University Quantum Mechanics II · topic UQMII-06.03
The first-order state and the energy-denominator condition
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 first-order state and the energy-denominator condition: predict a finite change
The first-order state is the part of the corrected eigenstate linear in H′: every other unperturbed state mixed in with amplitude ⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩/(Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾), trustworthy only while each amplitude is much smaller than 1. Use the supplied topic model Y_out = exp(-(0.023846*f/hw^1.5)^2), where Y_out is P(STAY IN |0⟩), EXACT (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. hw is Level spacing ħω, measured in meV. Keep f (Field F) = 100 kV m⁻¹ fixed. For this exercise use only 2.85 meV ≤ hw ≤ 9.15 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 hw = 4.236 meV and 5.118 meV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δhw across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
P(STAY IN |0⟩), EXACT against hw
Predictions from the supplied The first-order state and the energy-denominator condition 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): P(STAY IN |0⟩), EXACT
Horizontal axis (x): hw: Level spacing ħω / meV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| hw: Level spacing ħω / meV | P(STAY IN |0⟩), EXACT |
|---|---|
| 2.85 | 0.782204913244314 |
| 4.425 | 0.9364788747347376 |
| 6 | 0.974017953841428 |
| 7.574999999999999 | 0.9870029291772693 |
| 9.149999999999999 | 0.9926046802986603 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of hw separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At hw = 4.236 meV, substitute into Y_out = exp(-(0.023846*f/hw^1.5)^2): Y₀ = 0.927919.
- At hw = 5.118 meV, the same fixed-parameter model gives Y₁ = 0.958471.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.0305517; the input change is Δhw = 0.882 meV.
- Average sensitivity is ΔY_out/Δhw = 0.0346391 1 per meV. 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 = 0.927919, Y₁ = 0.958471; ΔY_out = 0.0305517; average sensitivity = 0.0346391 1 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 first-order state and the energy-denominator condition: infer an input with uncertainty
The first-order state is the part of the corrected eigenstate linear in H′: every other unperturbed state mixed in with amplitude ⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩/(Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾), trustworthy only while each amplitude is much smaller than 1. Use the supplied topic model Y_out = exp(-(0.023846*f/hw^1.5)^2), where Y_out is P(STAY IN |0⟩), EXACT (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. hw is Level spacing ħω, measured in meV. Keep f (Field F) = 100 kV m⁻¹ fixed. For this exercise use only 2.85 meV ≤ hw ≤ 9.15 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 = 0.98121303, with a bounded reading uncertainty ±0.0017087. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown hw by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00063 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.
P(STAY IN |0⟩), EXACT against hw
Predictions from the supplied The first-order state and the energy-denominator condition 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): P(STAY IN |0⟩), EXACT
Horizontal axis (x): hw: Level spacing ħω / meV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| hw: Level spacing ħω / meV | P(STAY IN |0⟩), EXACT |
|---|---|
| 2.85 | 0.782204913244314 |
| 4.425 | 0.9364788747347376 |
| 6 | 0.974017953841428 |
| 7.574999999999999 | 0.9870029291772693 |
| 9.149999999999999 | 0.9926046802986603 |
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(hw) = 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(2.85) = 0.782205 and Y_out(9.15) = 0.992605. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives hw_line = 2.85 + (0.981213 − 0.782205)(9.15 − 2.85)/(0.992605 − 0.782205) = 8.8089 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 hw = 6.693 meV; substituting back gives Y_out = 0.98121303.
- The interpolation error in the input is |hw_line − hw| = 2.1159 meV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.97950433 and 0.98292173. Sort the roots even if the response decreases: 6.4997 meV ≤ hw ≤ 6.91116 meV.
- The lower and upper input deviations are 0.1933 meV and 0.218156 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 hw ≈ 6.693 meV; compatible interval [6.4997, 6.91116] meV. Endpoint interpolation gives 8.8089 meV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.