University Advanced Quantum Mechanics · unit UAQM-09
Quantised Free Fields and the Spin-Statistics Connection
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.
Lagrangian field theory and Noether's theorem: infer an input with uncertainty
A Lagrangian density ℒ(φ, ∂_μφ) is the Lagrangian per unit volume of a field, a system with a degree of freedom at every point; making ∫dt∫d³x ℒ stationary gives the field equation, and each continuous symmetry of it gives a conserved current. Use the supplied topic model Y_out = 4*kw/sqrt(w0^2+4*kw^2), where Y_out is FIELD ENERGY SPEED v²K/ω (m s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. kw is Marked wavenumber K, measured in rad m⁻¹. Keep w0 (On-site frequency ω₀ (mass term)) = 8 s⁻¹; a (Bead spacing a) = 0.2 m fixed. For this exercise use only 1.50937 rad m⁻¹ ≤ kw ≤ 8.99062 rad m⁻¹ 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 = 1.6702472 m s⁻¹, with a bounded reading uncertainty ±0.023559 m s⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown kw by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000748125 rad m⁻¹, 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.
FIELD ENERGY SPEED v²K/ω against kw
Predictions from the supplied Lagrangian field theory and Noether's theorem 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): FIELD ENERGY SPEED v²K/ω / m s⁻¹
Horizontal axis (x): kw: Marked wavenumber K / rad m⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| kw: Marked wavenumber K / rad m⁻¹ | FIELD ENERGY SPEED v²K/ω / m s⁻¹ |
|---|---|
| 1.509375 | 0.7060903602492159 |
| 3.3796874999999997 | 1.2907873497081457 |
| 5.25 | 1.5908634400649975 |
| 7.120312499999999 | 1.7436924091671628 |
| 8.990625 | 1.8273086624719388 |
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(kw) = 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(1.50937) = 0.70609 m s⁻¹ and Y_out(8.99062) = 1.82731 m s⁻¹. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives kw_line = 1.50937 + (1.67025 − 0.70609)(8.99062 − 1.50937)/(1.82731 − 0.70609) = 7.94264 rad m⁻¹. 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 kw = 6.07294 rad m⁻¹; substituting back gives Y_out = 1.6702472 m s⁻¹.
- The interpolation error in the input is |kw_line − kw| = 1.86971 rad m⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.6466882 m s⁻¹ and 1.6938062 m s⁻¹. Sort the roots even if the response decreases: 5.80287 rad m⁻¹ ≤ kw ≤ 6.37072 rad m⁻¹.
- The lower and upper input deviations are 0.270066 rad m⁻¹ and 0.297781 rad m⁻¹. 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 kw ≈ 6.07294 rad m⁻¹; compatible interval [5.80287, 6.37072] rad m⁻¹. Endpoint interpolation gives 7.94264 rad m⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Canonical quantisation of the real Klein-Gordon field: infer an input with uncertainty
Canonical quantisation of a real field imposes [φ(x), π(y)] = iħ δ³(x − y) at equal times, which makes each normal mode of the field an independent quantum oscillator whose quanta are the field's particles. Use the supplied topic model Y_out = sqrt((197.327*k)^2+m^2), where Y_out is ONE QUANTUM ħωₖ (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. k is Mode wavenumber k, measured in fm⁻¹. Keep m (Mass of the quantum m) = 135 MeV/c²; n (Quanta in mode k, n) = 1 fixed. For this exercise use only 0.0525 fm⁻¹ ≤ k ≤ 0.7875 fm⁻¹ 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 = 167.30993 MeV, with a bounded reading uncertainty ±4.787 MeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown k 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 fm⁻¹, 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.
ONE QUANTUM ħωₖ against k
Predictions from the supplied Canonical quantisation of the real Klein-Gordon field 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): ONE QUANTUM ħωₖ / MeV
Horizontal axis (x): k: Mode wavenumber k / fm⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: Mode wavenumber k / fm⁻¹ | ONE QUANTUM ħωₖ / MeV |
|---|---|
| 0.05249999999999999 | 135.39690805446983 |
| 0.23624999999999996 | 142.82256436532978 |
| 0.41999999999999993 | 158.40976448904783 |
| 0.6037499999999999 | 180.05118297714975 |
| 0.7874999999999999 | 205.84608305691694 |
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(k) = 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) = 135.397 MeV and Y_out(0.7875) = 205.846 MeV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives k_line = 0.0525 + (167.31 − 135.397)(0.7875 − 0.0525)/(205.846 − 135.397) = 0.38545 fm⁻¹. 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 k = 0.50085 fm⁻¹; substituting back gives Y_out = 167.30993 MeV.
- The interpolation error in the input is |k_line − k| = 0.1154 fm⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 162.52293 MeV and 172.09693 MeV. Sort the roots even if the response decreases: 0.458586 fm⁻¹ ≤ k ≤ 0.540904 fm⁻¹.
