University Advanced Quantum Mechanics · topic UAQM-03.04
Minimal coupling and gauge invariance of the Klein-Gordon equation
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.
Klein-Gordon minimal coupling and gauge invariance: predict a finite change
Minimal coupling puts a charge q into the Klein-Gordon equation by replacing every four-momentum p^μ with the kinetic four-momentum p^μ − qA^μ (iħ∂^μ − qA^μ as an operator), which leaves the equation unchanged when the potentials are gauge-transformed and φ gains the phase e^(iqΛ/ħ). Use the supplied topic model Y_out = sqrt(max((en-v0)^2-1,0))/(en-v0), where Y_out is VELOCITY v = c²P/(E − V₀) (c). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. en is Particle energy E, measured in mc². Keep v0 (Plateau potential energy V₀ = qΦ₀) = 0.2 mc²; qa (Vector potential qA₀ (pure gauge)) = 0.4 mc fixed. For this exercise use only 1.2165 mc² ≤ en ≤ 1.85437 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 en = 1.35683 mc² and 1.44613 mc². Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δen across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
VELOCITY v = c²P/(E − V₀) against en
Predictions from the supplied Klein-Gordon minimal coupling and gauge invariance 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): VELOCITY v = c²P/(E − V₀) / c
Horizontal axis (x): en: Particle energy E / mc²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: Particle energy E / mc² | VELOCITY v = c²P/(E − V₀) / c |
|---|---|
| 1.2165 | 0.17944596707347446 |
| 1.37596875 | 0.5261969427641913 |
| 1.5354375 | 0.6627753742922543 |
| 1.6949062499999998 | 0.7433179612685342 |
| 1.8543749999999999 | 0.7966371037831561 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of en separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At en = 1.35683 mc², substitute into Y_out = sqrt(max((en-v0)^2-1,0))/(en-v0): Y₀ = 0.502754 c.
- At en = 1.44613 mc², the same fixed-parameter model gives Y₁ = 0.596677 c.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.093923 c; the input change is Δen = 0.0893025 mc².
- Average sensitivity is ΔY_out/Δen = 1.05174 c per mc². Its positive sign means that this response increases 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 = 0.502754 c, Y₁ = 0.596677 c; ΔY_out = 0.093923 c; average sensitivity = 1.05174 c 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.
Klein-Gordon minimal coupling and gauge invariance: infer an input with uncertainty
Minimal coupling puts a charge q into the Klein-Gordon equation by replacing every four-momentum p^μ with the kinetic four-momentum p^μ − qA^μ (iħ∂^μ − qA^μ as an operator), which leaves the equation unchanged when the potentials are gauge-transformed and φ gains the phase e^(iqΛ/ħ). Use the supplied topic model Y_out = sqrt(max((en-v0)^2-1,0))/(en-v0), where Y_out is VELOCITY v = c²P/(E − V₀) (c). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. en is Particle energy E, measured in mc². Keep v0 (Plateau potential energy V₀ = qΦ₀) = 0.2 mc²; qa (Vector potential qA₀ (pure gauge)) = 0.4 mc fixed. For this exercise use only 1.2165 mc² ≤ en ≤ 1.85437 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 = 0.70274881 c, with a bounded reading uncertainty ±0.014083 c. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown en by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000637875 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.
VELOCITY v = c²P/(E − V₀) against en
Predictions from the supplied Klein-Gordon minimal coupling and gauge invariance 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): VELOCITY v = c²P/(E − V₀) / c
Horizontal axis (x): en: Particle energy E / mc²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: Particle energy E / mc² | VELOCITY v = c²P/(E − V₀) / c |
|---|---|
| 1.2165 | 0.17944596707347446 |
| 1.37596875 | 0.5261969427641913 |
| 1.5354375 | 0.6627753742922543 |
| 1.6949062499999998 | 0.7433179612685342 |
| 1.8543749999999999 | 0.7966371037831561 |
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(en) = 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(1.2165) = 0.179446 c and Y_out(1.85437) = 0.796637 c. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives en_line = 1.2165 + (0.702749 − 0.179446)(1.85437 − 1.2165)/(0.796637 − 0.179446) = 1.75734 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 en = 1.6056 mc²; substituting back gives Y_out = 0.70274881 c.
- The interpolation error in the input is |en_line − en| = 0.151736 mc². This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.68866581 c and 0.71683181 c. Sort the roots even if the response decreases: 1.57916 mc² ≤ en ≤ 1.63421 mc².
- The lower and upper input deviations are 0.0264435 mc² and 0.02861 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 en ≈ 1.6056 mc²; compatible interval [1.57916, 1.63421] mc². Endpoint interpolation gives 1.75734 mc².
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.