University Advanced Quantum Mechanics · unit UAQM-11
The Quantised Radiation 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.
Maxwell's equations in Coulomb gauge: infer an input with uncertainty
Coulomb gauge is the choice ∇⋅A = 0, under which the scalar potential is the instantaneous Coulomb potential of the charges and the vector potential is a purely transverse field driven by the transverse part of the current. Use the supplied topic model Y_out = (clamp(t-d,0,1) + d - d*cos(2*pi*clamp(t-d,0,1)) + (2*pi*d^2 - 1/(2*pi))*sin(2*pi*clamp(t-d,0,1))) - (clamp(t,0,1) - sin(2*pi*clamp(t,0,1))/(2*pi)), where Y_out is TRANSVERSE PART E⊥/E₀ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; clamp(x,a,b) confines x to [a,b]; unmarked trigonometric functions use radians. d is Distance r, in units of cτ, dimensionless. Keep t (Time t, in units of τ) = 0.4 fixed. For this exercise use only 0.258125 ≤ d ≤ 0.389375 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 = -0.007591162, with a bounded reading uncertainty ±0.021125. 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 0.000013125, 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.
TRANSVERSE PART E⊥/E₀ against d
Predictions from the supplied Maxwell's equations in Coulomb gauge 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): TRANSVERSE PART E⊥/E₀
Horizontal axis (x): d: Distance r, in units of cτ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Distance r, in units of cτ | TRANSVERSE PART E⊥/E₀ |
|---|---|
| 0.258125 | 0.13323914030165523 |
| 0.2909375 | 0.10415067669368522 |
| 0.32375 | 0.03646460630413573 |
| 0.3565625 | -0.07739295427613299 |
| 0.389375 | -0.2420277904365249 |
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
- Endpoint predictions are Y_out(0.258125) = 0.133239 and Y_out(0.389375) = -0.242028. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 0.258125 + (-0.00759116 − 0.133239)(0.389375 − 0.258125)/(-0.242028 − 0.133239) = 0.307381. 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 d = 0.338187; substituting back gives Y_out = -0.007591162.
- The interpolation error in the input is |d_line − d| = 0.030807. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.028716162 and 0.013533838. Sort the roots even if the response decreases: 0.33164 ≤ d ≤ 0.344203.
- The lower and upper input deviations are 0.00654728 and 0.0060156. 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 ≈ 0.338187; compatible interval [0.33164, 0.344203]. Endpoint interpolation gives 0.307381.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The transverse field as a set of oscillators: infer an input with uncertainty
A mode of the free transverse field is one allowed wavevector k together with one of the two polarisation directions perpendicular to it; its amplitude evolves as an independent harmonic oscillator of angular frequency ω = c|k|. Use the supplied topic model Y_out = 1+b^2, where Y_out is TOTAL U ÷ (ε₀ω²A₀²V/4) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b is Initial E as a fraction of ωA₀, β, dimensionless. Keep ph (Phase of the oscillation ωt) = 0 °; n (Wavelengths across the box n) = 1 fixed. For this exercise use only 0.0525 ≤ b ≤ 0.7875 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 = 1.2508507, with a bounded reading uncertainty ±0.037214. 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 0.0000735, 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.
TOTAL U ÷ (ε₀ω²A₀²V/4) against b
Predictions from the supplied The transverse field as a set of oscillators 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): TOTAL U ÷ (ε₀ω²A₀²V/4)
Horizontal axis (x): b: Initial E as a fraction of ωA₀, β
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Initial E as a fraction of ωA₀, β | TOTAL U ÷ (ε₀ω²A₀²V/4) |
|---|---|
| 0.05249999999999999 | 1.00275625 |
| 0.23624999999999996 | 1.0558140625 |
| 0.41999999999999993 | 1.1764 |
| 0.6037499999999999 | 1.3645140624999998 |
| 0.7874999999999999 | 1.6201562499999997 |
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
- Endpoint predictions are Y_out(0.0525) = 1.00276 and Y_out(0.7875) = 1.62016. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives b_line = 0.0525 + (1.25085 − 1.00276)(0.7875 − 0.0525)/(1.62016 − 1.00276) = 0.347851. 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 b = 0.50085; substituting back gives Y_out = 1.2508507.
- The interpolation error in the input is |b_line − b| = 0.152999. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.2136367 and 1.2880647. Sort the roots even if the response decreases: 0.462209 ≤ b ≤ 0.536717.
