University Physics V · unit UPV-10
Quantum 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.
Intrinsic angular momentum and su(2): infer an input with uncertainty
Spin is an intrinsic angular momentum defined by its algebra alone — three Hermitian operators with [S_i, S_j] = iħ ε_ijk S_k acting on an internal space — with no r × p realisation and nothing rotating in space behind it. Use the supplied topic model Y_out = deg(acos(clamp((s-min(k,floor(2*s)))/sqrt(max(s*(s+1),0.0001)),-1,1))), where Y_out is TILT FROM z (°). 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; clamp(x,a,b) confines x to [a,b]; deg converts radians to degrees; unmarked trigonometric functions use radians. s is Trial spin quantum number s, dimensionless. Keep k (Rungs stepped down from the top) = 0 fixed. For this exercise use only 0.2125 ≤ s ≤ 1.7875 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 = 42.713784 °, with a bounded reading uncertainty ±0.88783 °. 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 0.0001575, 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.
TILT FROM z against s
Predictions from the supplied Intrinsic angular momentum and su(2) 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): TILT FROM z / °
Horizontal axis (x): s: Trial spin quantum number s
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Trial spin quantum number s | TILT FROM z / ° |
|---|---|
| 0.2125 | 65.25137468480418 |
| 0.60625 | 52.09494104166421 |
| 1 | 45.00000000000001 |
| 1.3937499999999998 | 40.266170564834134 |
| 1.7874999999999999 | 36.794930524972195 |
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(0.2125) = 65.2514 ° and Y_out(1.7875) = 36.7949 °. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives s_line = 0.2125 + (42.7138 − 65.2514)(1.7875 − 0.2125)/(36.7949 − 65.2514) = 1.4599. 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 = 1.17325; substituting back gives Y_out = 42.713784 °.
- The interpolation error in the input is |s_line − s| = 0.286655. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 41.825954 ° and 43.601614 °. Sort the roots even if the response decreases: 1.10259 ≤ s ≤ 1.24863.
- The lower and upper input deviations are 0.0706571 and 0.0753775. 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 ≈ 1.17325; compatible interval [1.10259, 1.24863]. Endpoint interpolation gives 1.4599.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Spin-half states in C²: infer an input with uncertainty
A spin-half state is a normalised vector |ψ⟩ = α|↑⟩ + β|↓⟩ in the two-dimensional complex space C², where |↑⟩ and |↓⟩ are the S_z eigenvectors, |α|² + |β|² = 1, and vectors differing only by an overall phase e^(iγ) are the same physical state. Use the supplied topic model Y_out = sind(th)*cosd(ph)/2, where Y_out is ⟨S_x⟩/ħ = Re(α*β) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. th is Polar angle θ, measured in °. Keep ph (Relative phase φ) = 90 °; gam (Global phase γ) = 0 ° fixed. For this exercise use only 94.725 ° ≤ th ≤ 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 = 2.1619977e-17, with a bounded reading uncertainty ±1.3338e-18. 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 0.006615 °, 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.
⟨S_x⟩/ħ = Re(α*β) against th
Predictions from the supplied Spin-half states in C² 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): ⟨S_x⟩/ħ = Re(α*β)
Horizontal axis (x): th: Polar angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Polar angle θ / ° | ⟨S_x⟩/ħ = Re(α*β) |
|---|---|
| 94.725 | 3.0512122135186783e-17 |
| 111.26249999999999 | 2.853208979535164e-17 |
| 127.8 | 2.4191520168403165e-17 |
| 144.3375 | 1.7849520369355162e-17 |
| 160.875 | 1.003078126760752e-17 |
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(94.725) = 3.05121e-17 and Y_out(160.875) = 1.00308e-17. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives th_line = 94.725 + (2.162e-17 − 3.05121e-17)(160.875 − 94.725)/(1.00308e-17 − 3.05121e-17) = 123.445 °. 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 = 135.077 °; substituting back gives Y_out = 2.1619977e-17.
- The interpolation error in the input is |th_line − th| = 11.6319 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.0286177e-17 and 2.2953777e-17. Sort the roots even if the response decreases: 131.433 ° ≤ th ≤ 138.502 °.
