| 3.5962499999999995 | -0.22193256864789707 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(k) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.60375) = 1.15631 and Y_out(3.59625) = -0.221933. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives k_line = 0.60375 + (-0.0883376 − 1.15631)(3.59625 − 0.60375)/(-0.221933 − 1.15631) = 3.30618 T₁. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives k = 2.42917 T₁; substituting back gives Y_out = -0.088337604.
- The interpolation error in the input is |k_line − k| = 0.877009 T₁. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.1083766 and -0.068298604. Sort the roots even if the response decreases: 2.31642 T₁ ≤ k ≤ 2.55347 T₁.
- The lower and upper input deviations are 0.112759 T₁ and 0.124299 T₁. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer k ≈ 2.42917 T₁; compatible interval [2.31642, 2.55347] T₁. Endpoint interpolation gives 3.30618 T₁.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The Born-Markov master equation for an atom in vacuum: infer an input with uncertainty
The Born-Markov master equation is the closed equation of motion for an atom's density matrix once the vacuum field has been traced out: unitary motion at the observed frequency, plus a dissipator that empties |e⟩ into |g⟩ at the rate Γ. Use the supplied topic model Y_out = p*exp(-t), where Y_out is EXCITED POPULATION ρₑₑ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. t is Time Γt, dimensionless. Keep p (Initial excited population ρₑₑ(0)) = 0.75; f (Coherence, fraction of its maximum) = 1 fixed. For this exercise use only 0.53125 ≤ t ≤ 4.46875 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.039922825, with a bounded reading uncertainty ±0.004699. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00039375, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
EXCITED POPULATION ρₑₑ against t
Predictions from the supplied The Born-Markov master equation for an atom in vacuum model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): EXCITED POPULATION ρₑₑ
Horizontal axis (x): t: Time Γt
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Time Γt | EXCITED POPULATION ρₑₑ |
|---|---|
| 0.53125 | 0.4409022548417599 |
| 1.515625 | 0.1647531357409315 |
| 2.5 | 0.0615637489679241 |
| 3.484375 | 0.023004692262400192 |
| 4.46875 | 0.00859622545669639 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(t) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.53125) = 0.440902 and Y_out(4.46875) = 0.00859623. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives t_line = 0.53125 + (0.0399228 − 0.440902)(4.46875 − 0.53125)/(0.00859623 − 0.440902) = 4.18342. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives t = 2.93312; substituting back gives Y_out = 0.039922825.
- The interpolation error in the input is |t_line − t| = 1.2503. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.035223825 and 0.044621825. Sort the roots even if the response decreases: 2.82185 ≤ t ≤ 3.05835.
- The lower and upper input deviations are 0.111275 and 0.125226. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer t ≈ 2.93312; compatible interval [2.82185, 3.05835]. Endpoint interpolation gives 4.18342.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The Lindblad form, the GKSL theorem and a thermal reservoir: infer an input with uncertainty
A Lindblad master equation is a first-order equation for the density operator, built from a Hamiltonian and jump operators Lₖ, whose solution at every later time is a completely positive, trace-preserving map of the initial state. Use the supplied topic model Y_out = (exp(1/temp)-1)/(exp(1/temp)+1), where Y_out is T₁ IN UNITS OF 1/Γ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. temp is Temperature k_B T/ħω₀, dimensionless. Keep p0 (Initial excited population pₑ(0)) = 1 fixed. For this exercise use only 0.408125 ≤ temp ≤ 2.69188 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.27067384, with a bounded reading uncertainty ±0.013056. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown temp by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000228375, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
T₁ IN UNITS OF 1/Γ against temp
Predictions from the supplied The Lindblad form, the GKSL theorem and a thermal reservoir model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): T₁ IN UNITS OF 1/Γ
Horizontal axis (x): temp: Temperature k_B T/ħω₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| temp: Temperature k_B T/ħω₀ | T₁ IN UNITS OF 1/Γ |
|---|---|
| 0.40812499999999996 | 0.8411564947415912 |
| 0.9790625 | 0.4704846845216812 |
| 1.5499999999999998 | 0.31183848721073015 |
| 2.1209375 | 0.2314725340747788 |
| 2.691875 | 0.1836370982174719 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(temp) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.408125) = 0.841156 and Y_out(2.69188) = 0.183637. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives temp_line = 0.408125 + (0.270674 − 0.841156)(2.69188 − 0.408125)/(0.183637 − 0.841156) = 2.38957. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives temp = 1.80121; substituting back gives Y_out = 0.27067384.
- The interpolation error in the input is |temp_line − temp| = 0.588359. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.25761784 and 0.28372984. Sort the roots even if the response decreases: 1.71389 ≤ temp ≤ 1.89714.
- The lower and upper input deviations are 0.0873214 and 0.0959231. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer temp ≈ 1.80121; compatible interval [1.71389, 1.89714]. Endpoint interpolation gives 2.38957.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.