University Quantum Mechanics I · unit UQMI-03
Operators and Observables
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.
Operators: linear instructions that act on a wavefunction: infer an input with uncertainty
An operator  is a rule that turns a function f(x) into a new function Âf(x); it is linear when Â(af + bg) = aÂf + bÂg for every pair of functions f, g it can act on and every pair of complex numbers a, b. Use the supplied topic model Y_out = sin(2*pi*m/n)/(2*pi*m/n), where Y_out is MATRIX ÷ EXACT sin(kh)/(kh) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; unmarked trigonometric functions use radians. n is Grid points N, dimensionless. Keep m (Wavelengths across the window m) = 2 fixed. For this exercise use only 8 ≤ n ≤ 64, at the specified test settings separated by 4. 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 = 0.98646084, with a bounded reading uncertainty ±0.028557. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of n consistent with that bound. Compare with a straight-line interpolation between the two endpoint predictions, and explain why a fractional setting may be inadmissible. Do not assume that the supplied calibration points prove monotonicity outside this branch.
MATRIX ÷ EXACT sin(kh)/(kh) against n
Predictions from the supplied Operators: linear instructions that act on a wavefunction 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): MATRIX ÷ EXACT sin(kh)/(kh)
Horizontal axis (x): n: Grid points N
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| n: Grid points N | MATRIX ÷ EXACT sin(kh)/(kh) |
|---|---|
| 8 | 0.6366197723675814 |
| 24 | 0.954929658551372 |
| 36 | 0.9798155360510165 |
| 52 | 0.99029504419782 |
| 64 | 0.9935868511442058 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Enumerate the permitted settings in the stated range, evaluate Y_out at each, and retain every setting satisfying |Y_out − Y_measured| ≤ u. The plotted predictions are supplied calibration evidence.
Reveal the worked solution
- Endpoint predictions are Y_out(8) = 0.63662 and Y_out(64) = 0.993587. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives n_line = 8 + (0.986461 − 0.63662)(64 − 8)/(0.993587 − 0.63662) = 62.8821. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Evaluate the model at each permitted setting. The central reading corresponds to n = 44; only the allowed settings 28, 32, 36, 40, 44, 48, 52, 56, 60, 64 lie inside the stated reading bound.
- The interpolation error in the input is |n_line − n| = 18.8821.
- The compatible allowed values span 28 to 64. Values between specified test settings are outside this exercise's input domain even if an interpolation passes through them.
- The quoted output bound restricts which specified settings remain compatible. It does not change physical restrictions on quantum numbers or counts, and the chosen grid alone does not establish physical quantisation.
Answer Compatible n: 28, 32, 36, 40, 44, 48, 52, 56, 60, 64. Endpoint interpolation predicts 62.8821; check admissibility against the discrete settings.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The position operator x̂ and the momentum operator p̂: infer an input with uncertainty
The position operator multiplies a wavefunction by x, x̂ψ = xψ, and the momentum operator differentiates it and multiplies by −iħ, p̂ψ = −iħ dψ/dx; each takes a function and returns a new function, not a number. Use the supplied topic model Y_out = 1/(2*k*sig), where Y_out is IM(p̂ψ/ψ)/ħk AT x = σ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. sig is Packet width σ, measured in nm. Keep k (Carrier wave number k) = 5 nm⁻¹; th (Prefactor phase θ in Ô = ħ e^(−iθ) d/dx) = 90 ° fixed. For this exercise use only 0.7825 nm ≤ sig ≤ 3.6175 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 = 0.039811294, with a bounded reading uncertainty ±0.0018252. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown sig by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0002835 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.
IM(p̂ψ/ψ)/ħk AT x = σ against sig
Predictions from the supplied The position operator x̂ and the momentum operator p̂ 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): IM(p̂ψ/ψ)/ħk AT x = σ
Horizontal axis (x): sig: Packet width σ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| sig: Packet width σ / nm | IM(p̂ψ/ψ)/ħk AT x = σ |
|---|---|
| 0.7825 | 0.12779552715654954 |
| 1.49125 | 0.06705783738474434 |
| 2.2 | 0.045454545454545456 |
| 2.9087499999999995 | 0.03437902879243662 |
| 3.6174999999999997 | 0.027643400138217002 |
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(sig) = 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(0.7825) = 0.127796 and Y_out(3.6175) = 0.0276434. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives sig_line = 0.7825 + (0.0398113 − 0.127796)(3.6175 − 0.7825)/(0.0276434 − 0.127796) = 3.27306 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 sig = 2.51185 nm; substituting back gives Y_out = 0.039811294.
