University Physics V · topic UPV-11.04
Radial wavefunctions: the Laguerre solution and its matrix route
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.
Radial wavefunctions and the Laguerre solution: predict a finite change
The radial wavefunction u_nl = r R_nl is the half-line eigenfunction of the radial Hamiltonian at fixed l: an r^(l+1) rise at the origin, an e^(−Zr/na₀) tail, and between them an associated Laguerre polynomial of degree n − l − 1. Use the supplied topic model Y_out = abs((2/(n*sqrt(n)))*rb*(1+(1-n)*rb/n+(1-n)*(2-n)*rb^2/(3*n^2)+(1-n)*(2-n)*(3-n)*rb^3/(18*n^3))*exp(-rb/n)), where Y_out is WALL AMPLITUDE |u(R)| (a_0^(-1/2)). 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; abs(x) means |x|. rb is Grid box radius R, measured in a₀. Keep n (Principal quantum number n) = 3 fixed. For this exercise use only 14.5725 a₀ ≤ rb ≤ 54.2625 a₀ 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 rb = 23.3043 a₀ and 28.8609 a₀. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δrb across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
WALL AMPLITUDE |u(R)| against rb
Predictions from the supplied Radial wavefunctions and the Laguerre solution 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): WALL AMPLITUDE |u(R)| / a_0^(-1/2)
Horizontal axis (x): rb: Grid box radius R / a₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rb: Grid box radius R / a₀ | WALL AMPLITUDE |u(R)| / a_0^(-1/2) |
|---|---|
| 14.5725 | 0.30572928930086346 |
| 24.494999999999997 | 0.07807736936718887 |
| 34.4175 | 0.009076391329130584 |
| 44.339999999999996 | 0.0007615972842707709 |
| 54.262499999999996 | 0.000053312819680487595 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of rb separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At rb = 23.3043 a₀, substitute into Y_out = abs((2/(n*sqrt(n)))*rb*(1+(1-n)*rb/n+(1-n)*(2-n)*rb^2/(3*n^2)+(1-n)*(2-n)*(3-n)*rb^3/(18*n^3))*exp(-rb/n)): Y₀ = 0.097488 a_0^(-1/2).
- At rb = 28.8609 a₀, the same fixed-parameter model gives Y₁ = 0.0320405 a_0^(-1/2).
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0654475 a_0^(-1/2); the input change is Δrb = 5.5566 a₀.
- Average sensitivity is ΔY_out/Δrb = -0.0117783 a_0^(-1/2) per a₀. 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 = 0.097488 a_0^(-1/2), Y₁ = 0.0320405 a_0^(-1/2); ΔY_out = -0.0654475 a_0^(-1/2); average sensitivity = -0.0117783 a_0^(-1/2) per a₀.
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.
Radial wavefunctions and the Laguerre solution: infer an input with uncertainty
The radial wavefunction u_nl = r R_nl is the half-line eigenfunction of the radial Hamiltonian at fixed l: an r^(l+1) rise at the origin, an e^(−Zr/na₀) tail, and between them an associated Laguerre polynomial of degree n − l − 1. Use the supplied topic model Y_out = abs((2/(n*sqrt(n)))*rb*(1+(1-n)*rb/n+(1-n)*(2-n)*rb^2/(3*n^2)+(1-n)*(2-n)*(3-n)*rb^3/(18*n^3))*exp(-rb/n)), where Y_out is WALL AMPLITUDE |u(R)| (a_0^(-1/2)). 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; abs(x) means |x|. rb is Grid box radius R, measured in a₀. Keep n (Principal quantum number n) = 3 fixed. For this exercise use only 14.5725 a₀ ≤ rb ≤ 54.2625 a₀ 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 = 0.0031394565 a_0^(-1/2), with a bounded reading uncertainty ±0.00046292 a_0^(-1/2). Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rb by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.003969 a₀, 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.
WALL AMPLITUDE |u(R)| against rb
Predictions from the supplied Radial wavefunctions and the Laguerre solution 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): WALL AMPLITUDE |u(R)| / a_0^(-1/2)
Horizontal axis (x): rb: Grid box radius R / a₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rb: Grid box radius R / a₀ | WALL AMPLITUDE |u(R)| / a_0^(-1/2) |
|---|---|
| 14.5725 | 0.30572928930086346 |
| 24.494999999999997 | 0.07807736936718887 |
| 34.4175 | 0.009076391329130584 |
| 44.339999999999996 | 0.0007615972842707709 |
| 54.262499999999996 | 0.000053312819680487595 |
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(rb) = 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(14.5725) = 0.305729 a_0^(-1/2) and Y_out(54.2625) = 0.0000533128 a_0^(-1/2). The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives rb_line = 14.5725 + (0.00313946 − 0.305729)(54.2625 − 14.5725)/(0.0000533128 − 0.305729) = 53.8618 a₀. 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 rb = 38.7834 a₀; substituting back gives Y_out = 0.0031394565 a_0^(-1/2).
- The interpolation error in the input is |rb_line − rb| = 15.0784 a₀. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.0026765365 a_0^(-1/2) and 0.0036023765 a_0^(-1/2). Sort the roots even if the response decreases: 38.2292 a₀ ≤ rb ≤ 39.4224 a₀.
- The lower and upper input deviations are 0.554158 a₀ and 0.639047 a₀. 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 rb ≈ 38.7834 a₀; compatible interval [38.2292, 39.4224] a₀. Endpoint interpolation gives 53.8618 a₀.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.