University Quantum Mechanics II · topic UQMII-06.04
Second-order energy, and why a ground state always goes down
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.
Second-order energy, and why a ground state always goes down: predict a finite change
The second-order energy shift Eₙ⁽²⁾ is the part of a level's energy quadratic in the perturbation: the sum, over every other unperturbed state m, of |⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩|² divided by the energy gap Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾. Use the supplied topic model Y_out = -(n+1)*d^2/2, where Y_out is PUSH FROM LEVEL n + 1 (ħω). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Field as displacement d/x₀ = qF/(mω²x₀), dimensionless. Keep n (Level n examined) = 0 fixed. For this exercise use only 0.14875 ≤ d ≤ 1.25125 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 = 0.3913 and 0.54565. 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 FROM LEVEL n + 1 against d
Predictions from the supplied Second-order energy, and why a ground state always goes down 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 FROM LEVEL n + 1 / ħω
Horizontal axis (x): d: Field as displacement d/x₀ = qF/(mω²x₀)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Field as displacement d/x₀ = qF/(mω²x₀) | PUSH FROM LEVEL n + 1 / ħω |
|---|---|
| 0.14875 | -0.01106328125 |
| 0.42437499999999995 | -0.09004707031249998 |
| 0.7 | -0.24499999999999997 |
| 0.9756249999999997 | -0.4759220703124997 |
| 1.2512499999999998 | -0.7828132812499997 |
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 = 0.3913, substitute into Y_out = -(n+1)*d^2/2: Y₀ = -0.0765578 ħω.
- At d = 0.54565, the same fixed-parameter model gives Y₁ = -0.148867 ħω.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0723091 ħω; the input change is Δd = 0.15435.
- Average sensitivity is ΔY_out/Δd = -0.468475 ħω per unit input. 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 = -0.0765578 ħω, Y₁ = -0.148867 ħω; ΔY_out = -0.0723091 ħω; average sensitivity = -0.468475 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.
Second-order energy, and why a ground state always goes down: infer an input with uncertainty
The second-order energy shift Eₙ⁽²⁾ is the part of a level's energy quadratic in the perturbation: the sum, over every other unperturbed state m, of |⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩|² divided by the energy gap Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾. Use the supplied topic model Y_out = -(n+1)*d^2/2, where Y_out is PUSH FROM LEVEL n + 1 (ħω). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Field as displacement d/x₀ = qF/(mω²x₀), dimensionless. Keep n (Level n examined) = 0 fixed. For this exercise use only 0.14875 ≤ d ≤ 1.25125 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 = -0.33724631 ħω, with a bounded reading uncertainty ±0.048927 ħω. 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 0.00011025, 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 FROM LEVEL n + 1 against d
Predictions from the supplied Second-order energy, and why a ground state always goes down 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 FROM LEVEL n + 1 / ħω
Horizontal axis (x): d: Field as displacement d/x₀ = qF/(mω²x₀)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Field as displacement d/x₀ = qF/(mω²x₀) | PUSH FROM LEVEL n + 1 / ħω |
|---|---|
| 0.14875 | -0.01106328125 |
| 0.42437499999999995 | -0.09004707031249998 |
| 0.7 | -0.24499999999999997 |
| 0.9756249999999997 | -0.4759220703124997 |
| 1.2512499999999998 | -0.7828132812499997 |
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(0.14875) = -0.0110633 ħω and Y_out(1.25125) = -0.782813 ħω. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 0.14875 + (-0.337246 − -0.0110633)(1.25125 − 0.14875)/(-0.782813 − -0.0110633) = 0.614726. 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 = 0.821275; substituting back gives Y_out = -0.33724631 ħω.
- The interpolation error in the input is |d_line − d| = 0.206549. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.38617331 ħω and -0.28831931 ħω. Sort the roots even if the response decreases: 0.759367 ≤ d ≤ 0.878833.
- The lower and upper input deviations are 0.0619077 and 0.0575575. 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 ≈ 0.821275; compatible interval [0.759367, 0.878833]. Endpoint interpolation gives 0.614726.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.