University Physics IV · topic UPIV-06.04
Stationary states and the evolution of superpositions
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.
Stationary states and superpositions: predict a finite change
A stationary state is a single energy eigenfunction carrying the phase e^(−iEt/ħ); its probability density and all its expectation values are frozen, so only a superposition of different energies can make anything move. Use the supplied topic model Y_out = 0.5*(0.5+2*sqrt(p*(1-p))*cos(2*pi*t*(n^2-1)/2.7496)*((cos(pi*(n-1))-1)/(pi*(n-1))^2-(cos(pi*(n+1))-1)/(pi*(n+1))^2)), where Y_out is MEAN POSITION ⟨x⟩ (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; unmarked trigonometric functions use radians. t is Time t since preparation, measured in fs. Keep p (Weight |c₂|² on the upper level) = 0.5; n (Upper level n) = 2 fixed. For this exercise use only 0.10625 fs ≤ t ≤ 0.42125 fs 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 t = 0.17555 fs and 0.21965 fs. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δt across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
MEAN POSITION ⟨x⟩ against t
Predictions from the supplied Stationary states and superpositions 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): MEAN POSITION ⟨x⟩ / nm
Horizontal axis (x): t: Time t since preparation / fs
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Time t since preparation / fs | MEAN POSITION ⟨x⟩ / nm |
|---|---|
| 0.10625 | 0.1827901979436082 |
| 0.185 | 0.223165094958426 |
| 0.26375 | 0.2711729122140766 |
| 0.3425 | 0.3131583064372691 |
| 0.42125 | 0.33717895232858247 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of t separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At t = 0.17555 fs, substitute into Y_out = 0.5*(0.5+2*sqrt(p*(1-p))*cos(2*pi*t*(n^2-1)/2.7496)*((cos(pi*(n-1))-1)/(pi*(n-1))^2-(cos(pi*(n+1))-1)/(pi*(n+1))^2)): Y₀ = 0.217656 nm.
- At t = 0.21965 fs, the same fixed-parameter model gives Y₁ = 0.244149 nm.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.0264933 nm; the input change is Δt = 0.0441 fs.
- Average sensitivity is ΔY_out/Δt = 0.600754 nm per fs. 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 = 0.217656 nm, Y₁ = 0.244149 nm; ΔY_out = 0.0264933 nm; average sensitivity = 0.600754 nm per fs.
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.
Stationary states and superpositions: infer an input with uncertainty
A stationary state is a single energy eigenfunction carrying the phase e^(−iEt/ħ); its probability density and all its expectation values are frozen, so only a superposition of different energies can make anything move. Use the supplied topic model Y_out = 0.5*(0.5+2*sqrt(p*(1-p))*cos(2*pi*t*(n^2-1)/2.7496)*((cos(pi*(n-1))-1)/(pi*(n-1))^2-(cos(pi*(n+1))-1)/(pi*(n+1))^2)), where Y_out is MEAN POSITION ⟨x⟩ (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; unmarked trigonometric functions use radians. t is Time t since preparation, measured in fs. Keep p (Weight |c₂|² on the upper level) = 0.5; n (Upper level n) = 2 fixed. For this exercise use only 0.10625 fs ≤ t ≤ 0.42125 fs 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 = 0.29117734 nm, with a bounded reading uncertainty ±0.0069002 nm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000315 fs, 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.
MEAN POSITION ⟨x⟩ against t
Predictions from the supplied Stationary states and superpositions 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): MEAN POSITION ⟨x⟩ / nm
Horizontal axis (x): t: Time t since preparation / fs
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Time t since preparation / fs | MEAN POSITION ⟨x⟩ / nm |
|---|---|
| 0.10625 | 0.1827901979436082 |
| 0.185 | 0.223165094958426 |
| 0.26375 | 0.2711729122140766 |
| 0.3425 | 0.3131583064372691 |
| 0.42125 | 0.33717895232858247 |
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(t) = 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(0.10625) = 0.18279 nm and Y_out(0.42125) = 0.337179 nm. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = 0.10625 + (0.291177 − 0.18279)(0.42125 − 0.10625)/(0.337179 − 0.18279) = 0.327393 fs. 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 t = 0.2984 fs; substituting back gives Y_out = 0.29117734 nm.
- The interpolation error in the input is |t_line − t| = 0.0289927 fs. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.28427714 nm and 0.29807754 nm. Sort the roots even if the response decreases: 0.286086 fs ≤ t ≤ 0.311275 fs.
- The lower and upper input deviations are 0.0123138 fs and 0.0128748 fs. 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 t ≈ 0.2984 fs; compatible interval [0.286086, 0.311275] fs. Endpoint interpolation gives 0.327393 fs.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.