- The lower and upper input deviations are 0.0422638 fm⁻¹ and 0.040054 fm⁻¹. 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 k ≈ 0.50085 fm⁻¹; compatible interval [0.458586, 0.540904] fm⁻¹. Endpoint interpolation gives 0.38545 fm⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Complex scalar field: antiparticles and the charge operator: infer an input with uncertainty
The complex scalar field is a spin-0 field with φ ≠ φ†, whose quanta come in two kinds, particles created by a_k† and antiparticles created by b_k†, with the same mass and positive energy and opposite values of the conserved charge Q. Use the supplied topic model Y_out = 0.13298*(na-nb)*exp(-d*d/72), where Y_out is CHARGE DENSITY AT x = 0 (q fm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. d is Packet separation d, measured in fm. Keep na (Particles N_a in the left packet) = 2; nb (Antiparticles N_b in the right packet) = 1 fixed. For this exercise use only 1.4875 fm ≤ d ≤ 12.5125 fm 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 = 0.052112426 q fm⁻¹, with a bounded reading uncertainty ±0.0055495 q fm⁻¹. 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 0.0011025 fm, 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.
CHARGE DENSITY AT x = 0 against d
Predictions from the supplied Complex scalar field: antiparticles and the charge operator 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): CHARGE DENSITY AT x = 0 / q fm⁻¹
Horizontal axis (x): d: Packet separation d / fm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Packet separation d / fm | CHARGE DENSITY AT x = 0 / q fm⁻¹ |
|---|---|
| 1.4875 | 0.12895550265135175 |
| 4.2437499999999995 | 0.10355138781380546 |
| 6.999999999999999 | 0.06733251030186442 |
| 9.75625 | 0.035452472267395146 |
| 12.5125 | 0.015115450069993373 |
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(1.4875) = 0.128956 q fm⁻¹ and Y_out(12.5125) = 0.0151155 q fm⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 1.4875 + (0.0521124 − 0.128956)(12.5125 − 1.4875)/(0.0151155 − 0.128956) = 8.92948 fm. 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 = 8.21275 fm; substituting back gives Y_out = 0.052112426 q fm⁻¹.
- The interpolation error in the input is |d_line − d| = 0.716726 fm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.046562926 q fm⁻¹ and 0.057661926 q fm⁻¹. Sort the roots even if the response decreases: 7.7565 fm ≤ d ≤ 8.69232 fm.
- The lower and upper input deviations are 0.456248 fm and 0.479567 fm. 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 ≈ 8.21275 fm; compatible interval [7.7565, 8.69232] fm. Endpoint interpolation gives 8.92948 fm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Microcausality: [φ(x), φ†(y)] = 0 at spacelike separation: infer an input with uncertainty
Microcausality is the requirement that every local observable commute with every other at spacelike separation, which for a scalar field means the field operators themselves commute there, so that nothing done at one event can change the statistics of measurements at an event outside its light cone. Use the supplied topic model Y_out = 1-w, where Y_out is WEIGHT OF Re D IN Re C (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. w is Antiparticle weight w, dimensionless. Keep sig (Probe √|x²|, spacelike if negative) = -1.5 ħ/mc fixed. For this exercise use only 0.10625 ≤ w ≤ 0.89375 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 = 0.413375, with a bounded reading uncertainty ±0.046069. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown w 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, 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.
WEIGHT OF Re D IN Re C against w
Predictions from the supplied Microcausality: [φ(x), φ†(y)] = 0 at spacelike separation 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): WEIGHT OF Re D IN Re C
Horizontal axis (x): w: Antiparticle weight w
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Antiparticle weight w | WEIGHT OF Re D IN Re C |
|---|---|
| 0.10625 | 0.89375 |
| 0.303125 | 0.696875 |
| 0.5 | 0.5 |
| 0.6968749999999999 | 0.3031250000000001 |
| 0.8937499999999999 | 0.10625000000000007 |
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(w) = 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(0.10625) = 0.89375 and Y_out(0.89375) = 0.10625. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 0.10625 + (0.413375 − 0.89375)(0.89375 − 0.10625)/(0.10625 − 0.89375) = 0.586625. 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 w = 0.586625; substituting back gives Y_out = 0.413375.
- The interpolation error in the input is |w_line − w| = 1.11022e-16. This is within the requested numerical tolerance.
- Invert both limits Y_out = 0.367306 and 0.459444. Sort the roots even if the response decreases: 0.540556 ≤ w ≤ 0.632694.
- The lower and upper input deviations are 0.046069 and 0.046069. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer w ≈ 0.586625; compatible interval [0.540556, 0.632694]. Endpoint interpolation gives 0.586625.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.