University Quantum Mechanics I · topic UQMI-03.09
Commutators, [x-hat, p-hat] = i hbar, and compatibility
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.
Commutators, [x̂, p̂] = iħ, and compatibility: predict a finite change
The commutator [Â, B̂] = ÂB̂ − B̂Â is the operator that measures how much the order of two operators matters, found by applying both orders to an arbitrary test function f and subtracting. Use the supplied topic model Y_out = 6.626*n*sind(2*th)*sind(ph), where Y_out is MEAN MOMENTUM ⟨p⟩ (10⁻²⁵ kg m s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. th is Mixing angle θ, measured in °. Keep ph (Relative phase φ) = 60 °; n (Ring level n) = 1 fixed. For this exercise use only 47.3625 ° ≤ th ≤ 80.4375 ° 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 th = 54.639 ° and 59.2695 °. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δth across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
MEAN MOMENTUM ⟨p⟩ against th
Predictions from the supplied Commutators, [x̂, p̂] = iħ, and compatibility 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): MEAN MOMENTUM ⟨p⟩ / 10⁻²⁵ kg m s⁻¹
Horizontal axis (x): th: Mixing angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Mixing angle θ / ° | MEAN MOMENTUM ⟨p⟩ / 10⁻²⁵ kg m s⁻¹ |
|---|---|
| 47.3625 | 5.71878299301463 |
| 55.631249999999994 | 5.34767228427733 |
| 63.9 | 4.5341341222114755 |
| 72.16875 | 3.3454747286823427 |
| 80.4375 | 1.8800351239316528 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of th separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At th = 54.639 °, substitute into Y_out = 6.626*n*sind(2*th)*sind(ph): Y₀ = 5.41653 10⁻²⁵ kg m s⁻¹.
- At th = 59.2695 °, the same fixed-parameter model gives Y₁ = 5.04104 10⁻²⁵ kg m s⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.375488 10⁻²⁵ kg m s⁻¹; the input change is Δth = 4.6305 °.
- Average sensitivity is ΔY_out/Δth = -0.0810903 10⁻²⁵ kg m s⁻¹ per °. 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 = 5.41653 10⁻²⁵ kg m s⁻¹, Y₁ = 5.04104 10⁻²⁵ kg m s⁻¹; ΔY_out = -0.375488 10⁻²⁵ kg m s⁻¹; average sensitivity = -0.0810903 10⁻²⁵ kg m s⁻¹ 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.
Commutators, [x̂, p̂] = iħ, and compatibility: infer an input with uncertainty
The commutator [Â, B̂] = ÂB̂ − B̂Â is the operator that measures how much the order of two operators matters, found by applying both orders to an arbitrary test function f and subtracting. Use the supplied topic model Y_out = 6.626*n*sind(2*th)*sind(ph), where Y_out is MEAN MOMENTUM ⟨p⟩ (10⁻²⁵ kg m s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. th is Mixing angle θ, measured in °. Keep ph (Relative phase φ) = 60 °; n (Ring level n) = 1 fixed. For this exercise use only 47.3625 ° ≤ th ≤ 80.4375 ° 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 = 4.0521586 10⁻²⁵ kg m s⁻¹, with a bounded reading uncertainty ±0.24999 10⁻²⁵ kg m s⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown th by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0033075 °, 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.
MEAN MOMENTUM ⟨p⟩ against th
Predictions from the supplied Commutators, [x̂, p̂] = iħ, and compatibility 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): MEAN MOMENTUM ⟨p⟩ / 10⁻²⁵ kg m s⁻¹
Horizontal axis (x): th: Mixing angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Mixing angle θ / ° | MEAN MOMENTUM ⟨p⟩ / 10⁻²⁵ kg m s⁻¹ |
|---|---|
| 47.3625 | 5.71878299301463 |
| 55.631249999999994 | 5.34767228427733 |
| 63.9 | 4.5341341222114755 |
| 72.16875 | 3.3454747286823427 |
| 80.4375 | 1.8800351239316528 |
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(th) = 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(47.3625) = 5.71878 10⁻²⁵ kg m s⁻¹ and Y_out(80.4375) = 1.88004 10⁻²⁵ kg m s⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives th_line = 47.3625 + (4.05216 − 5.71878)(80.4375 − 47.3625)/(1.88004 − 5.71878) = 61.7223 °. 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 th = 67.5382 °; substituting back gives Y_out = 4.0521586 10⁻²⁵ kg m s⁻¹.
- The interpolation error in the input is |th_line − th| = 5.81596 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.8021686 10⁻²⁵ kg m s⁻¹ and 4.3021486 10⁻²⁵ kg m s⁻¹. Sort the roots even if the response decreases: 65.7166 ° ≤ th ≤ 69.2509 °.
- The lower and upper input deviations are 1.82163 ° and 1.71264 °. 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 th ≈ 67.5382 °; compatible interval [65.7166, 69.2509] °. Endpoint interpolation gives 61.7223 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.