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University Quantum Mechanics I · unit UQMI-04

Measurement and the Uncertainty Principle

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

Observables, Hermitian operators, and real eigenvalues: infer an input with uncertainty

An observable is represented by a Hermitian operator Q̂: one for which ∫ ψ*(Q̂ψ) dx is real for every allowed state ψ, which is the same as ∫ f*(Q̂g) dx = ∫ (Q̂f)* g dx for every allowed pair f and g. Use the supplied topic model Y_out = sqrt(max(-1-b*c,0)), where Y_out is IM PART OF EIGENVALUES, ± (q₀). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. c is Reverse coupling Q₂₁, measured in q₀. Keep b (Coupling Q₁₂) = 1 q₀; phi (Relative phase φ) = 90 ° fixed. For this exercise use only -1.575 q₀ ≤ c ≤ -1.05 q₀ 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 = 0.50472765 q₀, with a bounded reading uncertainty ±0.038034 q₀. 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 0.0000525 q₀, 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

IM PART OF EIGENVALUES, ± against c

Predictions from the supplied Observables, Hermitian operators, and real eigenvalues 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): IM PART OF EIGENVALUES, ± / q₀

IM PART OF EIGENVALUES, ± against cPredictions from the supplied Observables, Hermitian operators, and real eigenvalues 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: Reverse coupling Q₂₁ / q₀. Vertical axis: IM PART OF EIGENVALUES, ± / q₀. Exact values are available in the data table below.0.20220.34660.49090.63530.7797-1.596-1.454-1.313-1.171-1.029xy

Horizontal axis (x): c: Reverse coupling Q₂₁ / q₀

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

Reading 1c: Reverse coupling Q₂₁ / q₀: -1.575IM PART OF EIGENVALUES, ± / q₀: 0.7583
Read the exact data as a table
IM PART OF EIGENVALUES, ± against c — supplied values
c: Reverse coupling Q₂₁ / q₀IM PART OF EIGENVALUES, ± / q₀
-1.5750.758287544405155
-1.443750.6661456297237115
-1.31250.5590169943749475
-1.181250.42573465914815994
-1.050.22360679774997907

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(-1.575) = 0.758288 q₀ and Y_out(-1.05) = 0.223607 q₀. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives c_line = -1.575 + (0.504728 − 0.758288)(-1.05 − -1.575)/(0.223607 − 0.758288) = -1.32603 q₀. 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 = -1.25475 q₀; substituting back gives Y_out = 0.50472765 q₀.
  4. The interpolation error in the input is |c_line − c| = 0.071281 q₀. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.46669365 q₀ and 0.54276165 q₀. Sort the roots even if the response decreases: -1.29459 q₀ ≤ c ≤ -1.2178 q₀.
  6. The lower and upper input deviations are 0.0398402 q₀ and 0.036947 q₀. 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 ≈ -1.25475 q₀; compatible interval [-1.29459, -1.2178] q₀. Endpoint interpolation gives -1.32603 q₀.

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

Extra challenging · 02

Eigenvalue equations and determinate states: infer an input with uncertainty

A determinate state of an observable Q is a normalisable state in which every ideal measurement of Q returns the same value q; such states are exactly the allowed eigenfunctions of Q̂, and q is the eigenvalue. Use the supplied topic model Y_out = (1/r^2 + r^2)/4 + d^2/2, where Y_out is MEAN ⟨E⟩ (ħω). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r is Trial width s, measured in b. Keep d (Centre offset x₀) = 0 b fixed. For this exercise use only 0.659375 b ≤ r ≤ 0.974375 b 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 = 0.5260562 ħω, with a bounded reading uncertainty ±0.0038073 ħω. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r 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 b, 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 ⟨E⟩ against r

Predictions from the supplied Eigenvalue equations and determinate states 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 ⟨E⟩ / ħω

MEAN ⟨E⟩ against rPredictions from the supplied Eigenvalue equations and determinate states 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: r: Trial width s / b. Vertical axis: MEAN ⟨E⟩ / ħω. Exact values are available in the data table below.0.49340.54280.59220.64160.6910.64680.73180.81690.90190.987xy

Horizontal axis (x): r: Trial width s / b

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

Reading 1r: Trial width s / b: 0.6594MEAN ⟨E⟩ / ħω: 0.6837
Read the exact data as a table
MEAN ⟨E⟩ against r — supplied values
r: Trial width s / bMEAN ⟨E⟩ / ħω
0.6593750.6837033937131669
0.7381250.5950671056334456
0.8168750.5414738327951809
0.8956250.512200719018691
0.9743750.5006740230959315

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(r) = 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(0.659375) = 0.683703 ħω and Y_out(0.974375) = 0.500674 ħω. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives r_line = 0.659375 + (0.526056 − 0.683703)(0.974375 − 0.659375)/(0.500674 − 0.683703) = 0.930691 b. 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 r = 0.851525 b; substituting back gives Y_out = 0.5260562 ħω.
  4. The interpolation error in the input is |r_line − r| = 0.0791664 b. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.5222489 ħω and 0.5298635 ħω. Sort the roots even if the response decreases: 0.842011 b ≤ r ≤ 0.861903 b.
  6. The lower and upper input deviations are 0.00951368 b and 0.0103776 b. 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 r ≈ 0.851525 b; compatible interval [0.842011, 0.861903] b. Endpoint interpolation gives 0.930691 b.

