University Physics V · topic UPV-06.02
The finite square well: parity reduction and transcendental roots
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 finite well, split by parity: predict a finite change
The bound levels of a symmetric finite well are the energies at which the interior logarithmic derivative matches the exterior decay — k tan(ka) = κ for even states, −k cot(ka) = κ for odd — with k and κ tied by k² + κ² = 2mV₀/ℏ². Use the supplied topic model Y_out = clamp(zc^2/max(z0^2,0.01)-1,-1,0), where Y_out is LEVEL AT CURSOR E/V₀ (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; clamp(x,a,b) confines x to [a,b]. z0 is Well depth z₀ = a√(2mV₀)/ℏ, dimensionless. Keep zc (Cursor z = ka) = 1.4 fixed. For this exercise use only 1.4925 ≤ z0 ≤ 4.51125 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 z0 = 2.15662 and 2.57925. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δz0 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
LEVEL AT CURSOR E/V₀ against z0
Predictions from the supplied The finite well, split by parity 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): LEVEL AT CURSOR E/V₀
Horizontal axis (x): z0: Well depth z₀ = a√(2mV₀)/ℏ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| z0: Well depth z₀ = a√(2mV₀)/ℏ | LEVEL AT CURSOR E/V₀ |
|---|---|
| 1.4924999999999997 | -0.12011200615023732 |
| 2.2471875 | -0.6118697870940322 |
| 3.0018749999999996 | -0.7824941894486186 |
| 3.7565624999999994 | -0.8611087668924998 |
| 4.5112499999999995 | -0.9036920183762139 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of z0 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At z0 = 2.15662, substitute into Y_out = clamp(zc^2/max(z0^2,0.01)-1,-1,0): Y₀ = -0.578588.
- At z0 = 2.57925, the same fixed-parameter model gives Y₁ = -0.705375.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.126787; the input change is Δz0 = 0.422625.
- Average sensitivity is ΔY_out/Δz0 = -0.299999 1 per unit input. 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 = -0.578588, Y₁ = -0.705375; ΔY_out = -0.126787; average sensitivity = -0.299999 1 per unit input.
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 finite well, split by parity: infer an input with uncertainty
The bound levels of a symmetric finite well are the energies at which the interior logarithmic derivative matches the exterior decay — k tan(ka) = κ for even states, −k cot(ka) = κ for odd — with k and κ tied by k² + κ² = 2mV₀/ℏ². Use the supplied topic model Y_out = clamp(zc^2/max(z0^2,0.01)-1,-1,0), where Y_out is LEVEL AT CURSOR E/V₀ (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; clamp(x,a,b) confines x to [a,b]. z0 is Well depth z₀ = a√(2mV₀)/ℏ, dimensionless. Keep zc (Cursor z = ka) = 1.4 fixed. For this exercise use only 1.4925 ≤ z0 ≤ 4.51125 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 = -0.82366393, with a bounded reading uncertainty ±0.012004. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown z0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000301875, 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.
LEVEL AT CURSOR E/V₀ against z0
Predictions from the supplied The finite well, split by parity 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): LEVEL AT CURSOR E/V₀
Horizontal axis (x): z0: Well depth z₀ = a√(2mV₀)/ℏ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| z0: Well depth z₀ = a√(2mV₀)/ℏ | LEVEL AT CURSOR E/V₀ |
|---|---|
| 1.4924999999999997 | -0.12011200615023732 |
| 2.2471875 | -0.6118697870940322 |
| 3.0018749999999996 | -0.7824941894486186 |
| 3.7565624999999994 | -0.8611087668924998 |
| 4.5112499999999995 | -0.9036920183762139 |
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(z0) = 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(1.4925) = -0.120112 and Y_out(4.51125) = -0.903692. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives z0_line = 1.4925 + (-0.823664 − -0.120112)(4.51125 − 1.4925)/(-0.903692 − -0.120112) = 4.20294. 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 z0 = 3.33394; substituting back gives Y_out = -0.82366393.
- The interpolation error in the input is |z0_line − z0| = 0.869003. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.83566793 and -0.81165993. Sort the roots even if the response decreases: 3.22594 ≤ z0 ≤ 3.45356.
- The lower and upper input deviations are 0.107995 and 0.119621. 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 z0 ≈ 3.33394; compatible interval [3.22594, 3.45356]. Endpoint interpolation gives 4.20294.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.