University Advanced Quantum Mechanics · topic UAQM-03.01
Covariant notation and 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.
Covariant notation and the Klein-Gordon equation: predict a finite change
The Klein-Gordon equation, (1/c²)∂²φ/∂t² − ∇²φ + (mc/ħ)²φ = 0, is the wave equation obtained by making E²/c² − p² = m²c² an operator identity with E = iħ∂/∂t and p = −iħ∇; written as (□ + (mc/ħ)²)φ = 0 it has the same form in every inertial frame. Use the supplied topic model Y_out = 197.327/m, where Y_out is COMPTON LENGTH ħ/mc (fm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. m is Rest energy mc², measured in MeV. Keep p (Momentum pc) = 100 MeV fixed. For this exercise use only 30.1875 MeV ≤ m ≤ 179.813 MeV 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 m = 63.105 MeV and 84.0525 MeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δm across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
COMPTON LENGTH ħ/mc against m
Predictions from the supplied Covariant notation and the Klein-Gordon equation 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): COMPTON LENGTH ħ/mc / fm
Horizontal axis (x): m: Rest energy mc² / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Rest energy mc² / MeV | COMPTON LENGTH ħ/mc / fm |
|---|---|
| 30.1875 | 6.536712215320911 |
| 67.59375 | 2.919308368007397 |
| 105 | 1.8793047619047618 |
| 142.40625 | 1.3856624972569673 |
| 179.8125 | 1.0974042405283282 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of m separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At m = 63.105 MeV, substitute into Y_out = 197.327/m: Y₀ = 3.12696 fm.
- At m = 84.0525 MeV, the same fixed-parameter model gives Y₁ = 2.34766 fm.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.779299 fm; the input change is Δm = 20.9475 MeV.
- Average sensitivity is ΔY_out/Δm = -0.0372025 fm per MeV. 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 = 3.12696 fm, Y₁ = 2.34766 fm; ΔY_out = -0.779299 fm; average sensitivity = -0.0372025 fm per MeV.
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.
Covariant notation and the Klein-Gordon equation: infer an input with uncertainty
The Klein-Gordon equation, (1/c²)∂²φ/∂t² − ∇²φ + (mc/ħ)²φ = 0, is the wave equation obtained by making E²/c² − p² = m²c² an operator identity with E = iħ∂/∂t and p = −iħ∇; written as (□ + (mc/ħ)²)φ = 0 it has the same form in every inertial frame. Use the supplied topic model Y_out = 197.327/m, where Y_out is COMPTON LENGTH ħ/mc (fm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. m is Rest energy mc², measured in MeV. Keep p (Momentum pc) = 100 MeV fixed. For this exercise use only 30.1875 MeV ≤ m ≤ 179.813 MeV 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 = 1.6246421 fm, with a bounded reading uncertainty ±0.079086 fm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown m by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0149625 MeV, 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.
COMPTON LENGTH ħ/mc against m
Predictions from the supplied Covariant notation and the Klein-Gordon equation 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): COMPTON LENGTH ħ/mc / fm
Horizontal axis (x): m: Rest energy mc² / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Rest energy mc² / MeV | COMPTON LENGTH ħ/mc / fm |
|---|---|
| 30.1875 | 6.536712215320911 |
| 67.59375 | 2.919308368007397 |
| 105 | 1.8793047619047618 |
| 142.40625 | 1.3856624972569673 |
| 179.8125 | 1.0974042405283282 |
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(m) = 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(30.1875) = 6.53671 fm and Y_out(179.813) = 1.0974 fm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives m_line = 30.1875 + (1.62464 − 6.53671)(179.813 − 30.1875)/(1.0974 − 6.53671) = 165.309 MeV. 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 m = 121.459 MeV; substituting back gives Y_out = 1.6246421 fm.
- The interpolation error in the input is |m_line − m| = 43.8504 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.5455561 fm and 1.7037281 fm. Sort the roots even if the response decreases: 115.821 MeV ≤ m ≤ 127.674 MeV.
- The lower and upper input deviations are 5.63804 MeV and 6.21504 MeV. 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 m ≈ 121.459 MeV; compatible interval [115.821, 127.674] MeV. Endpoint interpolation gives 165.309 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.