University Quantum Mechanics II · topic UQMII-14.05
The geometric phase for a cyclic change of parameters
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 geometric phase for a cyclic change of parameters: predict a finite change
Berry's phase is the phase γₙ = i∮⟨n|∇_R n⟩⋅dR, over and above the dynamic phase −(1/ħ)∫Eₙ dt, that an eigenstate acquires when the parameters R of its Hamiltonian are carried slowly once round a closed loop; it depends on the loop, not on how long the trip takes. Use the supplied topic model Y_out = deg(((sign(rho-c)+1)/2*(rad(phi)+atan2(-c*sind(phi),rho+c*cosd(phi)))+(1-sign(rho-c))/2*atan2(rho*sind(phi),c+rho*cosd(phi))))+1e-9, where Y_out is MIXING ANGLE θ (°). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees; deg converts radians to degrees; rad converts degrees to radians; atan2(y,x) is the angle of the vector (x,y), in radians. rho is Loop radius ρ, measured in meV. Keep c (Loop centre, X = c) = 0.5 meV; phi (Progress round the loop φ) = 300 ° fixed. For this exercise use only 0.909375 meV ≤ rho ≤ 2.09063 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 rho = 1.16925 meV and 1.33462 meV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δrho across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
MIXING ANGLE θ against rho
Predictions from the supplied The geometric phase for a cyclic change of parameters 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): MIXING ANGLE θ / °
Horizontal axis (x): rho: Loop radius ρ / meV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rho: Loop radius ρ / meV | MIXING ANGLE θ / ° |
|---|---|
| 0.909375 | 320.4800590847286 |
| 1.2046875000000001 | 316.5765402383228 |
| 1.5 | 313.897886249014 |
| 1.7953125 | 311.95357339735745 |
| 2.090625 | 310.48114895873823 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of rho separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At rho = 1.16925 meV, substitute into Y_out = deg(((sign(rho-c)+1)/2*(rad(phi)+atan2(-c*sind(phi),rho+c*cosd(phi)))+(1-sign(rho-c))/2*atan2(rho*sind(phi),c+rho*cosd(phi))))+1e-9: Y₀ = 316.967 °.
- At rho = 1.33462 meV, the same fixed-parameter model gives Y₁ = 315.283 °.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.68347 °; the input change is Δrho = 0.165375 meV.
- Average sensitivity is ΔY_out/Δrho = -10.1797 ° 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 = 316.967 °, Y₁ = 315.283 °; ΔY_out = -1.68347 °; average sensitivity = -10.1797 ° 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 geometric phase for a cyclic change of parameters: infer an input with uncertainty
Berry's phase is the phase γₙ = i∮⟨n|∇_R n⟩⋅dR, over and above the dynamic phase −(1/ħ)∫Eₙ dt, that an eigenstate acquires when the parameters R of its Hamiltonian are carried slowly once round a closed loop; it depends on the loop, not on how long the trip takes. Use the supplied topic model Y_out = deg(((sign(rho-c)+1)/2*(rad(phi)+atan2(-c*sind(phi),rho+c*cosd(phi)))+(1-sign(rho-c))/2*atan2(rho*sind(phi),c+rho*cosd(phi))))+1e-9, where Y_out is MIXING ANGLE θ (°). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees; deg converts radians to degrees; rad converts degrees to radians; atan2(y,x) is the angle of the vector (x,y), in radians. rho is Loop radius ρ, measured in meV. Keep c (Loop centre, X = c) = 0.5 meV; phi (Progress round the loop φ) = 300 ° fixed. For this exercise use only 0.909375 meV ≤ rho ≤ 2.09063 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 = 312.97091 °, with a bounded reading uncertainty ±0.37346 °. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rho by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000118125 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.
MIXING ANGLE θ against rho
Predictions from the supplied The geometric phase for a cyclic change of parameters 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): MIXING ANGLE θ / °
Horizontal axis (x): rho: Loop radius ρ / meV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rho: Loop radius ρ / meV | MIXING ANGLE θ / ° |
|---|---|
| 0.909375 | 320.4800590847286 |
| 1.2046875000000001 | 316.5765402383228 |
| 1.5 | 313.897886249014 |
| 1.7953125 | 311.95357339735745 |
| 2.090625 | 310.48114895873823 |
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(rho) = 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(0.909375) = 320.48 ° and Y_out(2.09063) = 310.481 °. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives rho_line = 0.909375 + (312.971 − 320.48)(2.09063 − 0.909375)/(310.481 − 320.48) = 1.79649 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 rho = 1.62994 meV; substituting back gives Y_out = 312.97091 °.
- The interpolation error in the input is |rho_line − rho| = 0.166552 meV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 312.59745 ° and 313.34437 °. Sort the roots even if the response decreases: 1.57546 meV ≤ rho ≤ 1.68759 meV.
- The lower and upper input deviations are 0.054481 meV and 0.0576543 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 rho ≈ 1.62994 meV; compatible interval [1.57546, 1.68759] meV. Endpoint interpolation gives 1.79649 meV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.