University Advanced Quantum Mechanics · topic UAQM-09.08
The scalar field with anticommutators violates microcausality
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.
The anticommuting scalar field violates microcausality: predict a finite change
An anticommuting scalar violates microcausality: with [a, a†]₊ = [b, b†]₊ = 1 its field anticommutator is D(x − y) + D(y − x), equal to 2D(x − y) ≠ 0 at spacelike separation, so the density φ†φ at two spacelike-separated points fails to commute. Use the supplied topic model Y_out = exp(-m/197.327)*(1+7.802*m/197.327+18.002*(m/197.327)^2+18.393*(m/197.327)^3+6.0881*(m/197.327)^4)^(1/8)/0.0394784, where Y_out is PARTICLE TERM AT 1 fm (10⁻³ fm⁻²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. m is Mass m, measured in MeV. Keep stat (Rule: −1 commutator, +1 anticommutator) = 1; w (Antiparticle norm w (negative forbidden)) = 1 fixed. For this exercise use only 106.25 MeV ≤ m ≤ 893.75 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. Calculate Y_out at m = 279.5 MeV and 389.75 MeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δm across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
PARTICLE TERM AT 1 fm against m
Predictions from the supplied The anticommuting scalar field violates microcausality 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): PARTICLE TERM AT 1 fm / 10⁻³ fm⁻²
Horizontal axis (x): m: Mass m / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Mass m / MeV | PARTICLE TERM AT 1 fm / 10⁻³ fm⁻² |
|---|---|
| 106.25 | 20.52550128656849 |
| 303.125 | 10.248295268415639 |
| 499.99999999999994 | 4.547581568237899 |
| 696.8749999999999 | 1.9177970060427734 |
| 893.7499999999999 | 0.785851643390852 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of m separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At m = 279.5 MeV, substitute into Y_out = exp(-m/197.327)*(1+7.802*m/197.327+18.002*(m/197.327)^2+18.393*(m/197.327)^3+6.0881*(m/197.327)^4)^(1/8)/0.0394784: Y₀ = 11.2352 10⁻³ fm⁻².
- At m = 389.75 MeV, the same fixed-parameter model gives Y₁ = 7.23039 10⁻³ fm⁻².
- Signed change: ΔY_out = Y₁ − Y₀ = -4.00481 10⁻³ fm⁻²; the input change is Δm = 110.25 MeV.
- Average sensitivity is ΔY_out/Δm = -0.0363248 10⁻³ fm⁻² per MeV. Its negative sign means that this response decreases 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 = 11.2352 10⁻³ fm⁻², Y₁ = 7.23039 10⁻³ fm⁻²; ΔY_out = -4.00481 10⁻³ fm⁻²; average sensitivity = -0.0363248 10⁻³ fm⁻² per MeV.
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.
The anticommuting scalar field violates microcausality: infer an input with uncertainty
An anticommuting scalar violates microcausality: with [a, a†]₊ = [b, b†]₊ = 1 its field anticommutator is D(x − y) + D(y − x), equal to 2D(x − y) ≠ 0 at spacelike separation, so the density φ†φ at two spacelike-separated points fails to commute. Use the supplied topic model Y_out = exp(-m/197.327)*(1+7.802*m/197.327+18.002*(m/197.327)^2+18.393*(m/197.327)^3+6.0881*(m/197.327)^4)^(1/8)/0.0394784, where Y_out is PARTICLE TERM AT 1 fm (10⁻³ fm⁻²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. m is Mass m, measured in MeV. Keep stat (Rule: −1 commutator, +1 anticommutator) = 1; w (Antiparticle norm w (negative forbidden)) = 1 fixed. For this exercise use only 106.25 MeV ≤ m ≤ 893.75 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 = 3.1243299 10⁻³ fm⁻², with a bounded reading uncertainty ±0.35077 10⁻³ fm⁻². 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 0.07875 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.
PARTICLE TERM AT 1 fm against m
Predictions from the supplied The anticommuting scalar field violates microcausality 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): PARTICLE TERM AT 1 fm / 10⁻³ fm⁻²
Horizontal axis (x): m: Mass m / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Mass m / MeV | PARTICLE TERM AT 1 fm / 10⁻³ fm⁻² |
|---|---|
| 106.25 | 20.52550128656849 |
| 303.125 | 10.248295268415639 |
| 499.99999999999994 | 4.547581568237899 |
| 696.8749999999999 | 1.9177970060427734 |
| 893.7499999999999 | 0.785851643390852 |
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(106.25) = 20.5255 10⁻³ fm⁻² and Y_out(893.75) = 0.785852 10⁻³ fm⁻². The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives m_line = 106.25 + (3.12433 − 20.5255)(893.75 − 106.25)/(0.785852 − 20.5255) = 800.458 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 = 586.625 MeV; substituting back gives Y_out = 3.1243299 10⁻³ fm⁻².
- The interpolation error in the input is |m_line − m| = 213.833 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.7735599 10⁻³ fm⁻² and 3.4750999 10⁻³ fm⁻². Sort the roots even if the response decreases: 562.258 MeV ≤ m ≤ 613.744 MeV.
- The lower and upper input deviations are 24.3671 MeV and 27.1188 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 ≈ 586.625 MeV; compatible interval [562.258, 613.744] MeV. Endpoint interpolation gives 800.458 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.