University Quantum Mechanics I · topic UQMI-12.07
The quantum numbers l and m
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 quantum numbers l and m: where the ladder must end: predict a finite change
The quantum numbers l and m label a state that is an eigenstate of both L² and L_z, with L² = ħ²l(l + 1) and L_z = mħ; m runs from −l to l in whole steps, and for orbital motion l is 0, 1, 2, …. Use the supplied topic model Y_out = (sqrt(1+4*lam)-1)/2, where Y_out is l NEEDED FOR THIS λ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. lam is L² eigenvalue λ, measured in ħ². Keep m0 (Starting rung m₀) = 0 fixed. For this exercise use only 0.6375 ħ² ≤ lam ≤ 5.3625 ħ² 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 lam = 1.677 ħ² and 2.3385 ħ². Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δlam across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
l NEEDED FOR THIS λ against lam
Predictions from the supplied The quantum numbers l and m: where the ladder must end 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): l NEEDED FOR THIS λ
Horizontal axis (x): lam: L² eigenvalue λ / ħ²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lam: L² eigenvalue λ / ħ² | l NEEDED FOR THIS λ |
|---|---|
| 0.6375 | 0.4420721840708386 |
| 1.8187499999999999 | 0.9383149863642524 |
| 3 | 1.3027756377319946 |
| 4.1812499999999995 | 1.6050534435020882 |
| 5.3625 | 1.8690715480964268 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of lam separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At lam = 1.677 ħ², substitute into Y_out = (sqrt(1+4*lam)-1)/2: Y₀ = 0.888164.
- At lam = 2.3385 ħ², the same fixed-parameter model gives Y₁ = 1.10888.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.220717; the input change is Δlam = 0.6615 ħ².
- Average sensitivity is ΔY_out/Δlam = 0.333662 1 per ħ². 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 = 0.888164, Y₁ = 1.10888; ΔY_out = 0.220717; average sensitivity = 0.333662 1 per ħ².
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 quantum numbers l and m: where the ladder must end: infer an input with uncertainty
The quantum numbers l and m label a state that is an eigenstate of both L² and L_z, with L² = ħ²l(l + 1) and L_z = mħ; m runs from −l to l in whole steps, and for orbital motion l is 0, 1, 2, …. Use the supplied topic model Y_out = (sqrt(1+4*lam)-1)/2, where Y_out is l NEEDED FOR THIS λ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. lam is L² eigenvalue λ, measured in ħ². Keep m0 (Starting rung m₀) = 0 fixed. For this exercise use only 0.6375 ħ² ≤ lam ≤ 5.3625 ħ² 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 = 1.4415844, with a bounded reading uncertainty ±0.064123. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lam by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0004725 ħ², 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.
l NEEDED FOR THIS λ against lam
Predictions from the supplied The quantum numbers l and m: where the ladder must end 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): l NEEDED FOR THIS λ
Horizontal axis (x): lam: L² eigenvalue λ / ħ²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lam: L² eigenvalue λ / ħ² | l NEEDED FOR THIS λ |
|---|---|
| 0.6375 | 0.4420721840708386 |
| 1.8187499999999999 | 0.9383149863642524 |
| 3 | 1.3027756377319946 |
| 4.1812499999999995 | 1.6050534435020882 |
| 5.3625 | 1.8690715480964268 |
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(lam) = 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(0.6375) = 0.442072 and Y_out(5.3625) = 1.86907. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives lam_line = 0.6375 + (1.44158 − 0.442072)(5.3625 − 0.6375)/(1.86907 − 0.442072) = 3.94703 ħ². 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 lam = 3.51975 ħ²; substituting back gives Y_out = 1.4415844.
- The interpolation error in the input is |lam_line − lam| = 0.427279 ħ². This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.3774614 and 1.5057074. Sort the roots even if the response decreases: 3.27486 ħ² ≤ lam ≤ 3.77286 ħ².
- The lower and upper input deviations are 0.244889 ħ² and 0.253112 ħ². 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 lam ≈ 3.51975 ħ²; compatible interval [3.27486, 3.77286] ħ². Endpoint interpolation gives 3.94703 ħ².
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.