University Quantum Mechanics II · unit UQMII-05
Rotations and the Addition of Angular Momentum
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.
Rotation operators and angular momentum as their generator: infer an input with uncertainty
The angular momentum J of any system is the generator of its rotations: the Hermitian operator for which a turn by a small angle ε about the unit vector n̂ acts on every state as D = 1 − iεJ⋅n̂/ħ. Use the supplied topic model Y_out = m*phi-n*deg(atan(m*phi*pi/(180*n))), where Y_out is N-STEP PHASE SHORTFALL (°). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; deg converts radians to degrees; unmarked trigonometric functions use radians. phi is Rotation angle φ about z, measured in °. Keep m (Azimuthal number m of cos mϕ) = 1; n (First-order steps N) = 4 fixed. For this exercise use only 9.5625 ° ≤ phi ≤ 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 = 0.90529253 °, with a bounded reading uncertainty ±0.13496 °. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown phi 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.
N-STEP PHASE SHORTFALL against phi
Predictions from the supplied Rotation operators and angular momentum as their generator 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): N-STEP PHASE SHORTFALL / °
Horizontal axis (x): phi: Rotation angle φ about z / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| phi: Rotation angle φ about z / ° | N-STEP PHASE SHORTFALL / ° |
|---|---|
| 9.5625 | 0.005543378424674117 |
| 27.28125 | 0.12777193501861106 |
| 45 | 0.5652776993588517 |
| 62.71875 | 1.4988929073741986 |
| 80.4375 | 3.078347870138188 |
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(phi) = 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) = 0.00554338 ° and Y_out(80.4375) = 3.07835 °. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives phi_line = 9.5625 + (0.905293 − 0.00554338)(80.4375 − 9.5625)/(3.07835 − 0.00554338) = 30.3154 °. 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 phi = 52.7963 °; substituting back gives Y_out = 0.90529253 °.
- The interpolation error in the input is |phi_line − phi| = 22.4808 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.77033253 ° and 1.0402525 °. Sort the roots even if the response decreases: 49.9773 ° ≤ phi ≤ 55.355 °.
- The lower and upper input deviations are 2.81894 ° and 2.55879 °. 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 phi ≈ 52.7963 °; compatible interval [49.9773, 55.355] °. Endpoint interpolation gives 30.3154 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Rotating a spin-half, and the sign change under 2π: infer an input with uncertainty
The spin-½ rotation operator D(n̂, φ) = cos(φ/2) I − i sin(φ/2) σ⋅n̂ is the 2 × 2 unitary matrix that turns a spinor through the angle φ about the unit axis n̂; it rotates ⟨S⟩ through φ, yet equals −I, not I, at φ = 2π. Use the supplied topic model Y_out = (1+cos(phi*pi/360))/2, where Y_out is FRINGE I/I₀ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; unmarked trigonometric functions use radians. phi is Rotation angle φ, measured in °. Keep theta (Spin tilt from axis θ) = 60 ° fixed. For this exercise use only 378.9 ° ≤ phi ≤ 643.5 ° 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 = 0.50133518, with a bounded reading uncertainty ±0.058698. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown phi by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.02646 °, 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.
FRINGE I/I₀ against phi
Predictions from the supplied Rotating a spin-half, and the sign change under 2π 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): FRINGE I/I₀
Horizontal axis (x): phi: Rotation angle φ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| phi: Rotation angle φ / ° | FRINGE I/I₀ |
|---|---|
| 378.9 | 0.006785371411752228 |
| 445.04999999999995 | 0.13150875760102443 |
| 511.2 | 0.37565505641757235 |
| 577.35 | 0.6600998308136561 |
| 643.5 | 0.8926584654403723 |
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(phi) = 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(378.9) = 0.00678537 and Y_out(643.5) = 0.892658. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives phi_line = 378.9 + (0.501335 − 0.00678537)(643.5 − 378.9)/(0.892658 − 0.00678537) = 526.616 °. 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 phi = 540.306 °; substituting back gives Y_out = 0.50133518.
- The interpolation error in the input is |phi_line − phi| = 13.6897 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.44263718 and 0.56003318. Sort the roots even if the response decreases: 526.824 ° ≤ phi ≤ 553.792 °.
