University Quantum Mechanics I · topic UQMI-12.04
L^2 and L_z: a compatible pair
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.
L² and L_z: a compatible pair: predict a finite change
L² and L_z are compatible because [L², L_z] = 0: a complete set of states exists in which both have definite values, and measuring one never disturbs a value already found for the other. Use the supplied topic model Y_out = sqrt(max(m*m*(1+cosd(beta)^2)/2+(1-m*m)*sind(beta)^2-(m*cosd(beta))^2, 0)), where Y_out is SPREAD ΔL_n (ħ). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value; sind/cosd/tand use degrees. beta is Tilt β of the measuring axis n from z, measured in °. Keep m (Prepared state m, with L_z = mħ) = 1 fixed. For this exercise use only 19.125 ° ≤ beta ≤ 85.275 ° 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 beta = 33.678 ° and 42.939 °. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δbeta across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
SPREAD ΔL_n against beta
Predictions from the supplied L² and L_z: a compatible pair 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): SPREAD ΔL_n / ħ
Horizontal axis (x): beta: Tilt β of the measuring axis n from z / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| beta: Tilt β of the measuring axis n from z / ° | SPREAD ΔL_n / ħ |
|---|---|
| 19.125 | 0.2316695216894395 |
| 35.6625 | 0.4122500267958414 |
| 52.2 | 0.558723967439391 |
| 68.73750000000001 | 0.6589731566607913 |
| 85.275 | 0.7047037067407294 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of beta separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At beta = 33.678 °, substitute into Y_out = sqrt(max(m*m*(1+cosd(beta)^2)/2+(1-m*m)*sind(beta)^2-(m*cosd(beta))^2, 0)): Y₀ = 0.392108 ħ.
- At beta = 42.939 °, the same fixed-parameter model gives Y₁ = 0.481695 ħ.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.0895865 ħ; the input change is Δbeta = 9.261 °.
- Average sensitivity is ΔY_out/Δbeta = 0.00967352 ħ per °. 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 = 0.392108 ħ, Y₁ = 0.481695 ħ; ΔY_out = 0.0895865 ħ; average sensitivity = 0.00967352 ħ per °.
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.
L² and L_z: a compatible pair: infer an input with uncertainty
L² and L_z are compatible because [L², L_z] = 0: a complete set of states exists in which both have definite values, and measuring one never disturbs a value already found for the other. Use the supplied topic model Y_out = sqrt(max(m*m*(1+cosd(beta)^2)/2+(1-m*m)*sind(beta)^2-(m*cosd(beta))^2, 0)), where Y_out is SPREAD ΔL_n (ħ). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value; sind/cosd/tand use degrees. beta is Tilt β of the measuring axis n from z, measured in °. Keep m (Prepared state m, with L_z = mħ) = 1 fixed. For this exercise use only 19.125 ° ≤ beta ≤ 85.275 ° 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 = 0.60911657 ħ, with a bounded reading uncertainty ±0.014338 ħ. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown beta by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.006615 °, 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.
SPREAD ΔL_n against beta
Predictions from the supplied L² and L_z: a compatible pair 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): SPREAD ΔL_n / ħ
Horizontal axis (x): beta: Tilt β of the measuring axis n from z / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| beta: Tilt β of the measuring axis n from z / ° | SPREAD ΔL_n / ħ |
|---|---|
| 19.125 | 0.2316695216894395 |
| 35.6625 | 0.4122500267958414 |
| 52.2 | 0.558723967439391 |
| 68.73750000000001 | 0.6589731566607913 |
| 85.275 | 0.7047037067407294 |
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(beta) = 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(19.125) = 0.23167 ħ and Y_out(85.275) = 0.704704 ħ. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives beta_line = 19.125 + (0.609117 − 0.23167)(85.275 − 19.125)/(0.704704 − 0.23167) = 71.9079 °. 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 beta = 59.4765 °; substituting back gives Y_out = 0.60911657 ħ.
- The interpolation error in the input is |beta_line − beta| = 12.4314 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.59477857 ħ and 0.62345457 ħ. Sort the roots even if the response decreases: 57.2611 ° ≤ beta ≤ 61.8479 °.
- The lower and upper input deviations are 2.21539 ° and 2.37136 °. 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 beta ≈ 59.4765 °; compatible interval [57.2611, 61.8479] °. Endpoint interpolation gives 71.9079 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.