University Physics V · topic UPV-07.06
Hermite-Gaussian wavefunctions in the position basis
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.
Hermite–Gaussians in the position basis: predict a finite change
The Hermite–Gaussian function ψ_n(x) = ⟨x|n⟩ is the n-th oscillator eigenvector written out in the position basis: a degree-n Hermite polynomial in ξ = x/x₀ multiplying one universal Gaussian exp(−ξ²/2), with x₀ = √(ħ/mω). Use the supplied topic model Y_out = 116.3/max(s,0.05)^2, where Y_out is TRAP ω/2π (⁸⁷Rb) (Hz). 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 Length scale x₀ = √(ħ/mω), measured in µm. Keep n (Quantum number n) = 2 fixed. For this exercise use only 0.685 µm ≤ s ≤ 1.315 µm 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 = 0.8236 µm and 0.9118 µm. 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.
TRAP ω/2π (⁸⁷Rb) against s
Predictions from the supplied Hermite–Gaussians in the position basis 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): TRAP ω/2π (⁸⁷Rb) / Hz
Horizontal axis (x): s: Length scale x₀ = √(ħ/mω) / µm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Length scale x₀ = √(ħ/mω) / µm | TRAP ω/2π (⁸⁷Rb) / Hz |
|---|---|
| 0.6849999999999999 | 247.85550642016094 |
| 0.8424999999999999 | 163.8475288150816 |
| 1 | 116.3 |
| 1.1575 | 86.80359566914993 |
| 1.315 | 67.25556246295307 |
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 = 0.8236 µm, substitute into Y_out = 116.3/max(s,0.05)^2: Y₀ = 171.454 Hz.
- At s = 0.9118 µm, the same fixed-parameter model gives Y₁ = 139.888 Hz.
- Signed change: ΔY_out = Y₁ − Y₀ = -31.5657 Hz; the input change is Δs = 0.0882 µm.
- Average sensitivity is ΔY_out/Δs = -357.888 Hz per µm. 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 = 171.454 Hz, Y₁ = 139.888 Hz; ΔY_out = -31.5657 Hz; average sensitivity = -357.888 Hz per µm.
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.
Hermite–Gaussians in the position basis: infer an input with uncertainty
The Hermite–Gaussian function ψ_n(x) = ⟨x|n⟩ is the n-th oscillator eigenvector written out in the position basis: a degree-n Hermite polynomial in ξ = x/x₀ multiplying one universal Gaussian exp(−ξ²/2), with x₀ = √(ħ/mω). Use the supplied topic model Y_out = 116.3/max(s,0.05)^2, where Y_out is TRAP ω/2π (⁸⁷Rb) (Hz). 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 Length scale x₀ = √(ħ/mω), measured in µm. Keep n (Quantum number n) = 2 fixed. For this exercise use only 0.685 µm ≤ s ≤ 1.315 µm 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 = 101.71396 Hz, with a bounded reading uncertainty ±5.1688 Hz. 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 0.000063 µm, 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.
TRAP ω/2π (⁸⁷Rb) against s
Predictions from the supplied Hermite–Gaussians in the position basis 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): TRAP ω/2π (⁸⁷Rb) / Hz
Horizontal axis (x): s: Length scale x₀ = √(ħ/mω) / µm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Length scale x₀ = √(ħ/mω) / µm | TRAP ω/2π (⁸⁷Rb) / Hz |
|---|---|
| 0.6849999999999999 | 247.85550642016094 |
| 0.8424999999999999 | 163.8475288150816 |
| 1 | 116.3 |
| 1.1575 | 86.80359566914993 |
| 1.315 | 67.25556246295307 |
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(0.685) = 247.856 Hz and Y_out(1.315) = 67.2556 Hz. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives s_line = 0.685 + (101.714 − 247.856)(1.315 − 0.685)/(67.2556 − 247.856) = 1.1948 µm. 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 = 1.0693 µm; substituting back gives Y_out = 101.71396 Hz.
- The interpolation error in the input is |s_line − s| = 0.125496 µm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 96.54516 Hz and 106.88276 Hz. Sort the roots even if the response decreases: 1.04312 µm ≤ s ≤ 1.09755 µm.
- The lower and upper input deviations are 0.0261758 µm and 0.0282507 µm. 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 ≈ 1.0693 µm; compatible interval [1.04312, 1.09755] µm. Endpoint interpolation gives 1.1948 µm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.