- The interpolation error in the input is |sig_line − sig| = 0.761214 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.037986094 and 0.041636494. Sort the roots even if the response decreases: 2.40174 nm ≤ sig ≤ 2.63254 nm.
- The lower and upper input deviations are 0.110111 nm and 0.120692 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 sig ≈ 2.51185 nm; compatible interval [2.40174, 2.63254] nm. Endpoint interpolation gives 3.27306 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Kinetic energy, potential energy and the Hamiltonian: infer an input with uncertainty
The Hamiltonian operator Ĥ = −(ħ²/2m) d²/dx² + V(x) is the energy written as an instruction on ψ: differentiate twice and scale by −ħ²/2m for the kinetic part, multiply by V(x) for the potential part, and add the two results. Use the supplied topic model Y_out = 1/(2*b^2), where Y_out is Ĥψ/ψ AT x = 0 (ħω₀). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b is Trial width b, measured in ℓ₀. Keep w (Spring frequency ω) = 1 ω₀ fixed. For this exercise use only 0.70625 ℓ₀ ≤ b ≤ 1.49375 ℓ₀ 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 = 0.35509374 ħω₀, with a bounded reading uncertainty ±0.019651 ħω₀. 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 0.00007875 ℓ₀, 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 = 0 against b
Predictions from the supplied Kinetic energy, potential energy and the Hamiltonian 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 = 0 / ħω₀
Horizontal axis (x): b: Trial width b / ℓ₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Trial width b / ℓ₀ | Ĥψ/ψ AT x = 0 / ħω₀ |
|---|---|
| 0.7062499999999999 | 1.002427754718459 |
| 0.903125 | 0.6130194801307456 |
| 1.0999999999999999 | 0.41322314049586784 |
| 1.296875 | 0.29728552765277977 |
| 1.49375 | 0.22408571278514033 |
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(0.70625) = 1.00243 ħω₀ and Y_out(1.49375) = 0.224086 ħω₀. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives b_line = 0.70625 + (0.355094 − 1.00243)(1.49375 − 0.70625)/(0.224086 − 1.00243) = 1.3612 ℓ₀. 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 = 1.18663 ℓ₀; substituting back gives Y_out = 0.35509374 ħω₀.
- The interpolation error in the input is |b_line − b| = 0.174576 ℓ₀. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.33544274 ħω₀ and 0.37474474 ħω₀. Sort the roots even if the response decreases: 1.15509 ℓ₀ ≤ b ≤ 1.22089 ℓ₀.
- The lower and upper input deviations are 0.0315313 ℓ₀ and 0.0342629 ℓ₀. 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 ≈ 1.18663 ℓ₀; compatible interval [1.15509, 1.22089] ℓ₀. Endpoint interpolation gives 1.3612 ℓ₀.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Eigenvalue equations: solving Âψ = aψ: infer an input with uncertainty
An eigenfunction of an operator  is a nonzero function that  returns multiplied by a single constant a, the eigenvalue, while also meeting the boundary and normalisability conditions the problem imposes. Use the supplied topic model Y_out = 1.504*q^2, where Y_out is ħ²k²/2m (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. q is Trial wave number k, measured in π/a. Keep c (Admixture c of sin 2kx) = 0.2 fixed. For this exercise use only 0.81875 π/a ≤ q ≤ 3.18125 π/a 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 = 7.6809807 eV, with a bounded reading uncertainty ±1.0009 eV. 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 0.00023625 π/a, 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.
ħ²k²/2m against q
Predictions from the supplied Eigenvalue equations: solving Âψ = aψ 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): ħ²k²/2m / eV
Horizontal axis (x): q: Trial wave number k / π/a
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q: Trial wave number k / π/a | ħ²k²/2m / eV |
|---|---|
| 0.81875 | 1.0082087499999999 |
| 1.4093749999999998 | 2.9874521874999993 |
| 2 | 6.016 |
| 2.5906249999999997 | 10.093852187499998 |
| 3.18125 | 15.22100875 |
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(0.81875) = 1.00821 eV and Y_out(3.18125) = 15.221 eV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives q_line = 0.81875 + (7.68098 − 1.00821)(3.18125 − 0.81875)/(15.221 − 1.00821) = 1.92792 π/a. 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 = 2.25988 π/a; substituting back gives Y_out = 7.6809807 eV.
- The interpolation error in the input is |q_line − q| = 0.331954 π/a. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 6.6800807 eV and 8.6818807 eV. Sort the roots even if the response decreases: 2.1075 π/a ≤ q ≤ 2.40261 π/a.
- The lower and upper input deviations are 0.152378 π/a and 0.142733 π/a. 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 ≈ 2.25988 π/a; compatible interval [2.1075, 2.40261] π/a. Endpoint interpolation gives 1.92792 π/a.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.