University Physics V · topic UPV-06.03
Transfer matrices for piecewise-constant potentials
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.
Transfer matrices for piecewise-constant potentials: predict a finite change
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. Calculate Y_out at v = 2.3975 eV and 2.94875 eV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δv across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
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
Substitute each value of v separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At v = 2.3975 eV, substitute into Y_out = 2*5.1231*sqrt(v)*w/ln(10): Y₀ = 3.44506 of 15.65.
- At v = 2.94875 eV, the same fixed-parameter model gives Y₁ = 3.82064 of 15.65.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.375583 of 15.65; the input change is Δv = 0.55125 eV.
- Average sensitivity is ΔY_out/Δv = 0.681329 of 15.65 per eV. Its positive sign means that this response increases 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 = 3.44506 of 15.65, Y₁ = 3.82064 of 15.65; ΔY_out = 0.375583 of 15.65; average sensitivity = 0.681329 of 15.65 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.
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.