University Quantum Mechanics II · topic UQMII-13.07
Natural linewidth and the Lorentzian line shape
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.
Natural linewidth and the Lorentzian line shape: predict a finite change
The natural linewidth is the spread in frequency a spectral line has because its upper level decays: an amplitude falling as e^(−t/2τ) radiates a Lorentzian whose full width at half maximum is Δω = 1/τ, or Δν = 1/(2πτ). Use the supplied topic model Y_out = (30.47*sqrt(temp/300))/(0.159155/tau+0.09606*p*sqrt(300/temp)), where Y_out is DOPPLER ÷ LORENTZIAN FWHM (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. p is Buffer-gas pressure p, measured in kPa. Keep tau (Upper-level lifetime τ) = 1.6 ns; temp (Gas temperature T) = 300 K fixed. For this exercise use only 10.625 kPa ≤ p ≤ 89.375 kPa 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 p = 27.95 kPa and 38.975 kPa. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δp across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
DOPPLER ÷ LORENTZIAN FWHM against p
Predictions from the supplied Natural linewidth and the Lorentzian line shape 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): DOPPLER ÷ LORENTZIAN FWHM
Horizontal axis (x): p: Buffer-gas pressure p / kPa
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Buffer-gas pressure p / kPa | DOPPLER ÷ LORENTZIAN FWHM |
|---|---|
| 10.625 | 27.202700629123825 |
| 30.3125 | 10.118584950597386 |
| 50 | 6.2152319843752295 |
| 69.6875 | 4.4850685543354105 |
| 89.375 | 3.5084146322594663 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of p separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At p = 27.95 kPa, substitute into Y_out = (30.47*sqrt(temp/300))/(0.159155/tau+0.09606*p*sqrt(300/temp)): Y₀ = 10.9433.
- At p = 38.975 kPa, the same fixed-parameter model gives Y₁ = 7.92785.
- Signed change: ΔY_out = Y₁ − Y₀ = -3.01546; the input change is Δp = 11.025 kPa.
- Average sensitivity is ΔY_out/Δp = -0.273511 1 per kPa. 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 = 10.9433, Y₁ = 7.92785; ΔY_out = -3.01546; average sensitivity = -0.273511 1 per kPa.
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.
Natural linewidth and the Lorentzian line shape: infer an input with uncertainty
The natural linewidth is the spread in frequency a spectral line has because its upper level decays: an amplitude falling as e^(−t/2τ) radiates a Lorentzian whose full width at half maximum is Δω = 1/τ, or Δν = 1/(2πτ). Use the supplied topic model Y_out = (30.47*sqrt(temp/300))/(0.159155/tau+0.09606*p*sqrt(300/temp)), where Y_out is DOPPLER ÷ LORENTZIAN FWHM (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. p is Buffer-gas pressure p, measured in kPa. Keep tau (Upper-level lifetime τ) = 1.6 ns; temp (Gas temperature T) = 300 K fixed. For this exercise use only 10.625 kPa ≤ p ≤ 89.375 kPa 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 = 5.3133688, with a bounded reading uncertainty ±0.27074. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown p by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.007875 kPa, 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.
DOPPLER ÷ LORENTZIAN FWHM against p
Predictions from the supplied Natural linewidth and the Lorentzian line shape 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): DOPPLER ÷ LORENTZIAN FWHM
Horizontal axis (x): p: Buffer-gas pressure p / kPa
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Buffer-gas pressure p / kPa | DOPPLER ÷ LORENTZIAN FWHM |
|---|---|
| 10.625 | 27.202700629123825 |
| 30.3125 | 10.118584950597386 |
| 50 | 6.2152319843752295 |
| 69.6875 | 4.4850685543354105 |
| 89.375 | 3.5084146322594663 |
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(p) = 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(10.625) = 27.2027 and Y_out(89.375) = 3.50841. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives p_line = 10.625 + (5.31337 − 27.2027)(89.375 − 10.625)/(3.50841 − 27.2027) = 83.3761 kPa. 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 p = 58.6625 kPa; substituting back gives Y_out = 5.3133688.
- The interpolation error in the input is |p_line − p| = 24.7136 kPa. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 5.0426288 and 5.5841088. Sort the roots even if the response decreases: 55.7681 kPa ≤ p ≤ 61.8677 kPa.
- The lower and upper input deviations are 2.8944 kPa and 3.2052 kPa. 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 p ≈ 58.6625 kPa; compatible interval [55.7681, 61.8677] kPa. Endpoint interpolation gives 83.3761 kPa.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.