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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.

Extra challenging · 01

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.

Explore the question data

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

SPACING ÷ SPOT WIDTH against gPredictions 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. Horizontal axis: g: Moment factor g. Vertical axis: SPACING ÷ SPOT WIDTH. Exact values are available in the data table below.1.1832.032.8773.7244.5710.53460.91731.31.6832.065xy

Horizontal axis (x): g: Moment factor g

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1g: Moment factor g: 0.5913SPACING ÷ SPOT WIDTH: 1.309
Read the exact data as a table
SPACING ÷ SPOT WIDTH against g — supplied values
g: Moment factor gSPACING ÷ SPOT WIDTH
0.591251.3086177095130238
0.9456252.092958345130238
1.32.877298980747452
1.6543753.6616396163646656
2.008754.445980251981879

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |g_line − g| = 4.44089e-16. This is within the requested numerical tolerance.
  5. Invert both limits Y_out = 3.0388689 and 3.4059489. Sort the roots even if the response decreases: 1.373 ≤ g ≤ 1.53885.
  6. 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.

Extra challenging · 02

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.

Explore the question data

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

2π|ψ|² AT φ = 2π against pPredictions 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. Horizontal axis: p: Weight p of the trial state. Vertical axis: 2π|ψ|² AT φ = 2π. Exact values are available in the data table below.-0.013910.089310.19250.29580.3990.091550.19080.290.38920.4885xy

Horizontal axis (x): p: Weight p of the trial state

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1p: Weight p of the trial state: 0.10622π|ψ|² AT φ = 2π: 0.3837
Read the exact data as a table
2π|ψ|² AT φ = 2π against p — supplied values
p: Weight p of the trial state2π|ψ|² AT φ = 2π
0.106250.3836853482189475
0.1981250.20282628147937543
0.290.09247589563692582
0.3818750.028307694020375584
0.473750.0013790759252036988

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |p_line − p| = 0.0876783. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.050584391 and 0.067950991. Sort the roots even if the response decreases: 0.318834 ≤ p ≤ 0.342989.
  6. 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.

Extra challenging · 03

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.

Explore the question data

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)

OVERLAP e^(−Δ²/8σ²) against sepPredictions 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. Horizontal axis: sep: Separation Δ of the components / σ. Vertical axis: OVERLAP e^(−Δ²/8σ²). Exact values are available in the data table below.-0.0094470.23980.4890.73820.98740.44851.72434.2765.551xy

Horizontal axis (x): sep: Separation Δ of the components / σ

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1sep: Separation Δ of the components / σ: 0.6375OVERLAP e^(−Δ²/8σ²): 0.9505
Read the exact data as a table
OVERLAP e^(−Δ²/8σ²) against sep — supplied values
sep: Separation Δ of the components / σOVERLAP e^(−Δ²/8σ²)
0.63750.9504680027199568
1.81874999999999990.6613438052942324
30.32465246735834974
4.18124999999999950.11243764863785047
5.36250.027473021800949925

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |sep_line − sep| = 0.895305 σ. This is larger than the requested numerical tolerance.
  5. 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.
  6. 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.

Extra challenging · 04

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.

Explore the question data

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)

I PART a⋅b (before display rounding) against alphaPredictions 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. Horizontal axis: alpha: Axis a (solid), α from +z / °. Vertical axis: I PART a⋅b (before display rounding). Exact values are available in the data table below.-1.028-0.7986-0.5689-0.3392-0.1095243.1263.5283.9304.4324.8xy

Horizontal axis (x): alpha: Axis a (solid), α from +z / °

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1alpha: Axis a (solid), α from +z / °: 246.1I PART a⋅b (before display rounding): -0.9942
Read the exact data as a table
I PART a⋅b (before display rounding) against alpha — supplied values
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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |alpha_line − alpha| = 12.1484 °. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = -0.66933345 and -0.55465945. Sort the roots even if the response decreases: 287.984 ° ≤ alpha ≤ 296.313 °.
  6. 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.

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