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.
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.
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
Horizontal axis (x): th: Angle θ between ε and d_ba / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Angle θ between ε and d_ba / ° | EMISSION RABI 2g√(n+1)/2π / kHz |
|---|---|
| 9.5625 | 65.54446244073816 |
| 27.28125 | 59.07461668780822 |
| 45 | 47.000000000000014 |
| 62.71875 | 30.46620526400575 |
| 80.4375 | 11.04189491684587 |
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
- 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.
- 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.
- 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.
- The interpolation error in the input is |th_line − th| = 10.2628 °. This is larger than the requested numerical tolerance.
- 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 °.
- 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.
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.
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
Horizontal axis (x): th: Viewing angle θ from z / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Viewing angle θ from z / ° | POLARISATION, + IN z–k PLANE |
|---|---|
| 19.125 | 0.0969362488532642 |
| 35.6625 | 0.25367574167316426 |
| 52.2 | 0.38436721587316286 |
| 68.73750000000001 | 0.4648088375752355 |
| 85.275 | 0.4982978823970005 |
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
- 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.
- 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.
- 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.
- The interpolation error in the input is |th_line − th| = 13.8766 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.41511235 and 0.43681235. Sort the roots even if the response decreases: 57.4003 ° ≤ th ≤ 61.7252 °.
- 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.
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.
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
Horizontal axis (x): nf: Field photons per mode n̄
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| nf: Field photons per mode n̄ | STIMULATED SHARE OF EMISSION |
|---|---|
| 0.4696875 | 0.31958324473740163 |
| 1.24734375 | 0.5550302440381006 |
| 2.025 | 0.6694214876033058 |
| 2.80265625 | 0.7370259275999507 |
| 3.5803125 | 0.7816742853244184 |
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
- 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.
- 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.
- 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.
- The interpolation error in the input is |nf_line − nf| = 0.683636. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.69121559 and 0.71481359. Sort the roots even if the response decreases: 2.23851 ≤ nf ≤ 2.50648.
- 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.
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.
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
Horizontal axis (x): mu: Transition moment |⟨f|μ|i⟩| / μ_B
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| mu: Transition moment |⟨f|μ|i⟩| / μ_B | E1 / M1 RATE AT EQUAL ω |
|---|---|
| 0.301875 | 824275.4694820587 |
| 0.6759375 | 164404.45785880258 |
| 1.0499999999999998 | 68131.51927437645 |
| 1.4240624999999998 | 37039.79471689857 |
| 1.7981249999999998 | 23232.01680196117 |
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
- 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.
- 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.
- 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.
- The interpolation error in the input is |mu_line − mu| = 0.531824. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 46764.833 and 55070.633. Sort the roots even if the response decreases: 1.16789 ≤ mu ≤ 1.26737.
- 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.