University Quantum Mechanics I · topic UQMI-01.01
Where the classical account fails
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.
Where the classical account fails: predict a finite change
The classical account fails wherever a motion's characteristic energy hν is not small beside k_BT, or its action is not large beside Planck's constant h: there its predictions miss the measurements by factors, not percent. Use the supplied topic model Y_out = 0.531977*r0^1.5*(1-sqrt(max(1-t/(1.05016e-4*r0^3),0))), where Y_out is ORBITS SO FAR (thousand). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. r0 is Starting orbit radius r₀, measured in pm. Keep t (Time since start t) = 8 ps fixed. For this exercise use only 42.65 pm ≤ r0 ≤ 65.75 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 r0 = 47.732 pm and 50.966 pm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δr0 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
ORBITS SO FAR against r0
Predictions from the supplied Where the classical account fails 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): ORBITS SO FAR / thousand
Horizontal axis (x): r0: Starting orbit radius r₀ / pm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r0: Starting orbit radius r₀ / pm | ORBITS SO FAR / thousand |
|---|---|
| 42.65 | 128.25169035237255 |
| 48.425 | 76.4181735058303 |
| 54.2 | 58.97240809055958 |
| 59.975 | 48.35776019659553 |
| 65.75 | 40.96448004401035 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of r0 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At r0 = 47.732 pm, substitute into Y_out = 0.531977*r0^1.5*(1-sqrt(max(1-t/(1.05016e-4*r0^3),0))): Y₀ = 79.4231 thousand.
- At r0 = 50.966 pm, the same fixed-parameter model gives Y₁ = 67.4379 thousand.
- Signed change: ΔY_out = Y₁ − Y₀ = -11.9852 thousand; the input change is Δr0 = 3.234 pm.
- Average sensitivity is ΔY_out/Δr0 = -3.70599 thousand 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 = 79.4231 thousand, Y₁ = 67.4379 thousand; ΔY_out = -11.9852 thousand; average sensitivity = -3.70599 thousand 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.
Where the classical account fails: infer an input with uncertainty
The classical account fails wherever a motion's characteristic energy hν is not small beside k_BT, or its action is not large beside Planck's constant h: there its predictions miss the measurements by factors, not percent. Use the supplied topic model Y_out = 0.531977*r0^1.5*(1-sqrt(max(1-t/(1.05016e-4*r0^3),0))), where Y_out is ORBITS SO FAR (thousand). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. r0 is Starting orbit radius r₀, measured in pm. Keep t (Time since start t) = 8 ps fixed. For this exercise use only 42.65 pm ≤ r0 ≤ 65.75 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 = 53.76463 thousand, with a bounded reading uncertainty ±1.92 thousand. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00231 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.
ORBITS SO FAR against r0
Predictions from the supplied Where the classical account fails 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): ORBITS SO FAR / thousand
Horizontal axis (x): r0: Starting orbit radius r₀ / pm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r0: Starting orbit radius r₀ / pm | ORBITS SO FAR / thousand |
|---|---|
| 42.65 | 128.25169035237255 |
| 48.425 | 76.4181735058303 |
| 54.2 | 58.97240809055958 |
| 59.975 | 48.35776019659553 |
| 65.75 | 40.96448004401035 |
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(r0) = 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(42.65) = 128.252 thousand and Y_out(65.75) = 40.9645 thousand. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r0_line = 42.65 + (53.7646 − 128.252)(65.75 − 42.65)/(40.9645 − 128.252) = 62.3625 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 r0 = 56.741 pm; substituting back gives Y_out = 53.76463 thousand.
- The interpolation error in the input is |r0_line − r0| = 5.62152 pm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 51.84463 thousand and 55.68463 thousand. Sort the roots even if the response decreases: 55.7463 pm ≤ r0 ≤ 57.8115 pm.
- The lower and upper input deviations are 0.994737 pm and 1.07051 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 r0 ≈ 56.741 pm; compatible interval [55.7463, 57.8115] pm. Endpoint interpolation gives 62.3625 pm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.