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University Physics V · course UPV

University Physics V

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

Complex amplitudes and the phase that survives: infer an input with uncertainty

A complex amplitude z = r e^(iφ) carries a modulus, which fixes the probability through |z|² = z*z = r², and a phase, which is measurable only as a difference — when two amplitudes for the same outcome are added before the modulus is squared. Use the supplied topic model Y_out = 2*r2/max(1 + r2*r2, 0.01), where Y_out is VISIBILITY (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. r2 is Modulus r₂ of z₂, dimensionless. Keep d (Relative phase δ) = 60 °; a (Global phase α) = 0 ° fixed. For this exercise use only 0.10625 ≤ r2 ≤ 0.89375 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 = 0.87287016, with a bounded reading uncertainty ±0.018128. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r2 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00007875, 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

VISIBILITY against r2

Predictions from the supplied Complex amplitudes and the phase that survives 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): VISIBILITY

VISIBILITY against r2Predictions from the supplied Complex amplitudes and the phase that survives 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: r2: Modulus r₂ of z₂. Vertical axis: VISIBILITY. Exact values are available in the data table below.0.17880.39040.60190.81351.0250.074750.28740.50.71260.9252xy

Horizontal axis (x): r2: Modulus r₂ of z₂

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

Reading 1r2: Modulus r₂ of z₂: 0.1062VISIBILITY: 0.2101
Read the exact data as a table
VISIBILITY against r2 — supplied values
r2: Modulus r₂ of z₂VISIBILITY
0.106250.21012785352852564
0.3031250.5552325841390228
0.50.8
0.69687499999999990.93815117433231
0.89374999999999990.9937240765271774

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(r2) = 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) = 0.210128 and Y_out(0.89375) = 0.993724. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives r2_line = 0.10625 + (0.87287 − 0.210128)(0.89375 − 0.10625)/(0.993724 − 0.210128) = 0.772294. 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 r2 = 0.586625; substituting back gives Y_out = 0.87287016.
  4. The interpolation error in the input is |r2_line − r2| = 0.185669. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.85474216 and 0.89099816. Sort the roots even if the response decreases: 0.562681 ≤ r2 ≤ 0.612788.
  6. The lower and upper input deviations are 0.023944 and 0.0261631. 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 r2 ≈ 0.586625; compatible interval [0.562681, 0.612788]. Endpoint interpolation gives 0.772294.

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

Extra challenging · 02

Four quantitative failures of the classical account: infer an input with uncertainty

The classical failure inventory is the list of sharp wrong predictions — atomic collapse in picoseconds, harmonic spectra, temperature-blind heat capacities, an undefined entropy constant — that classical mechanics, electrodynamics and statistical mechanics make about ordinary matter. Use the supplied topic model Y_out = 1.5+(tr/t)^2*exp(-tr/t)/max((1-exp(-tr/t))^2,0.000000001)+(tv/t)^2*exp(-tv/t)/max((1-exp(-tv/t))^2,0.000000001), where Y_out is C_V AT T (R). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; max(a,b) selects the larger value. tv is Vibration quantum θ_vib, measured in K. Keep tr (Rotation quantum θ_rot) = 85 K; t (Gas temperature T) = 300 K fixed. For this exercise use only 1384.38 K ≤ tv ≤ 5715.63 K 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 = 2.4936041 R, with a bounded reading uncertainty ±0.000039783 R. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown tv by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.433125 K, 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

C_V AT T against tv

Predictions from the supplied Four quantitative failures of the classical account 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): C_V AT T / R

C_V AT T against tvPredictions from the supplied Four quantitative failures of the classical account 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: tv: Vibration quantum θ_vib / K. Vertical axis: C_V AT T / R. Exact values are available in the data table below.2.48472.54282.60092.6592.717112112381355047195889xy

Horizontal axis (x): tv: Vibration quantum θ_vib / K

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

Reading 1tv: Vibration quantum θ_vib / K: 1384C_V AT T / R: 2.709
Read the exact data as a table
C_V AT T against tv — supplied values
tv: Vibration quantum θ_vib / KC_V AT T / R
1384.3752.7085280607810933
2467.18752.511482706259243
35502.49435336253227
4632.81252.4933838079367443
5715.6252.4933388825594234

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(tv) = 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(1384.38) = 2.70853 R and Y_out(5715.63) = 2.49334 R. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives tv_line = 1384.38 + (2.4936 − 2.70853)(5715.63 − 1384.38)/(2.49334 − 2.70853) = 5710.29 K. 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 tv = 4026.42 K; substituting back gives Y_out = 2.4936041 R.
  4. The interpolation error in the input is |tv_line − tv| = 1683.86 K. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 2.4935643 R and 2.4936439 R. Sort the roots even if the response decreases: 3977.43 K ≤ tv ≤ 4083.2 K.
  6. The lower and upper input deviations are 48.9911 K and 56.775 K. 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 tv ≈ 4026.42 K; compatible interval [3977.43, 4083.2] K. Endpoint interpolation gives 5710.29 K.

