University Quantum Mechanics I · unit UQMI-02
The Schrödinger Equation
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 time-dependent Schrödinger equation: infer an input with uncertainty
The time-dependent Schrödinger equation, iħ ∂Ψ/∂t = −(ħ²/2m) ∂²Ψ/∂x² + VΨ, is the postulated law of motion for a wavefunction: given V and Ψ at one instant, it fixes Ψ at every later instant. Use the supplied topic model Y_out = 1.994711/(2*(1+exp(-d^2/0.08)))*4*exp(-d^2/(0.08*(1+(1.447095*t)^2)))/sqrt((1+(1.447095*t)^2)), where Y_out is DENSITY AT x = 0 (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; exp(x) means eˣ. t is Time since release t, measured in fs. Keep d (Starting offset of each packet d) = 1 nm fixed. For this exercise use only 1.29688 fs ≤ t ≤ 3.26563 fs 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 = 0.43739029 nm⁻¹, with a bounded reading uncertainty ±0.0068939 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 t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000196875 fs, 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.
DENSITY AT x = 0 against t
Predictions from the supplied The time-dependent Schrödinger equation 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): DENSITY AT x = 0 / nm⁻¹
Horizontal axis (x): t: Time since release t / fs
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Time since release t / fs | DENSITY AT x = 0 / nm⁻¹ |
|---|---|
| 1.296875 | 0.11823345080310765 |
| 1.7890625 | 0.2836651117021689 |
| 2.28125 | 0.4044822834139529 |
| 2.7734375 | 0.46450257606941586 |
| 3.265625 | 0.483349519507742 |
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(t) = 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) = 0.118233 nm⁻¹ and Y_out(3.26563) = 0.48335 nm⁻¹. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = 1.29688 + (0.43739 − 0.118233)(3.26563 − 1.29688)/(0.48335 − 0.118233) = 3.01781 fs. 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 t = 2.49781 fs; substituting back gives Y_out = 0.43739029 nm⁻¹.
- The interpolation error in the input is |t_line − t| = 0.519995 fs. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.43049639 nm⁻¹ and 0.44428419 nm⁻¹. Sort the roots even if the response decreases: 2.44571 fs ≤ t ≤ 2.55516 fs.
- The lower and upper input deviations are 0.0520975 fs and 0.0573523 fs. 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 t ≈ 2.49781 fs; compatible interval [2.44571, 2.55516] fs. Endpoint interpolation gives 3.01781 fs.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The Hamiltonian operator: infer an input with uncertainty
The Hamiltonian operator is the classical energy p²/2m + V(x) with momentum replaced by −iħ ∂/∂x, giving Ĥ = −(ħ²/2m) ∂²/∂x² + V(x): the instruction that turns a wavefunction into the function driving its time evolution. Use the supplied topic model Y_out = 1/(2*s^2) + 0.5*(kappa - 1/s^4), where Y_out is Ĥψ/ψ AT x = ℓ (ε). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s is Width s of ψ = e^(−x²/2s²), measured in ℓ. Keep kappa (Stiffness κ in V = ½κx²) = 1 ε/ℓ² fixed. For this exercise use only 0.70625 ℓ ≤ s ≤ 1.38875 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 = 0.5819169 ε, with a bounded reading uncertainty ±0.0064368 ε. 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 0.00006825 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.
Ĥψ/ψ AT x = ℓ against s
Predictions from the supplied The Hamiltonian operator 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): Ĥψ/ψ AT x = ℓ / ε
Horizontal axis (x): s: Width s of ψ = e^(−x²/2s²) / ℓ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Width s of ψ = e^(−x²/2s²) / ℓ | Ĥψ/ψ AT x = ℓ / ε |
|---|---|
| 0.7062499999999999 | -0.5072950521413226 |
| 0.876875 | 0.30456567466656403 |
| 1.0474999999999999 | 0.5403897716514907 |
| 1.218125 | 0.6098737733284264 |
| 1.38875 | 0.6248288066254211 |
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(0.70625) = -0.507295 ε and Y_out(1.38875) = 0.624829 ε. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives s_line = 0.70625 + (0.581917 − -0.507295)(1.38875 − 0.70625)/(0.624829 − -0.507295) = 1.36288 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 = 1.12258 ℓ; substituting back gives Y_out = 0.5819169 ε.
