University Quantum Mechanics II · course UQMII
University Quantum Mechanics II
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 associated Legendre equation and its poles: infer an input with uncertainty
The associated Legendre equation is the polar part of the angular equation written in x = cos θ; its only singular points on the sphere are the poles x = ±1, and at each pole exactly one of its two solutions is regular. Use the supplied topic model Y_out = 57.29578*10^(-n), where Y_out is CAP HALF-ANGLE θ₀ (°). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. n is Cap size n, with θ₀ = 10^(−n) rad, dimensionless. Keep m (|m|, azimuthal quantum number) = 0 fixed. For this exercise use only 0.978125 ≤ n ≤ 4.52187 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 = 0.041525047 °, with a bounded reading uncertainty ±0.0059703 °. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown n by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000354375, 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.
CAP HALF-ANGLE θ₀ against n
Predictions from the supplied The associated Legendre equation and its poles 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): CAP HALF-ANGLE θ₀ / °
Horizontal axis (x): n: Cap size n, with θ₀ = 10^(−n) rad
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| n: Cap size n, with θ₀ = 10^(−n) rad | CAP HALF-ANGLE θ₀ / ° |
|---|---|
| 0.9781249999999999 | 6.025563063026317 |
| 1.8640624999999997 | 0.7835381305946377 |
| 2.75 | 0.10188790585611991 |
| 3.6359375 | 0.013249062112480949 |
| 4.521875 | 0.0017228506699142692 |
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(n) = 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(0.978125) = 6.02556 ° and Y_out(4.52187) = 0.00172285 °. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives n_line = 0.978125 + (0.041525 − 6.02556)(4.52187 − 0.978125)/(0.00172285 − 6.02556) = 4.49846. 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 n = 3.13981; substituting back gives Y_out = 0.041525047 °.
- The interpolation error in the input is |n_line − n| = 1.35865. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.035554747 ° and 0.047495347 °. Sort the roots even if the response decreases: 3.08147 ≤ n ≤ 3.20723.
- The lower and upper input deviations are 0.0583409 and 0.0674125. 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 n ≈ 3.13981; compatible interval [3.08147, 3.20723]. Endpoint interpolation gives 4.49846.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Centre of mass, relative coordinates, and the reduced mass: infer an input with uncertainty
The reduced mass μ = m₁m₂/(m₁ + m₂) is the inertia of the separation r₁ − r₂ of two interacting particles: once the free centre-of-mass motion is split off, the pair's internal problem is exactly a one-body problem of mass μ in the potential V(r). Use the supplied topic model Y_out = 1/(1+10^(-q)), where Y_out is REDUCED MASS μ/mₑ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. q is Partner mass, log₁₀(m₂/mₑ), dimensionless. Keep no other adjustable input fixed. For this exercise use only 0.40375 ≤ q ≤ 3.39625 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 = 0.99413497, with a bounded reading uncertainty ±0.00081954. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown q by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00029925, 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.
REDUCED MASS μ/mₑ against q
Predictions from the supplied Centre of mass, relative coordinates, and the reduced mass 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): REDUCED MASS μ/mₑ
Horizontal axis (x): q: Partner mass, log₁₀(m₂/mₑ)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q: Partner mass, log₁₀(m₂/mₑ) | REDUCED MASS μ/mₑ |
|---|---|
| 0.40375 | 0.7170080734221427 |
| 1.151875 | 0.9341520057683236 |
| 1.9 | 0.9875672647455576 |
| 2.648125 | 0.9977566366257081 |
| 3.3962499999999998 | 0.9995986015973534 |
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(q) = 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(0.40375) = 0.717008 and Y_out(3.39625) = 0.999599. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives q_line = 0.40375 + (0.994135 − 0.717008)(3.39625 − 0.40375)/(0.999599 − 0.717008) = 3.33839. 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 q = 2.22918; substituting back gives Y_out = 0.99413497.
- The interpolation error in the input is |q_line − q| = 1.10922. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.99331543 and 0.99495451. Sort the roots even if the response decreases: 2.17201 ≤ q ≤ 2.2949.
