University Advanced Quantum Mechanics · topic UAQM-09.02
Canonical quantisation of the real Klein-Gordon field
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.
Canonical quantisation of the real Klein-Gordon field: predict a finite change
Canonical quantisation of a real field imposes [φ(x), π(y)] = iħ δ³(x − y) at equal times, which makes each normal mode of the field an independent quantum oscillator whose quanta are the field's particles. Use the supplied topic model Y_out = sqrt((197.327*k)^2+m^2), where Y_out is ONE QUANTUM ħωₖ (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. k is Mode wavenumber k, measured in fm⁻¹. Keep m (Mass of the quantum m) = 135 MeV/c²; n (Quanta in mode k, n) = 1 fixed. For this exercise use only 0.0525 fm⁻¹ ≤ k ≤ 0.7875 fm⁻¹ 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 k = 0.2142 fm⁻¹ and 0.3171 fm⁻¹. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δk across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
ONE QUANTUM ħωₖ against k
Predictions from the supplied Canonical quantisation of the real Klein-Gordon field 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): ONE QUANTUM ħωₖ / MeV
Horizontal axis (x): k: Mode wavenumber k / fm⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: Mode wavenumber k / fm⁻¹ | ONE QUANTUM ħωₖ / MeV |
|---|---|
| 0.05249999999999999 | 135.39690805446983 |
| 0.23624999999999996 | 142.82256436532978 |
| 0.41999999999999993 | 158.40976448904783 |
| 0.6037499999999999 | 180.05118297714975 |
| 0.7874999999999999 | 205.84608305691694 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of k separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At k = 0.2142 fm⁻¹, substitute into Y_out = sqrt((197.327*k)^2+m^2): Y₀ = 141.462 MeV.
- At k = 0.3171 fm⁻¹, the same fixed-parameter model gives Y₁ = 148.796 MeV.
- Signed change: ΔY_out = Y₁ − Y₀ = 7.33404 MeV; the input change is Δk = 0.1029 fm⁻¹.
- Average sensitivity is ΔY_out/Δk = 71.2735 MeV per fm⁻¹. 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 = 141.462 MeV, Y₁ = 148.796 MeV; ΔY_out = 7.33404 MeV; average sensitivity = 71.2735 MeV per fm⁻¹.
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.
Canonical quantisation of the real Klein-Gordon field: infer an input with uncertainty
Canonical quantisation of a real field imposes [φ(x), π(y)] = iħ δ³(x − y) at equal times, which makes each normal mode of the field an independent quantum oscillator whose quanta are the field's particles. Use the supplied topic model Y_out = sqrt((197.327*k)^2+m^2), where Y_out is ONE QUANTUM ħωₖ (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. k is Mode wavenumber k, measured in fm⁻¹. Keep m (Mass of the quantum m) = 135 MeV/c²; n (Quanta in mode k, n) = 1 fixed. For this exercise use only 0.0525 fm⁻¹ ≤ k ≤ 0.7875 fm⁻¹ 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 = 167.30993 MeV, with a bounded reading uncertainty ±4.787 MeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown k by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000735 fm⁻¹, 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.
ONE QUANTUM ħωₖ against k
Predictions from the supplied Canonical quantisation of the real Klein-Gordon field 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): ONE QUANTUM ħωₖ / MeV
Horizontal axis (x): k: Mode wavenumber k / fm⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: Mode wavenumber k / fm⁻¹ | ONE QUANTUM ħωₖ / MeV |
|---|---|
| 0.05249999999999999 | 135.39690805446983 |
| 0.23624999999999996 | 142.82256436532978 |
| 0.41999999999999993 | 158.40976448904783 |
| 0.6037499999999999 | 180.05118297714975 |
| 0.7874999999999999 | 205.84608305691694 |
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(k) = 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(0.0525) = 135.397 MeV and Y_out(0.7875) = 205.846 MeV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives k_line = 0.0525 + (167.31 − 135.397)(0.7875 − 0.0525)/(205.846 − 135.397) = 0.38545 fm⁻¹. 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 k = 0.50085 fm⁻¹; substituting back gives Y_out = 167.30993 MeV.
- The interpolation error in the input is |k_line − k| = 0.1154 fm⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 162.52293 MeV and 172.09693 MeV. Sort the roots even if the response decreases: 0.458586 fm⁻¹ ≤ k ≤ 0.540904 fm⁻¹.
- The lower and upper input deviations are 0.0422638 fm⁻¹ and 0.040054 fm⁻¹. 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 k ≈ 0.50085 fm⁻¹; compatible interval [0.458586, 0.540904] fm⁻¹. Endpoint interpolation gives 0.38545 fm⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.