University Quantum Mechanics I · unit UQMI-09
The Quantum Harmonic Oscillator
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.
Why every stable minimum is an oscillator: infer an input with uncertainty
The harmonic approximation replaces a potential near a stable minimum x₀ by the parabola with the same curvature, V ≈ V(x₀) + ½V″(x₀)(x − x₀)², so the motion there is an oscillator with ω = √(V″(x₀)/m). Use the supplied topic model Y_out = (depth-(min(n+0.5,2*depth)-2*depth)^2/(4*depth)), where Y_out is MORSE Eₙ (ħω). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; min(a,b) selects the smaller value. depth is Well depth Dₑ, measured in ħω. Keep n (Level n) = 2 fixed. For this exercise use only 2.22187 ħω ≤ depth ≤ 11.2781 ħω 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 = 2.2982879 ħω, with a bounded reading uncertainty ±0.0094754 ħω. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown depth by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000905625 ħω, 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.
MORSE Eₙ against depth
Predictions from the supplied Why every stable minimum is an oscillator 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): MORSE Eₙ / ħω
Horizontal axis (x): depth: Well depth Dₑ / ħω
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| depth: Well depth Dₑ / ħω | MORSE Eₙ / ħω |
|---|---|
| 2.221875 | 1.7967651195499297 |
| 4.4859374999999995 | 2.1516893068617207 |
| 6.749999999999999 | 2.268518518518519 |
| 9.014062499999998 | 2.3266597330559886 |
| 11.278125 | 2.3614574674425057 |
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(depth) = 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(2.22187) = 1.79677 ħω and Y_out(11.2781) = 2.36146 ħω. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives depth_line = 2.22187 + (2.29829 − 1.79677)(11.2781 − 2.22187)/(2.36146 − 1.79677) = 10.265 ħω. 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 depth = 7.74619 ħω; substituting back gives Y_out = 2.2982879 ħω.
- The interpolation error in the input is |depth_line − depth| = 2.51885 ħω. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.2888125 ħω and 2.3077633 ħω. Sort the roots even if the response decreases: 7.39864 ħω ≤ depth ≤ 8.128 ħω.
- The lower and upper input deviations are 0.34755 ħω and 0.381812 ħω. 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 depth ≈ 7.74619 ħω; compatible interval [7.39864, 8.128] ħω. Endpoint interpolation gives 10.265 ħω.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Dimensionless form and the Gaussian that survives at large x: infer an input with uncertainty
Putting the oscillator in dimensionless form means measuring x in units of b = √(ħ/mω) and E in units of ħω/2, so the Schrödinger equation becomes ψ″ = (ξ² − ε)ψ, whose bound states must fall off as e^(−ξ²/2), times at most a power of ξ, far from the centre. Use the supplied topic model Y_out = (c^2*s^2*100^(2*s-2)-c*s*(s-1)*100^(s-2))/(10000-eps), where Y_out is SUPPLIED ÷ DEMANDED AT ξ = 100 (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s is Exponent s in exp(−c ξˢ), dimensionless. Keep c (Coefficient c in exp(−c ξˢ)) = 1; eps (Scaled energy ε = 2E/ħω) = 3 fixed. For this exercise use only 1.2125 ≤ s ≤ 2.7875 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 = 23.299099, with a bounded reading uncertainty ±3.4947. 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 0.0001575, 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.
SUPPLIED ÷ DEMANDED AT ξ = 100 against s
Predictions from the supplied Dimensionless form and the Gaussian that survives at large x 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): SUPPLIED ÷ DEMANDED AT ξ = 100
Horizontal axis (x): s: Exponent s in exp(−c ξˢ)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Exponent s in exp(−c ξˢ) | SUPPLIED ÷ DEMANDED AT ξ = 100 |
|---|---|
| 1.2125 | 0.001040417492380017 |
| 1.6062499999999997 | 0.06865245744238777 |
| 1.9999999999999998 | 4.001000300090018 |
| 2.39375 | 215.41888110513614 |
| 2.7874999999999996 | 10978.912377737515 |
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(1.2125) = 0.00104042 and Y_out(2.7875) = 10978.9. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives s_line = 1.2125 + (23.2991 − 0.00104042)(2.7875 − 1.2125)/(10978.9 − 0.00104042) = 1.21584. 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 = 2.17325; substituting back gives Y_out = 23.299099.
- The interpolation error in the input is |s_line − s| = 0.957408. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 19.804399 and 26.793799. Sort the roots even if the response decreases: 2.15721 ≤ s ≤ 2.18705.
