University Quantum Mechanics II · topic UQMII-06.02
The first-order energy shift E^(1) = <n|H'|n>
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 energy shift E⁽¹⁾ = ⟨n|H′|n⟩: predict a finite change
The first-order shift of a non-degenerate level is the expectation value of the perturbation in the unperturbed state, Eₙ⁽¹⁾ = ⟨n⁽⁰⁾|H′|n⁽⁰⁾⟩; for a potential it is H′ averaged over the unperturbed probability density. Use the supplied topic model Y_out = 0.37603*n^2+0.001*20/max(w,0.05)*(w-cos(2*pi*n*x0)*sin(pi*n*w)/(pi*n)), where Y_out is LEVEL Eₙ⁽⁰⁾ + E⁽¹⁾ (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value; unmarked trigonometric functions use radians. x0 is Bump centre x₀, measured in nm. Keep n (Level n) = 1; w (Bump width w) = 0.1 nm fixed. For this exercise use only 0.513125 nm ≤ x0 ≤ 0.696875 nm 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 x0 = 0.55355 nm and 0.579275 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δx0 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
LEVEL Eₙ⁽⁰⁾ + E⁽¹⁾ against x0
Predictions from the supplied The first-order energy shift E⁽¹⁾ = ⟨n|H′|n⟩ 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): LEVEL Eₙ⁽⁰⁾ + E⁽¹⁾ / eV
Horizontal axis (x): x0: Bump centre x₀ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| x0: Bump centre x₀ / nm | LEVEL Eₙ⁽⁰⁾ + E⁽¹⁾ / eV |
|---|---|
| 0.513125 | 0.4156357761964956 |
| 0.5590625 | 0.4143634926514627 |
| 0.605 | 0.41157442946227474 |
| 0.6509375 | 0.40749933340217587 |
| 0.6968749999999999 | 0.40247534823692 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of x0 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At x0 = 0.55355 nm, substitute into Y_out = 0.37603*n^2+0.001*20/max(w,0.05)*(w-cos(2*pi*n*x0)*sin(pi*n*w)/(pi*n)): Y₀ = 0.4146 eV.
- At x0 = 0.579275 nm, the same fixed-parameter model gives Y₁ = 0.413312 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.00128729 eV; the input change is Δx0 = 0.025725 nm.
- Average sensitivity is ΔY_out/Δx0 = -0.0500405 eV per nm. 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 = 0.4146 eV, Y₁ = 0.413312 eV; ΔY_out = -0.00128729 eV; average sensitivity = -0.0500405 eV per nm.
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 energy shift E⁽¹⁾ = ⟨n|H′|n⟩: infer an input with uncertainty
The first-order shift of a non-degenerate level is the expectation value of the perturbation in the unperturbed state, Eₙ⁽¹⁾ = ⟨n⁽⁰⁾|H′|n⁽⁰⁾⟩; for a potential it is H′ averaged over the unperturbed probability density. Use the supplied topic model Y_out = 0.37603*n^2+0.001*20/max(w,0.05)*(w-cos(2*pi*n*x0)*sin(pi*n*w)/(pi*n)), where Y_out is LEVEL Eₙ⁽⁰⁾ + E⁽¹⁾ (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value; unmarked trigonometric functions use radians. x0 is Bump centre x₀, measured in nm. Keep n (Level n) = 1; w (Bump width w) = 0.1 nm fixed. For this exercise use only 0.513125 nm ≤ x0 ≤ 0.696875 nm 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 = 0.40992207 eV, with a bounded reading uncertainty ±0.00085706 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown x0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000018375 nm, 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.
LEVEL Eₙ⁽⁰⁾ + E⁽¹⁾ against x0
Predictions from the supplied The first-order energy shift E⁽¹⁾ = ⟨n|H′|n⟩ 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): LEVEL Eₙ⁽⁰⁾ + E⁽¹⁾ / eV
Horizontal axis (x): x0: Bump centre x₀ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| x0: Bump centre x₀ / nm | LEVEL Eₙ⁽⁰⁾ + E⁽¹⁾ / eV |
|---|---|
| 0.513125 | 0.4156357761964956 |
| 0.5590625 | 0.4143634926514627 |
| 0.605 | 0.41157442946227474 |
| 0.6509375 | 0.40749933340217587 |
| 0.6968749999999999 | 0.40247534823692 |
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(x0) = 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(0.513125) = 0.415636 eV and Y_out(0.696875) = 0.402475 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives x0_line = 0.513125 + (0.409922 − 0.415636)(0.696875 − 0.513125)/(0.402475 − 0.415636) = 0.592902 nm. 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 x0 = 0.625212 nm; substituting back gives Y_out = 0.40992207 eV.
- The interpolation error in the input is |x0_line − x0| = 0.0323109 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.40906501 eV and 0.41077913 eV. Sort the roots even if the response decreases: 0.615092 nm ≤ x0 ≤ 0.634727 nm.
- The lower and upper input deviations are 0.0101204 nm and 0.00951485 nm. 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 x0 ≈ 0.625212 nm; compatible interval [0.615092, 0.634727] nm. Endpoint interpolation gives 0.592902 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.