University Quantum Mechanics I · unit UQMI-14
Introduction to Three-Dimensional Quantum Mechanics
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 Schrödinger equation in three dimensions: infer an input with uncertainty
The three-dimensional time-independent Schrödinger equation, −(ħ²/2m)∇²ψ + V(r)ψ = Eψ, is the energy-eigenvalue equation for one spinless, non-relativistic particle in a static potential, with the Laplacian ∇² in place of d²/dx² and |ψ|² read as probability per unit volume. Use the supplied topic model Y_out = d/4*(s^2+1/s^2), where Y_out is ⟨H⟩ IN THIS ψ (ħω). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s is Trial width a, measured in b. Keep d (Dimensions d) = 3 fixed. For this exercise use only 0.847813 b ≤ s ≤ 0.989563 b 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 = 1.5138848 ħω, with a bounded reading uncertainty ±0.0020332 ħω. 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 0.000014175 b, 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.
⟨H⟩ IN THIS ψ against s
Predictions from the supplied The Schrödinger equation in three dimensions 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): ⟨H⟩ IN THIS ψ / ħω
Horizontal axis (x): s: Trial width a / b
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Trial width a / b | ⟨H⟩ IN THIS ψ / ħω |
|---|---|
| 0.8478125000000001 | 1.5825154741695326 |
| 0.8832500000000001 | 1.5464754605225774 |
| 0.9186875000000001 | 1.5216296139981913 |
| 0.954125 | 1.5066207554839042 |
| 0.9895625 | 1.5003302805167014 |
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(0.847813) = 1.58252 ħω and Y_out(0.989563) = 1.50033 ħω. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives s_line = 0.847813 + (1.51388 − 1.58252)(0.989563 − 0.847813)/(1.50033 − 1.58252) = 0.966184 b. 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 = 0.93428 b; substituting back gives Y_out = 1.5138848 ħω.
- The interpolation error in the input is |s_line − s| = 0.0319041 b. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.5118516 ħω and 1.515918 ħω. Sort the roots even if the response decreases: 0.929807 b ≤ s ≤ 0.93912 b.
- The lower and upper input deviations are 0.0044729 b and 0.00483995 b. 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 ≈ 0.93428 b; compatible interval [0.929807, 0.93912] b. Endpoint interpolation gives 0.966184 b.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Additive Cartesian potentials and the three-dimensional box: infer an input with uncertainty
A potential is additive when it splits as V = Vₓ(x) + V_y(y) + V_z(z); its stationary states can then be taken as products X(x)Y(y)Z(z) of one-dimensional solutions, and each energy is the sum Eₓ + E_y + E_z of three one-dimensional eigenvalues. Use the supplied topic model Y_out = 0.376*(1+1/b^2+1/c^2), where Y_out is E (1,1,1) (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b is Edge b (a fixed at 1.00 nm), measured in nm. Keep c (Edge c) = 1 nm fixed. For this exercise use only 0.86375 nm ≤ b ≤ 1.33625 nm 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 = 1.035336 eV, with a bounded reading uncertainty ±0.010914 eV. 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 0.00004725 nm, 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.
E (1,1,1) against b
Predictions from the supplied Additive Cartesian potentials and the three-dimensional box 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): E (1,1,1) / eV
Horizontal axis (x): b: Edge b (a fixed at 1.00 nm) / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Edge b (a fixed at 1.00 nm) / nm | E (1,1,1) / eV |
|---|---|
| 0.86375 | 1.2559781687648304 |
| 0.981875 | 1.1420097283635076 |
| 1.1 | 1.0627438016528925 |
| 1.218125 | 1.0053985538360044 |
| 1.33625 | 0.9625777148502618 |
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
- Endpoint predictions are Y_out(0.86375) = 1.25598 eV and Y_out(1.33625) = 0.962578 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives b_line = 0.86375 + (1.03534 − 1.25598)(1.33625 − 0.86375)/(0.962578 − 1.25598) = 1.21908 nm. 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 b = 1.15197 nm; substituting back gives Y_out = 1.035336 eV.
- The interpolation error in the input is |b_line − b| = 0.0671031 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.024422 eV and 1.04625 eV. Sort the roots even if the response decreases: 1.13041 nm ≤ b ≤ 1.17482 nm.
