University Quantum Mechanics I · topic UQMI-05.02
Plane waves, and why no plane wave is a state
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.
Plane waves, and why no plane wave is a state: predict a finite change
A plane wave is the free-particle solution ψ(x, t) = A e^(i(kx − ωt)) with ω = ħk²/(2m); it has the sharp momentum ħk, but its probability density |A|² is the same everywhere, so it cannot be normalised and is not a physical state. Use the supplied topic model Y_out = 1/(s*sqrt(2*pi)), where Y_out is PEAK DENSITY |ψ(0)|² (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. s is Envelope width σ, measured in nm. Keep k (Wavenumber k) = 1.5 nm⁻¹ fixed. For this exercise use only 1.74375 nm ≤ s ≤ 7.25625 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 s = 2.9565 nm and 3.72825 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δs across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
PEAK DENSITY |ψ(0)|² against s
Predictions from the supplied Plane waves, and why no plane wave is a state 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): PEAK DENSITY |ψ(0)|² / nm⁻¹
Horizontal axis (x): s: Envelope width σ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Envelope width σ / nm | PEAK DENSITY |ψ(0)|² / nm⁻¹ |
|---|---|
| 1.74375 | 0.2287841034560188 |
| 3.1218749999999997 | 0.12778931904750598 |
| 4.5 | 0.08865384008920726 |
| 5.878124999999999 | 0.06786896848934529 |
| 7.25625 | 0.054979125636717696 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of s separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At s = 2.9565 nm, substitute into Y_out = 1/(s*sqrt(2*pi)): Y₀ = 0.134937 nm⁻¹.
- At s = 3.72825 nm, the same fixed-parameter model gives Y₁ = 0.107005 nm⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0279321 nm⁻¹; the input change is Δs = 0.77175 nm.
- Average sensitivity is ΔY_out/Δs = -0.0361932 nm⁻¹ per nm. 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 = 0.134937 nm⁻¹, Y₁ = 0.107005 nm⁻¹; ΔY_out = -0.0279321 nm⁻¹; average sensitivity = -0.0361932 nm⁻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.
Plane waves, and why no plane wave is a state: infer an input with uncertainty
A plane wave is the free-particle solution ψ(x, t) = A e^(i(kx − ωt)) with ω = ħk²/(2m); it has the sharp momentum ħk, but its probability density |A|² is the same everywhere, so it cannot be normalised and is not a physical state. Use the supplied topic model Y_out = 1/(s*sqrt(2*pi)), where Y_out is PEAK DENSITY |ψ(0)|² (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. s is Envelope width σ, measured in nm. Keep k (Wavenumber k) = 1.5 nm⁻¹ fixed. For this exercise use only 1.74375 nm ≤ s ≤ 7.25625 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 = 0.078126319 nm⁻¹, with a bounded reading uncertainty ±0.0034721 nm⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown s by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00055125 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.
PEAK DENSITY |ψ(0)|² against s
Predictions from the supplied Plane waves, and why no plane wave is a state 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): PEAK DENSITY |ψ(0)|² / nm⁻¹
Horizontal axis (x): s: Envelope width σ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Envelope width σ / nm | PEAK DENSITY |ψ(0)|² / nm⁻¹ |
|---|---|
| 1.74375 | 0.2287841034560188 |
| 3.1218749999999997 | 0.12778931904750598 |
| 4.5 | 0.08865384008920726 |
| 5.878124999999999 | 0.06786896848934529 |
| 7.25625 | 0.054979125636717696 |
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(s) = 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(1.74375) = 0.228784 nm⁻¹ and Y_out(7.25625) = 0.0549791 nm⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives s_line = 1.74375 + (0.0781263 − 0.228784)(7.25625 − 1.74375)/(0.0549791 − 0.228784) = 6.5221 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 s = 5.10637 nm; substituting back gives Y_out = 0.078126319 nm⁻¹.
- The interpolation error in the input is |s_line − s| = 1.41573 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.074654219 nm⁻¹ and 0.081598419 nm⁻¹. Sort the roots even if the response decreases: 4.88909 nm ≤ s ≤ 5.34387 nm.
- The lower and upper input deviations are 0.217282 nm and 0.237493 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 s ≈ 5.10637 nm; compatible interval [4.88909, 5.34387] nm. Endpoint interpolation gives 6.5221 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.