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University Physics V · unit UPV-03

The Quantum Wavefunction

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

The state vector and its position representation: infer an input with uncertainty

The state of the particle is a unit vector |ψ⟩ in the Hilbert space L²(ℝ), defined only up to a global phase; its position representation ψ(x) = ⟨x|ψ⟩ lists its components along the delta-normalised position basis. Use the supplied topic model Y_out = (2*s1*s2/(s1^2+s2^2))*exp(-d^2/(s1^2+s2^2)), where Y_out is PROBABILITY |⟨φ|ψ⟩|² (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. d is Separation d, dimensionless. Keep s1 (Width σ₁ of φ) = 1; s2 (Width σ₂ of ψ) = 1 fixed. For this exercise use only 0.425 ≤ d ≤ 3.575 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 = 0.063734381, with a bounded reading uncertainty ±0.0093085. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000315, 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

PROBABILITY |⟨φ|ψ⟩|² against d

Predictions from the supplied The state vector and its position representation 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): PROBABILITY |⟨φ|ψ⟩|²

PROBABILITY |⟨φ|ψ⟩|² against dPredictions from the supplied The state vector and its position representation 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: d: Separation d. Vertical axis: PROBABILITY |⟨φ|ψ⟩|². Exact values are available in the data table below.-0.03480.21140.45770.70390.95010.2991.14922.853.701xy

Horizontal axis (x): d: Separation d

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

Reading 1d: Separation d: 0.425PROBABILITY |⟨φ|ψ⟩|²: 0.9136
Read the exact data as a table
PROBABILITY |⟨φ|ψ⟩|² against d — supplied values
d: Separation dPROBABILITY |⟨φ|ψ⟩|²
0.4250.913645626396729
1.21250.4794679990742081
20.1353352832366127
2.78749999999999960.020546223525013892
3.57499999999999970.0016777318282638824

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(d) = 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(0.425) = 0.913646 and Y_out(3.575) = 0.00167773. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives d_line = 0.425 + (0.0637344 − 0.913646)(3.575 − 0.425)/(0.00167773 − 0.913646) = 3.36065. 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 d = 2.3465; substituting back gives Y_out = 0.063734381.
  4. The interpolation error in the input is |d_line − d| = 1.01415. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.054425881 and 0.073042881. Sort the roots even if the response decreases: 2.28767 ≤ d ≤ 2.41285.
  6. The lower and upper input deviations are 0.0588337 and 0.0663471. 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 d ≈ 2.3465; compatible interval [2.28767, 2.41285]. Endpoint interpolation gives 3.36065.

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

Extra challenging · 02

Born's rule, density and normalisation: infer an input with uncertainty

Born's rule reads |ψ(x)|² as a probability density: integrating it across an interval gives the chance of finding the particle there, and normalisation sets the whole-line integral to one. Use the supplied topic model Y_out = (1-w)*(c-sin(2*pi*c)/(2*pi))+w*(c-sin(4*pi*c)/(4*pi))+2*sqrt(w*(1-w))*cosd(phi)*(sin(pi*c)/pi-sin(3*pi*c)/(3*pi)), where Y_out is P(0 < x < cL) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; sind/cosd/tand use degrees; unmarked trigonometric functions use radians. c is Interval edge c, fraction of L, dimensionless. Keep w (Weight w of state |2⟩) = 0.5; phi (Relative phase φ) = 0 ° fixed. For this exercise use only 0.145625 ≤ c ≤ 0.854375 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 = 0.96990184, with a bounded reading uncertainty ±0.0033865. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown c by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000070875, 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

P(0 < x < cL) against c

Predictions from the supplied Born's rule, density and normalisation 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): P(0 < x < cL)

P(0 < x < cL) against cPredictions from the supplied Born's rule, density and normalisation 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: c: Interval edge c, fraction of L. Vertical axis: P(0 < x < cL). Exact values are available in the data table below.0.04420.29040.53660.78281.0290.11730.30860.50.69140.8827xy

Horizontal axis (x): c: Interval edge c, fraction of L

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

Reading 1c: Interval edge c, fraction of L: 0.1456P(0 < x < cL): 0.08067
Read the exact data as a table
P(0 < x < cL) against c — supplied values
c: Interval edge c, fraction of LP(0 < x < cL)
0.1456250.08066989588094756
0.322812499999999950.5427042339877804
0.499999999999999940.9244131815783875
0.67718749999999990.9767974191455719
0.85437499999999990.9924786602830681

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(c) = 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(0.145625) = 0.0806699 and Y_out(0.854375) = 0.992479. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives c_line = 0.145625 + (0.969902 − 0.0806699)(0.854375 − 0.145625)/(0.992479 − 0.0806699) = 0.836826. 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 c = 0.577962; substituting back gives Y_out = 0.96990184.
  4. The interpolation error in the input is |c_line − c| = 0.258864. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.96651534 and 0.97328834. Sort the roots even if the response decreases: 0.566231 ≤ c ≤ 0.59484.
  6. The lower and upper input deviations are 0.0117317 and 0.0168774. 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 c ≈ 0.577962; compatible interval [0.566231, 0.59484]. Endpoint interpolation gives 0.836826.

