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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.

Extra challenging · 01

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.

Explore the question data

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

COMPTON LENGTH ħ/mc against mPredictions 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. Horizontal axis: m: Rest energy mc² / MeV. Vertical axis: COMPTON LENGTH ħ/mc / fm. Exact values are available in the data table below.0.87982.3483.8175.2866.75424.264.6105145.4185.8xy

Horizontal axis (x): m: Rest energy mc² / MeV

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1m: Rest energy mc² / MeV: 30.19COMPTON LENGTH ħ/mc / fm: 6.537
Read the exact data as a table
COMPTON LENGTH ħ/mc against m — supplied values
m: Rest energy mc² / MeVCOMPTON LENGTH ħ/mc / fm
30.18756.536712215320911
67.593752.919308368007397
1051.8793047619047618
142.406251.3856624972569673
179.81251.0974042405283282

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |m_line − m| = 43.8504 MeV. This is larger than the requested numerical tolerance.
  5. 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.
  6. 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.

Extra challenging · 02

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.

Explore the question data

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|

THRESHOLD r = E₁/|E₂| against qPredictions 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. Horizontal axis: q: Momentum of wave 2, q = p₂/mc. Vertical axis: THRESHOLD r = E₁/|E₂|. Exact values are available in the data table below.0.24620.3780.50970.64150.77330.76161.5062.252.9943.738xy

Horizontal axis (x): q: Momentum of wave 2, q = p₂/mc

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1q: Momentum of wave 2, q = p₂/mc: 0.8719THRESHOLD r = E₁/|E₂|: 0.7537
Read the exact data as a table
THRESHOLD r = E₁/|E₂| against q — supplied values
q: Momentum of wave 2, q = p₂/mcTHRESHOLD r = E₁/|E₂|
0.8718750.7537428131110318
1.56093749999999990.5394363283035202
2.250.40613846605344767
2.93906249999999950.32211025216730715
3.6281250.2657161251333265

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |q_line − q| = 0.515945. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.34984639 and 0.37953839. Sort the roots even if the response decreases: 2.43763 ≤ q ≤ 2.67777.
  6. 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.

Extra challenging · 03

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.

Explore the question data

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 / %

TRUNCATION ERROR against qPredictions 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. Horizontal axis: q: Momentum q = p/mc. Vertical axis: TRUNCATION ERROR / %. Exact values are available in the data table below.-35.1-26-16.9-7.8061.2930.21210.53110.851.1691.488xy

Horizontal axis (x): q: Momentum q = p/mc

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1q: Momentum q = p/mc: 0.2594TRUNCATION ERROR / %: -0.0552
Read the exact data as a table
TRUNCATION ERROR against q — supplied values
q: Momentum q = p/mcTRUNCATION 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
  1. 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.
  2. 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.
  3. 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 %.
  4. The interpolation error in the input is |q_line − q| = 0.413863. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = -10.11696 % and -7.4921595 %. Sort the roots even if the response decreases: 0.936915 ≤ q ≤ 1.01878.
  6. 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.

Extra challenging · 04

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.

Explore the question data

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

VELOCITY v = c²P/(E − V₀) against enPredictions 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. Horizontal axis: en: Particle energy E / mc². Vertical axis: VELOCITY v = c²P/(E − V₀) / c. Exact values are available in the data table below.0.15480.32140.4880.65470.82131.1911.3631.5351.7081.88xy

Horizontal axis (x): en: Particle energy E / mc²

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1en: Particle energy E / mc²: 1.216VELOCITY v = c²P/(E − V₀) / c: 0.1794
Read the exact data as a table
VELOCITY v = c²P/(E − V₀) against en — supplied values
en: Particle energy E / mc²VELOCITY v = c²P/(E − V₀) / c
1.21650.17944596707347446
1.375968750.5261969427641913
1.53543750.6627753742922543
1.69490624999999980.7433179612685342
1.85437499999999990.7966371037831561

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |en_line − en| = 0.151736 mc². This is larger than the requested numerical tolerance.
  5. 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.
  6. 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.

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