University Advanced Quantum Mechanics · topic UAQM-05.07
The Gordon decomposition: a g = 2 moment in the free current
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 Gordon decomposition: a g = 2 moment in the free current: predict a finite change
The Gordon decomposition is the exact identity that splits the Dirac current between free positive-energy spinors into a convection term, set by the sum of the two momenta, and a spin term, set by σ^μν and the momentum transfer. Use the supplied topic model Y_out = 0.4965*sz/a, where Y_out is SPIN TERM AT x = a/√2 (10⁵ m s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. a is Packet width a, measured in nm. Keep v (Drift speed v along +y) = 0.5 10⁵ m s⁻¹; sz (Spin polarisation ⟨σ_z⟩) = 1 fixed. For this exercise use only 0.57 nm ≤ a ≤ 1.83 nm 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 a = 0.8472 nm and 1.0236 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δa across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
SPIN TERM AT x = a/√2 against a
Predictions from the supplied The Gordon decomposition: a g = 2 moment in the free current 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): SPIN TERM AT x = a/√2 / 10⁵ m s⁻¹
Horizontal axis (x): a: Packet width a / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Packet width a / nm | SPIN TERM AT x = a/√2 / 10⁵ m s⁻¹ |
|---|---|
| 0.5700000000000001 | 0.8710526315789473 |
| 0.885 | 0.5610169491525424 |
| 1.2000000000000002 | 0.41374999999999995 |
| 1.5150000000000001 | 0.3277227722772277 |
| 1.83 | 0.2713114754098361 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of a separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At a = 0.8472 nm, substitute into Y_out = 0.4965*sz/a: Y₀ = 0.586048 10⁵ m s⁻¹.
- At a = 1.0236 nm, the same fixed-parameter model gives Y₁ = 0.485053 10⁵ m s⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.100995 10⁵ m s⁻¹; the input change is Δa = 0.1764 nm.
- Average sensitivity is ΔY_out/Δa = -0.572536 10⁵ m s⁻¹ per nm. 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 = 0.586048 10⁵ m s⁻¹, Y₁ = 0.485053 10⁵ m s⁻¹; ΔY_out = -0.100995 10⁵ m s⁻¹; average sensitivity = -0.572536 10⁵ m s⁻¹ per nm.
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 Gordon decomposition: a g = 2 moment in the free current: infer an input with uncertainty
The Gordon decomposition is the exact identity that splits the Dirac current between free positive-energy spinors into a convection term, set by the sum of the two momenta, and a spin term, set by σ^μν and the momentum transfer. Use the supplied topic model Y_out = 0.4965*sz/a, where Y_out is SPIN TERM AT x = a/√2 (10⁵ m s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. a is Packet width a, measured in nm. Keep v (Drift speed v along +y) = 0.5 10⁵ m s⁻¹; sz (Spin polarisation ⟨σ_z⟩) = 1 fixed. For this exercise use only 0.57 nm ≤ a ≤ 1.83 nm 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 = 0.37090991 10⁵ m s⁻¹, with a bounded reading uncertainty ±0.01494 10⁵ m s⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000126 nm, 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.
SPIN TERM AT x = a/√2 against a
Predictions from the supplied The Gordon decomposition: a g = 2 moment in the free current 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): SPIN TERM AT x = a/√2 / 10⁵ m s⁻¹
Horizontal axis (x): a: Packet width a / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Packet width a / nm | SPIN TERM AT x = a/√2 / 10⁵ m s⁻¹ |
|---|---|
| 0.5700000000000001 | 0.8710526315789473 |
| 0.885 | 0.5610169491525424 |
| 1.2000000000000002 | 0.41374999999999995 |
| 1.5150000000000001 | 0.3277227722772277 |
| 1.83 | 0.2713114754098361 |
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(a) = 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(0.57) = 0.871053 10⁵ m s⁻¹ and Y_out(1.83) = 0.271311 10⁵ m s⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives a_line = 0.57 + (0.37091 − 0.871053)(1.83 − 0.57)/(0.271311 − 0.871053) = 1.62075 nm. 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 a = 1.3386 nm; substituting back gives Y_out = 0.37090991 10⁵ m s⁻¹.
- The interpolation error in the input is |a_line − a| = 0.282153 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.35596991 10⁵ m s⁻¹ and 0.38584991 10⁵ m s⁻¹. Sort the roots even if the response decreases: 1.28677 nm ≤ a ≤ 1.39478 nm.
- The lower and upper input deviations are 0.0518302 nm and 0.0561808 nm. 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 a ≈ 1.3386 nm; compatible interval [1.28677, 1.39478] nm. Endpoint interpolation gives 1.62075 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.