University Advanced Quantum Mechanics · topic UAQM-01.08
Permanent electric dipole moments violate both P and T
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.
Permanent electric dipole moments violate both P and T: predict a finite change
A permanent electric dipole moment is a nonzero ⟨d⟩ in a stationary state of an isolated system with no applied field, seen as an energy shift −⟨d⟩⋅E that stays linear in E however weak the field is made. Use the supplied topic model Y_out = sqrt((split/2)^2+(1.2795*w*field)^2)-split/2, where Y_out is STARK SHIFT OF EACH LEVEL (MHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. split is Opposite-parity splitting Δ/h, measured in MHz. Keep w (Dipole matrix element w) = 1.73 e a₀; field (Applied field E) = 100 V cm⁻¹ fixed. For this exercise use only 212.5 MHz ≤ split ≤ 1787.5 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 split = 559 MHz and 779.5 MHz. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δsplit across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
STARK SHIFT OF EACH LEVEL against split
Predictions from the supplied Permanent electric dipole moments violate both P and T 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): STARK SHIFT OF EACH LEVEL / MHz
Horizontal axis (x): split: Opposite-parity splitting Δ/h / MHz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| split: Opposite-parity splitting Δ/h / MHz | STARK SHIFT OF EACH LEVEL / MHz |
|---|---|
| 212.5 | 139.28296003235496 |
| 606.25 | 72.21769353119157 |
| 999.9999999999999 | 46.806521506693514 |
| 1393.7499999999998 | 34.31043310657526 |
| 1787.4999999999998 | 27.003188678839138 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of split separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At split = 559 MHz, substitute into Y_out = sqrt((split/2)^2+(1.2795*w*field)^2)-split/2: Y₀ = 77.0356 MHz.
- At split = 779.5 MHz, the same fixed-parameter model gives Y₁ = 58.4714 MHz.
- Signed change: ΔY_out = Y₁ − Y₀ = -18.5642 MHz; the input change is Δsplit = 220.5 MHz.
- Average sensitivity is ΔY_out/Δsplit = -0.0841913 MHz 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 = 77.0356 MHz, Y₁ = 58.4714 MHz; ΔY_out = -18.5642 MHz; average sensitivity = -0.0841913 MHz 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.
Permanent electric dipole moments violate both P and T: infer an input with uncertainty
A permanent electric dipole moment is a nonzero ⟨d⟩ in a stationary state of an isolated system with no applied field, seen as an energy shift −⟨d⟩⋅E that stays linear in E however weak the field is made. Use the supplied topic model Y_out = sqrt((split/2)^2+(1.2795*w*field)^2)-split/2, where Y_out is STARK SHIFT OF EACH LEVEL (MHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. split is Opposite-parity splitting Δ/h, measured in MHz. Keep w (Dipole matrix element w) = 1.73 e a₀; field (Applied field E) = 100 V cm⁻¹ fixed. For this exercise use only 212.5 MHz ≤ split ≤ 1787.5 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 = 40.372817 MHz, with a bounded reading uncertainty ±2.0054 MHz. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown split by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.1575 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.
STARK SHIFT OF EACH LEVEL against split
Predictions from the supplied Permanent electric dipole moments violate both P and T 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): STARK SHIFT OF EACH LEVEL / MHz
Horizontal axis (x): split: Opposite-parity splitting Δ/h / MHz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| split: Opposite-parity splitting Δ/h / MHz | STARK SHIFT OF EACH LEVEL / MHz |
|---|---|
| 212.5 | 139.28296003235496 |
| 606.25 | 72.21769353119157 |
| 999.9999999999999 | 46.806521506693514 |
| 1393.7499999999998 | 34.31043310657526 |
| 1787.4999999999998 | 27.003188678839138 |
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(split) = 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(212.5) = 139.283 MHz and Y_out(1787.5) = 27.0032 MHz. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives split_line = 212.5 + (40.3728 − 139.283)(1787.5 − 212.5)/(27.0032 − 139.283) = 1599.96 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 split = 1173.25 MHz; substituting back gives Y_out = 40.372817 MHz.
- The interpolation error in the input is |split_line − split| = 426.708 MHz. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 38.367417 MHz and 42.378217 MHz. Sort the roots even if the response decreases: 1113.81 MHz ≤ split ≤ 1238.69 MHz.
- The lower and upper input deviations are 59.4358 MHz and 65.4394 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 split ≈ 1173.25 MHz; compatible interval [1113.81, 1238.69] MHz. Endpoint interpolation gives 1599.96 MHz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.