- The lower and upper input deviations are 0.016036 and 0.0137992. 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 ≈ 2.17325; compatible interval [2.15721, 2.18705]. Endpoint interpolation gives 1.21584.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The power series, its recursion, and why it must terminate: infer an input with uncertainty
The series method writes h(ξ) = Σ aₖξᵏ and generates every coefficient from the one two places below it; an energy is allowed only if that chain can be made to stop, because an unstopped chain grows so fast that ψ cannot be normalised. Use the supplied topic model Y_out = sqrt(ep), where Y_out is TURNING POINTS ±√ε (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. ep is Dimensionless energy ε = 2E/ħω, dimensionless. Keep a0 (Even seed a₀) = 1; a1 (Odd seed a₁) = 0 fixed. For this exercise use only 1.24375 ≤ ep ≤ 6.75625 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 = 2.1462467, with a bounded reading uncertainty ±0.067955. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown ep 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, 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.
TURNING POINTS ±√ε against ep
Predictions from the supplied The power series, its recursion, and why it must terminate 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): TURNING POINTS ±√ε
Horizontal axis (x): ep: Dimensionless energy ε = 2E/ħω
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| ep: Dimensionless energy ε = 2E/ħω | TURNING POINTS ±√ε |
|---|---|
| 1.24375 | 1.1152354011597732 |
| 2.6218749999999997 | 1.6192204914711275 |
| 3.9999999999999996 | 1.9999999999999998 |
| 5.378124999999999 | 2.319078480776362 |
| 6.75625 | 2.599278746114006 |
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(ep) = 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(1.24375) = 1.11524 and Y_out(6.75625) = 2.59928. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives ep_line = 1.24375 + (2.14625 − 1.11524)(6.75625 − 1.24375)/(2.59928 − 1.11524) = 5.07346. 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 ep = 4.60637; substituting back gives Y_out = 2.1462467.
- The interpolation error in the input is |ep_line − ep| = 0.467081. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.0782917 and 2.2142017. Sort the roots even if the response decreases: 4.3193 ≤ ep ≤ 4.90269.
- The lower and upper input deviations are 0.287079 and 0.296314. 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 ep ≈ 4.60637; compatible interval [4.3193, 4.90269]. Endpoint interpolation gives 5.07346.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Eₙ = (n + ½)ħω and the energy that will not go away: infer an input with uncertainty
The zero-point energy of a harmonic oscillator is its lowest allowed energy, E₀ = ½ħω: the least ⟨H⟩ any state can have, because position and momentum cannot both be sharp. Every higher level sits a whole number of steps ħω above it. Use the supplied topic model Y_out = 1674340/(mu*k*sig^4), where Y_out is KINETIC ÷ POTENTIAL (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. sig is Position spread σ, measured in pm. Keep mu (Reduced mass μ) = 0.5 u; k (Spring constant k) = 575 N m⁻¹ fixed. For this exercise use only 5.3375 pm ≤ sig ≤ 22.6625 pm 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 = 0.090989064, with a bounded reading uncertainty ±0.010337. 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 0.0017325 pm, 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.
KINETIC ÷ POTENTIAL against sig
Predictions from the supplied Eₙ = (n + ½)ħω and the energy that will not go away 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): KINETIC ÷ POTENTIAL
Horizontal axis (x): sig: Position spread σ / pm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| sig: Position spread σ / pm | KINETIC ÷ POTENTIAL |
|---|---|
| 5.3375 | 7.175534411306192 |
| 9.66875 | 0.6663838748315171 |
| 13.999999999999998 | 0.15159806602321504 |
| 18.331249999999997 | 0.05157479166595816 |
| 22.662499999999994 | 0.02207875270848293 |
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(5.3375) = 7.17553 and Y_out(22.6625) = 0.0220788. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives sig_line = 5.3375 + (0.0909891 − 7.17553)(22.6625 − 5.3375)/(0.0220788 − 7.17553) = 22.4956 pm. 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 = 15.9057 pm; substituting back gives Y_out = 0.090989064.
- The interpolation error in the input is |sig_line − sig| = 6.58986 pm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.080652064 and 0.10132606. Sort the roots even if the response decreases: 15.4836 pm ≤ sig ≤ 16.3926 pm.
- The lower and upper input deviations are 0.422178 pm and 0.48684 pm. 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 ≈ 15.9057 pm; compatible interval [15.4836, 16.3926] pm. Endpoint interpolation gives 22.4956 pm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.