University Advanced Quantum Mechanics · topic UAQM-04.03
Gamma matrices, the Clifford algebra and trace identities
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.
Gamma matrices, the Clifford algebra and trace identities: predict a finite change
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. Calculate Y_out at a3 = -0.924 mc and -0.777 mc. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δa3 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
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
Substitute each value of a3 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At a3 = -0.924 mc, substitute into Y_out = sqrt(max(a0^2-a3^2,0)): Y₀ = 0.841857 mc.
- At a3 = -0.777 mc, the same fixed-parameter model gives Y₁ = 0.979169 mc.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.137311 mc; the input change is Δa3 = 0.147 mc.
- Average sensitivity is ΔY_out/Δa3 = 0.934089 mc per mc. Its positive sign means that this response increases 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 = 0.841857 mc, Y₁ = 0.979169 mc; ΔY_out = 0.137311 mc; average sensitivity = 0.934089 mc per mc.
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.
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.