University Quantum Mechanics I · topic UQMI-09.04
E_n = (n + 1/2) h-bar omega and the energy that will not go away
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.
Eₙ = (n + ½)ħω and the energy that will not go away: predict a finite change
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. Calculate Y_out at sig = 9.149 pm and 11.5745 pm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δsig across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
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
Substitute each value of sig separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At sig = 9.149 pm, substitute into Y_out = 1674340/(mu*k*sig^4): Y₀ = 0.831211.
- At sig = 11.5745 pm, the same fixed-parameter model gives Y₁ = 0.324487.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.506724; the input change is Δsig = 2.4255 pm.
- Average sensitivity is ΔY_out/Δsig = -0.208915 1 per pm. 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 = 0.831211, Y₁ = 0.324487; ΔY_out = -0.506724; average sensitivity = -0.208915 1 per pm.
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.
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.