University Advanced Quantum Mechanics · unit UAQM-03
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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
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.
The Klein-Gordon current and its indefinite density: infer an input with uncertainty
The Klein-Gordon current j^μ = (iħ/2m)(φ*∂^μφ − φ∂^μφ*) is the four-vector the equation conserves; its time component, the density ρ, depends on ∂φ/∂t as well as on φ and can be positive, negative or zero. Use the supplied topic model Y_out = 1/sqrt(1+q^2), where Y_out is THRESHOLD r = E₁/|E₂| (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. q is Momentum of wave 2, q = p₂/mc, dimensionless. Keep r (Amplitude ratio r = a₂/a₁) = 0.7; s (Sign of wave 2's energy) = 1 fixed. For this exercise use only 0.871875 ≤ q ≤ 3.62812 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 = 0.36469239, with a bounded reading uncertainty ±0.014846. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown q by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000275625, 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.
THRESHOLD r = E₁/|E₂| against q
Predictions from the supplied The Klein-Gordon current and its indefinite density 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): THRESHOLD r = E₁/|E₂|
Horizontal axis (x): q: Momentum of wave 2, q = p₂/mc
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q: Momentum of wave 2, q = p₂/mc | THRESHOLD r = E₁/|E₂| |
|---|---|
| 0.871875 | 0.7537428131110318 |
| 1.5609374999999999 | 0.5394363283035202 |
| 2.25 | 0.40613846605344767 |
| 2.9390624999999995 | 0.32211025216730715 |
| 3.628125 | 0.2657161251333265 |
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(q) = 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(0.871875) = 0.753743 and Y_out(3.62812) = 0.265716. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives q_line = 0.871875 + (0.364692 − 0.753743)(3.62812 − 0.871875)/(0.265716 − 0.753743) = 3.06913. 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 q = 2.55319; substituting back gives Y_out = 0.36469239.
- The interpolation error in the input is |q_line − q| = 0.515945. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.34984639 and 0.37953839. Sort the roots even if the response decreases: 2.43763 ≤ q ≤ 2.67777.
- The lower and upper input deviations are 0.115553 and 0.124579. 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 q ≈ 2.55319; compatible interval [2.43763, 2.67777]. Endpoint interpolation gives 3.06913.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The non-relativistic limit: Schrödinger plus −p⁴/(8m³c²): infer an input with uncertainty
The non-relativistic limit of the Klein-Gordon equation is what remains once the rest-energy oscillation e^(−imc²t/ħ) is factored out: Schrödinger's equation plus a correction smaller by T/2mc², whose leading term is −p⁴/(8m³c²). Use the supplied topic model Y_out = 100*((q^2/2 - clamp(n,0,1)*q^4/8 + clamp(n-1,0,1)*q^6/16 - clamp(n-2,0,1)*5*q^8/128)-(sqrt(1+q^2)-1))/(sqrt(1+q^2)-1), where Y_out is TRUNCATION ERROR (%). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; clamp(x,a,b) confines x to [a,b]. q is Momentum q = p/mc, dimensionless. Keep n (Correction terms kept n) = 1 fixed. For this exercise use only 0.259375 ≤ q ≤ 1.44063 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 = -8.8045595 %, with a bounded reading uncertainty ±1.3124 %. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown q by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000118125, 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.
TRUNCATION ERROR against q
Predictions from the supplied The non-relativistic limit: Schrödinger plus −p⁴/(8m³c²) 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): TRUNCATION ERROR / %
Horizontal axis (x): q: Momentum q = p/mc
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q: Momentum q = p/mc | TRUNCATION ERROR / % |
|---|---|
| 0.259375 | -0.05520102041122308 |
| 0.5546875 | -1.0671183898104144 |
| 0.85 | -5.262204296371332 |
| 1.1453125 | -15.305014970123983 |
| 1.4406249999999998 | -33.753309872752574 |
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(q) = 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(0.259375) = -0.055201 % and Y_out(1.44063) = -33.7533 %. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives q_line = 0.259375 + (-8.80456 − -0.055201)(1.44063 − 0.259375)/(-33.7533 − -0.055201) = 0.566074. 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 q = 0.979938; substituting back gives Y_out = -8.8045595 %.
- The interpolation error in the input is |q_line − q| = 0.413863. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -10.11696 % and -7.4921595 %. Sort the roots even if the response decreases: 0.936915 ≤ q ≤ 1.01878.
- The lower and upper input deviations are 0.0430224 and 0.0388376. 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 q ≈ 0.979938; compatible interval [0.936915, 1.01878]. Endpoint interpolation gives 0.566074.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.