University Quantum Mechanics I · topic UQMI-03.03
Kinetic energy, potential energy, and the Hamiltonian operator
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.
Further practice
Work through a fresh case before tackling the extension.
Kinetic energy, potential energy and the Hamiltonian: predict a finite change
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. Calculate Y_out at b = 0.8795 ℓ₀ and 0.98975 ℓ₀. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δb across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
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
Substitute each value of b separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At b = 0.8795 ℓ₀, substitute into Y_out = 1/(2*b^2): Y₀ = 0.646395 ħω₀.
- At b = 0.98975 ℓ₀, the same fixed-parameter model gives Y₁ = 0.51041 ħω₀.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.135986 ħω₀; the input change is Δb = 0.11025 ℓ₀.
- Average sensitivity is ΔY_out/Δb = -1.23343 ħω₀ per ℓ₀. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 0.646395 ħω₀, Y₁ = 0.51041 ħω₀; ΔY_out = -0.135986 ħω₀; average sensitivity = -1.23343 ħω₀ per ℓ₀.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
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.