University Quantum Mechanics I · topic UQMI-08.02
The potential step, above V0 and below it
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 potential step, above V₀ and below it: predict a finite change
A potential step is a potential that jumps from 0 to V₀ at one point; a particle sent at it is reflected with probability R = |(k₁ − k₂)/(k₁ + k₂)|², below 1 when E > V₀ and exactly 1 when E < V₀. Use the supplied topic model Y_out = ((sqrt(en)-sqrt(max(en-v0,0)))/(sqrt(en)+sqrt(max(en-v0,0))))^2, where Y_out is REFLECTED R (dimensionless). 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. en is Particle energy E, measured in eV. Keep v0 (Step height V₀) = 1 eV; m (Particle mass m) = 1 mₑ fixed. For this exercise use only 1.012 eV ≤ en ≤ 2.7025 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 en = 1.38391 eV and 1.62058 eV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δen across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
REFLECTED R against en
Predictions from the supplied The potential step, above V₀ and below it 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): REFLECTED R
Horizontal axis (x): en: Particle energy E / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: Particle energy E / eV | REFLECTED R |
|---|---|
| 1.012 | 0.6457735433762928 |
| 1.434625 | 0.08408806445568348 |
| 1.8572499999999998 | 0.036446343512241706 |
| 2.2798749999999997 | 0.020547917480208264 |
| 2.7024999999999997 | 0.013227010567968522 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of en separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At en = 1.38391 eV, substitute into Y_out = ((sqrt(en)-sqrt(max(en-v0,0)))/(sqrt(en)+sqrt(max(en-v0,0))))^2: Y₀ = 0.096111.
- At en = 1.62058 eV, the same fixed-parameter model gives Y₁ = 0.0554454.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0406655; the input change is Δen = 0.23667 eV.
- Average sensitivity is ΔY_out/Δen = -0.171824 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 = 0.096111, Y₁ = 0.0554454; ΔY_out = -0.0406655; average sensitivity = -0.171824 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 potential step, above V₀ and below it: infer an input with uncertainty
A potential step is a potential that jumps from 0 to V₀ at one point; a particle sent at it is reflected with probability R = |(k₁ − k₂)/(k₁ + k₂)|², below 1 when E > V₀ and exactly 1 when E < V₀. Use the supplied topic model Y_out = ((sqrt(en)-sqrt(max(en-v0,0)))/(sqrt(en)+sqrt(max(en-v0,0))))^2, where Y_out is REFLECTED R (dimensionless). 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. en is Particle energy E, measured in eV. Keep v0 (Step height V₀) = 1 eV; m (Particle mass m) = 1 mₑ fixed. For this exercise use only 1.012 eV ≤ en ≤ 2.7025 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 = 0.027719269, with a bounded reading uncertainty ±0.0021738. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown en by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00016905 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.
REFLECTED R against en
Predictions from the supplied The potential step, above V₀ and below it 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): REFLECTED R
Horizontal axis (x): en: Particle energy E / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: Particle energy E / eV | REFLECTED R |
|---|---|
| 1.012 | 0.6457735433762928 |
| 1.434625 | 0.08408806445568348 |
| 1.8572499999999998 | 0.036446343512241706 |
| 2.2798749999999997 | 0.020547917480208264 |
| 2.7024999999999997 | 0.013227010567968522 |
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(en) = 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(1.012) = 0.645774 and Y_out(2.7025) = 0.013227. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives en_line = 1.012 + (0.0277193 − 0.645774)(2.7025 − 1.012)/(0.013227 − 0.645774) = 2.66377 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 en = 2.0432 eV; substituting back gives Y_out = 0.027719269.
- The interpolation error in the input is |en_line − en| = 0.620564 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.025545469 and 0.029893069. Sort the roots even if the response decreases: 1.98918 eV ≤ en ≤ 2.10412 eV.
- The lower and upper input deviations are 0.054026 eV and 0.0609192 eV. 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 en ≈ 2.0432 eV; compatible interval [1.98918, 2.10412] eV. Endpoint interpolation gives 2.66377 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.