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.
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.
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
Horizontal axis (x): d: Separation d
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Separation d | PROBABILITY |⟨φ|ψ⟩|² |
|---|---|
| 0.425 | 0.913645626396729 |
| 1.2125 | 0.4794679990742081 |
| 2 | 0.1353352832366127 |
| 2.7874999999999996 | 0.020546223525013892 |
| 3.5749999999999997 | 0.0016777318282638824 |
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
- 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.
- 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.
- 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.
- The interpolation error in the input is |d_line − d| = 1.01415. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.054425881 and 0.073042881. Sort the roots even if the response decreases: 2.28767 ≤ d ≤ 2.41285.
- 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.
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.
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)
Horizontal axis (x): c: Interval edge c, fraction of L
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| c: Interval edge c, fraction of L | P(0 < x < cL) |
|---|---|
| 0.145625 | 0.08066989588094756 |
| 0.32281249999999995 | 0.5427042339877804 |
| 0.49999999999999994 | 0.9244131815783875 |
| 0.6771874999999999 | 0.9767974191455719 |
| 0.8543749999999999 | 0.9924786602830681 |
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
- 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.
- 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.
- 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.
- The interpolation error in the input is |c_line − c| = 0.258864. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.96651534 and 0.97328834. Sort the roots even if the response decreases: 0.566231 ≤ c ≤ 0.59484.
- 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.
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.
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
Horizontal axis (x): w: Box length L / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Box length L / nm | LEVEL SPACING h / L / ×10⁻²⁵ kg m/s |
|---|---|
| 1.765625 | 3.752778761061947 |
| 2.2578125 | 2.9346989619377166 |
| 2.75 | 2.4094545454545457 |
| 3.2421875 | 2.0436819277108436 |
| 3.734375 | 1.774326359832636 |
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
- 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.
- 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.
- 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.
- The interpolation error in the input is |w_line − w| = 0.310829 nm. This is larger than the requested numerical tolerance.
- 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.
- 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.
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.
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
Horizontal axis (x): th: Mixing angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Mixing angle θ / ° | MEAN <x> / L |
|---|---|
| 9.5625 | 0.4409851224611806 |
| 17.83125 | 0.3949845250712554 |
| 26.1 | 0.3576721431633102 |
| 34.368750000000006 | 0.332134931439803 |
| 42.6375 | 0.3204856527903301 |
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
- 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.
- 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.
- 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.
- The interpolation error in the input is |th_line − th| = 6.21571 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.34118286 and 0.34848766. Sort the roots even if the response decreases: 28.6306 ° ≤ th ≤ 30.9239 °.
- 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.