University Physics IV · unit UPIV-09
Angular Momentum and Spin
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.
Angular-momentum operators and commutators: infer an input with uncertainty
The angular-momentum commutation relations say that the three components of L = r × p do not commute with one another, [L_x, L_y] = iℏL_z and cyclically, while L² commutes with each of them. Use the supplied topic model Y_out = sqrt((l*(l+1)-clamp(m,-l,l)^2)/2), where Y_out is ΔL_x = ΔL_y (ℏ). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; clamp(x,a,b) confines x to [a,b]. m is Projection m (clamped to |m| ≤ l), dimensionless. Keep l (Angular momentum number l) = 3 fixed. For this exercise use only 0 ≤ m ≤ 3, 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 = 2 ℏ, with a bounded reading uncertainty ±0.09798 ℏ. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of m 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.
ΔL_x = ΔL_y against m
Predictions from the supplied Angular-momentum operators and commutators model, with all other inputs fixed as stated in the question. 4 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): ΔL_x = ΔL_y / ℏ
Horizontal axis (x): m: Projection m (clamped to |m| ≤ l)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Projection m (clamped to |m| ≤ l) | ΔL_x = ΔL_y / ℏ |
|---|---|
| 0 | 2.449489742783178 |
| 1 | 2.345207879911715 |
| 2 | 2 |
| 3 | 1.224744871391589 |
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(0) = 2.44949 ℏ and Y_out(3) = 1.22474 ℏ. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives m_line = 0 + (2 − 2.44949)(3 − 0)/(1.22474 − 2.44949) = 1.10102. 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 m = 2; only the allowed settings 2 lie inside the stated reading bound.
- The interpolation error in the input is |m_line − m| = 0.898979.
- The compatible allowed values span 2 to 2. 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 m: 2. Endpoint interpolation predicts 1.10102; check admissibility against the discrete settings.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Ladder operators and the angular-momentum spectrum: infer an input with uncertainty
The ladder operators J± = Jx ± iJy raise or lower the Jz eigenvalue by one unit of ħ without changing J², and demanding that every stepped state keep a non-negative norm forces the eigenvalues J² = j(j+1)ħ² and Jz = mħ with m running from −j to +j. Use the supplied topic model Y_out = sqrt(max(j*(j+1)-min(k-j,j)^2,0)/2), where Y_out is TRANSVERSE RMS ΔJx (ħ). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; min(a,b) selects the smaller value; max(a,b) selects the larger value. k is Raising steps from the bottom rung, dimensionless. Keep j (Quantum number j) = 2 fixed. For this exercise use only 0 ≤ k ≤ 2, 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 = 1.5811388 ħ, with a bounded reading uncertainty ±0.058564 ħ. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of k 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.
TRANSVERSE RMS ΔJx against k
Predictions from the supplied Ladder operators and the angular-momentum spectrum model, with all other inputs fixed as stated in the question. 3 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): TRANSVERSE RMS ΔJx / ħ
Horizontal axis (x): k: Raising steps from the bottom rung
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: Raising steps from the bottom rung | TRANSVERSE RMS ΔJx / ħ |
|---|---|
| 0 | 1 |
| 1 | 1.5811388300841898 |
| 2 | 1.7320508075688772 |
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(0) = 1 ħ and Y_out(2) = 1.73205 ħ. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives k_line = 0 + (1.58114 − 1)(2 − 0)/(1.73205 − 1) = 1.5877. 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 k = 1; only the allowed settings 1 lie inside the stated reading bound.
- The interpolation error in the input is |k_line − k| = 0.587701.