- The lower and upper input deviations are 13.4816 ° and 13.4859 °. 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 phi ≈ 540.306 °; compatible interval [526.824, 553.792] °. Endpoint interpolation gives 526.616 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Total angular momentum J = J₁ + J₂ and its two bases: infer an input with uncertainty
The total angular momentum J = J₁ + J₂ of a two-part system is itself an angular momentum, and its states can be labelled by m₁ and m₂ (the uncoupled basis) or by j and M (the coupled basis): two orthonormal bases of one space, needed because J² and J₁z do not commute. Use the supplied topic model Y_out = 0.5*(1+(m+f)/sqrt(4+2*m*f+f*f)), where Y_out is P(m₂ = +½), UPPER LEVEL (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. f is Field-to-coupling ratio b/a, dimensionless. Keep m (Block M = m₁ + m₂) = 0 fixed. For this exercise use only 0.53125 ≤ f ≤ 4.46875 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 = 0.91310436, with a bounded reading uncertainty ±0.006491. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown f by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00039375, 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.
P(m₂ = +½), UPPER LEVEL against f
Predictions from the supplied Total angular momentum J = J₁ + J₂ and its two bases 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): P(m₂ = +½), UPPER LEVEL
Horizontal axis (x): f: Field-to-coupling ratio b/a
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Field-to-coupling ratio b/a | P(m₂ = +½), UPPER LEVEL |
|---|---|
| 0.53125 | 0.6283613059199252 |
| 1.515625 | 0.8019887813363895 |
| 2.5 | 0.8904344047215151 |
| 3.484375 | 0.9336420962135578 |
| 4.46875 | 0.9563778137705827 |
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(f) = 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(0.53125) = 0.628361 and Y_out(4.46875) = 0.956378. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives f_line = 0.53125 + (0.913104 − 0.628361)(4.46875 − 0.53125)/(0.956378 − 0.628361) = 3.9493. 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 f = 2.93313; substituting back gives Y_out = 0.91310436.
- The interpolation error in the input is |f_line − f| = 1.01617. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.90661336 and 0.91959536. Sort the roots even if the response decreases: 2.79485 ≤ f ≤ 3.08618.
- The lower and upper input deviations are 0.138277 and 0.153058. 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 f ≈ 2.93313; compatible interval [2.79485, 3.08618]. Endpoint interpolation gives 3.9493.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Two spin-halves: the triplet and the singlet: infer an input with uncertainty
The triplet and the singlet are the four simultaneous eigenstates of S² and S_z for two spin-½ particles: three states with s = 1, symmetric under exchange, and one state with s = 0, antisymmetric. Use the supplied topic model Y_out = (1+cosd(dl))/2, where Y_out is SWAP ⟨P₁₂⟩ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. dl is Angle from spin 1 to spin 2, measured in °. Keep th (Spin 1 direction θ₁ from +z) = 30 ° fixed. For this exercise use only 19.125 ° ≤ dl ≤ 160.875 ° 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 = 0.36560313, with a bounded reading uncertainty ±0.050701. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown dl by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.014175 °, 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.
SWAP ⟨P₁₂⟩ against dl
Predictions from the supplied Two spin-halves: the triplet and the singlet 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): SWAP ⟨P₁₂⟩
Horizontal axis (x): dl: Angle from spin 1 to spin 2 / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dl: Angle from spin 1 to spin 2 / ° | SWAP ⟨P₁₂⟩ |
|---|---|
| 19.125 | 0.9724030232334391 |
| 54.5625 | 0.7899072740632569 |
| 90 | 0.5 |
| 125.4375 | 0.21009272593674322 |
| 160.875 | 0.027596976766561032 |
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(dl) = 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) = 0.972403 and Y_out(160.875) = 0.027597. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives dl_line = 19.125 + (0.365603 − 0.972403)(160.875 − 19.125)/(0.027597 − 0.972403) = 110.164 °. 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 dl = 105.592 °; substituting back gives Y_out = 0.36560313.
- The interpolation error in the input is |dl_line − dl| = 4.57117 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.31490213 and 0.41630413. Sort the roots even if the response decreases: 99.6362 ° ≤ dl ≤ 111.728 °.
- The lower and upper input deviations are 5.9563 ° and 6.13519 °. 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 dl ≈ 105.592 °; compatible interval [99.6362, 111.728] °. Endpoint interpolation gives 110.164 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.