University Advanced Quantum Mechanics · topic UAQM-08.08
What hole theory cannot do: the case for quantised fields
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 hole theory cannot do: the case for quantised fields: predict a finite change
Hole theory's limits are the assumptions it cannot justify: it subtracts the infinite charge and energy of a sea of real electrons by hand, breaks the symmetry charge conjugation demands, has no sea for bosons, and treats number-changing processes with a fixed-number equation. Use the supplied topic model Y_out = ((1-sqrt(max((en-v)^2-1,0))*(en+1)/(sqrt(en^2-1)*max(abs(en-v+1),1e-12)))/(1+sqrt(max((en-v)^2-1,0))*(en+1)/(sqrt(en^2-1)*max(abs(en-v+1),1e-12))))^2, where Y_out is R, SPIN ½ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; abs(x) means |x|; max(a,b) selects the larger value. v is Step height V₀, measured in mc². Keep en (Total energy E) = 1.5 mc² fixed. For this exercise use only 2.56675 mc² ≤ v ≤ 7.07125 mc² 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 v = 3.55774 mc² and 4.18837 mc². Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δv across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
R, SPIN ½ against v
Predictions from the supplied What hole theory cannot do: the case for quantised fields 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): R, SPIN ½
Horizontal axis (x): v: Step height V₀ / mc²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: Step height V₀ / mc² | R, SPIN ½ |
|---|---|
| 2.5667500000000003 | 0.7245700150118838 |
| 3.6928750000000004 | 0.3256507623310945 |
| 4.819000000000001 | 0.25640721096005836 |
| 5.945125000000001 | 0.2258399110774177 |
| 7.071250000000001 | 0.20854251196817356 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of v separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At v = 3.55774 mc², substitute into Y_out = ((1-sqrt(max((en-v)^2-1,0))*(en+1)/(sqrt(en^2-1)*max(abs(en-v+1),1e-12)))/(1+sqrt(max((en-v)^2-1,0))*(en+1)/(sqrt(en^2-1)*max(abs(en-v+1),1e-12))))^2: Y₀ = 0.340466.
- At v = 4.18837 mc², the same fixed-parameter model gives Y₁ = 0.286676.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0537897; the input change is Δv = 0.63063 mc².
- Average sensitivity is ΔY_out/Δv = -0.0852952 1 per mc². 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 = 0.340466, Y₁ = 0.286676; ΔY_out = -0.0537897; average sensitivity = -0.0852952 1 per mc².
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 hole theory cannot do: the case for quantised fields: infer an input with uncertainty
Hole theory's limits are the assumptions it cannot justify: it subtracts the infinite charge and energy of a sea of real electrons by hand, breaks the symmetry charge conjugation demands, has no sea for bosons, and treats number-changing processes with a fixed-number equation. Use the supplied topic model Y_out = ((1-sqrt(max((en-v)^2-1,0))*(en+1)/(sqrt(en^2-1)*max(abs(en-v+1),1e-12)))/(1+sqrt(max((en-v)^2-1,0))*(en+1)/(sqrt(en^2-1)*max(abs(en-v+1),1e-12))))^2, where Y_out is R, SPIN ½ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; abs(x) means |x|; max(a,b) selects the larger value. v is Step height V₀, measured in mc². Keep en (Total energy E) = 1.5 mc² fixed. For this exercise use only 2.56675 mc² ≤ v ≤ 7.07125 mc² 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 = 0.24048076, with a bounded reading uncertainty ±0.0047907. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown v by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00045045 mc², 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.
R, SPIN ½ against v
Predictions from the supplied What hole theory cannot do: the case for quantised fields 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): R, SPIN ½
Horizontal axis (x): v: Step height V₀ / mc²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: Step height V₀ / mc² | R, SPIN ½ |
|---|---|
| 2.5667500000000003 | 0.7245700150118838 |
| 3.6928750000000004 | 0.3256507623310945 |
| 4.819000000000001 | 0.25640721096005836 |
| 5.945125000000001 | 0.2258399110774177 |
| 7.071250000000001 | 0.20854251196817356 |
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(v) = 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(2.56675) = 0.72457 and Y_out(7.07125) = 0.208543. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives v_line = 2.56675 + (0.240481 − 0.72457)(7.07125 − 2.56675)/(0.208543 − 0.72457) = 6.79246 mc². 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 v = 5.3145 mc²; substituting back gives Y_out = 0.24048076.
- The interpolation error in the input is |v_line − v| = 1.47796 mc². This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.23569006 and 0.24527146. Sort the roots even if the response decreases: 5.14865 mc² ≤ v ≤ 5.49813 mc².
- The lower and upper input deviations are 0.165847 mc² and 0.183631 mc². 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 v ≈ 5.3145 mc²; compatible interval [5.14865, 5.49813] mc². Endpoint interpolation gives 6.79246 mc².
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.