- The lower and upper input deviations are 3.64326 ° and 3.42528 °. 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 ≈ 135.077 °; compatible interval [131.433, 138.502] °. Endpoint interpolation gives 123.445 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Pauli matrices and the spin operator: infer an input with uncertainty
The Pauli matrices are the three Hermitian, traceless 2×2 matrices that square to the identity and represent spin-half in the S_z eigenbasis, the spin operator itself being S = (ħ/2)σ. Use the supplied topic model Y_out = -(ax*ax+ay*ay+az*az), where Y_out is det A (meV²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. ay is Component a_y, measured in meV. Keep ax (Component a_x) = 1 meV; az (Component a_z) = 2 meV fixed. For this exercise use only 0.105 meV ≤ ay ≤ 1.575 meV 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 = -6.0034029 meV², with a bounded reading uncertainty ±0.14886 meV². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown ay by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000147 meV, 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.
det A against ay
Predictions from the supplied Pauli matrices and the spin operator 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): det A / meV²
Horizontal axis (x): ay: Component a_y / meV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| ay: Component a_y / meV | det A / meV² |
|---|---|
| 0.10499999999999998 | -5.011025 |
| 0.4724999999999999 | -5.22325625 |
| 0.8399999999999999 | -5.7056 |
| 1.2074999999999998 | -6.458056249999999 |
| 1.5749999999999997 | -7.480624999999999 |
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(ay) = 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(0.105) = -5.01103 meV² and Y_out(1.575) = -7.48062 meV². The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives ay_line = 0.105 + (-6.0034 − -5.01103)(1.575 − 0.105)/(-7.48062 − -5.01103) = 0.695701 meV. 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 ay = 1.0017 meV; substituting back gives Y_out = -6.0034029 meV².
- The interpolation error in the input is |ay_line − ay| = 0.305999 meV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -6.1522629 meV² and -5.8545429 meV². Sort the roots even if the response decreases: 0.924415 meV ≤ ay ≤ 1.07344 meV.
- The lower and upper input deviations are 0.0772851 meV and 0.0717351 meV. 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 ay ≈ 1.0017 meV; compatible interval [0.924415, 1.07344] meV. Endpoint interpolation gives 0.695701 meV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Spin states and eigenspinors along any axis: infer an input with uncertainty
The eigenspinors along a unit vector n̂ are the two orthonormal vectors in C² that n̂⋅σ sends to ±1 times themselves — |n,+⟩ = (cos(θ/2), e^(iφ) sin(θ/2))ᵀ and its orthogonal partner — the states with spin definitely ±ħ/2 along n̂. Use the supplied topic model Y_out = (cosd((t-s)/2))^2, where Y_out is P(+ħ/2 ALONG n̂) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. t is Analyser axis n̂: θ from +z, measured in °. Keep s (Prepared state m̂: angle from +z) = 0 ° fixed. For this exercise use only 189.45 ° ≤ t ≤ 321.75 ° 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 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 0.01323 °, 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(+ħ/2 ALONG n̂) against t
Predictions from the supplied Spin states and eigenspinors along any axis 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(+ħ/2 ALONG n̂)
Horizontal axis (x): t: Analyser axis n̂: θ from +z / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Analyser axis n̂: θ from +z / ° | P(+ħ/2 ALONG n̂) |
|---|---|
| 189.45 | 0.006785371411752245 |
| 222.52499999999998 | 0.13150875760102443 |
| 255.6 | 0.3756550564175723 |
| 288.675 | 0.6600998308136561 |
| 321.75 | 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(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(189.45) = 0.00678537 and Y_out(321.75) = 0.892658. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = 189.45 + (0.501335 − 0.00678537)(321.75 − 189.45)/(0.892658 − 0.00678537) = 263.308 °. 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 = 270.153 °; substituting back gives Y_out = 0.50133518.
- The interpolation error in the input is |t_line − t| = 6.84486 °. 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: 263.412 ° ≤ t ≤ 276.896 °.
- The lower and upper input deviations are 6.7408 ° and 6.74293 °. 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 ≈ 270.153 °; compatible interval [263.412, 276.896] °. Endpoint interpolation gives 263.308 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.