University Advanced Quantum Mechanics · topic UAQM-12.08
Emission depends on the modes: the Purcell factor
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.
Further practice
Work through a fresh case before tackling the extension.
Emission depends on the modes: the Purcell factor: predict a finite change
The Purcell factor F_P is the rate at which an emitter radiates into one resonant cavity mode divided by its free-space rate Γ₀: F_P = (3/(4π²))(λ³/V)Q for a dipole aligned with the mode's field at its maximum. Use the supplied topic model Y_out = 1000/(2*pi*(gam+4*gc^2/kap)), where Y_out is GOLDEN-RULE LIFETIME (ns). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. gc is Coupling g/2π, measured in MHz. Keep kap (Cavity linewidth κ/2π) = 60 MHz; gam (Free-space rate Γ₀/2π) = 0.4 MHz fixed. For this exercise use only 2.125 MHz ≤ gc ≤ 17.875 MHz 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. Calculate Y_out at gc = 5.59 MHz and 7.795 MHz. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δgc across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
GOLDEN-RULE LIFETIME against gc
Predictions from the supplied Emission depends on the modes: the Purcell factor 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): GOLDEN-RULE LIFETIME / ns
Horizontal axis (x): gc: Coupling g/2π / MHz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| gc: Coupling g/2π / MHz | GOLDEN-RULE LIFETIME / ns |
|---|---|
| 2.125 | 227.0263675605045 |
| 6.0625 | 55.83873745754939 |
| 10 | 22.52192590923047 |
| 13.9375 | 11.921486032827037 |
| 17.875 | 7.33397712130848 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of gc separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At gc = 5.59 MHz, substitute into Y_out = 1000/(2*pi*(gam+4*gc^2/kap)): Y₀ = 64.0925 ns.
- At gc = 7.795 MHz, the same fixed-parameter model gives Y₁ = 35.7587 ns.
- Signed change: ΔY_out = Y₁ − Y₀ = -28.3338 ns; the input change is Δgc = 2.205 MHz.
- Average sensitivity is ΔY_out/Δgc = -12.8498 ns per MHz. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 64.0925 ns, Y₁ = 35.7587 ns; ΔY_out = -28.3338 ns; average sensitivity = -12.8498 ns per MHz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Emission depends on the modes: the Purcell factor: infer an input with uncertainty
The Purcell factor F_P is the rate at which an emitter radiates into one resonant cavity mode divided by its free-space rate Γ₀: F_P = (3/(4π²))(λ³/V)Q for a dipole aligned with the mode's field at its maximum. Use the supplied topic model Y_out = 1000/(2*pi*(gam+4*gc^2/kap)), where Y_out is GOLDEN-RULE LIFETIME (ns). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. gc is Coupling g/2π, measured in MHz. Keep kap (Cavity linewidth κ/2π) = 60 MHz; gam (Free-space rate Γ₀/2π) = 0.4 MHz fixed. For this exercise use only 2.125 MHz ≤ gc ≤ 17.875 MHz 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 = 16.618853 ns, with a bounded reading uncertainty ±1.3927 ns. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown gc by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.001575 MHz, 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.
GOLDEN-RULE LIFETIME against gc
Predictions from the supplied Emission depends on the modes: the Purcell factor 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): GOLDEN-RULE LIFETIME / ns
Horizontal axis (x): gc: Coupling g/2π / MHz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| gc: Coupling g/2π / MHz | GOLDEN-RULE LIFETIME / ns |
|---|---|
| 2.125 | 227.0263675605045 |
| 6.0625 | 55.83873745754939 |
| 10 | 22.52192590923047 |
| 13.9375 | 11.921486032827037 |
| 17.875 | 7.33397712130848 |
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(gc) = 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(2.125) = 227.026 ns and Y_out(17.875) = 7.33398 ns. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives gc_line = 2.125 + (16.6189 − 227.026)(17.875 − 2.125)/(7.33398 − 227.026) = 17.2094 MHz. 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 gc = 11.7325 MHz; substituting back gives Y_out = 16.618853 ns.
- The interpolation error in the input is |gc_line − gc| = 5.47686 MHz. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 15.226153 ns and 18.011553 ns. Sort the roots even if the response decreases: 11.2492 MHz ≤ gc ≤ 12.2797 MHz.
- The lower and upper input deviations are 0.48332 MHz and 0.5472 MHz. 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 gc ≈ 11.7325 MHz; compatible interval [11.2492, 12.2797] MHz. Endpoint interpolation gives 17.2094 MHz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.