Skip to main content
Course and syllabus guidance

University Advanced Quantum Mechanics · unit UAQM-12

Spontaneous Emission and Radiative Decay from the Quantised Field

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

Emission and absorption matrix elements: √n and √(n + 1): infer an input with uncertainty

The emission and absorption matrix elements are the first-order amplitudes of the A⋅p coupling that take an atom with n photons in one mode to its other level with n + 1 or n − 1 photons: √(n + 1) and √n times atomic factors that are complex conjugates of each other. Use the supplied topic model Y_out = 47*sqrt(n+1)*cosd(th), where Y_out is EMISSION RABI 2g√(n+1)/2π (kHz). 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. th is Angle θ between ε and d_ba, measured in °. Keep n (Photons in the mode n) = 1 fixed. For this exercise use only 9.5625 ° ≤ th ≤ 80.4375 ° 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 = 40.189982 kHz, with a bounded reading uncertainty ±3.8032 kHz. 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 0.0070875 °, 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

EMISSION RABI 2g√(n+1)/2π against th

Predictions from the supplied Emission and absorption matrix elements: √n and √(n + 1) 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): EMISSION RABI 2g√(n+1)/2π / kHz

EMISSION RABI 2g√(n+1)/2π against thPredictions from the supplied Emission and absorption matrix elements: √n and √(n + 1) 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: Angle θ between ε and d_ba / °. Vertical axis: EMISSION RABI 2g√(n+1)/2π / kHz. Exact values are available in the data table below.8.86223.5838.2953.0167.726.72825.864564.1483.27xy

Horizontal axis (x): th: Angle θ between ε and d_ba / °

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

Reading 1th: Angle θ between ε and d_ba / °: 9.563EMISSION RABI 2g√(n+1)/2π / kHz: 65.54
Read the exact data as a table
EMISSION RABI 2g√(n+1)/2π against th — supplied values
th: Angle θ between ε and d_ba / °EMISSION RABI 2g√(n+1)/2π / kHz
9.562565.54446244073816
27.2812559.07461668780822
4547.000000000000014
62.7187530.46620526400575
80.437511.04189491684587

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) = 65.5445 kHz and Y_out(80.4375) = 11.0419 kHz. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives th_line = 9.5625 + (40.19 − 65.5445)(80.4375 − 9.5625)/(11.0419 − 65.5445) = 42.5334 °. 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 = 52.7962 °; substituting back gives Y_out = 40.189982 kHz.
  4. The interpolation error in the input is |th_line − th| = 10.2628 °. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 36.386782 kHz and 43.993182 kHz. Sort the roots even if the response decreases: 48.5574 ° ≤ th ≤ 56.8089 °.
  6. The lower and upper input deviations are 4.23888 ° and 4.01268 °. 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 ≈ 52.7962 °; compatible interval [48.5574, 56.8089] °. Endpoint interpolation gives 42.5334 °.

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

Extra challenging · 02

The A coefficient from the golden rule over empty modes: infer an input with uncertainty

The A coefficient is the rate at which an excited atom alone in the photon vacuum emits, obtained here as the golden-rule rate into every empty mode of the field: for an electric-dipole line, A = ω₀³|d|²/(3πε₀ħc³). Use the supplied topic model Y_out = (n0-n1)*sind(th)^2/max((n0*sind(th)^2+n1*(1+cosd(th)^2)),0.000001), where Y_out is POLARISATION, + IN z–k PLANE (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; sind/cosd/tand use degrees. th is Viewing angle θ from z, measured in °. Keep n0 (Atoms in m = 0, n₀) = 6; n1 (Atoms in each of m = ±1, n₁) = 2 fixed. For this exercise use only 19.125 ° ≤ th ≤ 85.275 ° 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 = 0.42596235, with a bounded reading uncertainty ±0.01085. 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 0.006615 °, 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

POLARISATION, + IN z–k PLANE against th

Predictions from the supplied The A coefficient from the golden rule over empty modes 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): POLARISATION, + IN z–k PLANE

POLARISATION, + IN z–k PLANE against thPredictions from the supplied The A coefficient from the golden rule over empty modes 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: Viewing angle θ from z / °. Vertical axis: POLARISATION, + IN z–k PLANE. Exact values are available in the data table below.0.080880.18920.29760.4060.514416.4834.3452.270.0687.92xy

Horizontal axis (x): th: Viewing angle θ from z / °

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

Reading 1th: Viewing angle θ from z / °: 19.13POLARISATION, + IN z–k PLANE: 0.09694
Read the exact data as a table
POLARISATION, + IN z–k PLANE against th — supplied values
th: Viewing angle θ from z / °POLARISATION, + IN z–k PLANE
19.1250.0969362488532642
35.66250.25367574167316426
52.20.38436721587316286
68.737500000000010.4648088375752355
85.2750.4982978823970005

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(19.125) = 0.0969362 and Y_out(85.275) = 0.498298. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives th_line = 19.125 + (0.425962 − 0.0969362)(85.275 − 19.125)/(0.498298 − 0.0969362) = 73.3531 °. 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 = 59.4765 °; substituting back gives Y_out = 0.42596235.
  4. The interpolation error in the input is |th_line − th| = 13.8766 °. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.41511235 and 0.43681235. Sort the roots even if the response decreases: 57.4003 ° ≤ th ≤ 61.7252 °.
  6. The lower and upper input deviations are 2.07625 ° and 2.24868 °. 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 ≈ 59.4765 °; compatible interval [57.4003, 61.7252] °. Endpoint interpolation gives 73.3531 °.