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

Extra challenging · 03

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 · 04

The Hamiltonian as a self-adjoint operator: infer an input with uncertainty

The Hamiltonian is the self-adjoint energy operator built by substituting x̂ and p̂ = −iℏ d/dx into the classical energy, together with the domain of states on which its boundary terms vanish and D(Ĥ†) = D(Ĥ). Use the supplied topic model Y_out = b, where Y_out is ψ AT LEFT WALL β (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b is Wall leak β = ψ(0), dimensionless. Keep a (Interior amplitude α) = 1 fixed. For this exercise use only -0.4725 ≤ b ≤ 0.4725 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 = 0.10395, with a bounded reading uncertainty ±0.055282. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000945, 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

ψ AT LEFT WALL β against b

Predictions from the supplied The Hamiltonian as a self-adjoint operator 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): ψ AT LEFT WALL β

ψ AT LEFT WALL β against bPredictions from the supplied The Hamiltonian as a self-adjoint operator 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: b: Wall leak β = ψ(0). Vertical axis: ψ AT LEFT WALL β. Exact values are available in the data table below.-0.5103-0.255100.25510.5103-0.5103-0.255100.25510.5103xy

Horizontal axis (x): b: Wall leak β = ψ(0)

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

Reading 1b: Wall leak β = ψ(0): -0.4725ψ AT LEFT WALL β: -0.4725
Read the exact data as a table
ψ AT LEFT WALL β against b — supplied values
b: Wall leak β = ψ(0)ψ AT LEFT WALL β
-0.4725-0.4725
-0.23625000000000002-0.23625000000000002
-5.551115123125783e-17-5.551115123125783e-17
0.23624999999999990.2362499999999999
0.472499999999999860.47249999999999986

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(b) = 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(-0.4725) = -0.4725 and Y_out(0.4725) = 0.4725. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives b_line = -0.4725 + (0.10395 − -0.4725)(0.4725 − -0.4725)/(0.4725 − -0.4725) = 0.10395. 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 b = 0.10395; substituting back gives Y_out = 0.10395.
  4. The interpolation error in the input is |b_line − b| = 5.55112e-17. This is within the requested numerical tolerance.
  5. Invert both limits Y_out = 0.048668 and 0.159232. Sort the roots even if the response decreases: 0.048668 ≤ b ≤ 0.159232.
  6. The lower and upper input deviations are 0.055282 and 0.055282. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.

Answer b ≈ 0.10395; compatible interval [0.048668, 0.159232]. Endpoint interpolation gives 0.10395.

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

Choose a unit or pathway

UPV-01 · unitMathematical Foundations of Quantum Physics4 selected questions · Explore →UPV-02 · unitFoundations of Quantum Mechanics4 selected questions · Explore →UPV-03 · unitThe Quantum Wavefunction4 selected questions · Explore →UPV-04 · unitThe Schrödinger Equation4 selected questions · Explore →UPV-05 · unitParticle in a Box4 selected questions · Explore →UPV-06 · unitFinite Potential Wells and Quantum Tunneling4 selected questions · Explore →UPV-07 · unitThe Quantum Harmonic Oscillator4 selected questions · Explore →UPV-08 · unitQuantum Uncertainty and Commutation Relations4 selected questions · Explore →UPV-09 · unitAngular Momentum in Quantum Mechanics4 selected questions · Explore →UPV-10 · unitQuantum Spin4 selected questions · Explore →UPV-11 · unitThe Hydrogen Atom4 selected questions · Explore →UPV-12 · unitMulti-Electron Atoms4 selected questions · Explore →UPV-13 · unitAtomic Spectroscopy4 selected questions · Explore →UPV-14 · unitNuclear Physics4 selected questions · Explore →UPV-15 · unitRadioactive Decay4 selected questions · Explore →UPV-16 · unitNuclear Fission and Fusion4 selected questions · Explore →UPV-17 · unitElementary Particle Physics4 selected questions · Explore →UPV-18 · unitThe Standard Model4 selected questions · Explore →UPV-19 · unitConservation Laws and Particle Reactions4 selected questions · Explore →UPV-20 · unitIntroduction to Quantum Information4 selected questions · Explore →