University Quantum Mechanics I · unit UQMI-13
Spin-½
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 Stern-Gerlach experiment and the two-beam result: infer an input with uncertainty
The Stern-Gerlach experiment sends neutral atoms through a magnetic field whose strength varies across the beam, so that each atom is pushed in proportion to its magnetic-moment component μ_z, and reads the allowed values of μ_z off the spots on a screen. Use the supplied topic model Y_out = sign(j)*0.0013029*b*g/0.3532, where Y_out is SPACING ÷ SPOT WIDTH (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. g is Moment factor g, dimensionless. Keep j (Angular momentum j) = 0.5; b (Gradient ∂B_z/∂z) = 600 T m⁻¹ fixed. For this exercise use only 0.59125 ≤ g ≤ 2.00875 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 = 3.2224089, with a bounded reading uncertainty ±0.18354. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown g 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, 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.
SPACING ÷ SPOT WIDTH against g
Predictions from the supplied The Stern-Gerlach experiment and the two-beam result 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): SPACING ÷ SPOT WIDTH
Horizontal axis (x): g: Moment factor g
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| g: Moment factor g | SPACING ÷ SPOT WIDTH |
|---|---|
| 0.59125 | 1.3086177095130238 |
| 0.945625 | 2.092958345130238 |
| 1.3 | 2.877298980747452 |
| 1.654375 | 3.6616396163646656 |
| 2.00875 | 4.445980251981879 |
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(g) = 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(0.59125) = 1.30862 and Y_out(2.00875) = 4.44598. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives g_line = 0.59125 + (3.22241 − 1.30862)(2.00875 − 0.59125)/(4.44598 − 1.30862) = 1.45593. 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 g = 1.45593; substituting back gives Y_out = 3.2224089.
- The interpolation error in the input is |g_line − g| = 4.44089e-16. This is within the requested numerical tolerance.
- Invert both limits Y_out = 3.0388689 and 3.4059489. Sort the roots even if the response decreases: 1.373 ≤ g ≤ 1.53885.
- The lower and upper input deviations are 0.0829257 and 0.0829257. 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 g ≈ 1.45593; compatible interval [1.373, 1.53885]. Endpoint interpolation gives 1.45593.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Intrinsic angular momentum: spin is not rotation: infer an input with uncertainty
Spin is angular momentum that belongs to a particle itself rather than to its motion: operators S that obey the same commutators as L but act on an extra discrete label instead of on the spatial wavefunction, and may take the half-integer values that L cannot. Use the supplied topic model Y_out = 1+2*sqrt(max(p*(1-p),0))*cos(2*pi*m), where Y_out is 2π|ψ|² AT φ = 2π (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value; unmarked trigonometric functions use radians. p is Weight p of the trial state, dimensionless. Keep m (Trial L_z quantum number m) = 0.5 fixed. For this exercise use only 0.10625 ≤ p ≤ 0.47375 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 = 0.059267691, with a bounded reading uncertainty ±0.0086833. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown p by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00003675, 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.
2π|ψ|² AT φ = 2π against p
Predictions from the supplied Intrinsic angular momentum: spin is not rotation 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): 2π|ψ|² AT φ = 2π
Horizontal axis (x): p: Weight p of the trial state
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Weight p of the trial state | 2π|ψ|² AT φ = 2π |
|---|---|
| 0.10625 | 0.3836853482189475 |
| 0.198125 | 0.20282628147937543 |
| 0.29 | 0.09247589563692582 |
| 0.381875 | 0.028307694020375584 |
| 0.47375 | 0.0013790759252036988 |
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(p) = 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(0.10625) = 0.383685 and Y_out(0.47375) = 0.00137908. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives p_line = 0.10625 + (0.0592677 − 0.383685)(0.47375 − 0.10625)/(0.00137908 − 0.383685) = 0.418103. 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 p = 0.330425; substituting back gives Y_out = 0.059267691.
- The interpolation error in the input is |p_line − p| = 0.0876783. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.050584391 and 0.067950991. Sort the roots even if the response decreases: 0.318834 ≤ p ≤ 0.342989.
