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University Quantum Mechanics I · unit UQMI-12

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.

Extra challenging · 01

From L = r × p to three Hermitian operators: infer an input with uncertainty

The orbital angular momentum operators are the components of L = r × p with p = −iħ∇, such as L_z = −iħ(x ∂/∂y − y ∂/∂x); each is Hermitian, and each is measured about a chosen origin. Use the supplied topic model Y_out = (100*s*cosd(theta))/100, where Y_out is ⟨x⟩ FROM O (before display rounding) (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. theta is Direction of k, θ, measured in °. Keep a (Origin O raised by a) = 0 nm; s (Packet position along its path s) = -1 nm fixed. For this exercise use only -23.625 ° ≤ theta ≤ -1.575 ° 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 = -0.98427432 nm, with a bounded reading uncertainty ±0.0023022 nm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown theta by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.002205 °, 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

⟨x⟩ FROM O (before display rounding) against theta

Predictions from the supplied From L = r × p to three Hermitian operators 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): ⟨x⟩ FROM O (before display rounding) / nm

⟨x⟩ FROM O (before display rounding) against thetaPredictions from the supplied From L = r × p to three Hermitian operators 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: theta: Direction of k, θ / °. Vertical axis: ⟨x⟩ FROM O (before display rounding) / nm. Exact values are available in the data table below.-1.003-0.98043-0.95791-0.93538-0.91285-24.51-18.55-12.6-6.646-0.693xy

Horizontal axis (x): theta: Direction of k, θ / °

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

Reading 1theta: Direction of k, θ / °: -23.63⟨x⟩ FROM O (before display rounding) / nm: -0.9162
Read the exact data as a table
⟨x⟩ FROM O (before display rounding) against theta — supplied values
theta: Direction of k, θ / °⟨x⟩ FROM O (before display rounding) / nm
-23.625-0.9161879571171361
-18.1125-0.9504479299251564
-12.6-0.9759167619387474
-7.087499999999999-0.9923588799286435
-1.5749999999999993-0.9996222029973468

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(theta) = 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(-23.625) = -0.916188 nm and Y_out(-1.575) = -0.999622 nm. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives theta_line = -23.625 + (-0.984274 − -0.916188)(-1.575 − -23.625)/(-0.999622 − -0.916188) = -5.63114 °. 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 theta = -10.1745 °; substituting back gives Y_out = -0.98427432 nm.
  4. The interpolation error in the input is |theta_line − theta| = 4.54336 °. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = -0.98657652 nm and -0.98197212 nm. Sort the roots even if the response decreases: -10.8959 ° ≤ theta ≤ -9.39847 °.
  6. The lower and upper input deviations are 0.721436 ° and 0.77603 °. 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 theta ≈ -10.1745 °; compatible interval [-10.8959, -9.39847] °. Endpoint interpolation gives -5.63114 °.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 02

The components in spherical coordinates: infer an input with uncertainty

In spherical coordinates the orbital angular-momentum operators act on the angles alone: L_z = −iħ ∂/∂φ, and L² is −ħ² times the angular part of r²∇², with no r and no ∂/∂r anywhere in either. Use the supplied topic model Y_out = 2*pi*(2/(2*l+1)+b^2*(1.6449*max(0,1-abs(l))+1.2150*max(0,1-abs(l-1))+0.8290*max(0,1-abs(l-2)))), where Y_out is NORM ∫f² dΩ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; abs(x) means |x|; max(a,b) selects the larger value. b is Singular admixture B, dimensionless. Keep l (Angular number l) = 1 fixed. For this exercise use only -0.7875 ≤ b ≤ -0.0525 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 = 5.0668817, with a bounded reading uncertainty ±0.12856. 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 0.0000735, 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

NORM ∫f² dΩ against b

Predictions from the supplied The components in spherical coordinates 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): NORM ∫f² dΩ

NORM ∫f² dΩ against bPredictions from the supplied The components in spherical coordinates 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: b: Singular admixture B. Vertical axis: NORM ∫f² dΩ. Exact values are available in the data table below.4.0215.2946.5667.8399.112-0.8169-0.6184-0.42-0.2216-0.0231xy

Horizontal axis (x): b: Singular admixture B

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

Reading 1b: Singular admixture B: -0.7875NORM ∫f² dΩ: 8.923
Read the exact data as a table
NORM ∫f² dΩ against b — supplied values
b: Singular admixture BNORM ∫f² dΩ
-0.78758.923106520145433
-0.603756.971516127925206
-0.425.535440178932963
-0.236249999999999964.614878673168704
-0.052499999999999994.209831610632431

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
  1. Endpoint predictions are Y_out(-0.7875) = 8.92311 and Y_out(-0.0525) = 4.20983. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives b_line = -0.7875 + (5.06688 − 8.92311)(-0.0525 − -0.7875)/(4.20983 − 8.92311) = -0.186151. 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 b = -0.33915; substituting back gives Y_out = 5.0668817.
  4. The interpolation error in the input is |b_line − b| = 0.152999. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 4.9383217 and 5.1954417. Sort the roots even if the response decreases: -0.363129 ≤ b ≤ -0.313341.
  6. The lower and upper input deviations are 0.0239795 and 0.0258092. 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 b ≈ -0.33915; compatible interval [-0.363129, -0.313341]. Endpoint interpolation gives -0.186151.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 03

