University Quantum Mechanics I · unit UQMI-08
Potential Barriers and Quantum Tunneling
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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Probability current and what R and T count: infer an input with uncertainty
The probability current j = (ħ/m) Im(ψ* ∂ψ/∂x) is the rate at which probability flows past a point, and the reflection and transmission coefficients R and T are the reflected and transmitted currents divided by the incident current. Use the supplied topic model Y_out = (2/(1+r))^2, where Y_out is DENSITY RATIO |C/A|² (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r is Wave-number ratio k₂/k₁, dimensionless. Keep wt (Time phase ωt) = 0 ° fixed. For this exercise use only 0.542188 ≤ r ≤ 2.70781 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 = 0.4879226, with a bounded reading uncertainty ±0.029545. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000216562, 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.
DENSITY RATIO |C/A|² against r
Predictions from the supplied Probability current and what R and T count 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): DENSITY RATIO |C/A|²
Horizontal axis (x): r: Wave-number ratio k₂/k₁
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Wave-number ratio k₂/k₁ | DENSITY RATIO |C/A|² |
|---|---|
| 0.5421875 | 1.6818437047370631 |
| 1.08359375 | 0.9213696431928011 |
| 1.6249999999999998 | 0.5804988662131518 |
| 2.1664062499999996 | 0.39895758208426246 |
| 2.7078124999999993 | 0.29095408753733054 |
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(r) = 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(0.542188) = 1.68184 and Y_out(2.70781) = 0.290954. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r_line = 0.542188 + (0.487923 − 1.68184)(2.70781 − 0.542188)/(0.290954 − 1.68184) = 2.40113. 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 r = 1.86322; substituting back gives Y_out = 0.4879226.
- The interpolation error in the input is |r_line − r| = 0.537912. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.4583776 and 0.5174676. Sort the roots even if the response decreases: 1.78028 ≤ r ≤ 1.95405.
- The lower and upper input deviations are 0.0829395 and 0.0908344. 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 r ≈ 1.86322; compatible interval [1.78028, 1.95405]. Endpoint interpolation gives 2.40113.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.
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.
The thick-barrier limit and T ~ e^(−2κa): infer an input with uncertainty
The thick-barrier limit is the form the exact rectangular-barrier transmission takes when κa ≫ 1, T ≈ [16E(V₀ − E)/V₀²] e^(−2κa), with corrections that are powers of e^(−2κa); dropping the bracket leaves the estimate T ~ e^(−2κa). Use the supplied topic model Y_out = 1-(2-16*u*(1-u))*exp(-2*b*sqrt(1-u))+exp(-4*b*sqrt(1-u)), where Y_out is THICK FORM ÷ EXACT (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. b is Barrier strength κ₀a = √(2mV₀)a/ħ, dimensionless. Keep u (Marker energy E/V₀) = 0.5 fixed. For this exercise use only 1.53125 ≤ b ≤ 5.46875 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 = 1.0076948, with a bounded reading uncertainty ±0.0010229. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00039375, 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.
THICK FORM ÷ EXACT against b
Predictions from the supplied The thick-barrier limit and T ~ e^(−2κa) 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): THICK FORM ÷ EXACT
Horizontal axis (x): b: Barrier strength κ₀a = √(2mV₀)a/ħ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Barrier strength κ₀a = √(2mV₀)a/ħ | THICK FORM ÷ EXACT |
|---|---|
| 1.53125 | 1.2425358224897622 |
| 2.515625 | 1.057825160509003 |
| 3.5 | 1.0142205959102335 |
| 4.484375 | 1.003525134659196 |
| 5.46875 | 1.0008755884236942 |
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(b) = 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(1.53125) = 1.24254 and Y_out(5.46875) = 1.00088. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives b_line = 1.53125 + (1.00769 − 1.24254)(5.46875 − 1.53125)/(1.00088 − 1.24254) = 5.35764. 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 b = 3.93312; substituting back gives Y_out = 1.0076948.
- The interpolation error in the input is |b_line − b| = 1.42452. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.0066719 and 1.0087177. Sort the roots even if the response decreases: 3.84505 ≤ b ≤ 4.0338.
- The lower and upper input deviations are 0.0880747 and 0.100682. 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 b ≈ 3.93312; compatible interval [3.84505, 4.0338]. Endpoint interpolation gives 5.35764.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.