University Physics V · unit UPV-06
Finite Potential Wells 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.
Bound states and the continuum: infer an input with uncertainty
A bound state is a square-integrable eigenvector of Ĥ with energy below the potential's value at infinity, normalised by ⟨n|m⟩ = δ_nm; an unbound state has energy above that floor, is not square-integrable, and is normalised by ⟨k|k′⟩ = δ(k − k′). Use the supplied topic model Y_out = 1-exp(-2*13.123*a*w), where Y_out is BOUND WEIGHT IN ±X (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. a is Well strength α, measured in eV nm. Keep k (Continuum wavenumber k) = 8 nm⁻¹; w (Window half-width X) = 0.5 nm fixed. For this exercise use only 0.197813 eV nm ≤ a ≤ 0.552187 eV nm 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 = 0.99562862, with a bounded reading uncertainty ±0.00054879. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000354375 eV nm, 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.
BOUND WEIGHT IN ±X against a
Predictions from the supplied Bound states and the continuum 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): BOUND WEIGHT IN ±X
Horizontal axis (x): a: Well strength α / eV nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Well strength α / eV nm | BOUND WEIGHT IN ±X |
|---|---|
| 0.1978125 | 0.9254207855670302 |
| 0.28640625 | 0.9766815227754335 |
| 0.375 | 0.9927090760581645 |
| 0.46359374999999997 | 0.9977203669256057 |
| 0.5521874999999999 | 0.9992872334157742 |
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(a) = 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(0.197813) = 0.925421 and Y_out(0.552187) = 0.999287. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives a_line = 0.197813 + (0.995629 − 0.925421)(0.552187 − 0.197813)/(0.999287 − 0.925421) = 0.534635 eV nm. 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 a = 0.413981 eV nm; substituting back gives Y_out = 0.99562862.
- The interpolation error in the input is |a_line − a| = 0.120654 eV nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.99507983 and 0.99617741. Sort the roots even if the response decreases: 0.404969 eV nm ≤ a ≤ 0.424204 eV nm.
- The lower and upper input deviations are 0.00901199 eV nm and 0.0102226 eV nm. 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 a ≈ 0.413981 eV nm; compatible interval [0.404969, 0.424204] eV nm. Endpoint interpolation gives 0.534635 eV nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.
Transfer matrices for piecewise-constant potentials: infer an input with uncertainty
A transfer matrix is the 2×2 matrix that carries the pair of plane-wave amplitudes (A, B) from one constant-potential region across to the next, so an entire staircase potential is solved by one ordered product of such matrices. Use the supplied topic model Y_out = 2*5.1231*sqrt(v)*w/ln(10), where Y_out is DIGITS LOST (of 15.65). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; ln(x) means the natural logarithm. v is Barrier height above E, measured in eV. Keep w (Slab width L) = 0.5 nm; n (Slabs the barrier is cut into) = 1 fixed. For this exercise use only 1.53125 eV ≤ v ≤ 5.46875 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 = 4.4125132 of 15.65, with a bounded reading uncertainty ±0.11859 of 15.65. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown v 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 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.
DIGITS LOST against v
Predictions from the supplied Transfer matrices for piecewise-constant potentials 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): DIGITS LOST / of 15.65
Horizontal axis (x): v: Barrier height above E / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: Barrier height above E / eV | DIGITS LOST / of 15.65 |
|---|---|
| 1.53125 | 2.753215432953294 |
| 2.515625 | 3.5289060432155024 |
| 3.5 | 4.162470480788253 |
| 4.484375 | 4.711596666208206 |
| 5.46875 | 5.203088100985315 |
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(v) = 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) = 2.75322 of 15.65 and Y_out(5.46875) = 5.20309 of 15.65. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives v_line = 1.53125 + (4.41251 − 2.75322)(5.46875 − 1.53125)/(5.20309 − 2.75322) = 4.19812 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 v = 3.93312 eV; substituting back gives Y_out = 4.4125132 of 15.65.
- The interpolation error in the input is |v_line − v| = 0.264992 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 4.2939232 of 15.65 and 4.5311032 of 15.65. Sort the roots even if the response decreases: 3.72455 eV ≤ v ≤ 4.14738 eV.
- The lower and upper input deviations are 0.208571 eV and 0.214253 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 v ≈ 3.93312 eV; compatible interval [3.72455, 4.14738] eV. Endpoint interpolation gives 4.19812 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The scattering matrix and R + T = 1: infer an input with uncertainty
The scattering matrix S is the map from the flux-normalised amplitudes of the waves running into a scatterer to those running out of it, and its unitarity is nothing more than the statement that probability current is conserved. Use the supplied topic model Y_out = (1+b^2+w^2)/max((1+w)^2+b^2,0.01), where Y_out is R + T (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. w is Absorption w = mW/(ℏ²k), dimensionless. Keep b (Real strength β = mα/(ℏ²k)) = 1 fixed. For this exercise use only 0.2125 ≤ w ≤ 1.3675 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 = 0.60768592, with a bounded reading uncertainty ±0.0032644. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown w by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001155, 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.
R + T against w
Predictions from the supplied The scattering matrix and R + T = 1 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): R + T
Horizontal axis (x): w: Absorption w = mW/(ℏ²k)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Absorption w = mW/(ℏ²k) | R + T |
|---|---|
| 0.2125 | 0.8279461066481119 |
| 0.50125 | 0.6918941164549959 |
| 0.79 | 0.6241763992293238 |
| 1.0787499999999999 | 0.594546461986235 |
| 1.3675 | 0.5859232841506838 |
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(w) = 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(0.2125) = 0.827946 and Y_out(1.3675) = 0.585923. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 0.2125 + (0.607686 − 0.827946)(1.3675 − 0.2125)/(0.585923 − 0.827946) = 1.26364. 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 w = 0.91705; substituting back gives Y_out = 0.60768592.
- The interpolation error in the input is |w_line − w| = 0.346593. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.60442152 and 0.61095032. Sort the roots even if the response decreases: 0.887688 ≤ w ≤ 0.949488.
- The lower and upper input deviations are 0.0293618 and 0.0324377. 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 w ≈ 0.91705; compatible interval [0.887688, 0.949488]. Endpoint interpolation gives 1.26364.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.