University Quantum Mechanics II · topic UQMII-12.09
Second-order transitions through virtual intermediate states
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 transitions through virtual intermediate states: predict a finite change
A second-order transition is one whose amplitude comes from the perturbation acting twice, i → m → f, summed over every intermediate state m, each path weighted by 1/(Eᵢ + ħω − Eₘ), the distance of m from the virtual level that one photon reaches. Use the supplied topic model Y_out = (2*pi*v*v/d)^2, where Y_out is (Mt/ħ)² AT 1 μs (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Intermediate detuning Δ/h, measured in MHz. Keep v (Coupling per step V/h) = 1.5 MHz fixed. For this exercise use only 4.975 MHz ≤ d ≤ 27.025 MHz 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 = 9.826 MHz and 12.913 MHz. 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.
(Mt/ħ)² AT 1 μs against d
Predictions from the supplied Second-order transitions through virtual intermediate states 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): (Mt/ħ)² AT 1 μs
Horizontal axis (x): d: Intermediate detuning Δ/h / MHz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Intermediate detuning Δ/h / MHz | (Mt/ħ)² AT 1 μs |
|---|---|
| 4.975 | 8.074926961321562 |
| 10.487499999999999 | 1.8171083550301539 |
| 15.999999999999998 | 0.7807011293830451 |
| 21.512499999999996 | 0.4318600257006306 |
| 27.025 | 0.27364868637178136 |
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 = 9.826 MHz, substitute into Y_out = (2*pi*v*v/d)^2: Y₀ = 2.07.
- At d = 12.913 MHz, the same fixed-parameter model gives Y₁ = 1.19859.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.871415; the input change is Δd = 3.087 MHz.
- Average sensitivity is ΔY_out/Δd = -0.282285 1 per MHz. 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 = 2.07, Y₁ = 1.19859; ΔY_out = -0.871415; average sensitivity = -0.282285 1 per MHz.
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 transitions through virtual intermediate states: infer an input with uncertainty
A second-order transition is one whose amplitude comes from the perturbation acting twice, i → m → f, summed over every intermediate state m, each path weighted by 1/(Eᵢ + ħω − Eₘ), the distance of m from the virtual level that one photon reaches. Use the supplied topic model Y_out = (2*pi*v*v/d)^2, where Y_out is (Mt/ħ)² AT 1 μs (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Intermediate detuning Δ/h, measured in MHz. Keep v (Coupling per step V/h) = 1.5 MHz fixed. For this exercise use only 4.975 MHz ≤ d ≤ 27.025 MHz 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 = 0.58868939, with a bounded reading uncertainty ±0.047256. 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 0.002205 MHz, 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.
(Mt/ħ)² AT 1 μs against d
Predictions from the supplied Second-order transitions through virtual intermediate states 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): (Mt/ħ)² AT 1 μs
Horizontal axis (x): d: Intermediate detuning Δ/h / MHz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Intermediate detuning Δ/h / MHz | (Mt/ħ)² AT 1 μs |
|---|---|
| 4.975 | 8.074926961321562 |
| 10.487499999999999 | 1.8171083550301539 |
| 15.999999999999998 | 0.7807011293830451 |
| 21.512499999999996 | 0.4318600257006306 |
| 27.025 | 0.27364868637178136 |
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(4.975) = 8.07493 and Y_out(27.025) = 0.273649. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 4.975 + (0.588689 − 8.07493)(27.025 − 4.975)/(0.273649 − 8.07493) = 26.1346 MHz. 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 = 18.4255 MHz; substituting back gives Y_out = 0.58868939.
- The interpolation error in the input is |d_line − d| = 7.70905 MHz. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.54143339 and 0.63594539. Sort the roots even if the response decreases: 17.7277 MHz ≤ d ≤ 19.2128 MHz.
- The lower and upper input deviations are 0.697797 MHz and 0.787265 MHz. 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 ≈ 18.4255 MHz; compatible interval [17.7277, 19.2128] MHz. Endpoint interpolation gives 26.1346 MHz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.