University Quantum Mechanics I · topic UQMI-11.04
The 2x2 eigenvalue problem and the mixing angle
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 2×2 eigenvalue problem and the mixing angle: predict a finite change
The mixing angle χ of a two-state Hamiltonian is the angle of the rotation that diagonalises it: with the coupling made real, the upper eigenstate is cos χ|1⟩ + sin χ|2⟩, where tan 2χ = 2|V|/(E₁ − E₂) and 0 ≤ χ ≤ 90°. Use the supplied topic model Y_out = sqrt(d^2/4+v^2)-abs(d)/2, where Y_out is PUSH E₊ − max(E₁, E₂) (meV). 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 Uncoupled offset E₁ − E₂, measured in meV. Keep v (Coupling |V|) = 5 meV fixed. For this exercise use only 2.1 meV ≤ d ≤ 31.5 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 d = 8.568 meV and 12.684 meV. 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.
PUSH E₊ − max(E₁, E₂) against d
Predictions from the supplied The 2×2 eigenvalue problem and the mixing angle 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): PUSH E₊ − max(E₁, E₂) / meV
Horizontal axis (x): d: Uncoupled offset E₁ − E₂ / meV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Uncoupled offset E₁ − E₂ / meV | PUSH E₊ − max(E₁, E₂) / meV |
|---|---|
| 2.1000000000000014 | 4.059060579010587 |
| 9.450000000000001 | 2.1543622524184602 |
| 16.8 | 1.3754795278799499 |
| 24.15 | 0.9942626035289379 |
| 31.5 | 0.7746028696607397 |
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 = 8.568 meV, substitute into Y_out = sqrt(d^2/4+v^2)-abs(d)/2: Y₀ = 2.30027 meV.
- At d = 12.684 meV, the same fixed-parameter model gives Y₁ = 1.73395 meV.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.566324 meV; the input change is Δd = 4.116 meV.
- Average sensitivity is ΔY_out/Δd = -0.137591 meV 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 = 2.30027 meV, Y₁ = 1.73395 meV; ΔY_out = -0.566324 meV; average sensitivity = -0.137591 meV 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 2×2 eigenvalue problem and the mixing angle: infer an input with uncertainty
The mixing angle χ of a two-state Hamiltonian is the angle of the rotation that diagonalises it: with the coupling made real, the upper eigenstate is cos χ|1⟩ + sin χ|2⟩, where tan 2χ = 2|V|/(E₁ − E₂) and 0 ≤ χ ≤ 90°. Use the supplied topic model Y_out = sqrt(d^2/4+v^2)-abs(d)/2, where Y_out is PUSH E₊ − max(E₁, E₂) (meV). 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 Uncoupled offset E₁ − E₂, measured in meV. Keep v (Coupling |V|) = 5 meV fixed. For this exercise use only 2.1 meV ≤ d ≤ 31.5 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 = 1.1785477 meV, with a bounded reading uncertainty ±0.060592 meV. 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 0.00294 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.
PUSH E₊ − max(E₁, E₂) against d
Predictions from the supplied The 2×2 eigenvalue problem and the mixing angle 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): PUSH E₊ − max(E₁, E₂) / meV
Horizontal axis (x): d: Uncoupled offset E₁ − E₂ / meV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Uncoupled offset E₁ − E₂ / meV | PUSH E₊ − max(E₁, E₂) / meV |
|---|---|
| 2.1000000000000014 | 4.059060579010587 |
| 9.450000000000001 | 2.1543622524184602 |
| 16.8 | 1.3754795278799499 |
| 24.15 | 0.9942626035289379 |
| 31.5 | 0.7746028696607397 |
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(2.1) = 4.05906 meV and Y_out(31.5) = 0.774603 meV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 2.1 + (1.17855 − 4.05906)(31.5 − 2.1)/(0.774603 − 4.05906) = 27.8842 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 d = 20.034 meV; substituting back gives Y_out = 1.1785477 meV.
- The interpolation error in the input is |d_line − d| = 7.85019 meV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.1179557 meV and 1.2391397 meV. Sort the roots even if the response decreases: 18.9361 meV ≤ d ≤ 21.2443 meV.
- The lower and upper input deviations are 1.09785 meV and 1.21029 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 d ≈ 20.034 meV; compatible interval [18.9361, 21.2443] meV. Endpoint interpolation gives 27.8842 meV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.