- The lower and upper input deviations are 0.0386415 and 0.0358666. 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 b ≈ 0.50085; compatible interval [0.462209, 0.536717]. Endpoint interpolation gives 0.347851.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Quantising the field: A, E and B as operators: infer an input with uncertainty
Quantising the radiation field means promoting each transverse mode's amplitude to an oscillator ladder operator, so that the vector potential A and the fields E and B become operator-valued functions of position whose commutators all follow from [a, a†] = 1. Use the supplied topic model Y_out = 9.47802*(1-3*cosd(th)^2)*t0/r^3*(1+sign(3.33564*r-abs(t0)))/2, where Y_out is A–A COMMUTATOR / i AT PROBE (10⁻³⁴ T² m²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; abs(x) means |x|; sind/cosd/tand use degrees. r is Separation r, measured in m. Keep th (Angle θ between r and the x axis) = 0 °; t0 (Probe time t₀) = 2 ns fixed. For this exercise use only 1.10625 m ≤ r ≤ 1.89375 m 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 = -9.4919345 10⁻³⁴ T² m², with a bounded reading uncertainty ±0.58645 10⁻³⁴ T² m². 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 0.00007875 m, 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.
A–A COMMUTATOR / i AT PROBE against r
Predictions from the supplied Quantising the field: A, E and B as operators 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): A–A COMMUTATOR / i AT PROBE / 10⁻³⁴ T² m²
Horizontal axis (x): r: Separation r / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Separation r / m | A–A COMMUTATOR / i AT PROBE / 10⁻³⁴ T² m² |
|---|---|
| 1.10625 | -28.003851178120023 |
| 1.3031249999999999 | -17.132446415105385 |
| 1.5 | -11.233208888888889 |
| 1.696875 | -7.75939853172152 |
| 1.8937499999999998 | -5.582255041110426 |
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
- Endpoint predictions are Y_out(1.10625) = -28.0039 10⁻³⁴ T² m² and Y_out(1.89375) = -5.58226 10⁻³⁴ T² m². The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives r_line = 1.10625 + (-9.49193 − -28.0039)(1.89375 − 1.10625)/(-5.58226 − -28.0039) = 1.75643 m. 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 r = 1.58663 m; substituting back gives Y_out = -9.4919345 10⁻³⁴ T² m².
- The interpolation error in the input is |r_line − r| = 0.169808 m. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -10.078384 10⁻³⁴ T² m² and -8.9054845 10⁻³⁴ T² m². Sort the roots even if the response decreases: 1.55523 m ≤ r ≤ 1.62072 m.
- The lower and upper input deviations are 0.0313916 m and 0.0340901 m. 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 ≈ 1.58663 m; compatible interval [1.55523, 1.62072] m. Endpoint interpolation gives 1.75643 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Photons: energy, momentum, polarisation and helicity: infer an input with uncertainty
A photon is one quantum of a single transverse mode of the electromagnetic field, the state a†(k, λ)|0⟩, carrying energy ħω, momentum ħk with ω = c|k|, and, when λ is a circular polarisation, angular momentum ±ħ along k̂. Use the supplied topic model Y_out = abs(cosd(2*chi)), where Y_out is SPREAD OF S⋅k̂ (ħ). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; abs(x) means |x|; sind/cosd/tand use degrees. chi is Ellipticity angle χ, measured in °. Keep psi (Major-axis angle ψ from e₁) = 30 °; wt (Phase ωt) = 45 ° fixed. For this exercise use only -35.4375 ° ≤ chi ≤ -2.3625 ° 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 = 0.86142092 ħ, with a bounded reading uncertainty ±0.020277 ħ. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown chi by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0033075 °, 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.
SPREAD OF S⋅k̂ against chi
Predictions from the supplied Photons: energy, momentum, polarisation and helicity 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): SPREAD OF S⋅k̂ / ħ
Horizontal axis (x): chi: Ellipticity angle χ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| chi: Ellipticity angle χ / ° | SPREAD OF S⋅k̂ / ħ |
|---|---|
| -35.4375 | 0.32763017956169366 |
| -27.16875 | 0.5830095789833509 |
| -18.9 | 0.7901550123756904 |
| -10.631249999999998 | 0.9319287753895013 |
| -2.362499999999997 | 0.9966015395273318 |
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(chi) = 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(-35.4375) = 0.32763 ħ and Y_out(-2.3625) = 0.996602 ħ. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives chi_line = -35.4375 + (0.861421 − 0.32763)(-2.3625 − -35.4375)/(0.996602 − 0.32763) = -9.04604 °. 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 chi = -15.2617 °; substituting back gives Y_out = 0.86142092 ħ.
- The interpolation error in the input is |chi_line − chi| = 6.21571 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.84114392 ħ and 0.88169792 ħ. Sort the roots even if the response decreases: -16.3694 ° ≤ chi ≤ -14.0761 °.
- The lower and upper input deviations are 1.10769 ° and 1.18568 °. 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 chi ≈ -15.2617 °; compatible interval [-16.3694, -14.0761] °. Endpoint interpolation gives -9.04604 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.