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University Physics IV · unit UPIV-06

The Schrödinger Equation

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.

Extra challenging questions

A selection from this pathway. Open a topic below for its complete practice and challenge set.

Extra challenging · 01

Building the Schrödinger equation: infer an input with uncertainty

The time-dependent Schrödinger equation is the postulated law of motion for a quantum state: it fixes the rate of change of ψ in time, through a factor of i, from the curvature of ψ in space plus the potential energy at that point. Use the supplied topic model Y_out = exp(-5.7884*q^2*s*cosd(th)), where Y_out is AMPLITUDE |ψ|, 1 AT t = 0 (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; sind/cosd/tand use degrees. th is Phase θ of the coefficient C, measured in °. Keep q (Wavenumber k) = 1 10^10 m^-1; s (Elapsed time t) = 0.2 fs fixed. For this exercise use only 9.5625 ° ≤ th ≤ 80.4375 ° 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 = 0.49658895, with a bounded reading uncertainty ±0.026592. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown th by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0070875 °, 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.

Explore the question data

AMPLITUDE |ψ|, 1 AT t = 0 against th

Predictions from the supplied Building the Schrödinger equation 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): AMPLITUDE |ψ|, 1 AT t = 0

AMPLITUDE |ψ|, 1 AT t = 0 against thPredictions from the supplied Building the Schrödinger equation 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. Horizontal axis: th: Phase θ of the coefficient C / °. Vertical axis: AMPLITUDE |ψ|, 1 AT t = 0. Exact values are available in the data table below.0.29910.43560.57220.70870.84536.72825.864564.1483.27xy

Horizontal axis (x): th: Phase θ of the coefficient C / °

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1th: Phase θ of the coefficient C / °: 9.563AMPLITUDE |ψ|, 1 AT t = 0: 0.3193
Read the exact data as a table
AMPLITUDE |ψ|, 1 AT t = 0 against th — supplied values
th: Phase θ of the coefficient C / °AMPLITUDE |ψ|, 1 AT t = 0
9.56250.3193096337184467
27.281250.35739698859939883
450.4410472005910194
62.718750.5882326563568182
80.43750.8250446999053394

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(th) = 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
  1. Endpoint predictions are Y_out(9.5625) = 0.31931 and Y_out(80.4375) = 0.825045. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives th_line = 9.5625 + (0.496589 − 0.31931)(80.4375 − 9.5625)/(0.825045 − 0.31931) = 34.4069 °. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives th = 52.7962 °; substituting back gives Y_out = 0.49658895.
  4. The interpolation error in the input is |th_line − th| = 18.3894 °. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.46999695 and 0.52318095. Sort the roots even if the response decreases: 49.293 ° ≤ th ≤ 55.9724 °.
  6. The lower and upper input deviations are 3.50329 ° and 3.1762 °. 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 th ≈ 52.7962 °; compatible interval [49.293, 55.9724] °. Endpoint interpolation gives 34.4069 °.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 02

Normalisation and the probability current: infer an input with uncertainty

A wavefunction is admissible when it is single-valued, continuous, and square-integrable, and normalising it means dividing by √(∫|ψ|²dx) so that the probability of finding the particle somewhere is exactly one. Use the supplied topic model Y_out = pi/max(k,0.01), 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; max(a,b) selects the larger value. k is Wavenumber k, measured in nm⁻¹. Keep r (Reflected amplitude r = |B|/|A|) = 0.6 fixed. For this exercise use only 2.6375 nm⁻¹ ≤ k ≤ 7.3625 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 = 0.56915488 nm, with a bounded reading uncertainty ±0.021368 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 0.0004725 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.

Explore the question data

FRINGE SPACING π / k against k

Predictions from the supplied Normalisation and the 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

FRINGE SPACING π / k against kPredictions from the supplied Normalisation and the 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. Horizontal axis: k: Wavenumber k / nm⁻¹. Vertical axis: FRINGE SPACING π / k / nm. Exact values are available in the data table below.0.39610.60250.80891.0151.2222.4493.72456.2767.551xy

Horizontal axis (x): k: Wavenumber k / nm⁻¹

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1k: Wavenumber k / nm⁻¹: 2.638FRINGE SPACING π / k / nm: 1.191
Read the exact data as a table
FRINGE SPACING π / k against k — supplied values
k: Wavenumber k / nm⁻¹FRINGE SPACING π / k / nm
2.63751.1911251767165092
3.818750.8226756539678672
50.6283185307179586
6.181250.5082455253532526
7.36250.42670188843324863