- The lower and upper input deviations are 0.0215658 nm and 0.0228491 nm. 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 ≈ 1.15197 nm; compatible interval [1.13041, 1.17482] nm. Endpoint interpolation gives 1.21908 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Central potentials and why Cartesian coordinates fail: infer an input with uncertainty
A central potential depends only on the distance r from a fixed centre, V = V(r), so every direction is equivalent; it separates in x, y and z only if it is also a sum V₁(x) + V₂(y) + V₃(z), and among central potentials only V = ½κr² + constant is. Use the supplied topic model Y_out = (abs(cosd(alpha))^n+abs(sind(alpha))^n)^(-1/n), where Y_out is V_add = V₀ REACHED AT (a). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; abs(x) means |x|; sind/cosd/tand use degrees. n is Power n in V = V₀(r/a)ⁿ, dimensionless. Keep alpha (Direction α in the xy-plane) = 30 ° fixed. For this exercise use only 1.29688 ≤ n ≤ 7.20312 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 = 1.1393471 a, with a bounded reading uncertainty ±0.0018481 a. 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 0.000590625, 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.
V_add = V₀ REACHED AT against n
Predictions from the supplied Central potentials and why Cartesian coordinates fail 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): V_add = V₀ REACHED AT / a
Horizontal axis (x): n: Power n in V = V₀(r/a)ⁿ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| n: Power n in V = V₀(r/a)ⁿ | V_add = V₀ REACHED AT / a |
|---|---|
| 1.296875 | 0.8488314419121912 |
| 2.7734375 | 1.075459020993421 |
| 4.25 | 1.1298546857605525 |
| 5.726562499999999 | 1.1462348242905136 |
| 7.203124999999999 | 1.1516675201486648 |
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(1.29688) = 0.848831 a and Y_out(7.20312) = 1.15167 a. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives n_line = 1.29688 + (1.13935 − 0.848831)(7.20312 − 1.29688)/(1.15167 − 0.848831) = 6.96284. 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 = 4.89969; substituting back gives Y_out = 1.1393471 a.
- The interpolation error in the input is |n_line − n| = 2.06315. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.137499 a and 1.1411952 a. Sort the roots even if the response decreases: 4.74465 ≤ n ≤ 5.07577.
- The lower and upper input deviations are 0.155035 and 0.176081. 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 ≈ 4.89969; compatible interval [4.74465, 5.07577]. Endpoint interpolation gives 6.96284.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Spherical coordinates, the Laplacian, and the volume element: infer an input with uncertainty
Spherical coordinates label a point by its distance r from the origin, its polar angle θ from the +z axis and its azimuth φ around that axis; in them a small volume is dV = r² sin θ dr dθ dφ, not dr dθ dφ. Use the supplied topic model Y_out = r*sind(theta), where Y_out is DISTANCE FROM z AXIS r sin θ (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 Polar angle θ, measured in °. Keep r (Radius r) = 1.2 nm fixed. For this exercise use only 17.5313 ° ≤ theta ≤ 86.8312 ° 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 = 1.0371745 nm, with a bounded reading uncertainty ±0.024149 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 0.00693 °, 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.
DISTANCE FROM z AXIS r sin θ against theta
Predictions from the supplied Spherical coordinates, the Laplacian, and the volume element 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): DISTANCE FROM z AXIS r sin θ / nm
Horizontal axis (x): theta: Polar angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| theta: Polar angle θ / ° | DISTANCE FROM z AXIS r sin θ / nm |
|---|---|
| 17.53125 | 0.36147111198677334 |
| 34.85625 | 0.6858233446450177 |
| 52.18125 | 0.9479452760463911 |
| 69.50625 | 1.1240524624496506 |
| 86.83125 | 1.198165273897687 |
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
- Endpoint predictions are Y_out(17.5313) = 0.361471 nm and Y_out(86.8312) = 1.19817 nm. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives theta_line = 17.5313 + (1.03717 − 0.361471)(86.8312 − 17.5313)/(1.19817 − 0.361471) = 73.497 °. 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 theta = 59.8042 °; substituting back gives Y_out = 1.0371745 nm.
- The interpolation error in the input is |theta_line − theta| = 13.6928 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.0130255 nm and 1.0613235 nm. Sort the roots even if the response decreases: 57.585 ° ≤ theta ≤ 62.1822 °.
- The lower and upper input deviations are 2.21921 ° and 2.37798 °. 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 ≈ 59.8042 °; compatible interval [57.585, 62.1822] °. Endpoint interpolation gives 73.497 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.