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University Advanced Quantum Mechanics · unit UAQM-04

The Dirac Equation and Its Free Solutions

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

Dirac's requirements and the algebra of α and β: infer an input with uncertainty

Dirac's requirements are that the Hamiltonian H = cα⋅p + βmc² be Hermitian and square to p²c² + m²c⁴; they hold exactly when α₁, α₂, α₃ and β are Hermitian, square to 1 and anticommute in pairs, which no ordinary numbers can do. Use the supplied topic model Y_out = 1.022*m*sqrt(max(1-s^2,0)), where Y_out is SOLID GAP AT ITS MINIMUM (MeV). 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. s is Overlap s, where αβ + βα = 2s, dimensionless. Keep m (Mass m in electron masses) = 1 mₑ; p0 (Probe momentum p₀) = 1 MeV/c fixed. For this exercise use only 0.10625 ≤ s ≤ 0.89375 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 = 0.82767559 MeV, with a bounded reading uncertainty ±0.028281 MeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown s by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00007875, 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

SOLID GAP AT ITS MINIMUM against s

Predictions from the supplied Dirac's requirements and the algebra of α and β 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): SOLID GAP AT ITS MINIMUM / MeV

SOLID GAP AT ITS MINIMUM against sPredictions from the supplied Dirac's requirements and the algebra of α and β 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: s: Overlap s, where αβ + βα = 2s. Vertical axis: SOLID GAP AT ITS MINIMUM / MeV. Exact values are available in the data table below.0.43610.58670.73730.88791.0390.074750.28740.50.71260.9252xy

Horizontal axis (x): s: Overlap s, where αβ + βα = 2s

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

Reading 1s: Overlap s, where αβ + βα = 2s: 0.1062SOLID GAP AT ITS MINIMUM / MeV: 1.016
Read the exact data as a table
SOLID GAP AT ITS MINIMUM against s — supplied values
s: Overlap s, where αβ + βα = 2sSOLID GAP AT ITS MINIMUM / MeV
0.106251.0162149156766742
0.3031250.9739157214363764
0.50.8850779626676962
0.69687499999999990.7329708435271745
0.89374999999999990.45843386092625193

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(s) = 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(0.10625) = 1.01621 MeV and Y_out(0.89375) = 0.458434 MeV. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives s_line = 0.10625 + (0.827676 − 1.01621)(0.89375 − 0.10625)/(0.458434 − 1.01621) = 0.372438. 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 s = 0.586625; substituting back gives Y_out = 0.82767559 MeV.
  4. The interpolation error in the input is |s_line − s| = 0.214187. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.79939459 MeV and 0.85595659 MeV. Sort the roots even if the response decreases: 0.54639 ≤ s ≤ 0.623044.
  6. The lower and upper input deviations are 0.0402351 and 0.0364194. 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 s ≈ 0.586625; compatible interval [0.54639, 0.623044]. Endpoint interpolation gives 0.372438.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 02

Why four components: the Dirac representation constructed: infer an input with uncertainty

The Dirac representation is the choice β = [[1, 0], [0, −1]] and αᵢ = [[0, σᵢ], [σᵢ, 0]] in 2 × 2 blocks: four Hermitian 4 × 4 matrices that square to 1 and anticommute in pairs, the smallest matrices that can. Use the supplied topic model Y_out = (bl*sind(th)*cosd(ph))^2+(bl*sind(th)*sind(ph))^2+(bl*cosd(th))^2+(bl^2-1)^2, where Y_out is MISMATCH M (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. bl is Length |b|, dimensionless. Keep th (Polar angle θ of b) = 55 °; ph (Azimuth φ of b) = 45 ° fixed. For this exercise use only 0.75 ≤ bl ≤ 1.34062 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 = 1.2868851, with a bounded reading uncertainty ±0.079947. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown bl by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000590625, 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

MISMATCH M against bl

Predictions from the supplied Why four components: the Dirac representation constructed 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): MISMATCH M

MISMATCH M against blPredictions from the supplied Why four components: the Dirac representation constructed 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: bl: Length |b|. Vertical axis: MISMATCH M. Exact values are available in the data table below.0.68671.141.5932.0472.50.72640.88581.0451.2051.364xy

Horizontal axis (x): bl: Length |b|

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

Reading 1bl: Length |b|: 0.75MISMATCH M: 0.7539
Read the exact data as a table
MISMATCH M against bl — supplied values
bl: Length |b|MISMATCH M
0.750.7539062499999999
0.897656250.8435055322948843
1.04531251.1012674756109717
1.192968751.602251043856516
1.3406252.4329234391212458

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(bl) = 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(0.75) = 0.753906 and Y_out(1.34062) = 2.43292. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives bl_line = 0.75 + (1.28689 − 0.753906)(1.34062 − 0.75)/(2.43292 − 0.753906) = 0.937485. 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 bl = 1.11028; substituting back gives Y_out = 1.2868851.
  4. The interpolation error in the input is |bl_line − bl| = 0.172796. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 1.2069381 and 1.3668321. Sort the roots even if the response decreases: 1.08442 ≤ bl ≤ 1.13375.
  6. The lower and upper input deviations are 0.0258588 and 0.0234677. 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 bl ≈ 1.11028; compatible interval [1.08442, 1.13375]. Endpoint interpolation gives 0.937485.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 03

