University Quantum Mechanics II · topic UQMII-08.08
The Lamb shift: the 1947 measurement and Welton's estimate
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.
The Lamb shift: the 1947 measurement and Welton's estimate: predict a finite change
The Lamb shift is the displacement of a hydrogen level from its Dirac energy by the electron's coupling to the quantised electromagnetic field, seen most directly as a splitting of levels that share n and j but differ in l. Use the supplied topic model Y_out = 135.64*ln(510999*hi/(13.6057*lo)), where Y_out is 2S₁/₂ SHIFT, WELTON (MHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm. lo is Lower cutoff ħωₘᵢₙ, measured in Ry. Keep hi (Upper cutoff ħωₘₐₓ) = 1 mc² fixed. For this exercise use only 5.14375 Ry ≤ lo ≤ 35.8562 Ry 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 lo = 11.9005 Ry and 16.2002 Ry. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δlo across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
2S₁/₂ SHIFT, WELTON against lo
Predictions from the supplied The Lamb shift: the 1947 measurement and Welton's estimate 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): 2S₁/₂ SHIFT, WELTON / MHz
Horizontal axis (x): lo: Lower cutoff ħωₘᵢₙ / Ry
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lo: Lower cutoff ħωₘᵢₙ / Ry | 2S₁/₂ SHIFT, WELTON / MHz |
|---|---|
| 5.14375 | 1206.6333251590067 |
| 12.821874999999999 | 1082.7437761471374 |
| 20.499999999999996 | 1019.091696363907 |
| 28.178124999999998 | 975.9417528608684 |
| 35.856249999999996 | 943.2563214066469 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of lo separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At lo = 11.9005 Ry, substitute into Y_out = 135.64*ln(510999*hi/(13.6057*lo)): Y₀ = 1092.86 MHz.
- At lo = 16.2002 Ry, the same fixed-parameter model gives Y₁ = 1051.02 MHz.
- Signed change: ΔY_out = Y₁ − Y₀ = -41.8377 MHz; the input change is Δlo = 4.29975 Ry.
- Average sensitivity is ΔY_out/Δlo = -9.73025 MHz per Ry. 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 = 1092.86 MHz, Y₁ = 1051.02 MHz; ΔY_out = -41.8377 MHz; average sensitivity = -9.73025 MHz per Ry.
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.
The Lamb shift: the 1947 measurement and Welton's estimate: infer an input with uncertainty
The Lamb shift is the displacement of a hydrogen level from its Dirac energy by the electron's coupling to the quantised electromagnetic field, seen most directly as a splitting of levels that share n and j but differ in l. Use the supplied topic model Y_out = 135.64*ln(510999*hi/(13.6057*lo)), where Y_out is 2S₁/₂ SHIFT, WELTON (MHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm. lo is Lower cutoff ħωₘᵢₙ, measured in Ry. Keep hi (Upper cutoff ħωₘₐₓ) = 1 mc² fixed. For this exercise use only 5.14375 Ry ≤ lo ≤ 35.8562 Ry 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 = 998.40004 MHz, with a bounded reading uncertainty ±8.2716 MHz. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lo by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00307125 Ry, 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.
2S₁/₂ SHIFT, WELTON against lo
Predictions from the supplied The Lamb shift: the 1947 measurement and Welton's estimate 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): 2S₁/₂ SHIFT, WELTON / MHz
Horizontal axis (x): lo: Lower cutoff ħωₘᵢₙ / Ry
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lo: Lower cutoff ħωₘᵢₙ / Ry | 2S₁/₂ SHIFT, WELTON / MHz |
|---|---|
| 5.14375 | 1206.6333251590067 |
| 12.821874999999999 | 1082.7437761471374 |
| 20.499999999999996 | 1019.091696363907 |
| 28.178124999999998 | 975.9417528608684 |
| 35.856249999999996 | 943.2563214066469 |
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(lo) = 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(5.14375) = 1206.63 MHz and Y_out(35.8562) = 943.256 MHz. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives lo_line = 5.14375 + (998.4 − 1206.63)(35.8562 − 5.14375)/(943.256 − 1206.63) = 29.4259 Ry. 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 lo = 23.8784 Ry; substituting back gives Y_out = 998.40004 MHz.
- The interpolation error in the input is |lo_line − lo| = 5.54754 Ry. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 990.12844 MHz and 1006.6716 MHz. Sort the roots even if the response decreases: 22.4657 Ry ≤ lo ≤ 25.3798 Ry.
- The lower and upper input deviations are 1.41264 Ry and 1.50147 Ry. 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 lo ≈ 23.8784 Ry; compatible interval [22.4657, 25.3798] Ry. Endpoint interpolation gives 29.4259 Ry.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.