University Physics V · topic UPV-13.08
The Stark effect, polarisability, and level mixing
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 Stark effect, polarisability and level mixing: predict a finite change
The Stark effect is the shift and splitting of atomic levels caused by the odd-parity operator eFz, and the static polarisability α is the coefficient in the second-order shift ΔE = −½αF² it produces whenever no degeneracy blocks the sum. Use the supplied topic model Y_out = -(0.0012795*d*f)^2/max(s,0.5), where Y_out is QUADRATIC −½αF² (GHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. s is Zero-field splitting Δ, measured in GHz. Keep d (Matrix element ⟨2s|z|2p₀⟩) = 3 a₀; f (Field marker F) = 200 V cm⁻¹ fixed. For this exercise use only 1.72188 GHz ≤ s ≤ 10.7781 GHz 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 s = 3.71425 GHz and 4.98212 GHz. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δs across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
QUADRATIC −½αF² against s
Predictions from the supplied The Stark effect, polarisability and level mixing 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): QUADRATIC −½αF² / GHz
Horizontal axis (x): s: Zero-field splitting Δ / GHz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Zero-field splitting Δ / GHz | QUADRATIC −½αF² / GHz |
|---|---|
| 1.721875 | -0.3422799506352086 |
| 3.9859374999999995 | -0.14786064508036062 |
| 6.249999999999999 | -0.09429812639999999 |
| 8.5140625 | -0.069222335400991 |
| 10.778125 | -0.05468143021165554 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of s separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At s = 3.71425 GHz, substitute into Y_out = -(0.0012795*d*f)^2/max(s,0.5): Y₀ = -0.158676 GHz.
- At s = 4.98212 GHz, the same fixed-parameter model gives Y₁ = -0.118296 GHz.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.0403807 GHz; the input change is Δs = 1.26787 GHz.
- Average sensitivity is ΔY_out/Δs = 0.0318491 GHz per GHz. 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 = -0.158676 GHz, Y₁ = -0.118296 GHz; ΔY_out = 0.0403807 GHz; average sensitivity = 0.0318491 GHz per GHz.
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 Stark effect, polarisability and level mixing: infer an input with uncertainty
The Stark effect is the shift and splitting of atomic levels caused by the odd-parity operator eFz, and the static polarisability α is the coefficient in the second-order shift ΔE = −½αF² it produces whenever no degeneracy blocks the sum. Use the supplied topic model Y_out = -(0.0012795*d*f)^2/max(s,0.5), where Y_out is QUADRATIC −½αF² (GHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. s is Zero-field splitting Δ, measured in GHz. Keep d (Matrix element ⟨2s|z|2p₀⟩) = 3 a₀; f (Field marker F) = 200 V cm⁻¹ fixed. For this exercise use only 1.72188 GHz ≤ s ≤ 10.7781 GHz 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 = -0.081334259 GHz, with a bounded reading uncertainty ±0.0039979 GHz. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown s by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000905625 GHz, 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.
QUADRATIC −½αF² against s
Predictions from the supplied The Stark effect, polarisability and level mixing 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): QUADRATIC −½αF² / GHz
Horizontal axis (x): s: Zero-field splitting Δ / GHz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Zero-field splitting Δ / GHz | QUADRATIC −½αF² / GHz |
|---|---|
| 1.721875 | -0.3422799506352086 |
| 3.9859374999999995 | -0.14786064508036062 |
| 6.249999999999999 | -0.09429812639999999 |
| 8.5140625 | -0.069222335400991 |
| 10.778125 | -0.05468143021165554 |
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(s) = 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(1.72188) = -0.34228 GHz and Y_out(10.7781) = -0.0546814 GHz. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives s_line = 1.72188 + (-0.0813343 − -0.34228)(10.7781 − 1.72188)/(-0.0546814 − -0.34228) = 9.93885 GHz. 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 s = 7.24619 GHz; substituting back gives Y_out = -0.081334259 GHz.
- The interpolation error in the input is |s_line − s| = 2.69266 GHz. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.085332159 GHz and -0.077336359 GHz. Sort the roots even if the response decreases: 6.9067 GHz ≤ s ≤ 7.62078 GHz.
- The lower and upper input deviations are 0.339491 GHz and 0.374591 GHz. 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 s ≈ 7.24619 GHz; compatible interval [6.9067, 7.62078] GHz. Endpoint interpolation gives 9.93885 GHz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.