University Quantum Mechanics I · topic UQMI-01.08
Normalising psi and reading probabilities off it
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.
Normalising ψ and reading probabilities off it: predict a finite change
Normalising ψ means choosing the constant that multiplies it so that ∫|ψ|² dx over all x equals 1; after that, the integral of |ψ|² over any interval is the probability of finding the particle there. Use the supplied topic model Y_out = a*a*s*sqrt(pi)*(1-(0.3480242+(-0.0958798+0.7478556/(1+0.47047*w/s))/(1+0.47047*w/s))/(1+0.47047*w/s)*exp(-(w/s)^2)), where Y_out is ∫|ψ|² dx OVER |x| ≤ w (dimensionless). 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. w is Window half-width w, measured in nm. Keep s (Width σ of ψ) = 1 nm; a (Amplitude A) = 1 nm^(-1/2) fixed. For this exercise use only 0.4975 nm ≤ w ≤ 2.7025 nm 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 w = 0.9826 nm and 1.2913 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δw across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
∫|ψ|² dx OVER |x| ≤ w against w
Predictions from the supplied Normalising ψ and reading probabilities off it 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): ∫|ψ|² dx OVER |x| ≤ w
Horizontal axis (x): w: Window half-width w / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Window half-width w / nm | ∫|ψ|² dx OVER |x| ≤ w |
|---|---|
| 0.4975 | 0.9186408406526156 |
| 1.0487499999999998 | 1.5278187583409517 |
| 1.5999999999999999 | 1.7304938680680653 |
| 2.1512499999999997 | 1.768276595171887 |
| 2.7024999999999997 | 1.7722170697491633 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of w separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At w = 0.9826 nm, substitute into Y_out = a*a*s*sqrt(pi)*(1-(0.3480242+(-0.0958798+0.7478556/(1+0.47047*w/s))/(1+0.47047*w/s))/(1+0.47047*w/s)*exp(-(w/s)^2)): Y₀ = 1.48065.
- At w = 1.2913 nm, the same fixed-parameter model gives Y₁ = 1.65222.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.17157; the input change is Δw = 0.3087 nm.
- Average sensitivity is ΔY_out/Δw = 0.555782 1 per nm. 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 = 1.48065, Y₁ = 1.65222; ΔY_out = 0.17157; average sensitivity = 0.555782 1 per nm.
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.
Normalising ψ and reading probabilities off it: infer an input with uncertainty
Normalising ψ means choosing the constant that multiplies it so that ∫|ψ|² dx over all x equals 1; after that, the integral of |ψ|² over any interval is the probability of finding the particle there. Use the supplied topic model Y_out = a*a*s*sqrt(pi)*(1-(0.3480242+(-0.0958798+0.7478556/(1+0.47047*w/s))/(1+0.47047*w/s))/(1+0.47047*w/s)*exp(-(w/s)^2)), where Y_out is ∫|ψ|² dx OVER |x| ≤ w (dimensionless). 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. w is Window half-width w, measured in nm. Keep s (Width σ of ψ) = 1 nm; a (Amplitude A) = 1 nm^(-1/2) fixed. For this exercise use only 0.4975 nm ≤ w ≤ 2.7025 nm 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 = 1.7561734, with a bounded reading uncertainty ±0.0024066. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown w by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0002205 nm, 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.
∫|ψ|² dx OVER |x| ≤ w against w
Predictions from the supplied Normalising ψ and reading probabilities off it 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): ∫|ψ|² dx OVER |x| ≤ w
Horizontal axis (x): w: Window half-width w / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Window half-width w / nm | ∫|ψ|² dx OVER |x| ≤ w |
|---|---|
| 0.4975 | 0.9186408406526156 |
| 1.0487499999999998 | 1.5278187583409517 |
| 1.5999999999999999 | 1.7304938680680653 |
| 2.1512499999999997 | 1.768276595171887 |
| 2.7024999999999997 | 1.7722170697491633 |
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(w) = 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(0.4975) = 0.918641 and Y_out(2.7025) = 1.77222. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives w_line = 0.4975 + (1.75617 − 0.918641)(2.7025 − 0.4975)/(1.77222 − 0.918641) = 2.66106 nm. 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 w = 1.84255 nm; substituting back gives Y_out = 1.7561734.
- The interpolation error in the input is |w_line − w| = 0.818505 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.7537668 and 1.75858. Sort the roots even if the response decreases: 1.80886 nm ≤ w ≤ 1.88102 nm.
- The lower and upper input deviations are 0.0336862 nm and 0.038466 nm. 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 w ≈ 1.84255 nm; compatible interval [1.80886, 1.88102] nm. Endpoint interpolation gives 2.66106 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.