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University Quantum Mechanics I · course UQMI

University Quantum Mechanics I

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

Where the classical account fails: infer an input with uncertainty

The classical account fails wherever a motion's characteristic energy hν is not small beside k_BT, or its action is not large beside Planck's constant h: there its predictions miss the measurements by factors, not percent. Use the supplied topic model Y_out = 0.531977*r0^1.5*(1-sqrt(max(1-t/(1.05016e-4*r0^3),0))), where Y_out is ORBITS SO FAR (thousand). 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. r0 is Starting orbit radius r₀, measured in pm. Keep t (Time since start t) = 8 ps fixed. For this exercise use only 42.65 pm ≤ r0 ≤ 65.75 pm 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 = 53.76463 thousand, with a bounded reading uncertainty ±1.92 thousand. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00231 pm, 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

ORBITS SO FAR against r0

Predictions from the supplied Where the classical account fails 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): ORBITS SO FAR / thousand

ORBITS SO FAR against r0Predictions from the supplied Where the classical account fails 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: r0: Starting orbit radius r₀ / pm. Vertical axis: ORBITS SO FAR / thousand. Exact values are available in the data table below.37.4761.0484.61108.2131.741.7347.9654.260.4466.67xy

Horizontal axis (x): r0: Starting orbit radius r₀ / pm

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

Reading 1r0: Starting orbit radius r₀ / pm: 42.65ORBITS SO FAR / thousand: 128.3
Read the exact data as a table
ORBITS SO FAR against r0 — supplied values
r0: Starting orbit radius r₀ / pmORBITS SO FAR / thousand
42.65128.25169035237255
48.42576.4181735058303
54.258.97240809055958
59.97548.35776019659553
65.7540.96448004401035

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(r0) = 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(42.65) = 128.252 thousand and Y_out(65.75) = 40.9645 thousand. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives r0_line = 42.65 + (53.7646 − 128.252)(65.75 − 42.65)/(40.9645 − 128.252) = 62.3625 pm. 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 r0 = 56.741 pm; substituting back gives Y_out = 53.76463 thousand.
  4. The interpolation error in the input is |r0_line − r0| = 5.62152 pm. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 51.84463 thousand and 55.68463 thousand. Sort the roots even if the response decreases: 55.7463 pm ≤ r0 ≤ 57.8115 pm.
  6. The lower and upper input deviations are 0.994737 pm and 1.07051 pm. 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 r0 ≈ 56.741 pm; compatible interval [55.7463, 57.8115] pm. Endpoint interpolation gives 62.3625 pm.

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

Extra challenging · 02

The time-dependent Schrödinger equation: infer an input with uncertainty

The time-dependent Schrödinger equation, iħ ∂Ψ/∂t = −(ħ²/2m) ∂²Ψ/∂x² + VΨ, is the postulated law of motion for a wavefunction: given V and Ψ at one instant, it fixes Ψ at every later instant. Use the supplied topic model Y_out = 1.994711/(2*(1+exp(-d^2/0.08)))*4*exp(-d^2/(0.08*(1+(1.447095*t)^2)))/sqrt((1+(1.447095*t)^2)), where Y_out is DENSITY AT x = 0 (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; exp(x) means eˣ. t is Time since release t, measured in fs. Keep d (Starting offset of each packet d) = 1 nm fixed. For this exercise use only 1.29688 fs ≤ t ≤ 3.26563 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 = 0.43739029 nm⁻¹, with a bounded reading uncertainty ±0.0068939 nm⁻¹. 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 0.000196875 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

DENSITY AT x = 0 against t

Predictions from the supplied The time-dependent 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): DENSITY AT x = 0 / nm⁻¹

DENSITY AT x = 0 against tPredictions from the supplied The time-dependent 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: t: Time since release t / fs. Vertical axis: DENSITY AT x = 0 / nm⁻¹. Exact values are available in the data table below.0.10360.20220.30080.39940.4981.2181.752.2812.8133.344xy

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

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

Reading 1t: Time since release t / fs: 1.297DENSITY AT x = 0 / nm⁻¹: 0.1182
Read the exact data as a table
DENSITY AT x = 0 against t — supplied values
t: Time since release t / fsDENSITY AT x = 0 / nm⁻¹
1.2968750.11823345080310765
1.78906250.2836651117021689
2.281250.4044822834139529
2.77343750.46450257606941586
3.2656250.483349519507742

