University Physics IV · topic UPIV-08.04
The evenly spaced ladder and zero-point energy
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.
The ladder and zero-point energy: predict a finite change
The harmonic ladder is the spectrum Eₙ = (n + ½)ħω for n = 0, 1, 2, …: infinitely many levels separated by the same ħω, the lowest of them — the zero-point energy — sitting half a quantum above the bottom of the well rather than on it. Use the supplied topic model Y_out = (w-2*a*(n+1))/max(w,0.01), where Y_out is GAP AS A FRACTION OF ħω (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. a is Anharmonicity a, measured in eV. Keep w (Vibrational quantum ħω) = 0.27 eV; n (Marked level n) = 0 fixed. For this exercise use only 0.0017 eV ≤ a ≤ 0.0143 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. Calculate Y_out at a = 0.004472 eV and 0.006236 eV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δa across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
GAP AS A FRACTION OF ħω against a
Predictions from the supplied The ladder and zero-point energy 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): GAP AS A FRACTION OF ħω
Horizontal axis (x): a: Anharmonicity a / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Anharmonicity a / eV | GAP AS A FRACTION OF ħω |
|---|---|
| 0.0017 | 0.9874074074074074 |
| 0.004849999999999999 | 0.9640740740740741 |
| 0.007999999999999998 | 0.9407407407407407 |
| 0.011149999999999998 | 0.9174074074074074 |
| 0.014299999999999998 | 0.8940740740740741 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of a separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At a = 0.004472 eV, substitute into Y_out = (w-2*a*(n+1))/max(w,0.01): Y₀ = 0.966874.
- At a = 0.006236 eV, the same fixed-parameter model gives Y₁ = 0.953807.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0130667; the input change is Δa = 0.001764 eV.
- Average sensitivity is ΔY_out/Δa = -7.40741 1 per eV. 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 = 0.966874, Y₁ = 0.953807; ΔY_out = -0.0130667; average sensitivity = -7.40741 1 per eV.
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.
The ladder and zero-point energy: infer an input with uncertainty
The harmonic ladder is the spectrum Eₙ = (n + ½)ħω for n = 0, 1, 2, …: infinitely many levels separated by the same ħω, the lowest of them — the zero-point energy — sitting half a quantum above the bottom of the well rather than on it. Use the supplied topic model Y_out = (w-2*a*(n+1))/max(w,0.01), where Y_out is GAP AS A FRACTION OF ħω (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. a is Anharmonicity a, measured in eV. Keep w (Vibrational quantum ħω) = 0.27 eV; n (Marked level n) = 0 fixed. For this exercise use only 0.0017 eV ≤ a ≤ 0.0143 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 = 0.93047407, with a bounded reading uncertainty ±0.00546. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00000126 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.
GAP AS A FRACTION OF ħω against a
Predictions from the supplied The ladder and zero-point energy 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): GAP AS A FRACTION OF ħω
Horizontal axis (x): a: Anharmonicity a / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Anharmonicity a / eV | GAP AS A FRACTION OF ħω |
|---|---|
| 0.0017 | 0.9874074074074074 |
| 0.004849999999999999 | 0.9640740740740741 |
| 0.007999999999999998 | 0.9407407407407407 |
| 0.011149999999999998 | 0.9174074074074074 |
| 0.014299999999999998 | 0.8940740740740741 |
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(a) = 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(0.0017) = 0.987407 and Y_out(0.0143) = 0.894074. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives a_line = 0.0017 + (0.930474 − 0.987407)(0.0143 − 0.0017)/(0.894074 − 0.987407) = 0.009386 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 a = 0.009386 eV; substituting back gives Y_out = 0.93047407.
- The interpolation error in the input is |a_line − a| = 1.21431e-17 eV. This is within the requested numerical tolerance.
- Invert both limits Y_out = 0.92501407 and 0.93593407. Sort the roots even if the response decreases: 0.0086489 eV ≤ a ≤ 0.0101231 eV.
- The lower and upper input deviations are 0.0007371 eV and 0.0007371 eV. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer a ≈ 0.009386 eV; compatible interval [0.0086489, 0.0101231] eV. Endpoint interpolation gives 0.009386 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.