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

Extra challenging · 03

Self-adjoint operators and their spectra: infer an input with uncertainty

An observable is a self-adjoint operator: a rule A together with a dense domain D(A) on which ⟨φ|Aψ⟩ = ⟨Aφ|ψ⟩ for every pair, and for which the adjoint's domain D(A†) is no larger than D(A) itself. Use the supplied topic model Y_out = 6.626/w, where Y_out is LEVEL SPACING h / L (×10⁻²⁵ kg m/s). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. w is Box length L, measured in nm. Keep t (Boundary twist θ) = 0 °; n (Level index n) = 1 fixed. For this exercise use only 1.76563 nm ≤ w ≤ 3.73438 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 = 2.2335616 ×10⁻²⁵ kg m/s, with a bounded reading uncertainty ±0.068885 ×10⁻²⁵ kg m/s. 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 0.000196875 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.

Explore the question data

LEVEL SPACING h / L against w

Predictions from the supplied Self-adjoint operators and their spectra 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): LEVEL SPACING h / L / ×10⁻²⁵ kg m/s

LEVEL SPACING h / L against wPredictions from the supplied Self-adjoint operators and their spectra 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: w: Box length L / nm. Vertical axis: LEVEL SPACING h / L / ×10⁻²⁵ kg m/s. Exact values are available in the data table below.1.6952.2292.7643.2983.8321.6872.2182.753.2823.813xy

Horizontal axis (x): w: Box length L / nm

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

Reading 1w: Box length L / nm: 1.766LEVEL SPACING h / L / ×10⁻²⁵ kg m/s: 3.753
Read the exact data as a table
LEVEL SPACING h / L against w — supplied values
w: Box length L / nmLEVEL SPACING h / L / ×10⁻²⁵ kg m/s
1.7656253.752778761061947
2.25781252.9346989619377166
2.752.4094545454545457
3.24218752.0436819277108436
3.7343751.774326359832636

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
  1. Endpoint predictions are Y_out(1.76563) = 3.75278 ×10⁻²⁵ kg m/s and Y_out(3.73438) = 1.77433 ×10⁻²⁵ kg m/s. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives w_line = 1.76563 + (2.23356 − 3.75278)(3.73438 − 1.76563)/(1.77433 − 3.75278) = 3.27739 nm. 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 w = 2.96656 nm; substituting back gives Y_out = 2.2335616 ×10⁻²⁵ kg m/s.
  4. The interpolation error in the input is |w_line − w| = 0.310829 nm. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 2.1646766 ×10⁻²⁵ kg m/s and 2.3024466 ×10⁻²⁵ kg m/s. Sort the roots even if the response decreases: 2.87781 nm ≤ w ≤ 3.06097 nm.
  6. The lower and upper input deviations are 0.0887541 nm and 0.0944029 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 ≈ 2.96656 nm; compatible interval [2.87781, 3.06097] nm. Endpoint interpolation gives 3.27739 nm.

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

Extra challenging · 04

Expectation values, variances and moments: infer an input with uncertainty

The expectation value of an observable A in the state |ψ⟩ is the number ⟨A⟩ = ⟨ψ|A|ψ⟩: the mean of the outcomes A would give over a large ensemble of systems all prepared in |ψ⟩, weighted by the Born probabilities. Use the supplied topic model Y_out = 0.5-0.1801265*sind(2*th)*cosd(ph), where Y_out is MEAN <x> / L (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 Mixing angle θ, measured in °. Keep ph (Relative phase φ) = 0 ° fixed. For this exercise use only 9.5625 ° ≤ th ≤ 42.6375 ° 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 = 0.34483526, with a bounded reading uncertainty ±0.0036524. 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 0.0033075 °, 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 <x> / L against th

Predictions from the supplied Expectation values, variances and moments 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 <x> / L

MEAN <x> / L against thPredictions from the supplied Expectation values, variances and moments 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: Mixing angle θ / °. Vertical axis: MEAN <x> / L. Exact values are available in the data table below.0.31570.34820.38070.41330.44588.23917.1726.135.0343.96xy

Horizontal axis (x): th: Mixing angle θ / °

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

Reading 1th: Mixing angle θ / °: 9.563MEAN <x> / L: 0.441
Read the exact data as a table
MEAN <x> / L against th — supplied values
th: Mixing angle θ / °MEAN <x> / L
9.56250.4409851224611806
17.831250.3949845250712554
26.10.3576721431633102
34.3687500000000060.332134931439803
42.63750.3204856527903301

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) = 0.440985 and Y_out(42.6375) = 0.320486. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives th_line = 9.5625 + (0.344835 − 0.440985)(42.6375 − 9.5625)/(0.320486 − 0.440985) = 35.954 °. 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 = 29.7383 °; substituting back gives Y_out = 0.34483526.
  4. The interpolation error in the input is |th_line − th| = 6.21571 °. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.34118286 and 0.34848766. Sort the roots even if the response decreases: 28.6306 ° ≤ th ≤ 30.9239 °.
  6. The lower and upper input deviations are 1.10769 ° and 1.18567 °. 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 ≈ 29.7383 °; compatible interval [28.6306, 30.9239] °. Endpoint interpolation gives 35.954 °.

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

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