University Quantum Mechanics I · unit UQMI-05
The Free Particle
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 with V = 0: infer an input with uncertainty
The free-particle Schrödinger equation is the time-independent equation with V = 0 for every x, −(ħ²/2m)ψ″ = Eψ, whose solutions at each energy E > 0 are the plane waves e^(ikx) and e^(−ikx), with k = √(2mE)/ħ. Use the supplied topic model Y_out = 2*pi/k, where Y_out is WAVELENGTH 2π/k (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. k is Wavenumber k, measured in nm⁻¹. Keep r (Left-moving amplitude r = B/A) = 0; t (Time t, in periods T = h/E) = 0 T fixed. For this exercise use only 2.85 nm⁻¹ ≤ k ≤ 9.15 nm⁻¹ 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 = 0.93876966 nm, with a bounded reading uncertainty ±0.037812 nm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown k by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00063 nm⁻¹, 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.
WAVELENGTH 2π/k against k
Predictions from the supplied The Schrödinger equation with V = 0 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): WAVELENGTH 2π/k / nm
Horizontal axis (x): k: Wavenumber k / nm⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: Wavenumber k / nm⁻¹ | WAVELENGTH 2π/k / nm |
|---|---|
| 2.85 | 2.2046264235717845 |
| 4.425 | 1.4199288829784376 |
| 6 | 1.0471975511965976 |
| 7.574999999999999 | 0.8294634068883943 |
| 9.149999999999999 | 0.6866869188174413 |
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(k) = 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(2.85) = 2.20463 nm and Y_out(9.15) = 0.686687 nm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives k_line = 2.85 + (0.93877 − 2.20463)(9.15 − 2.85)/(0.686687 − 2.20463) = 8.10377 nm⁻¹. 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 k = 6.693 nm⁻¹; substituting back gives Y_out = 0.93876966 nm.
- The interpolation error in the input is |k_line − k| = 1.41077 nm⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.90095766 nm and 0.97658166 nm. Sort the roots even if the response decreases: 6.43386 nm⁻¹ ≤ k ≤ 6.9739 nm⁻¹.
- The lower and upper input deviations are 0.259144 nm⁻¹ and 0.280896 nm⁻¹. 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 k ≈ 6.693 nm⁻¹; compatible interval [6.43386, 6.9739] nm⁻¹. Endpoint interpolation gives 8.10377 nm⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Plane waves, and why no plane wave is a state: infer an input with uncertainty
A plane wave is the free-particle solution ψ(x, t) = A e^(i(kx − ωt)) with ω = ħk²/(2m); it has the sharp momentum ħk, but its probability density |A|² is the same everywhere, so it cannot be normalised and is not a physical state. Use the supplied topic model Y_out = 1/(s*sqrt(2*pi)), where Y_out is PEAK DENSITY |ψ(0)|² (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. s is Envelope width σ, measured in nm. Keep k (Wavenumber k) = 1.5 nm⁻¹ fixed. For this exercise use only 1.74375 nm ≤ s ≤ 7.25625 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 = 0.078126319 nm⁻¹, with a bounded reading uncertainty ±0.0034721 nm⁻¹. 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 0.00055125 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.
PEAK DENSITY |ψ(0)|² against s
Predictions from the supplied Plane waves, and why no plane wave is a state 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): PEAK DENSITY |ψ(0)|² / nm⁻¹
Horizontal axis (x): s: Envelope width σ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Envelope width σ / nm | PEAK DENSITY |ψ(0)|² / nm⁻¹ |
|---|---|
| 1.74375 | 0.2287841034560188 |
| 3.1218749999999997 | 0.12778931904750598 |
| 4.5 | 0.08865384008920726 |
| 5.878124999999999 | 0.06786896848934529 |
| 7.25625 | 0.054979125636717696 |
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(1.74375) = 0.228784 nm⁻¹ and Y_out(7.25625) = 0.0549791 nm⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives s_line = 1.74375 + (0.0781263 − 0.228784)(7.25625 − 1.74375)/(0.0549791 − 0.228784) = 6.5221 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 s = 5.10637 nm; substituting back gives Y_out = 0.078126319 nm⁻¹.
- The interpolation error in the input is |s_line − s| = 1.41573 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.074654219 nm⁻¹ and 0.081598419 nm⁻¹. Sort the roots even if the response decreases: 4.88909 nm ≤ s ≤ 5.34387 nm.
