University Quantum Mechanics I · topic UQMI-02.06
Probability density and probability current
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.
Probability density and probability current: predict a finite change
The probability density ρ = |Ψ|² is the probability per unit length of finding the particle near x, and the probability current j = (ħ/m) Im(Ψ* ∂Ψ/∂x) is the probability per unit time crossing x, counted positive towards +x. Use the supplied topic model Y_out = pi/k, where Y_out is FRINGE SPACING π/k (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. k is Wave number k, measured in nm⁻¹. Keep r (Reflected amplitude r = |B|/|A|) = 0.5; phi (Phase φ of B relative to A) = 0 ° fixed. For this exercise use only 2.85 nm⁻¹ ≤ k ≤ 9.15 nm⁻¹ 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 k = 4.236 nm⁻¹ and 5.118 nm⁻¹. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δk across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
FRINGE SPACING π/k against k
Predictions from the supplied Probability density and probability current 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): FRINGE SPACING π/k / nm
Horizontal axis (x): k: Wave number k / nm⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: Wave number k / nm⁻¹ | FRINGE SPACING π/k / nm |
|---|---|
| 2.85 | 1.1023132117858923 |
| 4.425 | 0.7099644414892188 |
| 6 | 0.5235987755982988 |
| 7.574999999999999 | 0.41473170344419713 |
| 9.149999999999999 | 0.34334345940872063 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of k separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At k = 4.236 nm⁻¹, substitute into Y_out = pi/k: Y₀ = 0.741641 nm.
- At k = 5.118 nm⁻¹, the same fixed-parameter model gives Y₁ = 0.613832 nm.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.127809 nm; the input change is Δk = 0.882 nm⁻¹.
- Average sensitivity is ΔY_out/Δk = -0.144908 nm per nm⁻¹. 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 = 0.741641 nm, Y₁ = 0.613832 nm; ΔY_out = -0.127809 nm; average sensitivity = -0.144908 nm per nm⁻¹.
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.
Probability density and probability current: infer an input with uncertainty
The probability density ρ = |Ψ|² is the probability per unit length of finding the particle near x, and the probability current j = (ħ/m) Im(Ψ* ∂Ψ/∂x) is the probability per unit time crossing x, counted positive towards +x. Use the supplied topic model Y_out = pi/k, where Y_out is FRINGE SPACING π/k (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. k is Wave number k, measured in nm⁻¹. Keep r (Reflected amplitude r = |B|/|A|) = 0.5; phi (Phase φ of B relative to A) = 0 ° fixed. For this exercise use only 2.85 nm⁻¹ ≤ k ≤ 9.15 nm⁻¹ 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 = 0.46938483 nm, with a bounded reading uncertainty ±0.018906 nm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown k by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00063 nm⁻¹, 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.
FRINGE SPACING π/k against k
Predictions from the supplied Probability density and probability current 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): FRINGE SPACING π/k / nm
Horizontal axis (x): k: Wave number k / nm⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: Wave number k / nm⁻¹ | FRINGE SPACING π/k / nm |
|---|---|
| 2.85 | 1.1023132117858923 |
| 4.425 | 0.7099644414892188 |
| 6 | 0.5235987755982988 |
| 7.574999999999999 | 0.41473170344419713 |
| 9.149999999999999 | 0.34334345940872063 |
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(k) = 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(2.85) = 1.10231 nm and Y_out(9.15) = 0.343343 nm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives k_line = 2.85 + (0.469385 − 1.10231)(9.15 − 2.85)/(0.343343 − 1.10231) = 8.10377 nm⁻¹. 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 k = 6.693 nm⁻¹; substituting back gives Y_out = 0.46938483 nm.
- The interpolation error in the input is |k_line − k| = 1.41077 nm⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.45047883 nm and 0.48829083 nm. Sort the roots even if the response decreases: 6.43386 nm⁻¹ ≤ k ≤ 6.9739 nm⁻¹.
- The lower and upper input deviations are 0.259144 nm⁻¹ and 0.280896 nm⁻¹. 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 k ≈ 6.693 nm⁻¹; compatible interval [6.43386, 6.9739] nm⁻¹. Endpoint interpolation gives 8.10377 nm⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.