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

Extra challenging · 03

Eigenbasis expansion and the probability of an eigenvalue: infer an input with uncertainty

The probability that a measurement of Q returns the eigenvalue qₙ is |cₙ|², where cₙ = ∫ ψₙ* ψ dx is the overlap of the prepared state ψ with the normalised eigenfunction ψₙ of Q̂. Use the supplied topic model Y_out = (1/r)*(sin(pi*((p0*r)/r-p0)+0.000001)/(pi*((p0*r)/r-p0)+0.000001))^2, where Y_out is |cₙ|² NEAREST p₀ (before display rounding) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; unmarked trigonometric functions use radians. r is Ring circumference L, measured in w. Keep p0 (Mean momentum p₀) = 0.25 h/w fixed. For this exercise use only 2.6375 w ≤ r ≤ 7.3625 w 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 = 0.18116763, with a bounded reading uncertainty ±0.0068016. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r 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 w, 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

|cₙ|² NEAREST p₀ (before display rounding) against r

Predictions from the supplied Eigenbasis expansion and the probability of an eigenvalue 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): |cₙ|² NEAREST p₀ (before display rounding)

|cₙ|² NEAREST p₀ (before display rounding) against rPredictions from the supplied Eigenbasis expansion and the probability of an eigenvalue 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: r: Ring circumference L / w. Vertical axis: |cₙ|² NEAREST p₀ (before display rounding). Exact values are available in the data table below.0.12610.19180.25750.32320.38892.4493.72456.2767.551xy

Horizontal axis (x): r: Ring circumference L / w

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

Reading 1r: Ring circumference L / w: 2.638|cₙ|² NEAREST p₀ (before display rounding): 0.3791
Read the exact data as a table
|cₙ|² NEAREST p₀ (before display rounding) against r — supplied values
r: Ring circumference L / w|cₙ|² NEAREST p₀ (before display rounding)
2.63750.37914691943115325
3.818750.26186579378060015
50.19999999999993334
6.181250.16177957532856083
7.36250.13582342954155066

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(r) = 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) = 0.379147 and Y_out(7.3625) = 0.135823. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives r_line = 2.6375 + (0.181168 − 0.379147)(7.3625 − 2.6375)/(0.135823 − 0.379147) = 6.48198 w. 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 r = 5.51975 w; substituting back gives Y_out = 0.18116763.
  4. The interpolation error in the input is |r_line − r| = 0.96223 w. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.17436603 and 0.18796923. Sort the roots even if the response decreases: 5.32002 w ≤ r ≤ 5.73506 w.
  6. The lower and upper input deviations are 0.19973 w and 0.215312 w. 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 r ≈ 5.51975 w; compatible interval [5.32002, 5.73506] w. Endpoint interpolation gives 6.48198 w.

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

Extra challenging · 04

Collapse, repeatability, and the state after measurement: infer an input with uncertainty

The projection rule: a measurement that returns qₙ leaves the system in the matching eigenfunction ψₙ, or, when the result cannot tell several eigenfunctions apart, in the part of ψ it cannot exclude, renormalised. That state is where every later calculation starts. Use the supplied topic model Y_out = 1/(b-a-(sin(2*pi*b)-sin(2*pi*a))/(2*pi)), where Y_out is DENSITY RESCALE 1/P (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; unmarked trigonometric functions use radians. a is Band starts at a, measured in nm. Keep b (Band ends at b) = 0.75 nm fixed. For this exercise use only 0.0425 nm ≤ a ≤ 0.3575 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 = 1.2005952, with a bounded reading uncertainty ±0.01501. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a 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 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

DENSITY RESCALE 1/P against a

Predictions from the supplied Collapse, repeatability, and the state after measurement 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): DENSITY RESCALE 1/P

DENSITY RESCALE 1/P against aPredictions from the supplied Collapse, repeatability, and the state after measurement 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: a: Band starts at a / nm. Vertical axis: DENSITY RESCALE 1/P. Exact values are available in the data table below.1.0851.1881.291.3921.4950.02990.11490.20.2850.3701xy

Horizontal axis (x): a: Band starts at a / nm

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

Reading 1a: Band starts at a / nm: 0.0425DENSITY RESCALE 1/P: 1.101
Read the exact data as a table
DENSITY RESCALE 1/P against a — supplied values
a: Band starts at a / nmDENSITY RESCALE 1/P
0.04251.1005317685018885
0.121251.1138811603477352
0.21.1620876497534212
0.278751.2706961182702252
0.35751.4795869522398626

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(a) = 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(0.0425) = 1.10053 and Y_out(0.3575) = 1.47959. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives a_line = 0.0425 + (1.2006 − 1.10053)(0.3575 − 0.0425)/(1.47959 − 1.10053) = 0.125654 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 a = 0.23465 nm; substituting back gives Y_out = 1.2005952.
  4. The interpolation error in the input is |a_line − a| = 0.108996 nm. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 1.1855852 and 1.2156052. Sort the roots even if the response decreases: 0.222469 nm ≤ a ≤ 0.245614 nm.
  6. The lower and upper input deviations are 0.0121807 nm and 0.0109639 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 a ≈ 0.23465 nm; compatible interval [0.222469, 0.245614] nm. Endpoint interpolation gives 0.125654 nm.

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

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