University Quantum Mechanics I · topic UQMI-07.09
Solving the well numerically by shooting
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.
Solving the well numerically by shooting: predict a finite change
The shooting method finds a bound-state energy by integrating the Schrödinger equation outward from fixed starting values at a trial energy, and adjusting that energy until the diverging far tail changes sign; the eigenvalue lies where it changes. Use the supplied topic model Y_out = 0.0976*max(ln(abs(cos((2.5616*sqrt(eps*v0)-p*pi/2)-asin(sqrt(eps))))/max(abs(cos((2.5616*sqrt(eps*v0)-p*pi/2)+asin(sqrt(eps)))),1e-12))/sqrt((1-eps)*v0),0), where Y_out is TAIL TURNS AT x − a (nm). 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; abs(x) means |x|; max(a,b) selects the larger value; unmarked trigonometric functions use radians. v0 is Well depth V₀, measured in eV. Keep eps (Trial energy E/V₀) = 0.05; p (Parity of the shot (0 even, 1 odd)) = 0 fixed. For this exercise use only 1.95625 eV ≤ v0 ≤ 5.26375 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 v0 = 2.6839 eV and 3.14695 eV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δv0 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
TAIL TURNS AT x − a against v0
Predictions from the supplied Solving the well numerically by shooting 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): TAIL TURNS AT x − a / nm
Horizontal axis (x): v0: Well depth V₀ / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v0: Well depth V₀ / eV | TAIL TURNS AT x − a / nm |
|---|---|
| 1.9562499999999998 | 0.03455598130889587 |
| 2.783125 | 0.04043185695448696 |
| 3.61 | 0.04951432628002489 |
| 4.436875 | 0.06615298034649969 |
| 5.26375 | 0.11788801536512739 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of v0 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At v0 = 2.6839 eV, substitute into Y_out = 0.0976*max(ln(abs(cos((2.5616*sqrt(eps*v0)-p*pi/2)-asin(sqrt(eps))))/max(abs(cos((2.5616*sqrt(eps*v0)-p*pi/2)+asin(sqrt(eps)))),1e-12))/sqrt((1-eps)*v0),0): Y₀ = 0.0395995 nm.
- At v0 = 3.14695 eV, the same fixed-parameter model gives Y₁ = 0.0438878 nm.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.00428823 nm; the input change is Δv0 = 0.46305 eV.
- Average sensitivity is ΔY_out/Δv0 = 0.00926083 nm 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 = 0.0395995 nm, Y₁ = 0.0438878 nm; ΔY_out = 0.00428823 nm; average sensitivity = 0.00926083 nm 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.
Solving the well numerically by shooting: infer an input with uncertainty
The shooting method finds a bound-state energy by integrating the Schrödinger equation outward from fixed starting values at a trial energy, and adjusting that energy until the diverging far tail changes sign; the eigenvalue lies where it changes. Use the supplied topic model Y_out = 0.0976*max(ln(abs(cos((2.5616*sqrt(eps*v0)-p*pi/2)-asin(sqrt(eps))))/max(abs(cos((2.5616*sqrt(eps*v0)-p*pi/2)+asin(sqrt(eps)))),1e-12))/sqrt((1-eps)*v0),0), where Y_out is TAIL TURNS AT x − a (nm). 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; abs(x) means |x|; max(a,b) selects the larger value; unmarked trigonometric functions use radians. v0 is Well depth V₀, measured in eV. Keep eps (Trial energy E/V₀) = 0.05; p (Parity of the shot (0 even, 1 odd)) = 0 fixed. For this exercise use only 1.95625 eV ≤ v0 ≤ 5.26375 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 = 0.055399704 nm, with a bounded reading uncertainty ±0.0031266 nm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown v0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00033075 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.
TAIL TURNS AT x − a against v0
Predictions from the supplied Solving the well numerically by shooting 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): TAIL TURNS AT x − a / nm
Horizontal axis (x): v0: Well depth V₀ / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v0: Well depth V₀ / eV | TAIL TURNS AT x − a / nm |
|---|---|
| 1.9562499999999998 | 0.03455598130889587 |
| 2.783125 | 0.04043185695448696 |
| 3.61 | 0.04951432628002489 |
| 4.436875 | 0.06615298034649969 |
| 5.26375 | 0.11788801536512739 |
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(v0) = 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(1.95625) = 0.034556 nm and Y_out(5.26375) = 0.117888 nm. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives v0_line = 1.95625 + (0.0553997 − 0.034556)(5.26375 − 1.95625)/(0.117888 − 0.034556) = 2.78355 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 v0 = 3.97382 eV; substituting back gives Y_out = 0.055399704 nm.
- The interpolation error in the input is |v0_line − v0| = 1.19027 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.052273104 nm and 0.058526304 nm. Sort the roots even if the response decreases: 3.79316 eV ≤ v0 ≤ 4.13076 eV.
- The lower and upper input deviations are 0.180669 eV and 0.156932 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 v0 ≈ 3.97382 eV; compatible interval [3.79316, 4.13076] eV. Endpoint interpolation gives 2.78355 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.