- The lower and upper input deviations are 0.0115911 and 0.0125639. 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 p ≈ 0.330425; compatible interval [0.318834, 0.342989]. Endpoint interpolation gives 0.418103.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Spin states in a two-dimensional state space: infer an input with uncertainty
A spin-½ state is a normalised vector α|+z⟩ + β|−z⟩ in a two-dimensional complex space, stored as the column (α, β)ᵀ; columns that differ only by an overall phase are the same state, so two real numbers fix it. Use the supplied topic model Y_out = exp(-sep^2/8), where Y_out is OVERLAP e^(−Δ²/8σ²) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. sep is Separation Δ of the components, measured in σ. Keep theta (Amplitude split θ) = 90 °; phi (Relative phase φ) = 0 ° fixed. For this exercise use only 0.6375 σ ≤ sep ≤ 5.3625 σ 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 = 0.21254958, with a bounded reading uncertainty ±0.027761. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown sep by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0004725 σ, 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.
OVERLAP e^(−Δ²/8σ²) against sep
Predictions from the supplied Spin states in a two-dimensional state space 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): OVERLAP e^(−Δ²/8σ²)
Horizontal axis (x): sep: Separation Δ of the components / σ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| sep: Separation Δ of the components / σ | OVERLAP e^(−Δ²/8σ²) |
|---|---|
| 0.6375 | 0.9504680027199568 |
| 1.8187499999999999 | 0.6613438052942324 |
| 3 | 0.32465246735834974 |
| 4.1812499999999995 | 0.11243764863785047 |
| 5.3625 | 0.027473021800949925 |
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(sep) = 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(0.6375) = 0.950468 and Y_out(5.3625) = 0.027473. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives sep_line = 0.6375 + (0.21255 − 0.950468)(5.3625 − 0.6375)/(0.027473 − 0.950468) = 4.41506 σ. 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 sep = 3.51975 σ; substituting back gives Y_out = 0.21254958.
- The interpolation error in the input is |sep_line − sep| = 0.895305 σ. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.18478858 and 0.24031058. Sort the roots even if the response decreases: 3.37736 σ ≤ sep ≤ 3.67537 σ.
- The lower and upper input deviations are 0.142386 σ and 0.15562 σ. 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 sep ≈ 3.51975 σ; compatible interval [3.37736, 3.67537] σ. Endpoint interpolation gives 4.41506 σ.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The three Pauli matrices: infer an input with uncertainty
The Pauli matrices σₓ, σ_y and σ_z are the three 2×2 Hermitian, traceless matrices that represent the x, y and z components of spin-½ in units of ħ/2, written in the basis |+z⟩, |−z⟩. Use the supplied topic model Y_out = (1000*cosd(beta-alpha))/1000, where Y_out is I PART a⋅b (before display rounding) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. alpha is Axis a (solid), α from +z, measured in °. Keep beta (Axis b (dashed), β from +z) = 60 ° fixed. For this exercise use only 246.15 ° ≤ alpha ≤ 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 = -0.61199645, with a bounded reading uncertainty ±0.057337. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown alpha by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00756 °, 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.
I PART a⋅b (before display rounding) against alpha
Predictions from the supplied The three Pauli matrices 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): I PART a⋅b (before display rounding)
Horizontal axis (x): alpha: Axis a (solid), α from +z / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| alpha: Axis a (solid), α from +z / ° | I PART a⋅b (before display rounding) |
|---|---|
| 246.14999999999998 | -0.9942448326331055 |
| 265.04999999999995 | -0.9059386379797364 |
| 283.95 | -0.7199457301444867 |
| 302.85 | -0.45632159090047586 |
| 321.75 | -0.1434926219911794 |
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(alpha) = 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(246.15) = -0.994245 and Y_out(321.75) = -0.143493. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives alpha_line = 246.15 + (-0.611996 − -0.994245)(321.75 − 246.15)/(-0.143493 − -0.994245) = 280.118 °. 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 alpha = 292.266 °; substituting back gives Y_out = -0.61199645.
- The interpolation error in the input is |alpha_line − alpha| = 12.1484 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.66933345 and -0.55465945. Sort the roots even if the response decreases: 287.984 ° ≤ alpha ≤ 296.313 °.
- The lower and upper input deviations are 4.28164 ° and 4.04674 °. 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 alpha ≈ 292.266 °; compatible interval [287.984, 296.313] °. Endpoint interpolation gives 280.118 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.