University Quantum Mechanics I · topic UQMI-14.05
Separating psi = R(r) Y(theta, phi) for a central potential
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.
Separating ψ = R(r)Y(θ, φ) for a central potential: predict a finite change
Separation in a central potential writes ψ = R(r)Y(θ, φ) and splits the Schrödinger equation into a radial equation holding V, E and the mass and an angular equation holding none of them, joined by one dimensionless constant λ = l(l+1). Use the supplied topic model Y_out = 2*en-2*l-3, where Y_out is RADIAL TILT 2E/ħω − 2l − 3 (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. en is Trial energy E, measured in ħω. Keep l (Angular quantum number l) = 1; kap (Squash κ in V = ½mω²(r² + κz²)) = 0 fixed. For this exercise use only 1.03125 ħω ≤ en ≤ 4.96875 ħω 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 en = 1.8975 ħω and 2.44875 ħω. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δen across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
RADIAL TILT 2E/ħω − 2l − 3 against en
Predictions from the supplied Separating ψ = R(r)Y(θ, φ) for a central potential 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): RADIAL TILT 2E/ħω − 2l − 3
Horizontal axis (x): en: Trial energy E / ħω
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: Trial energy E / ħω | RADIAL TILT 2E/ħω − 2l − 3 |
|---|---|
| 1.03125 | -2.9375 |
| 2.015625 | -0.96875 |
| 3 | 1 |
| 3.984375 | 2.96875 |
| 4.96875 | 4.9375 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of en separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At en = 1.8975 ħω, substitute into Y_out = 2*en-2*l-3: Y₀ = -1.205.
- At en = 2.44875 ħω, the same fixed-parameter model gives Y₁ = -0.1025.
- Signed change: ΔY_out = Y₁ − Y₀ = 1.1025; the input change is Δen = 0.55125 ħω.
- Average sensitivity is ΔY_out/Δen = 2 1 per ħω. 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 = -1.205, Y₁ = -0.1025; ΔY_out = 1.1025; average sensitivity = 2 1 per ħω.
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.
Separating ψ = R(r)Y(θ, φ) for a central potential: infer an input with uncertainty
Separation in a central potential writes ψ = R(r)Y(θ, φ) and splits the Schrödinger equation into a radial equation holding V, E and the mass and an angular equation holding none of them, joined by one dimensionless constant λ = l(l+1). Use the supplied topic model Y_out = 2*en-2*l-3, where Y_out is RADIAL TILT 2E/ħω − 2l − 3 (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. en is Trial energy E, measured in ħω. Keep l (Angular quantum number l) = 1; kap (Squash κ in V = ½mω²(r² + κz²)) = 0 fixed. For this exercise use only 1.03125 ħω ≤ en ≤ 4.96875 ħω 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 = 1.86625, with a bounded reading uncertainty ±0.46069. 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 0.00039375 ħω, 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.
RADIAL TILT 2E/ħω − 2l − 3 against en
Predictions from the supplied Separating ψ = R(r)Y(θ, φ) for a central potential 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): RADIAL TILT 2E/ħω − 2l − 3
Horizontal axis (x): en: Trial energy E / ħω
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: Trial energy E / ħω | RADIAL TILT 2E/ħω − 2l − 3 |
|---|---|
| 1.03125 | -2.9375 |
| 2.015625 | -0.96875 |
| 3 | 1 |
| 3.984375 | 2.96875 |
| 4.96875 | 4.9375 |
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(1.03125) = -2.9375 and Y_out(4.96875) = 4.9375. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives en_line = 1.03125 + (1.86625 − -2.9375)(4.96875 − 1.03125)/(4.9375 − -2.9375) = 3.43312 ħω. 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 = 3.43312 ħω; substituting back gives Y_out = 1.86625.
- The interpolation error in the input is |en_line − en| = 4.44089e-16 ħω. This is within the requested numerical tolerance.
- Invert both limits Y_out = 1.40556 and 2.32694. Sort the roots even if the response decreases: 3.20278 ħω ≤ en ≤ 3.66347 ħω.
- The lower and upper input deviations are 0.230345 ħω and 0.230345 ħω. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer en ≈ 3.43312 ħω; compatible interval [3.20278, 3.66347] ħω. Endpoint interpolation gives 3.43312 ħω.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.