University Quantum Mechanics II · topic UQMII-13.01
The electric-dipole Hamiltonian and its validity
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 electric-dipole Hamiltonian and its validity: predict a finite change
The electric-dipole Hamiltonian H′ = −d⋅E(t), with d = qr, is the coupling of an atom to a light wave whose field is treated as uniform across the atom: the leading term of an expansion in k a, the atom's size times the wave number, which is of order α. Use the supplied topic model Y_out = 10^lg, where Y_out is WAVELENGTH λ (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lg is Wavelength, log₁₀(λ / 1 nm), dimensionless. Keep wt (Phase of the wave ωt) = 60 ° fixed. For this exercise use only -0.575 ≤ lg ≤ 2.575 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 lg = 0.118 and 0.559. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δlg across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
WAVELENGTH λ against lg
Predictions from the supplied The electric-dipole Hamiltonian and its validity 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): WAVELENGTH λ / nm
Horizontal axis (x): lg: Wavelength, log₁₀(λ / 1 nm)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lg: Wavelength, log₁₀(λ / 1 nm) | WAVELENGTH λ / nm |
|---|---|
| -0.575 | 0.26607250597988097 |
| 0.2124999999999999 | 1.631172909227838 |
| 0.9999999999999998 | 9.999999999999995 |
| 1.7874999999999999 | 61.30557921498205 |
| 2.5749999999999993 | 375.83740428844357 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of lg separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At lg = 0.118, substitute into Y_out = 10^lg: Y₀ = 1.3122 nm.
- At lg = 0.559, the same fixed-parameter model gives Y₁ = 3.62243 nm.
- Signed change: ΔY_out = Y₁ − Y₀ = 2.31023 nm; the input change is Δlg = 0.441.
- Average sensitivity is ΔY_out/Δlg = 5.23862 nm 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 = 1.3122 nm, Y₁ = 3.62243 nm; ΔY_out = 2.31023 nm; average sensitivity = 5.23862 nm 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.
The electric-dipole Hamiltonian and its validity: infer an input with uncertainty
The electric-dipole Hamiltonian H′ = −d⋅E(t), with d = qr, is the coupling of an atom to a light wave whose field is treated as uniform across the atom: the leading term of an expansion in k a, the atom's size times the wave number, which is of order α. Use the supplied topic model Y_out = 10^lg, where Y_out is WAVELENGTH λ (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lg is Wavelength, log₁₀(λ / 1 nm), dimensionless. Keep wt (Phase of the wave ωt) = 60 ° fixed. For this exercise use only -0.575 ≤ lg ≤ 2.575 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 = 22.207517 nm, with a bounded reading uncertainty ±3.2912 nm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lg by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000315, 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.
WAVELENGTH λ against lg
Predictions from the supplied The electric-dipole Hamiltonian and its validity 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): WAVELENGTH λ / nm
Horizontal axis (x): lg: Wavelength, log₁₀(λ / 1 nm)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lg: Wavelength, log₁₀(λ / 1 nm) | WAVELENGTH λ / nm |
|---|---|
| -0.575 | 0.26607250597988097 |
| 0.2124999999999999 | 1.631172909227838 |
| 0.9999999999999998 | 9.999999999999995 |
| 1.7874999999999999 | 61.30557921498205 |
| 2.5749999999999993 | 375.83740428844357 |
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(lg) = 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(-0.575) = 0.266073 nm and Y_out(2.575) = 375.837 nm. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives lg_line = -0.575 + (22.2075 − 0.266073)(2.575 − -0.575)/(375.837 − 0.266073) = -0.390972. 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 lg = 1.3465; substituting back gives Y_out = 22.207517 nm.
- The interpolation error in the input is |lg_line − lg| = 1.73747. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 18.916317 nm and 25.498717 nm. Sort the roots even if the response decreases: 1.27684 ≤ lg ≤ 1.40652.
- The lower and upper input deviations are 0.0696634 and 0.0600183. 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 lg ≈ 1.3465; compatible interval [1.27684, 1.40652]. Endpoint interpolation gives -0.390972.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.