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
  1. Endpoint predictions are Y_out(2.6375) = 1.19113 nm and Y_out(7.3625) = 0.426702 nm. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives k_line = 2.6375 + (0.569155 − 1.19113)(7.3625 − 2.6375)/(0.426702 − 1.19113) = 6.48198 nm⁻¹. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives k = 5.51975 nm⁻¹; substituting back gives Y_out = 0.56915488 nm.
  4. The interpolation error in the input is |k_line − k| = 0.962229 nm⁻¹. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.54778688 nm and 0.59052288 nm. Sort the roots even if the response decreases: 5.32002 nm⁻¹ ≤ k ≤ 5.73506 nm⁻¹.
  6. The lower and upper input deviations are 0.199731 nm⁻¹ and 0.215314 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 ≈ 5.51975 nm⁻¹; compatible interval [5.32002, 5.73506] nm⁻¹. Endpoint interpolation gives 6.48198 nm⁻¹.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 03

Separation of variables and the time-independent equation: infer an input with uncertainty

Separation of variables writes a solution as a function of position alone multiplied by a function of time alone, which reduces the time-dependent Schrödinger equation to the eigenvalue equation Ĥψ = Eψ — but only when the potential has no explicit time dependence. Use the supplied topic model Y_out = cosd(th), where Y_out is REAL PART Re φ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. th is Phase θ = Eₙt/ħ, measured in °. Keep n (Level n in ψₙ) = 2 fixed. For this exercise use only 189.45 ° ≤ th ≤ 321.75 ° 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 = 0.0026703506, with a bounded reading uncertainty ±0.1174. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown th by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.01323 °, 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.

Explore the question data

REAL PART Re φ against th

Predictions from the supplied Separation of variables and the time-independent equation 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): REAL PART Re φ

REAL PART Re φ against thPredictions from the supplied Separation of variables and the time-independent equation 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. Horizontal axis: th: Phase θ = Eₙt/ħ / °. Vertical axis: REAL PART Re φ. Exact values are available in the data table below.-1.057-0.5789-0.10060.37780.8562184.2219.9255.6291.3327xy

Horizontal axis (x): th: Phase θ = Eₙt/ħ / °

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1th: Phase θ = Eₙt/ħ / °: 189.4REAL PART Re φ: -0.9864
Read the exact data as a table
REAL PART Re φ against th — supplied values
th: Phase θ = Eₙt/ħ / °REAL PART Re φ
189.45-0.9864292571764955
222.52499999999998-0.7369824847979511
255.6-0.2486898871648553
288.6750.32019966162731217
321.750.7853169308807446

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(th) = 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
  1. Endpoint predictions are Y_out(189.45) = -0.986429 and Y_out(321.75) = 0.785317. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives th_line = 189.45 + (0.00267035 − -0.986429)(321.75 − 189.45)/(0.785317 − -0.986429) = 263.308 °. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives th = 270.153 °; substituting back gives Y_out = 0.0026703506.
  4. The interpolation error in the input is |th_line − th| = 6.84486 °. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = -0.11472965 and 0.12007035. Sort the roots even if the response decreases: 263.412 ° ≤ th ≤ 276.896 °.
  6. The lower and upper input deviations are 6.74103 ° and 6.74316 °. 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 th ≈ 270.153 °; compatible interval [263.412, 276.896] °. Endpoint interpolation gives 263.308 °.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 04

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.

Explore the question data

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

MEAN POSITION ⟨x⟩ against tPredictions 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. Horizontal axis: t: Time t since preparation / fs. Vertical axis: MEAN POSITION ⟨x⟩ / nm. Exact values are available in the data table below.0.17660.21830.260.30170.34340.093650.17870.26370.34880.4339xy

Horizontal axis (x): t: Time t since preparation / fs

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1t: Time t since preparation / fs: 0.1062MEAN POSITION ⟨x⟩ / nm: 0.1828
Read the exact data as a table
MEAN POSITION ⟨x⟩ against t — supplied values
t: Time t since preparation / fsMEAN POSITION ⟨x⟩ / nm
0.106250.1827901979436082
0.1850.223165094958426
0.263750.2711729122140766
0.34250.3131583064372691
0.421250.33717895232858247

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |t_line − t| = 0.0289927 fs. This is larger than the requested numerical tolerance.
  5. 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.
  6. 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.

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