University Quantum Mechanics II · topic UQMII-14.07
Berry curvature and the monopole at a degeneracy
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.
Berry curvature and the monopole at a degeneracy: predict a finite change
The Berry curvature of a level is the curl, in parameter space, of its Berry connection i⟨n|∇_R n⟩; its flux through a surface bounded by a closed loop is the geometric phase the level picks up when the Hamiltonian is carried slowly around that loop. Use the supplied topic model Y_out = pi*(sign(-d-b1)+1)-pi*(1-(d+b1*cosd(th))/hypot(d+b1*cosd(th),b1*sind(th))), where Y_out is BERRY PHASE γ (CAP FLUX) (rad). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; hypot(a,b) means √(a²+b²); sind/cosd/tand use degrees. d is Offset field d along z, measured in mT. Keep b1 (Sphere radius B₁) = 1.75 mT; th (Loop latitude θ) = 60 ° fixed. For this exercise use only -1.47 mT ≤ d ≤ 2.75625 mT 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 d = -0.540225 mT and 0.05145 mT. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δd across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
BERRY PHASE γ (CAP FLUX) against d
Predictions from the supplied Berry curvature and the monopole at a degeneracy 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): BERRY PHASE γ (CAP FLUX) / rad
Horizontal axis (x): d: Offset field d along z / mT
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Offset field d along z / mT | BERRY PHASE γ (CAP FLUX) / rad |
|---|---|
| -1.4700000000000002 | -4.289667404628125 |
| -0.4134375000000001 | -2.2263191549463963 |
| 0.643125 | -0.9182623545182519 |
| 1.6996875 | -0.43421640595052485 |
| 2.75625 | -0.2423768231303443 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of d separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At d = -0.540225 mT, substitute into Y_out = pi*(sign(-d-b1)+1)-pi*(1-(d+b1*cosd(th))/hypot(d+b1*cosd(th),b1*sind(th))): Y₀ = -2.46397 rad.
- At d = 0.05145 mT, the same fixed-parameter model gives Y₁ = -1.50304 rad.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.960925 rad; the input change is Δd = 0.591675 mT.
- Average sensitivity is ΔY_out/Δd = 1.62408 rad per mT. 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 = -2.46397 rad, Y₁ = -1.50304 rad; ΔY_out = 0.960925 rad; average sensitivity = 1.62408 rad per mT.
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.
Berry curvature and the monopole at a degeneracy: infer an input with uncertainty
The Berry curvature of a level is the curl, in parameter space, of its Berry connection i⟨n|∇_R n⟩; its flux through a surface bounded by a closed loop is the geometric phase the level picks up when the Hamiltonian is carried slowly around that loop. Use the supplied topic model Y_out = pi*(sign(-d-b1)+1)-pi*(1-(d+b1*cosd(th))/hypot(d+b1*cosd(th),b1*sind(th))), where Y_out is BERRY PHASE γ (CAP FLUX) (rad). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; hypot(a,b) means √(a²+b²); sind/cosd/tand use degrees. d is Offset field d along z, measured in mT. Keep b1 (Sphere radius B₁) = 1.75 mT; th (Loop latitude θ) = 60 ° fixed. For this exercise use only -1.47 mT ≤ d ≤ 2.75625 mT 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 = -0.64551143 rad, with a bounded reading uncertainty ±0.06047 rad. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000422625 mT, 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.
BERRY PHASE γ (CAP FLUX) against d
Predictions from the supplied Berry curvature and the monopole at a degeneracy 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): BERRY PHASE γ (CAP FLUX) / rad
Horizontal axis (x): d: Offset field d along z / mT
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Offset field d along z / mT | BERRY PHASE γ (CAP FLUX) / rad |
|---|---|
| -1.4700000000000002 | -4.289667404628125 |
| -0.4134375000000001 | -2.2263191549463963 |
| 0.643125 | -0.9182623545182519 |
| 1.6996875 | -0.43421640595052485 |
| 2.75625 | -0.2423768231303443 |
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(d) = 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(-1.47) = -4.28967 rad and Y_out(2.75625) = -0.242377 rad. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives d_line = -1.47 + (-0.645511 − -4.28967)(2.75625 − -1.47)/(-0.242377 − -4.28967) = 2.33529 mT. 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 d = 1.10801 mT; substituting back gives Y_out = -0.64551143 rad.
- The interpolation error in the input is |d_line − d| = 1.22728 mT. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.70598143 rad and -0.58504143 rad. Sort the roots even if the response decreases: 0.985252 mT ≤ d ≤ 1.24709 mT.
- The lower and upper input deviations are 0.122761 mT and 0.139074 mT. 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 d ≈ 1.10801 mT; compatible interval [0.985252, 1.24709] mT. Endpoint interpolation gives 2.33529 mT.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.