Commutation relations: [L_x, L_y] = iħL_z: infer an input with uncertainty

The angular-momentum commutation relations [L̂_x, L̂_y] = iħL̂_z, [L̂_y, L̂_z] = iħL̂_x and [L̂_z, L̂_x] = iħL̂_y say that two components applied in opposite orders differ by iħ times the third, so no two components are sharp together unless all three are zero. Use the supplied topic model Y_out = abs(sind(theta))*sqrt(1-m*m/2), where Y_out is SPREAD ΔL_z (ħ). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; abs(x) means |x|; sind/cosd/tand use degrees. m is Sharp value of L along n̂, measured in ħ. Keep theta (Tilt θ of the sharp axis n̂ from z) = 30 ° fixed. For this exercise use only -0.7875 ħ ≤ m ≤ -0.0525 ħ 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 = 0.48540927 ħ, with a bounded reading uncertainty ±0.0021369 ħ. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown m by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000735 ħ, 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

SPREAD ΔL_z against m

Predictions from the supplied Commutation relations: [L_x, L_y] = iħL_z 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): SPREAD ΔL_z / ħ

SPREAD ΔL_z against mPredictions from the supplied Commutation relations: [L_x, L_y] = iħL_z 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: m: Sharp value of L along n̂ / ħ. Vertical axis: SPREAD ΔL_z / ħ. Exact values are available in the data table below.0.41190.43470.45750.48030.503-0.8169-0.6184-0.42-0.2216-0.0231xy

Horizontal axis (x): m: Sharp value of L along n̂ / ħ

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

Reading 1m: Sharp value of L along n̂ / ħ: -0.7875SPREAD ΔL_z / ħ: 0.4153
Read the exact data as a table
SPREAD ΔL_z against m — supplied values
m: Sharp value of L along n̂ / ħSPREAD ΔL_z / ħ
-0.78750.4153076796183764
-0.603750.45214570902254503
-0.420.47744109584324634
-0.236249999999999960.49297387576574475
-0.052499999999999990.49965534996635425

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(m) = 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(-0.7875) = 0.415308 ħ and Y_out(-0.0525) = 0.499655 ħ. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives m_line = -0.7875 + (0.485409 − 0.415308)(-0.0525 − -0.7875)/(0.499655 − 0.415308) = -0.176639 ħ. 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 m = -0.33915 ħ; substituting back gives Y_out = 0.48540927 ħ.
  4. The interpolation error in the input is |m_line − m| = 0.162511 ħ. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.48327237 ħ and 0.48754617 ħ. Sort the roots even if the response decreases: -0.362743 ħ ≤ m ≤ -0.313672 ħ.
  6. The lower and upper input deviations are 0.0235931 ħ and 0.0254784 ħ. 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 m ≈ -0.33915 ħ; compatible interval [-0.362743, -0.313672] ħ. Endpoint interpolation gives -0.176639 ħ.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 04

L² and L_z: a compatible pair: infer an input with uncertainty

L² and L_z are compatible because [L², L_z] = 0: a complete set of states exists in which both have definite values, and measuring one never disturbs a value already found for the other. Use the supplied topic model Y_out = sqrt(max(m*m*(1+cosd(beta)^2)/2+(1-m*m)*sind(beta)^2-(m*cosd(beta))^2, 0)), where Y_out is SPREAD ΔL_n (ħ). 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; sind/cosd/tand use degrees. beta is Tilt β of the measuring axis n from z, measured in °. Keep m (Prepared state m, with L_z = mħ) = 1 fixed. For this exercise use only 19.125 ° ≤ beta ≤ 85.275 ° 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 = 0.60911657 ħ, with a bounded reading uncertainty ±0.014338 ħ. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown beta 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.

Explore the question data

SPREAD ΔL_n against beta

Predictions from the supplied L² and L_z: a compatible pair 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): SPREAD ΔL_n / ħ

SPREAD ΔL_n against betaPredictions from the supplied L² and L_z: a compatible pair 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: beta: Tilt β of the measuring axis n from z / °. Vertical axis: SPREAD ΔL_n / ħ. Exact values are available in the data table below.0.21270.34050.46820.59590.723616.4834.3452.270.0687.92xy

Horizontal axis (x): beta: Tilt β of the measuring axis n from z / °

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

Reading 1beta: Tilt β of the measuring axis n from z / °: 19.13SPREAD ΔL_n / ħ: 0.2317
Read the exact data as a table
SPREAD ΔL_n against beta — supplied values
beta: Tilt β of the measuring axis n from z / °SPREAD ΔL_n / ħ
19.1250.2316695216894395
35.66250.4122500267958414
52.20.558723967439391
68.737500000000010.6589731566607913
85.2750.7047037067407294

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(beta) = 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(19.125) = 0.23167 ħ and Y_out(85.275) = 0.704704 ħ. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives beta_line = 19.125 + (0.609117 − 0.23167)(85.275 − 19.125)/(0.704704 − 0.23167) = 71.9079 °. 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 beta = 59.4765 °; substituting back gives Y_out = 0.60911657 ħ.
  4. The interpolation error in the input is |beta_line − beta| = 12.4314 °. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.59477857 ħ and 0.62345457 ħ. Sort the roots even if the response decreases: 57.2611 ° ≤ beta ≤ 61.8479 °.
  6. The lower and upper input deviations are 2.21539 ° and 2.37136 °. 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 beta ≈ 59.4765 °; compatible interval [57.2611, 61.8479] °. Endpoint interpolation gives 71.9079 °.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

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