- The compatible allowed values span 1 to 1. 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 k: 1. Endpoint interpolation predicts 1.5877; check admissibility against the discrete settings.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Spherical harmonics and integer l: infer an input with uncertainty
The spherical harmonics Y_l^m(θ, φ) are the simultaneous eigenfunctions of L̂² and L̂_z on the unit sphere, and they exist only for integer l ≥ 0, with m running in integer steps from −l to +l. Use the supplied topic model Y_out = -max(sind(t),0.02)^(j+0.5)+2*(j-1)*cosd(t)^2*max(sind(t),0.02)^(j-1.5), where Y_out is RUNG 2 AT θ₀ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value; sind/cosd/tand use degrees. t is Marker polar angle θ₀, measured in °. Keep j (Trial angular momentum j) = 0.5 ħ fixed. For this exercise use only 21.4875 ° ≤ t ≤ 85.4325 ° 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 = -1.1490242, with a bounded reading uncertainty ±0.021876. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0063945 °, 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.
RUNG 2 AT θ₀ against t
Predictions from the supplied Spherical harmonics and integer l 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): RUNG 2 AT θ₀
Horizontal axis (x): t: Marker polar angle θ₀ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Marker polar angle θ₀ / ° | RUNG 2 AT θ₀ |
|---|---|
| 21.4875 | -2.7300159039946137 |
| 37.47375 | -1.643661187108329 |
| 53.46000000000001 | -1.2446458445636992 |
| 69.44625 | -1.0679844255704887 |
| 85.4325 | -1.0031859082845505 |
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(t) = 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(21.4875) = -2.73002 and Y_out(85.4325) = -1.00319. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = 21.4875 + (-1.14902 − -2.73002)(85.4325 − 21.4875)/(-1.00319 − -2.73002) = 80.0321 °. 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 t = 60.4939 °; substituting back gives Y_out = -1.1490242.
- The interpolation error in the input is |t_line − t| = 19.5381 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -1.1709002 and -1.1271482. Sort the roots even if the response decreases: 58.6543 ° ≤ t ≤ 62.5228 °.
- The lower and upper input deviations are 1.8397 ° and 2.02888 °. 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 t ≈ 60.4939 °; compatible interval [58.6543, 62.5228] °. Endpoint interpolation gives 80.0321 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Orbital moments and the normal Zeeman effect: infer an input with uncertainty
An electron's orbital motion gives it a magnetic moment μ = −(e/2mₑ)L, so a field B shifts each sublevel by m_l μ_B B and splits every level into 2l+1 equally spaced rungs — the normal Zeeman effect is the three-line spectrum those rungs produce in singlet atoms. Use the supplied topic model Y_out = 13.996*b, where Y_out is π–σ SPACING μ_B B/h (GHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b is Magnetic field B, measured in T. Keep gu (Upper-level g factor) = 1 fixed. For this exercise use only 0.2125 T ≤ b ≤ 1.7875 T 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 = 16.420807 GHz, with a bounded reading uncertainty ±1.2896 GHz. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001575 T, 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.
π–σ SPACING μ_B B/h against b
Predictions from the supplied Orbital moments and the normal Zeeman effect 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): π–σ SPACING μ_B B/h / GHz
Horizontal axis (x): b: Magnetic field B / T
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Magnetic field B / T | π–σ SPACING μ_B B/h / GHz |
|---|---|
| 0.2125 | 2.97415 |
| 0.60625 | 8.485075 |
| 1 | 13.996 |
| 1.3937499999999998 | 19.506925 |
| 1.7874999999999999 | 25.01785 |
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(b) = 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(0.2125) = 2.97415 GHz and Y_out(1.7875) = 25.0178 GHz. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives b_line = 0.2125 + (16.4208 − 2.97415)(1.7875 − 0.2125)/(25.0178 − 2.97415) = 1.17325 T. 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 b = 1.17325 T; substituting back gives Y_out = 16.420807 GHz.
- The interpolation error in the input is |b_line − b| = 0 T. This is within the requested numerical tolerance.
- Invert both limits Y_out = 15.131207 GHz and 17.710407 GHz. Sort the roots even if the response decreases: 1.08111 T ≤ b ≤ 1.26539 T.
- The lower and upper input deviations are 0.0921406 T and 0.0921406 T. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer b ≈ 1.17325 T; compatible interval [1.08111, 1.26539] T. Endpoint interpolation gives 1.17325 T.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.