University Quantum Mechanics II · topic UQMII-09.05
The ground-state Stark effect: a second-order shift and a polarisability
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 ground-state Stark effect and the polarisability: predict a finite change
The static polarisability αₚₒₗ of hydrogen 1s is the constant in d = αₚₒₗF, the dipole a weak uniform field induces; because 1s has no permanent dipole, it fixes the whole leading shift, ΔE = −½αₚₒₗF². Use the supplied topic model Y_out = -5.59872*(f/100)^2, where Y_out is SHIFT ΔE/h, EXACT 9/2 (MHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. f is Field F, measured in kV/cm. Keep k (p terms kept exactly, k) = 1 fixed. For this exercise use only 42.5 kV/cm ≤ f ≤ 357.5 kV/cm 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 f = 111.8 kV/cm and 155.9 kV/cm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δf across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
SHIFT ΔE/h, EXACT 9/2 against f
Predictions from the supplied The ground-state Stark effect and the polarisability 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): SHIFT ΔE/h, EXACT 9/2 / MHz
Horizontal axis (x): f: Field F / kV/cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Field F / kV/cm | SHIFT ΔE/h, EXACT 9/2 / MHz |
|---|---|
| 42.5 | -1.0112687999999999 |
| 121.25 | -8.230993199999999 |
| 200 | -22.39488 |
| 278.75 | -43.502929200000004 |
| 357.5 | -71.5551408 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of f separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At f = 111.8 kV/cm, substitute into Y_out = -5.59872*(f/100)^2: Y₀ = -6.99797 MHz.
- At f = 155.9 kV/cm, the same fixed-parameter model gives Y₁ = -13.6076 MHz.
- Signed change: ΔY_out = Y₁ − Y₀ = -6.60961 MHz; the input change is Δf = 44.1 kV/cm.
- Average sensitivity is ΔY_out/Δf = -0.149878 MHz per kV/cm. 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 = -6.99797 MHz, Y₁ = -13.6076 MHz; ΔY_out = -6.60961 MHz; average sensitivity = -0.149878 MHz per kV/cm.
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 ground-state Stark effect and the polarisability: infer an input with uncertainty
The static polarisability αₚₒₗ of hydrogen 1s is the constant in d = αₚₒₗF, the dipole a weak uniform field induces; because 1s has no permanent dipole, it fixes the whole leading shift, ΔE = −½αₚₒₗF². Use the supplied topic model Y_out = -5.59872*(f/100)^2, where Y_out is SHIFT ΔE/h, EXACT 9/2 (MHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. f is Field F, measured in kV/cm. Keep k (p terms kept exactly, k) = 1 fixed. For this exercise use only 42.5 kV/cm ≤ f ≤ 357.5 kV/cm 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 = -30.826901 MHz, with a bounded reading uncertainty ±4.4723 MHz. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown f by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0315 kV/cm, 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.
SHIFT ΔE/h, EXACT 9/2 against f
Predictions from the supplied The ground-state Stark effect and the polarisability 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): SHIFT ΔE/h, EXACT 9/2 / MHz
Horizontal axis (x): f: Field F / kV/cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Field F / kV/cm | SHIFT ΔE/h, EXACT 9/2 / MHz |
|---|---|
| 42.5 | -1.0112687999999999 |
| 121.25 | -8.230993199999999 |
| 200 | -22.39488 |
| 278.75 | -43.502929200000004 |
| 357.5 | -71.5551408 |
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(f) = 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(42.5) = -1.01127 MHz and Y_out(357.5) = -71.5551 MHz. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives f_line = 42.5 + (-30.8269 − -1.01127)(357.5 − 42.5)/(-71.5551 − -1.01127) = 175.636 kV/cm. 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 f = 234.65 kV/cm; substituting back gives Y_out = -30.826901 MHz.
- The interpolation error in the input is |f_line − f| = 59.0141 kV/cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -35.299201 MHz and -26.354601 MHz. Sort the roots even if the response decreases: 216.962 kV/cm ≤ f ≤ 251.095 kV/cm.
- The lower and upper input deviations are 17.6879 kV/cm and 16.445 kV/cm. 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 f ≈ 234.65 kV/cm; compatible interval [216.962, 251.095] kV/cm. Endpoint interpolation gives 175.636 kV/cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.