University Physics V · unit UPV-08
Quantum Uncertainty and Commutation Relations
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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
The commutator as an algebraic object: infer an input with uncertainty
The commutator of two operators, [A, B] = AB − BA, is the operator measuring how much their order matters; it is linear in each argument, antisymmetric, and obeys the product rule [A, BC] = [A, B]C + B[A, C]. Use the supplied topic model Y_out = 100*(d-1)/max(d,1), where Y_out is RUNGS OBEYING THE ALGEBRA (%). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. d is States kept in the truncation d, measured in states. Keep n (Rung probed n) = 3 fixed. For this exercise use only 2 states ≤ d ≤ 10 states, at the specified test settings separated by 1. 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 = 85.714286 %, with a bounded reading uncertainty ±3.2 %. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of d consistent with that bound. Compare with a straight-line interpolation between the two endpoint predictions, and explain why a fractional setting may be inadmissible. Do not assume that the supplied calibration points prove monotonicity outside this branch.
RUNGS OBEYING THE ALGEBRA against d
Predictions from the supplied The commutator as an algebraic object 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): RUNGS OBEYING THE ALGEBRA / %
Horizontal axis (x): d: States kept in the truncation d / states
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: States kept in the truncation d / states | RUNGS OBEYING THE ALGEBRA / % |
|---|---|
| 2 | 50 |
| 4 | 75 |
| 6 | 83.33333333333333 |
| 8 | 87.5 |
| 10 | 90 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Enumerate the permitted settings in the stated range, evaluate Y_out at each, and retain every setting satisfying |Y_out − Y_measured| ≤ u. The plotted predictions are supplied calibration evidence.
Reveal the worked solution
- Endpoint predictions are Y_out(2) = 50 % and Y_out(10) = 90 %. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives d_line = 2 + (85.7143 − 50)(10 − 2)/(90 − 50) = 9.14286 states. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Evaluate the model at each permitted setting. The central reading corresponds to d = 7 states; only the allowed settings 6 states, 7 states, 8 states, 9 states lie inside the stated reading bound.
- The interpolation error in the input is |d_line − d| = 2.14286 states.
- The compatible allowed values span 6 states to 9 states. Values between specified test settings are outside this exercise's input domain even if an interpolation passes through them.
- The quoted output bound restricts which specified settings remain compatible. It does not change physical restrictions on quantum numbers or counts, and the chosen grid alone does not establish physical quantisation.
Answer Compatible d: 6 states, 7 states, 8 states, 9 states. Endpoint interpolation predicts 9.14286 states; check admissibility against the discrete settings.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The canonical commutation relation: infer an input with uncertainty
The canonical commutation relation says position and momentum fail to commute by a fixed amount: x̂p̂ − p̂x̂ is iħ times the identity operator — the same number in every state, in every representation, and at every time. Use the supplied topic model Y_out = (b*b/(a*a))*exp(-b*b/(2*a*a)) - (b*b/(a*a) - 1)*exp(-b*b/(2*a*a)), where Y_out is DIFFERENCE ÷ iħ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. a is Gaussian width a, measured in nm. Keep b (Probe position x) = 1 nm fixed. For this exercise use only 0.79125 nm ≤ a ≤ 2.20875 nm 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 = 0.83331561, with a bounded reading uncertainty ±0.010391. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00014175 nm, 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.
DIFFERENCE ÷ iħ against a
Predictions from the supplied The canonical commutation relation 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): DIFFERENCE ÷ iħ
Horizontal axis (x): a: Gaussian width a / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Gaussian width a / nm | DIFFERENCE ÷ iħ |
|---|---|
| 0.79125 | 0.44994750007515355 |
| 1.145625 | 0.6832018242185902 |
| 1.5 | 0.8007374029168081 |
| 1.8543749999999999 | 0.864673322251564 |
| 2.2087499999999998 | 0.9025881583348507 |
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(a) = 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(0.79125) = 0.449948 and Y_out(2.20875) = 0.902588. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives a_line = 0.79125 + (0.833316 − 0.449948)(2.20875 − 0.79125)/(0.902588 − 0.449948) = 1.99181 nm. 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 a = 1.65592 nm; substituting back gives Y_out = 0.83331561.
- The interpolation error in the input is |a_line − a| = 0.335889 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.82292461 and 0.84370661. Sort the roots even if the response decreases: 1.60173 nm ≤ a ≤ 1.71524 nm.
