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.
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.
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
Horizontal axis (x): s: Overlap s, where αβ + βα = 2s
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Overlap s, where αβ + βα = 2s | SOLID GAP AT ITS MINIMUM / MeV |
|---|---|
| 0.10625 | 1.0162149156766742 |
| 0.303125 | 0.9739157214363764 |
| 0.5 | 0.8850779626676962 |
| 0.6968749999999999 | 0.7329708435271745 |
| 0.8937499999999999 | 0.45843386092625193 |
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
- 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.
- 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.
- 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.
- The interpolation error in the input is |s_line − s| = 0.214187. This is larger than the requested numerical tolerance.
- 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.
- 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.
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.
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
Horizontal axis (x): bl: Length |b|
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| bl: Length |b| | MISMATCH M |
|---|---|
| 0.75 | 0.7539062499999999 |
| 0.89765625 | 0.8435055322948843 |
| 1.0453125 | 1.1012674756109717 |
| 1.19296875 | 1.602251043856516 |
| 1.340625 | 2.4329234391212458 |
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
- 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.
- 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.
- 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.
- The interpolation error in the input is |bl_line − bl| = 0.172796. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.2069381 and 1.3668321. Sort the roots even if the response decreases: 1.08442 ≤ bl ≤ 1.13375.
- 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.
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.
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
Horizontal axis (x): a3: Space component a³ / mc
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a3: Space component a³ / mc | REAL PARTS OF λ (±) / mc |
|---|---|
| -1.155 | 0.47799058568135 |
| -0.8925000000000001 | 0.8751821239033621 |
| -0.63 | 1.0796295661012623 |
| -0.36749999999999994 | 1.1947567744105911 |
| -0.10499999999999998 | 1.2455821931932072 |
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
- 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.
- 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.
- 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.
- The interpolation error in the input is |a3_line − a3| = 0.263986 mc. This is larger than the requested numerical tolerance.
- 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.
- 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.
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.
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)
Horizontal axis (x): p: Momentum p along z / mc
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Momentum p along z / mc | CURRENT j_z/cρ (before display rounding) |
|---|---|
| 0.2125 | 0.2078587558200447 |
| 0.60625 | 0.5184201595739372 |
| 1 | 0.7071067811865475 |
| 1.3937499999999998 | 0.8125007339595185 |
| 1.7874999999999999 | 0.8727140745172804 |
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
- 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.
- 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.
- 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.
- The interpolation error in the input is |p_line − p| = 0.349754 mc. This is larger than the requested numerical tolerance.
- 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.
- 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.