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

Extra challenging · 03

The Einstein relations and Planck's law as outputs: infer an input with uncertainty

The Einstein relations, A/B₂₁ = ħω³/(π²c³) and g₁B₁₂ = g₂B₂₁, fix the absorption and stimulated-emission coefficients B from the spontaneous rate A; with the quantised field they follow from the photon factors n̄ and n̄ + 1, and Planck's law follows from them. Use the supplied topic model Y_out = nf/(nf+s), where Y_out is STIMULATED SHARE OF EMISSION (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. nf is Field photons per mode n̄, dimensionless. Keep hw (Level spacing ħω/kT) = 0.7; s (Vacuum term (1 quantised, 0 classical)) = 1 fixed. For this exercise use only 0.469687 ≤ nf ≤ 3.58031 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 = 0.70301459, with a bounded reading uncertainty ±0.011799. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown nf by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000311062, 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

STIMULATED SHARE OF EMISSION against nf

Predictions from the supplied The Einstein relations and Planck's law as outputs 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): STIMULATED SHARE OF EMISSION

STIMULATED SHARE OF EMISSION against nfPredictions from the supplied The Einstein relations and Planck's law as outputs 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: nf: Field photons per mode n̄. Vertical axis: STIMULATED SHARE OF EMISSION. Exact values are available in the data table below.0.30110.42590.55060.67540.80020.34531.1852.0252.8653.705xy

Horizontal axis (x): nf: Field photons per mode n̄

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

Reading 1nf: Field photons per mode n̄: 0.4697STIMULATED SHARE OF EMISSION: 0.3196
Read the exact data as a table
STIMULATED SHARE OF EMISSION against nf — supplied values
nf: Field photons per mode n̄STIMULATED SHARE OF EMISSION
0.46968750.31958324473740163
1.247343750.5550302440381006
2.0250.6694214876033058
2.802656250.7370259275999507
3.58031250.7816742853244184

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(nf) = 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(0.469687) = 0.319583 and Y_out(3.58031) = 0.781674. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives nf_line = 0.469687 + (0.703015 − 0.319583)(3.58031 − 0.469687)/(0.781674 − 0.319583) = 3.0508. 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 nf = 2.36717; substituting back gives Y_out = 0.70301459.
  4. The interpolation error in the input is |nf_line − nf| = 0.683636. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.69121559 and 0.71481359. Sort the roots even if the response decreases: 2.23851 ≤ nf ≤ 2.50648.
  6. The lower and upper input deviations are 0.128663 and 0.13931. 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 nf ≈ 2.36717; compatible interval [2.23851, 2.50648]. Endpoint interpolation gives 3.0508.

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

Extra challenging · 04

Magnetic-dipole emission and the 21 cm rate computed: infer an input with uncertainty

Magnetic-dipole emission is spontaneous decay driven by the coupling −μ⋅B of an atom's magnetic moment to the vacuum fluctuations of the quantised magnetic field, at the rate μ₀ω³|⟨f|μ|i⟩|²/(3πħc³). Use the supplied topic model Y_out = 75115/mu^2, where Y_out is E1 / M1 RATE AT EQUAL ω (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. mu is Transition moment |⟨f|μ|i⟩| / μ_B, dimensionless. Keep dec (Frequency log₁₀(ν / 1420 MHz)) = 0; tb (Background temperature T) = 2.725 K fixed. For this exercise use only 0.301875 ≤ mu ≤ 1.79812 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 = 50917.733, with a bounded reading uncertainty ±4152.9. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown mu by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000149625, 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

E1 / M1 RATE AT EQUAL ω against mu

Predictions from the supplied Magnetic-dipole emission and the 21 cm rate computed 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): E1 / M1 RATE AT EQUAL ω

E1 / M1 RATE AT EQUAL ω against muPredictions from the supplied Magnetic-dipole emission and the 21 cm rate computed 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: mu: Transition moment |⟨f|μ|i⟩| / μ_B. Vertical axis: E1 / M1 RATE AT EQUAL ω. Exact values are available in the data table below.-88102.07e54.24e56.40e58.56e50.2420.6461.051.4541.858xy

Horizontal axis (x): mu: Transition moment |⟨f|μ|i⟩| / μ_B

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

Reading 1mu: Transition moment |⟨f|μ|i⟩| / μ_B: 0.3019E1 / M1 RATE AT EQUAL ω: 8.24e5
Read the exact data as a table
E1 / M1 RATE AT EQUAL ω against mu — supplied values
mu: Transition moment |⟨f|μ|i⟩| / μ_BE1 / M1 RATE AT EQUAL ω
0.301875824275.4694820587
0.6759375164404.45785880258
1.049999999999999868131.51927437645
1.424062499999999837039.79471689857
1.798124999999999823232.01680196117

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(mu) = 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(0.301875) = 824275 and Y_out(1.79812) = 23232. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives mu_line = 0.301875 + (50917.7 − 824275)(1.79812 − 0.301875)/(23232 − 824275) = 1.74641. 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 mu = 1.21459; substituting back gives Y_out = 50917.733.
  4. The interpolation error in the input is |mu_line − mu| = 0.531824. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 46764.833 and 55070.633. Sort the roots even if the response decreases: 1.16789 ≤ mu ≤ 1.26737.
  6. The lower and upper input deviations are 0.0466938 and 0.0527831. 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 mu ≈ 1.21459; compatible interval [1.16789, 1.26737]. Endpoint interpolation gives 1.74641.

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

Choose a topic