- The lower and upper input deviations are 0.0541944 nm and 0.0593108 nm. 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 a ≈ 1.65592 nm; compatible interval [1.60173, 1.71524] nm. Endpoint interpolation gives 1.99181 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The Robertson bound from Cauchy-Schwarz: infer an input with uncertainty
The Robertson relation says that for any two Hermitian observables and any normalised state, the product of their standard deviations is at least half the modulus of the expectation of their commutator, taken in that same state. Use the supplied topic model Y_out = min(hypot(0.5*sind(th)^2*sind(2*ph), cosd(th))/max(abs(cosd(th)), 0.0005), 99), where Y_out is PRODUCT / FLOOR (×). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; hypot(a,b) means √(a²+b²); abs(x) means |x|; min(a,b) selects the smaller value; max(a,b) selects the larger value; sind/cosd/tand use degrees. th is Polar angle θ from +z, measured in °. Keep ph (Azimuth φ in the xy-plane) = 35 ° fixed. For this exercise use only 9.5625 ° ≤ 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 = 1.1149035 ×, with a bounded reading uncertainty ±0.017223 ×. 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 0.0070875 °, 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.
PRODUCT / FLOOR against th
Predictions from the supplied The Robertson bound from Cauchy-Schwarz 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): PRODUCT / FLOOR / ×
Horizontal axis (x): th: Polar angle θ from +z / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Polar angle θ from +z / ° | PRODUCT / FLOOR / × |
|---|---|
| 9.5625 | 1.00008644494265 |
| 27.28125 | 1.0061488571797534 |
| 45 | 1.053744645393245 |
| 62.71875 | 1.286709366518848 |
| 80.4375 | 2.926405129562407 |
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(9.5625) = 1.00009 × and Y_out(80.4375) = 2.92641 ×. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives th_line = 9.5625 + (1.1149 − 1.00009)(80.4375 − 9.5625)/(2.92641 − 1.00009) = 13.787 °. 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 = 52.7963 °; substituting back gives Y_out = 1.1149035 ×.
- The interpolation error in the input is |th_line − th| = 39.0093 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.0976805 × and 1.1321265 ×. Sort the roots even if the response decreases: 51.0779 ° ≤ th ≤ 54.2914 °.
- The lower and upper input deviations are 1.71839 ° and 1.49516 °. 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 ≈ 52.7963 °; compatible interval [51.0779, 54.2914] °. Endpoint interpolation gives 13.787 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Δx Δp ≥ ħ/2 and the cost of confinement: infer an input with uncertainty
The position–momentum uncertainty relation says that for one prepared state, the standard deviation of x̂ and the standard deviation of p̂ — each read from its own ensemble of identical copies — satisfy σ_x σ_p ≥ ħ/2, whatever the state and whatever the apparatus. Use the supplied topic model Y_out = q*q/(2*s*s), where Y_out is KINETIC ⟨T⟩ (ħω). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s is Position spread σx / √(ħ/mω), dimensionless. Keep q (Uncertainty product σx σp / ħ) = 0.5 fixed. For this exercise use only 0.480625 ≤ s ≤ 1.81937 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 = 0.074276988 ħω, with a bounded reading uncertainty ±0.0054771 ħω. 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 0.000133875, 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.
KINETIC ⟨T⟩ against s
Predictions from the supplied Δx Δp ≥ ħ/2 and the cost of confinement 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): KINETIC ⟨T⟩ / ħω
Horizontal axis (x): s: Position spread σx / √(ħ/mω)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Position spread σx / √(ħ/mω) | KINETIC ⟨T⟩ / ħω |
|---|---|
| 0.48062499999999997 | 0.5411246260744283 |
| 0.8153125 | 0.18804500916058325 |
| 1.15 | 0.09451795841209831 |
| 1.4846875000000002 | 0.056707422616187185 |
| 1.819375 | 0.03776291990449286 |
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(0.480625) = 0.541125 ħω and Y_out(1.81937) = 0.0377629 ħω. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives s_line = 0.480625 + (0.074277 − 0.541125)(1.81937 − 0.480625)/(0.0377629 − 0.541125) = 1.72226. 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 = 1.29726; substituting back gives Y_out = 0.074276988 ħω.
- The interpolation error in the input is |s_line − s| = 0.424999. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.068799888 ħω and 0.079754088 ħω. Sort the roots even if the response decreases: 1.25193 ≤ s ≤ 1.34791.
- The lower and upper input deviations are 0.0453369 and 0.0506483. 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 ≈ 1.29726; compatible interval [1.25193, 1.34791]. Endpoint interpolation gives 1.72226.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.