- The interpolation error in the input is |s_line − s| = 0.240306 ℓ. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.5754801 ε and 0.5883537 ε. Sort the roots even if the response decreases: 1.1079 ℓ ≤ s ≤ 1.13907 ℓ.
- The lower and upper input deviations are 0.0146775 ℓ and 0.0164938 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 ≈ 1.12258 ℓ; compatible interval [1.1079, 1.13907] ℓ. Endpoint interpolation gives 1.36288 ℓ.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Separation of variables and the constant E: infer an input with uncertainty
Separation of variables looks for solutions of the Schrödinger equation of the form Ψ(x,t) = ψ(x) f(t); when V depends on x alone this forces f(t) = e^(−iEt/ħ), where the real constant E is the energy of the state. Use the supplied topic model Y_out = 0.5-0.36025*sqrt(max(p*(1-p),0))*cos(6*pi*t/(10.998*w^2)), where Y_out is ⟨x⟩ / L (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 of ψ₂, p = |c₂|², dimensionless. Keep t (Time t) = 2 fs; w (Well width L) = 1 nm 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 = 0.66255573, with a bounded reading uncertainty ±0.0015004. 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.
⟨x⟩ / L against p
Predictions from the supplied Separation of variables and the constant E 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⟩ / L
Horizontal axis (x): p: Weight of ψ₂, p = |c₂|²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Weight of ψ₂, p = |c₂|² | ⟨x⟩ / L |
|---|---|
| 0.10625 | 0.6064973295058715 |
| 0.198125 | 0.6377492356045297 |
| 0.29 | 0.6568174524126194 |
| 0.381875 | 0.6679055258368167 |
| 0.47375 | 0.6725587105471457 |
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
- Endpoint predictions are Y_out(0.10625) = 0.606497 and Y_out(0.47375) = 0.672559. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives p_line = 0.10625 + (0.662556 − 0.606497)(0.47375 − 0.10625)/(0.672559 − 0.606497) = 0.418103. 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 p = 0.330425; substituting back gives Y_out = 0.66255573.
- The interpolation error in the input is |p_line − p| = 0.0876783. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.66105533 and 0.66405613. Sort the roots even if the response decreases: 0.318834 ≤ p ≤ 0.342988.
- The lower and upper input deviations are 0.0115908 and 0.0125634. 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, 0.342988]. Endpoint interpolation gives 0.418103.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The time-independent Schrödinger equation: infer an input with uncertainty
The time-independent Schrödinger equation, −(ħ²/2m)ψ″ + V(x)ψ = Eψ, asks for the pairs of a number E and a function ψ for which the Hamiltonian returns ψ multiplied by E; the admissible pairs are the energy eigenvalues and eigenfunctions. Use the supplied topic model Y_out = 2*pi*sqrt(0.0381/en), where Y_out is LEFT WAVELENGTH (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. en is Electron energy E, measured in eV. Keep phi (Phase φ of the left-hand wave) = 45 ° fixed. For this exercise use only 0.29125 eV ≤ en ≤ 1.70875 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 = 1.140716 nm, with a bounded reading uncertainty ±0.030375 nm. 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 0.00014175 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.
LEFT WAVELENGTH against en
Predictions from the supplied The time-independent Schrödinger equation 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): LEFT WAVELENGTH / nm
Horizontal axis (x): en: Electron energy E / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: Electron energy E / eV | LEFT WAVELENGTH / nm |
|---|---|
| 0.29125 | 2.272528765410293 |
| 0.6456249999999999 | 1.5263431668161582 |
| 0.9999999999999999 | 1.226428844542568 |
| 1.3543749999999999 | 1.0538356638692952 |
| 1.7087499999999998 | 0.9382165561340096 |
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(0.29125) = 2.27253 nm and Y_out(1.70875) = 0.938217 nm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives en_line = 0.29125 + (1.14072 − 2.27253)(1.70875 − 0.29125)/(0.938217 − 2.27253) = 1.49363 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 = 1.15592 eV; substituting back gives Y_out = 1.140716 nm.
- The interpolation error in the input is |en_line − en| = 0.337701 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.110341 nm and 1.171091 nm. Sort the roots even if the response decreases: 1.09674 eV ≤ en ≤ 1.22003 eV.
- The lower and upper input deviations are 0.0591856 eV and 0.0641091 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 ≈ 1.15592 eV; compatible interval [1.09674, 1.22003] eV. Endpoint interpolation gives 1.49363 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.