University Quantum Field Theory I · topic UQFTI-15.09
What one loop leaves out, and where the next course begins
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.
What one loop leaves out, and where the next course begins: predict a finite change
Beyond one loop, hydrogen's Lamb shift is a series in four small numbers, α for loops, Zα for Coulomb exchanges, m/M for recoil and mrₚ for the proton's size, each term being (Zα)⁴m/n³ times powers of them and a computed or quoted coefficient. Use the supplied topic model Y_out = -5.64+7.14*min(step,1)-1.7173*rec*clamp(step-1,0,1)+0.3584*rec*clamp(step-2,0,1)+0.1955*rn*rn*clamp(step-3,0,1)+0.1*clamp(step-4,0,1)-0.41*clamp(step-5,0,1), where Y_out is MINUS MEASURED 1057.83 (MHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; min(a,b) selects the smaller value; clamp(x,a,b) confines x to [a,b]. rec is Recoil m/M, in units of hydrogen's, dimensionless. Keep step (Entries added past one loop) = 6; rn (Nuclear charge radius) = 0.84 fm fixed. For this exercise use only 0.2125 ≤ rec ≤ 1.7875 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 rec = 0.559 and 0.7795. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δrec across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
MINUS MEASURED 1057.83 against rec
Predictions from the supplied What one loop leaves out, and where the next course begins 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): MINUS MEASURED 1057.83 / MHz
Horizontal axis (x): rec: Recoil m/M, in units of hydrogen's
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rec: Recoil m/M, in units of hydrogen's | MINUS MEASURED 1057.83 / MHz |
|---|---|
| 0.2125 | 1.0391785500000004 |
| 0.60625 | 0.5041116750000001 |
| 1 | -0.030955200000000016 |
| 1.3937499999999998 | -0.5660220749999998 |
| 1.7874999999999999 | -1.1010889499999998 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of rec separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At rec = 0.559, substitute into Y_out = -5.64+7.14*min(step,1)-1.7173*rec*clamp(step-1,0,1)+0.3584*rec*clamp(step-2,0,1)+0.1955*rn*rn*clamp(step-3,0,1)+0.1*clamp(step-4,0,1)-0.41*clamp(step-5,0,1): Y₀ = 0.56832 MHz.
- At rec = 0.7795, the same fixed-parameter model gives Y₁ = 0.268682 MHz.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.299637 MHz; the input change is Δrec = 0.2205.
- Average sensitivity is ΔY_out/Δrec = -1.3589 MHz per unit input. 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 = 0.56832 MHz, Y₁ = 0.268682 MHz; ΔY_out = -0.299637 MHz; average sensitivity = -1.3589 MHz per unit input.
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.
What one loop leaves out, and where the next course begins: infer an input with uncertainty
Beyond one loop, hydrogen's Lamb shift is a series in four small numbers, α for loops, Zα for Coulomb exchanges, m/M for recoil and mrₚ for the proton's size, each term being (Zα)⁴m/n³ times powers of them and a computed or quoted coefficient. Use the supplied topic model Y_out = -5.64+7.14*min(step,1)-1.7173*rec*clamp(step-1,0,1)+0.3584*rec*clamp(step-2,0,1)+0.1955*rn*rn*clamp(step-3,0,1)+0.1*clamp(step-4,0,1)-0.41*clamp(step-5,0,1), where Y_out is MINUS MEASURED 1057.83 (MHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; min(a,b) selects the smaller value; clamp(x,a,b) confines x to [a,b]. rec is Recoil m/M, in units of hydrogen's, dimensionless. Keep step (Entries added past one loop) = 6; rn (Nuclear charge radius) = 0.84 fm fixed. For this exercise use only 0.2125 ≤ rec ≤ 1.7875 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 = -0.26638463 MHz, with a bounded reading uncertainty ±0.12521 MHz. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rec by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001575, 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.
MINUS MEASURED 1057.83 against rec
Predictions from the supplied What one loop leaves out, and where the next course begins 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): MINUS MEASURED 1057.83 / MHz
Horizontal axis (x): rec: Recoil m/M, in units of hydrogen's
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rec: Recoil m/M, in units of hydrogen's | MINUS MEASURED 1057.83 / MHz |
|---|---|
| 0.2125 | 1.0391785500000004 |
| 0.60625 | 0.5041116750000001 |
| 1 | -0.030955200000000016 |
| 1.3937499999999998 | -0.5660220749999998 |
| 1.7874999999999999 | -1.1010889499999998 |
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(rec) = 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(0.2125) = 1.03918 MHz and Y_out(1.7875) = -1.10109 MHz. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives rec_line = 0.2125 + (-0.266385 − 1.03918)(1.7875 − 0.2125)/(-1.10109 − 1.03918) = 1.17325. 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 rec = 1.17325; substituting back gives Y_out = -0.26638463 MHz.
- The interpolation error in the input is |rec_line − rec| = 2.22045e-16. This is within the requested numerical tolerance.
- Invert both limits Y_out = -0.39159463 MHz and -0.14117463 MHz. Sort the roots even if the response decreases: 1.08111 ≤ rec ≤ 1.26539.
- The lower and upper input deviations are 0.0921407 and 0.0921407. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer rec ≈ 1.17325; compatible interval [1.08111, 1.26539]. Endpoint interpolation gives 1.17325.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.