Gamma matrices, the Clifford algebra and trace identities: infer an input with uncertainty

The gamma matrices are the four fixed 4 × 4 matrices γ⁰ = β and γⁱ = βαᵢ, defined by the Clifford algebra (γ^μ, γ^ν) = 2g^(μν), in terms of which the free Dirac equation reads (iħγ^μ∂_μ − mc)ψ = 0. Use the supplied topic model Y_out = sqrt(max(a0^2-a3^2,0)), where Y_out is REAL PARTS OF λ (±) (mc). 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. a3 is Space component a³, measured in mc. Keep a0 (Time component a⁰) = 1.25 mc fixed. For this exercise use only -1.155 mc ≤ a3 ≤ -0.105 mc 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 = 1.1392058 mc, with a bounded reading uncertainty ±0.015956 mc. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a3 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000105 mc, 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

REAL PARTS OF λ (±) against a3

Predictions from the supplied Gamma matrices, the Clifford algebra and trace identities 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): REAL PARTS OF λ (±) / mc

REAL PARTS OF λ (±) against a3Predictions from the supplied Gamma matrices, the Clifford algebra and trace identities 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: a3: Space component a³ / mc. Vertical axis: REAL PARTS OF λ (±) / mc. Exact values are available in the data table below.0.44730.65450.86181.0691.276-1.197-0.9135-0.63-0.3465-0.063xy

Horizontal axis (x): a3: Space component a³ / mc

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

Reading 1a3: Space component a³ / mc: -1.155REAL PARTS OF λ (±) / mc: 0.478
Read the exact data as a table
REAL PARTS OF λ (±) against a3 — supplied values
a3: Space component a³ / mcREAL PARTS OF λ (±) / mc
-1.1550.47799058568135
-0.89250000000000010.8751821239033621
-0.631.0796295661012623
-0.367499999999999941.1947567744105911
-0.104999999999999981.2455821931932072

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(a3) = 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(-1.155) = 0.477991 mc and Y_out(-0.105) = 1.24558 mc. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives a3_line = -1.155 + (1.13921 − 0.477991)(-0.105 − -1.155)/(1.24558 − 0.477991) = -0.250514 mc. 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 a3 = -0.5145 mc; substituting back gives Y_out = 1.1392058 mc.
  4. The interpolation error in the input is |a3_line − a3| = 0.263986 mc. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 1.1232498 mc and 1.1551618 mc. Sort the roots even if the response decreases: -0.548461 mc ≤ a3 ≤ -0.477599 mc.
  6. The lower and upper input deviations are 0.0339615 mc and 0.0369005 mc. 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 a3 ≈ -0.5145 mc; compatible interval [-0.548461, -0.477599] mc. Endpoint interpolation gives -0.250514 mc.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 04

The conserved current and its positive density: infer an input with uncertainty

The Dirac density ρ = ψ†ψ and current j = cψ†αψ are the pair the Dirac equation itself conserves, ∂ρ/∂t + ∇⋅j = 0, and because ρ is a sum of four squared moduli it is never negative. Use the supplied topic model Y_out = (1000000*(p*cosd(2*th)-sind(2*th)*cosd(ph))/sqrt(1+p^2))/1000000, where Y_out is CURRENT j_z/cρ (before display rounding) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; sind/cosd/tand use degrees. p is Momentum p along z, measured in mc. Keep th (Branch mixing angle θ) = 0 °; ph (Phase 2Et/ħ) = 0 ° fixed. For this exercise use only 0.2125 mc ≤ p ≤ 1.7875 mc 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 = 0.76106234, with a bounded reading uncertainty ±0.016748. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown p by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001575 mc, 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

CURRENT j_z/cρ (before display rounding) against p

Predictions from the supplied The conserved current and its positive 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): CURRENT j_z/cρ (before display rounding)

CURRENT j_z/cρ (before display rounding) against pPredictions from the supplied The conserved current and its positive 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: p: Momentum p along z / mc. Vertical axis: CURRENT j_z/cρ (before display rounding). Exact values are available in the data table below.0.18130.36080.54030.71980.89930.14950.574711.4251.85xy

Horizontal axis (x): p: Momentum p along z / mc

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

Reading 1p: Momentum p along z / mc: 0.2125CURRENT j_z/cρ (before display rounding): 0.2079
Read the exact data as a table
CURRENT j_z/cρ (before display rounding) against p — supplied values
p: Momentum p along z / mcCURRENT j_z/cρ (before display rounding)
0.21250.2078587558200447
0.606250.5184201595739372
10.7071067811865475
1.39374999999999980.8125007339595185
1.78749999999999990.8727140745172804

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(p) = 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(0.2125) = 0.207859 and Y_out(1.7875) = 0.872714. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives p_line = 0.2125 + (0.761062 − 0.207859)(1.7875 − 0.2125)/(0.872714 − 0.207859) = 1.523 mc. 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 p = 1.17325 mc; substituting back gives Y_out = 0.76106234.
  4. The interpolation error in the input is |p_line − p| = 0.349754 mc. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.74431434 and 0.77781034. Sort the roots even if the response decreases: 1.11453 mc ≤ p ≤ 1.23757 mc.
  6. The lower and upper input deviations are 0.0587224 mc and 0.064318 mc. 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 p ≈ 1.17325 mc; compatible interval [1.11453, 1.23757] mc. Endpoint interpolation gives 1.523 mc.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

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