- The lower and upper input deviations are 0.0571614 and 0.0657248. 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 q ≈ 2.22918; compatible interval [2.17201, 2.2949]. Endpoint interpolation gives 3.33839.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The Heisenberg picture, conservation and Ehrenfest's theorem: infer an input with uncertainty
In the Heisenberg picture the state stays at |ψ(0)⟩ and each observable carries the time dependence, A_H(t) = U†(t)AU(t), evolving by dA_H/dt = (i/ħ)[H_H, A_H] + (∂A/∂t)_H, with every prediction the same as in the Schrödinger picture. Use the supplied topic model Y_out = -(x0+b*(x0^3+3*x0*s^2)), where Y_out is MEAN FORCE ⟨F⟩ = d⟨p⟩/dt (kℓ). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s is Packet width σ, measured in ℓ. Keep x0 (Packet centre ⟨x⟩) = 0.8 ℓ; b (Anharmonicity β = 4λℓ²/k) = 0.6 fixed. For this exercise use only 0.153125 ℓ ≤ s ≤ 0.546875 ℓ 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 = -1.3299604 kℓ, with a bounded reading uncertainty ±0.028349 kℓ. 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 0.000039375 ℓ, 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.
MEAN FORCE ⟨F⟩ = d⟨p⟩/dt against s
Predictions from the supplied The Heisenberg picture, conservation and Ehrenfest's theorem 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): MEAN FORCE ⟨F⟩ = d⟨p⟩/dt / kℓ
Horizontal axis (x): s: Packet width σ / ℓ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Packet width σ / ℓ | MEAN FORCE ⟨F⟩ = d⟨p⟩/dt / kℓ |
|---|---|
| 0.153125 | -1.1409640625000002 |
| 0.2515625 | -1.198328515625 |
| 0.35 | -1.2836 |
| 0.4484375 | -1.396778515625 |
| 0.546875 | -1.5378640625000002 |
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(0.153125) = -1.14096 kℓ and Y_out(0.546875) = -1.53786 kℓ. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives s_line = 0.153125 + (-1.32996 − -1.14096)(0.546875 − 0.153125)/(-1.53786 − -1.14096) = 0.340621 ℓ. 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 = 0.393312 ℓ; substituting back gives Y_out = -1.3299604 kℓ.
- The interpolation error in the input is |s_line − s| = 0.0526911 ℓ. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -1.3583094 kℓ and -1.3016114 kℓ. Sort the roots even if the response decreases: 0.367434 ℓ ≤ s ≤ 0.41759 ℓ.
- The lower and upper input deviations are 0.0258783 ℓ and 0.0242776 ℓ. 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 ≈ 0.393312 ℓ; compatible interval [0.367434, 0.41759] ℓ. Endpoint interpolation gives 0.340621 ℓ.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Tensor-product spaces for two subsystems: infer an input with uncertainty
The tensor product H_A ⊗ H_B is the state space of systems A and B taken together: it is spanned by the product kets |i⟩|j⟩ of basis kets of each part, so it has d_A d_B dimensions, and any superposition Σ cᵢⱼ|i⟩|j⟩ of them is a state. Use the supplied topic model Y_out = (200*(cosd(th1/2)*cosd(th2/2)*sind(th1/2)*sind(th2/2)+cosd(th1/2)*sind(th2/2)*sind(th1/2)*cosd(th2/2)))/100, where Y_out is ⟨σₓ ⊗ σₓ⟩ FROM THE COLUMN (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. th1 is Spin 1 polar angle θ₁, measured in °. Keep th2 (Spin 2 polar angle θ₂) = 150 ° fixed. For this exercise use only 283.95 ° ≤ th1 ≤ 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 = -0.39927574, with a bounded reading uncertainty ±0.012897. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown th1 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00378 °, 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.
⟨σₓ ⊗ σₓ⟩ FROM THE COLUMN (before display rounding) against th1
Predictions from the supplied Tensor-product spaces for two subsystems 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): ⟨σₓ ⊗ σₓ⟩ FROM THE COLUMN (before display rounding)
Horizontal axis (x): th1: Spin 1 polar angle θ₁ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th1: Spin 1 polar angle θ₁ / ° | ⟨σₓ ⊗ σₓ⟩ FROM THE COLUMN (before display rounding) |
|---|---|
| 283.95 | -0.4852532367342713 |
| 293.4 | -0.4588773128419906 |
| 302.85 | -0.4200467769494708 |
| 312.3 | -0.36981554748930484 |
| 321.75 | -0.30954697465491715 |
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(th1) = 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(283.95) = -0.485253 and Y_out(321.75) = -0.309547. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives th1_line = 283.95 + (-0.399276 − -0.485253)(321.75 − 283.95)/(-0.309547 − -0.485253) = 302.446 °. 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 th1 = 307.008 °; substituting back gives Y_out = -0.39927574.
- The interpolation error in the input is |th1_line − th1| = 4.56151 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.41217274 and -0.38637874. Sort the roots even if the response decreases: 304.478 ° ≤ th1 ≤ 309.398 °.
- The lower and upper input deviations are 2.53019 ° and 2.38984 °. 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 th1 ≈ 307.008 °; compatible interval [304.478, 309.398] °. Endpoint interpolation gives 302.446 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.