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University Advanced Quantum Mechanics · topic UAQM-04.09

Minimal coupling and the gauge invariance of the Dirac equation

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.

Further practice · 01

Minimal coupling and gauge invariance of the Dirac equation: predict a finite change

Minimal coupling puts the electromagnetic field into the Dirac equation only through the substitution p^μ → p^μ − qA^μ: H = cα⋅(p − qA) + βmc² + qφ, with no term that gauge invariance does not demand. Use the supplied topic model Y_out = (k-a)/sqrt(1+(k-a)^2), where Y_out is VELOCITY ⟨cα_z⟩ (c). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. a is Vector potential qA_z, measured in mc. Keep k (Canonical momentum p_z = ħk) = 1.5 mc fixed. For this exercise use only -0.7875 mc ≤ a ≤ 0.7875 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. Calculate Y_out at a = -0.441 mc and -0.2205 mc. 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.

Explore the question data

VELOCITY ⟨cα_z⟩ against a

Predictions from the supplied Minimal coupling and gauge invariance of the Dirac equation 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): VELOCITY ⟨cα_z⟩ / c

VELOCITY ⟨cα_z⟩ against aPredictions from the supplied Minimal coupling and gauge invariance of the Dirac equation 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: a: Vector potential qA_z / mc. Vertical axis: VELOCITY ⟨cα_z⟩ / c. Exact values are available in the data table below.0.56680.65760.74830.8390.9297-0.8505-0.4253-1.11e-160.42520.8505xy

Horizontal axis (x): a: Vector potential qA_z / mc

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

Reading 1a: Vector potential qA_z / mc: -0.7875VELOCITY ⟨cα_z⟩ / c: 0.9163
Read the exact data as a table
VELOCITY ⟨cα_z⟩ against a — supplied values
a: Vector potential qA_z / mcVELOCITY ⟨cα_z⟩ / c
-0.78750.9162722109020922
-0.393750000000000040.8842843398019031
-1.1102230246251565e-160.8320502943378437
0.393749999999999930.7418335808003366
0.78749999999999980.5802747891877948

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
  1. At a = -0.441 mc, substitute into Y_out = (k-a)/sqrt(1+(k-a)^2): Y₀ = 0.888958 c.
  2. At a = -0.2205 mc, the same fixed-parameter model gives Y₁ = 0.864571 c.
  3. Signed change: ΔY_out = Y₁ − Y₀ = -0.024387 c; the input change is Δa = 0.2205 mc.
  4. Average sensitivity is ΔY_out/Δa = -0.110598 c per mc. Its negative sign means that this response decreases over the specified interval.
  5. 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 = 0.888958 c, Y₁ = 0.864571 c; ΔY_out = -0.024387 c; average sensitivity = -0.110598 c 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.

Extra challenging · 01

Minimal coupling and gauge invariance of the Dirac equation: infer an input with uncertainty

Minimal coupling puts the electromagnetic field into the Dirac equation only through the substitution p^μ → p^μ − qA^μ: H = cα⋅(p − qA) + βmc² + qφ, with no term that gauge invariance does not demand. Use the supplied topic model Y_out = (k-a)/sqrt(1+(k-a)^2), where Y_out is VELOCITY ⟨cα_z⟩ (c). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. a is Vector potential qA_z, measured in mc. Keep k (Canonical momentum p_z = ħk) = 1.5 mc fixed. For this exercise use only -0.7875 mc ≤ a ≤ 0.7875 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 = 0.79857124 c, with a bounded reading uncertainty ±0.017655 c. 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 0.0001575 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

VELOCITY ⟨cα_z⟩ against a

Predictions from the supplied Minimal coupling and gauge invariance of the Dirac equation 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): VELOCITY ⟨cα_z⟩ / c

VELOCITY ⟨cα_z⟩ against aPredictions from the supplied Minimal coupling and gauge invariance of the Dirac equation 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: a: Vector potential qA_z / mc. Vertical axis: VELOCITY ⟨cα_z⟩ / c. Exact values are available in the data table below.0.56680.65760.74830.8390.9297-0.8505-0.4253-1.11e-160.42520.8505xy

Horizontal axis (x): a: Vector potential qA_z / mc

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

Reading 1a: Vector potential qA_z / mc: -0.7875VELOCITY ⟨cα_z⟩ / c: 0.9163
Read the exact data as a table
VELOCITY ⟨cα_z⟩ against a — supplied values
a: Vector potential qA_z / mcVELOCITY ⟨cα_z⟩ / c
-0.78750.9162722109020922
-0.393750000000000040.8842843398019031
-1.1102230246251565e-160.8320502943378437
0.393749999999999930.7418335808003366
0.78749999999999980.5802747891877948

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
  1. Endpoint predictions are Y_out(-0.7875) = 0.916272 c and Y_out(0.7875) = 0.580275 c. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives a_line = -0.7875 + (0.798571 − 0.916272)(0.7875 − -0.7875)/(0.580275 − 0.916272) = -0.235772 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 a = 0.17325 mc; substituting back gives Y_out = 0.79857124 c.
  4. The interpolation error in the input is |a_line − a| = 0.409022 mc. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.78091624 c and 0.81622624 c. Sort the roots even if the response decreases: 0.0871898 mc ≤ a ≤ 0.249805 mc.
  6. The lower and upper input deviations are 0.0860602 mc and 0.0765554 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 a ≈ 0.17325 mc; compatible interval [0.0871898, 0.249805] mc. Endpoint interpolation gives -0.235772 mc.

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