University Quantum Mechanics II · topic UQMII-06.06
Finding the good states from a commuting operator
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.
Finding the good states from a commuting operator: predict a finite change
A good state of a degenerate level can be read off as an eigenstate of any Hermitian operator that commutes with both H₀ and H′, provided no other state of the level shares its eigenvalue; shared eigenvalues leave blocks of W still to diagonalise. Use the supplied topic model Y_out = sqrt(1+hypot(eps,dlt)) - sqrt(1-hypot(eps,dlt)), where Y_out is EXACT SPLITTING (ħω). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; hypot(a,b) means √(a²+b²). dlt is Swap-breaking anisotropy δ, dimensionless. Keep eps (Swap-symmetric coupling ε) = 0.2 fixed. For this exercise use only 0.021 ≤ dlt ≤ 0.315 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 dlt = 0.08568 and 0.12684. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δdlt across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
EXACT SPLITTING against dlt
Predictions from the supplied Finding the good states from a commuting operator 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): EXACT SPLITTING / ħω
Horizontal axis (x): dlt: Swap-breaking anisotropy δ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dlt: Swap-breaking anisotropy δ | EXACT SPLITTING / ħω |
|---|---|
| 0.02100000000000002 | 0.2021344879431093 |
| 0.0945 | 0.22258458952558846 |
| 0.16799999999999998 | 0.2634940226666338 |
| 0.24149999999999996 | 0.3175936453184276 |
| 0.31499999999999995 | 0.3800536608165773 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of dlt separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At dlt = 0.08568, substitute into Y_out = sqrt(1+hypot(eps,dlt)) - sqrt(1-hypot(eps,dlt)): Y₀ = 0.218895 ħω.
- At dlt = 0.12684, the same fixed-parameter model gives Y₁ = 0.238532 ħω.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.0196374 ħω; the input change is Δdlt = 0.04116.
- Average sensitivity is ΔY_out/Δdlt = 0.4771 ħω per unit input. 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 = 0.218895 ħω, Y₁ = 0.238532 ħω; ΔY_out = 0.0196374 ħω; average sensitivity = 0.4771 ħω per unit input.
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.
Finding the good states from a commuting operator: infer an input with uncertainty
A good state of a degenerate level can be read off as an eigenstate of any Hermitian operator that commutes with both H₀ and H′, provided no other state of the level shares its eigenvalue; shared eigenvalues leave blocks of W still to diagonalise. Use the supplied topic model Y_out = sqrt(1+hypot(eps,dlt)) - sqrt(1-hypot(eps,dlt)), where Y_out is EXACT SPLITTING (ħω). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; hypot(a,b) means √(a²+b²). dlt is Swap-breaking anisotropy δ, dimensionless. Keep eps (Swap-symmetric coupling ε) = 0.2 fixed. For this exercise use only 0.021 ≤ dlt ≤ 0.315 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 = 0.28602326 ħω, with a bounded reading uncertainty ±0.012583 ħω. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown dlt by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000294, 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.
EXACT SPLITTING against dlt
Predictions from the supplied Finding the good states from a commuting operator 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): EXACT SPLITTING / ħω
Horizontal axis (x): dlt: Swap-breaking anisotropy δ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dlt: Swap-breaking anisotropy δ | EXACT SPLITTING / ħω |
|---|---|
| 0.02100000000000002 | 0.2021344879431093 |
| 0.0945 | 0.22258458952558846 |
| 0.16799999999999998 | 0.2634940226666338 |
| 0.24149999999999996 | 0.3175936453184276 |
| 0.31499999999999995 | 0.3800536608165773 |
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(dlt) = 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(0.021) = 0.202134 ħω and Y_out(0.315) = 0.380054 ħω. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives dlt_line = 0.021 + (0.286023 − 0.202134)(0.315 − 0.021)/(0.380054 − 0.202134) = 0.159621. 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 dlt = 0.20034; substituting back gives Y_out = 0.28602326 ħω.
- The interpolation error in the input is |dlt_line − dlt| = 0.0407192. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.27344026 ħω and 0.29860626 ħω. Sort the roots even if the response decreases: 0.18268 ≤ dlt ≤ 0.217205.
- The lower and upper input deviations are 0.0176601 and 0.0168651. 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 dlt ≈ 0.20034; compatible interval [0.18268, 0.217205]. Endpoint interpolation gives 0.159621.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.