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(1.29688) = 0.118233 nm⁻¹ and Y_out(3.26563) = 0.48335 nm⁻¹. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives t_line = 1.29688 + (0.43739 − 0.118233)(3.26563 − 1.29688)/(0.48335 − 0.118233) = 3.01781 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 = 2.49781 fs; substituting back gives Y_out = 0.43739029 nm⁻¹.
  4. The interpolation error in the input is |t_line − t| = 0.519995 fs. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.43049639 nm⁻¹ and 0.44428419 nm⁻¹. Sort the roots even if the response decreases: 2.44571 fs ≤ t ≤ 2.55516 fs.
  6. The lower and upper input deviations are 0.0520975 fs and 0.0573523 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 ≈ 2.49781 fs; compatible interval [2.44571, 2.55516] fs. Endpoint interpolation gives 3.01781 fs.

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

Extra challenging · 03

Operators: linear instructions that act on a wavefunction: infer an input with uncertainty

An operator  is a rule that turns a function f(x) into a new function Âf(x); it is linear when Â(af + bg) = aÂf + bÂg for every pair of functions f, g it can act on and every pair of complex numbers a, b. Use the supplied topic model Y_out = sin(2*pi*m/n)/(2*pi*m/n), where Y_out is MATRIX ÷ EXACT sin(kh)/(kh) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; unmarked trigonometric functions use radians. n is Grid points N, dimensionless. Keep m (Wavelengths across the window m) = 2 fixed. For this exercise use only 8 ≤ n ≤ 64, at the specified test settings separated by 4. 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 = 0.98646084, with a bounded reading uncertainty ±0.028557. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of n consistent with that bound. Compare with a straight-line interpolation between the two endpoint predictions, and explain why a fractional setting may be inadmissible. Do not assume that the supplied calibration points prove monotonicity outside this branch.

Explore the question data

MATRIX ÷ EXACT sin(kh)/(kh) against n

Predictions from the supplied Operators: linear instructions that act on a wavefunction 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): MATRIX ÷ EXACT sin(kh)/(kh)

MATRIX ÷ EXACT sin(kh)/(kh) against nPredictions from the supplied Operators: linear instructions that act on a wavefunction 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: n: Grid points N. Vertical axis: MATRIX ÷ EXACT sin(kh)/(kh). Exact values are available in the data table below.0.62230.71870.81510.91151.0085.7620.883651.1266.24xy

Horizontal axis (x): n: Grid points N

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

Reading 1n: Grid points N: 8MATRIX ÷ EXACT sin(kh)/(kh): 0.6366
Read the exact data as a table
MATRIX ÷ EXACT sin(kh)/(kh) against n — supplied values
n: Grid points NMATRIX ÷ EXACT sin(kh)/(kh)
80.6366197723675814
240.954929658551372
360.9798155360510165
520.99029504419782
640.9935868511442058

These are the supplied values, not a worked answer. No fitted trend is drawn.

Show a hint

Enumerate the permitted settings in the stated range, evaluate Y_out at each, and retain every setting satisfying |Y_out − Y_measured| ≤ u. The plotted predictions are supplied calibration evidence.

Reveal the worked solution
  1. Endpoint predictions are Y_out(8) = 0.63662 and Y_out(64) = 0.993587. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives n_line = 8 + (0.986461 − 0.63662)(64 − 8)/(0.993587 − 0.63662) = 62.8821. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Evaluate the model at each permitted setting. The central reading corresponds to n = 44; only the allowed settings 28, 32, 36, 40, 44, 48, 52, 56, 60, 64 lie inside the stated reading bound.
  4. The interpolation error in the input is |n_line − n| = 18.8821.
  5. The compatible allowed values span 28 to 64. Values between specified test settings are outside this exercise's input domain even if an interpolation passes through them.
  6. The quoted output bound restricts which specified settings remain compatible. It does not change physical restrictions on quantum numbers or counts, and the chosen grid alone does not establish physical quantisation.

Answer Compatible n: 28, 32, 36, 40, 44, 48, 52, 56, 60, 64. Endpoint interpolation predicts 62.8821; check admissibility against the discrete settings.

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

Extra challenging · 04

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.

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