- The lower and upper input deviations are 0.217282 nm and 0.237493 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 s ≈ 5.10637 nm; compatible interval [4.88909, 5.34387] nm. Endpoint interpolation gives 6.5221 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Momentum eigenstates and delta normalisation: infer an input with uncertainty
A momentum eigenfunction is fₖ(x) = e^(ikx)/√(2π), on which p̂ = −iħ d/dx returns exactly ħk; since it cannot be normalised to 1, it is normalised to a Dirac delta instead: ∫ fₖ*(x) fₖ′(x) dx = δ(k − k′). Use the supplied topic model Y_out = 1/(pi*q), where Y_out is BOUND 1/(π × GAP) (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. q is Wavenumber gap k′ − k, measured in nm⁻¹. Keep hw (Window half-length X) = 4 nm fixed. For this exercise use only 0.74 nm⁻¹ ≤ q ≤ 3.26 nm⁻¹ 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 = 0.13978126 nm, with a bounded reading uncertainty ±0.006321 nm. 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 0.000252 nm⁻¹, 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.
BOUND 1/(π × GAP) against q
Predictions from the supplied Momentum eigenstates and delta normalisation 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): BOUND 1/(π × GAP) / nm
Horizontal axis (x): q: Wavenumber gap k′ − k / nm⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q: Wavenumber gap k′ − k / nm⁻¹ | BOUND 1/(π × GAP) / nm |
|---|---|
| 0.74 | 0.43014849484296036 |
| 1.3699999999999999 | 0.23234298261590564 |
| 1.9999999999999998 | 0.15915494309189537 |
| 2.63 | 0.12103037497482537 |
| 3.26 | 0.0976410693815309 |
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(0.74) = 0.430148 nm and Y_out(3.26) = 0.0976411 nm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives q_line = 0.74 + (0.139781 − 0.430148)(3.26 − 0.74)/(0.0976411 − 0.430148) = 2.94063 nm⁻¹. 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 = 2.2772 nm⁻¹; substituting back gives Y_out = 0.13978126 nm.
- The interpolation error in the input is |q_line − q| = 0.663429 nm⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.13346026 nm and 0.14610226 nm. Sort the roots even if the response decreases: 2.17868 nm⁻¹ ≤ q ≤ 2.38505 nm⁻¹.
- The lower and upper input deviations are 0.0985213 nm⁻¹ and 0.107854 nm⁻¹. 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 ≈ 2.2772 nm⁻¹; compatible interval [2.17868, 2.38505] nm⁻¹. Endpoint interpolation gives 2.94063 nm⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A continuous spectrum, degenerate above zero: infer an input with uncertainty
The spectrum of a free particle is the set of energies at which −(ħ²/2m)ψ″ = Eψ has a solution that stays bounded as x → ±∞: every E ≥ 0, with two independent solutions at each E > 0 and only the constant at E = 0. Use the supplied topic model Y_out = sqrt(max(-en,0)/0.0381), where Y_out is GROWTH CONSTANT κ (nm⁻¹). 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. en is Energy E, measured in eV. Keep w (Window half-width W) = 6 nm fixed. For this exercise use only -0.225625 eV ≤ en ≤ -0.005125 eV 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 = 1.5464802 nm⁻¹, with a bounded reading uncertainty ±0.13305 nm⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown en by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00002205 eV, 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.
GROWTH CONSTANT κ against en
Predictions from the supplied A continuous spectrum, degenerate above zero 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): GROWTH CONSTANT κ / nm⁻¹
Horizontal axis (x): en: Energy E / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: Energy E / eV | GROWTH CONSTANT κ / nm⁻¹ |
|---|---|
| -0.225625 | 2.4334987179981598 |
| -0.17049999999999998 | 2.115435089242376 |
| -0.11537499999999999 | 1.7401767792661504 |
| -0.06025 | 1.2575232917907824 |
| -0.005124999999999991 | 0.3667620968632634 |
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(en) = 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(-0.225625) = 2.4335 nm⁻¹ and Y_out(-0.005125) = 0.366762 nm⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives en_line = -0.225625 + (1.54648 − 2.4335)(-0.005125 − -0.225625)/(0.366762 − 2.4335) = -0.130989 eV. 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 en = -0.09112 eV; substituting back gives Y_out = 1.5464802 nm⁻¹.
- The interpolation error in the input is |en_line − en| = 0.0398691 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.4134302 nm⁻¹ and 1.6795302 nm⁻¹. Sort the roots even if the response decreases: -0.107473 eV ≤ en ≤ -0.0761156 eV.
- The lower and upper input deviations are 0.0163533 eV and 0.0150044 eV. 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 en ≈ -0.09112 eV; compatible interval [-0.107473, -0.0761156] eV. Endpoint interpolation gives -0.130989 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.