University Advanced Quantum Mechanics · topic UAQM-01.07
Kramers degeneracy derived
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.
Kramers degeneracy derived: predict a finite change
Kramers' theorem: if the Hamiltonian commutes with time reversal and Θ² = −1, as for any odd number of spin-half particles, every energy level is at least twofold degenerate, |n⟩ and Θ|n⟩ being orthogonal states of one energy. Use the supplied topic model Y_out = abs(27.99*bt-sqrt((d+27.99*bt)^2+3*er^2)+sqrt((d-27.99*bt)^2+3*er^2)), where Y_out is S = 3/2 GROUND GAP AT CURSOR (GHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; abs(x) means |x|. d is Axial term D/h, measured in GHz. Keep er (Rhombic term E/h) = 1.5 GHz; bt (Field cursor B along z) = 0.05 T fixed. For this exercise use only -4.725 GHz ≤ d ≤ 1.26 GHz 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 = -3.4083 GHz and -2.5704 GHz. 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.
S = 3/2 GROUND GAP AT CURSOR against d
Predictions from the supplied Kramers degeneracy derived 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): S = 3/2 GROUND GAP AT CURSOR / GHz
Horizontal axis (x): d: Axial term D/h / GHz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Axial term D/h / GHz | S = 3/2 GROUND GAP AT CURSOR / GHz |
|---|---|
| -4.725 | 3.832216197025401 |
| -3.22875 | 3.529662748260444 |
| -1.7325 | 2.849497340988343 |
| -0.23625000000000007 | 1.6230233799098126 |
| 1.2599999999999998 | 0.2833971468521752 |
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 = -3.4083 GHz, substitute into Y_out = abs(27.99*bt-sqrt((d+27.99*bt)^2+3*er^2)+sqrt((d-27.99*bt)^2+3*er^2)): Y₀ = 3.58029 GHz.
- At d = -2.5704 GHz, the same fixed-parameter model gives Y₁ = 3.29424 GHz.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.286044 GHz; the input change is Δd = 0.8379 GHz.
- Average sensitivity is ΔY_out/Δd = -0.341382 GHz per GHz. 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 = 3.58029 GHz, Y₁ = 3.29424 GHz; ΔY_out = -0.286044 GHz; average sensitivity = -0.341382 GHz per GHz.
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.
Kramers degeneracy derived: infer an input with uncertainty
Kramers' theorem: if the Hamiltonian commutes with time reversal and Θ² = −1, as for any odd number of spin-half particles, every energy level is at least twofold degenerate, |n⟩ and Θ|n⟩ being orthogonal states of one energy. Use the supplied topic model Y_out = abs(27.99*bt-sqrt((d+27.99*bt)^2+3*er^2)+sqrt((d-27.99*bt)^2+3*er^2)), where Y_out is S = 3/2 GROUND GAP AT CURSOR (GHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; abs(x) means |x|. d is Axial term D/h, measured in GHz. Keep er (Rhombic term E/h) = 1.5 GHz; bt (Field cursor B along z) = 0.05 T fixed. For this exercise use only -4.725 GHz ≤ d ≤ 1.26 GHz 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 = 2.3684628 GHz, with a bounded reading uncertainty ±0.21956 GHz. 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 0.0005985 GHz, 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.
S = 3/2 GROUND GAP AT CURSOR against d
Predictions from the supplied Kramers degeneracy derived 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): S = 3/2 GROUND GAP AT CURSOR / GHz
Horizontal axis (x): d: Axial term D/h / GHz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Axial term D/h / GHz | S = 3/2 GROUND GAP AT CURSOR / GHz |
|---|---|
| -4.725 | 3.832216197025401 |
| -3.22875 | 3.529662748260444 |
| -1.7325 | 2.849497340988343 |
| -0.23625000000000007 | 1.6230233799098126 |
| 1.2599999999999998 | 0.2833971468521752 |
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(-4.725) = 3.83222 GHz and Y_out(1.26) = 0.283397 GHz. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = -4.725 + (2.36846 − 3.83222)(1.26 − -4.725)/(0.283397 − 3.83222) = -2.25641 GHz. 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 = -1.07415 GHz; substituting back gives Y_out = 2.3684628 GHz.
- The interpolation error in the input is |d_line − d| = 1.18226 GHz. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.1489028 GHz and 2.5880228 GHz. Sort the roots even if the response decreases: -1.3557 GHz ≤ d ≤ -0.81341 GHz.
- The lower and upper input deviations are 0.28155 GHz and 0.26074 GHz. 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 ≈ -1.07415 GHz; compatible interval [-1.3557, -0.81341] GHz. Endpoint interpolation gives -2.25641 GHz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.