University Quantum Mechanics I · topic UQMI-08.03
The rectangular barrier solved exactly
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 rectangular barrier solved exactly: predict a finite change
The exact transmission of a rectangular barrier is the flux ratio T = |F/A|² found by matching ψ and ψ′ at both of its faces; for E < V₀ it is T = [1 + V₀² sinh²(κa)/(4E(V₀ − E))]⁻¹. Use the supplied topic model Y_out = 1/(1+vb^2*((exp(2*a*sqrt(max(vb-en+1e-9,0)/0.0381))+exp(-2*a*sqrt(max(vb-en+1e-9,0)/0.0381)))/2+cos(2*a*sqrt(max(en-vb,0)/0.0381))-2)/(8*en*(vb-en+1e-9))), where Y_out is TRANSMISSION T = |F/A|² (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; exp(x) means eˣ; max(a,b) selects the larger value; unmarked trigonometric functions use radians. vb is Barrier height V₀, measured in eV. Keep en (Electron energy E) = 0.6 eV; a (Barrier width a) = 0.5 nm fixed. For this exercise use only 0.60625 eV ≤ vb ≤ 1.39375 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 vb = 0.7795 eV and 0.88975 eV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δvb across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
TRANSMISSION T = |F/A|² against vb
Predictions from the supplied The rectangular barrier solved exactly 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): TRANSMISSION T = |F/A|²
Horizontal axis (x): vb: Barrier height V₀ / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| vb: Barrier height V₀ / eV | TRANSMISSION T = |F/A|² |
|---|---|
| 0.60625 | 0.4953741933620302 |
| 0.8031249999999999 | 0.27025098922234125 |
| 0.9999999999999999 | 0.1400582436035499 |
| 1.196875 | 0.07356512525490828 |
| 1.3937499999999998 | 0.04005464452702559 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of vb separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At vb = 0.7795 eV, substitute into Y_out = 1/(1+vb^2*((exp(2*a*sqrt(max(vb-en+1e-9,0)/0.0381))+exp(-2*a*sqrt(max(vb-en+1e-9,0)/0.0381)))/2+cos(2*a*sqrt(max(en-vb,0)/0.0381))-2)/(8*en*(vb-en+1e-9))): Y₀ = 0.291972.
- At vb = 0.88975 eV, the same fixed-parameter model gives Y₁ = 0.202623.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0893493; the input change is Δvb = 0.11025 eV.
- Average sensitivity is ΔY_out/Δvb = -0.810425 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 = 0.291972, Y₁ = 0.202623; ΔY_out = -0.0893493; average sensitivity = -0.810425 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 rectangular barrier solved exactly: infer an input with uncertainty
The exact transmission of a rectangular barrier is the flux ratio T = |F/A|² found by matching ψ and ψ′ at both of its faces; for E < V₀ it is T = [1 + V₀² sinh²(κa)/(4E(V₀ − E))]⁻¹. Use the supplied topic model Y_out = 1/(1+vb^2*((exp(2*a*sqrt(max(vb-en+1e-9,0)/0.0381))+exp(-2*a*sqrt(max(vb-en+1e-9,0)/0.0381)))/2+cos(2*a*sqrt(max(en-vb,0)/0.0381))-2)/(8*en*(vb-en+1e-9))), where Y_out is TRANSMISSION T = |F/A|² (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; exp(x) means eˣ; max(a,b) selects the larger value; unmarked trigonometric functions use radians. vb is Barrier height V₀, measured in eV. Keep en (Electron energy E) = 0.6 eV; a (Barrier width a) = 0.5 nm fixed. For this exercise use only 0.60625 eV ≤ vb ≤ 1.39375 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 = 0.10511967, with a bounded reading uncertainty ±0.0097598. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown vb by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00007875 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.
TRANSMISSION T = |F/A|² against vb
Predictions from the supplied The rectangular barrier solved exactly 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): TRANSMISSION T = |F/A|²
Horizontal axis (x): vb: Barrier height V₀ / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| vb: Barrier height V₀ / eV | TRANSMISSION T = |F/A|² |
|---|---|
| 0.60625 | 0.4953741933620302 |
| 0.8031249999999999 | 0.27025098922234125 |
| 0.9999999999999999 | 0.1400582436035499 |
| 1.196875 | 0.07356512525490828 |
| 1.3937499999999998 | 0.04005464452702559 |
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(vb) = 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(0.60625) = 0.495374 and Y_out(1.39375) = 0.0400546. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives vb_line = 0.60625 + (0.10512 − 0.495374)(1.39375 − 0.60625)/(0.0400546 − 0.495374) = 1.28122 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 vb = 1.08663 eV; substituting back gives Y_out = 0.10511967.
- The interpolation error in the input is |vb_line − vb| = 0.194591 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.09535987 and 0.11487947. Sort the roots even if the response decreases: 1.05967 eV ≤ vb ≤ 1.1164 eV.
- The lower and upper input deviations are 0.0269557 eV and 0.0297789 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 vb ≈ 1.08663 eV; compatible interval [1.05967, 1.1164] eV. Endpoint interpolation gives 1.28122 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.