University Quantum Mechanics II · topic UQMII-13.02
Absorption and stimulated emission of incoherent light
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.
Absorption and stimulated emission of incoherent light: predict a finite change
In incoherent light, absorption and stimulated emission are constant rates, each proportional to the squared transition dipole and to the light's spectral energy density at the transition frequency, and equal to each other for non-degenerate levels. Use the supplied topic model Y_out = 1/(2*w), where Y_out is INCOHERENT RATE R = ω₁²/(2γ) (ω₁). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. w is Band half-width γ, measured in ω₁. Keep tm (Time marker ω₁t) = 3 fixed. For this exercise use only 1.50937 ω₁ ≤ w ≤ 8.99062 ω₁ 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 w = 3.15525 ω₁ and 4.20262 ω₁. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δw across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
INCOHERENT RATE R = ω₁²/(2γ) against w
Predictions from the supplied Absorption and stimulated emission of incoherent light 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): INCOHERENT RATE R = ω₁²/(2γ) / ω₁
Horizontal axis (x): w: Band half-width γ / ω₁
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Band half-width γ / ω₁ | INCOHERENT RATE R = ω₁²/(2γ) / ω₁ |
|---|---|
| 1.509375 | 0.33126293995859213 |
| 3.3796874999999997 | 0.14794267221451687 |
| 5.25 | 0.09523809523809523 |
| 7.120312499999999 | 0.07022163704191355 |
| 8.990625 | 0.05561348627042058 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of w separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At w = 3.15525 ω₁, substitute into Y_out = 1/(2*w): Y₀ = 0.158466 ω₁.
- At w = 4.20262 ω₁, the same fixed-parameter model gives Y₁ = 0.118973 ω₁.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0394928 ω₁; the input change is Δw = 1.04737 ω₁.
- Average sensitivity is ΔY_out/Δw = -0.0377064 ω₁ per ω₁. 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 = 0.158466 ω₁, Y₁ = 0.118973 ω₁; ΔY_out = -0.0394928 ω₁; average sensitivity = -0.0377064 ω₁ per ω₁.
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.
Absorption and stimulated emission of incoherent light: infer an input with uncertainty
In incoherent light, absorption and stimulated emission are constant rates, each proportional to the squared transition dipole and to the light's spectral energy density at the transition frequency, and equal to each other for non-degenerate levels. Use the supplied topic model Y_out = 1/(2*w), where Y_out is INCOHERENT RATE R = ω₁²/(2γ) (ω₁). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. w is Band half-width γ, measured in ω₁. Keep tm (Time marker ω₁t) = 3 fixed. For this exercise use only 1.50937 ω₁ ≤ w ≤ 8.99062 ω₁ 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 = 0.082332479 ω₁, with a bounded reading uncertainty ±0.0040078 ω₁. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown w by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000748125 ω₁, 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.
INCOHERENT RATE R = ω₁²/(2γ) against w
Predictions from the supplied Absorption and stimulated emission of incoherent light 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): INCOHERENT RATE R = ω₁²/(2γ) / ω₁
Horizontal axis (x): w: Band half-width γ / ω₁
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Band half-width γ / ω₁ | INCOHERENT RATE R = ω₁²/(2γ) / ω₁ |
|---|---|
| 1.509375 | 0.33126293995859213 |
| 3.3796874999999997 | 0.14794267221451687 |
| 5.25 | 0.09523809523809523 |
| 7.120312499999999 | 0.07022163704191355 |
| 8.990625 | 0.05561348627042058 |
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(w) = 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(1.50937) = 0.331263 ω₁ and Y_out(8.99062) = 0.0556135 ω₁. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 1.50937 + (0.0823325 − 0.331263)(8.99062 − 1.50937)/(0.0556135 − 0.331263) = 8.26546 ω₁. 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 w = 6.07294 ω₁; substituting back gives Y_out = 0.082332479 ω₁.
- The interpolation error in the input is |w_line − w| = 2.19252 ω₁. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.078324679 ω₁ and 0.086340279 ω₁. Sort the roots even if the response decreases: 5.79104 ω₁ ≤ w ≤ 6.38368 ω₁.
- The lower and upper input deviations are 0.281898 ω₁ and 0.310746 ω₁. 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 w ≈ 6.07294 ω₁; compatible interval [5.79104, 6.38368] ω₁. Endpoint interpolation gives 8.26546 ω₁.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.