University Advanced Quantum Mechanics · unit UAQM-01
Discrete Symmetries: Parity, Time Reversal and Kramers Degeneracy
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.
Wigner's theorem: every symmetry is unitary or antiunitary: infer an input with uncertainty
A symmetry is a one-to-one map of rays that keeps every transition probability |⟨a|b⟩|², and Wigner's theorem says it is carried out by an operator that is either unitary or antiunitary, fixed up to one overall phase. Use the supplied topic model Y_out = 3*sqrt(3)/8*sind(th)^2*sind(90-th), where Y_out is IM B (dimensionless). 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 Polar angle θ of a, b, c, measured in °. Keep chi (Rotation χ about z, applied after) = 45 °; s (Type s: +1 unitary, −1 antiunitary) = 1 fixed. For this exercise use only 56.8375 ° ≤ th ≤ 82.5625 ° 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 = 0.17741851, with a bounded reading uncertainty ±0.010735. 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 0.0025725 °, 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.
IM B against th
Predictions from the supplied Wigner's theorem: every symmetry is unitary or antiunitary 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 B
Horizontal axis (x): th: Polar angle θ of a, b, c / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Polar angle θ of a, b, c / ° | IM B |
|---|---|
| 56.8375 | 0.24898290403971113 |
| 63.26875 | 0.23304670803299413 |
| 69.7 | 0.19821825562703652 |
| 76.13125 | 0.14674362213060246 |
| 82.5625 | 0.08266796225957404 |
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(56.8375) = 0.248983 and Y_out(82.5625) = 0.082668. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives th_line = 56.8375 + (0.177419 − 0.248983)(82.5625 − 56.8375)/(0.082668 − 0.248983) = 67.9068 °. 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 = 72.5298 °; substituting back gives Y_out = 0.17741851.
- The interpolation error in the input is |th_line − th| = 4.62293 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.16668351 and 0.18815351. Sort the roots even if the response decreases: 71.1221 ° ≤ th ≤ 73.8505 °.
- The lower and upper input deviations are 1.40768 ° and 1.32075 °. 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 ≈ 72.5298 °; compatible interval [71.1221, 73.8505] °. Endpoint interpolation gives 67.9068 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Antiunitary operators and complex conjugation K: infer an input with uncertainty
An antiunitary operator Θ maps kets one-to-one, complex-conjugates every number it passes, Θ(c|a⟩) = c*Θ|a⟩, and conjugates every overlap, ⟨Θa|Θb⟩ = ⟨a|b⟩*, so every transition probability is kept. Use the supplied topic model Y_out = x0*cosd(th)+p0*sind(th), where Y_out is KEPT x cos θ + p sin θ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. th is Angle θ of the basis that K conjugates, measured in °. Keep x0 (Mean position ⟨x⟩ of ψ) = 1.5 √(ħ/mω); p0 (Mean momentum ⟨p⟩ of ψ) = 1 √(ħmω) fixed. For this exercise use only 38.025 ° ≤ th ≤ 160.875 ° 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 = 0.33553661, with a bounded reading uncertainty ±0.21377. 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 0.012285 °, 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.
KEPT x cos θ + p sin θ against th
Predictions from the supplied Antiunitary operators and complex conjugation K 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): KEPT x cos θ + p sin θ
Horizontal axis (x): th: Angle θ of the basis that K conjugates / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Angle θ of the basis that K conjugates / ° | KEPT x cos θ + p sin θ |
|---|---|
| 38.025 | 1.7976183196728561 |
| 68.7375 | 1.4758908187365583 |
| 99.44999999999999 | 0.7401489873232017 |
| 130.1625 | -0.20321815319141523 |
| 160.875 | -1.0895788901386232 |
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(38.025) = 1.79762 and Y_out(160.875) = -1.08958. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives th_line = 38.025 + (0.335537 − 1.79762)(160.875 − 38.025)/(-1.08958 − 1.79762) = 100.236 °. 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 = 112.964 °; substituting back gives Y_out = 0.33553661.
- The interpolation error in the input is |th_line − th| = 12.727 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.12176661 and 0.54930661. Sort the roots even if the response decreases: 105.95 ° ≤ th ≤ 119.817 °.
- The lower and upper input deviations are 7.01358 ° and 6.85363 °. 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 ≈ 112.964 °; compatible interval [105.95, 119.817] °. Endpoint interpolation gives 100.236 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Parity as a unitary symmetry and parity-odd terms: infer an input with uncertainty
Parity is the unitary operator P that inverts position and momentum through the origin while leaving every angular momentum, spin included, unchanged; with its phase chosen so that P² = 1, its eigenvalues are +1 and −1. Use the supplied topic model Y_out = 50*(1-gap/sqrt(gap*gap+4*v*v)), where Y_out is ODD-PARITY WEIGHT (%). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. gap is Gap Δ to the odd level, measured in meV. Keep v (Coupling εM) = 6 meV fixed. For this exercise use only 19.5625 meV ≤ gap ≤ 90.4375 meV 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 = 0.88865932 %, with a bounded reading uncertainty ±0.068135 %. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown gap 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 meV, 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.
ODD-PARITY WEIGHT against gap
Predictions from the supplied Parity as a unitary symmetry and parity-odd terms 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): ODD-PARITY WEIGHT / %
Horizontal axis (x): gap: Gap Δ to the odd level / meV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| gap: Gap Δ to the odd level / meV | ODD-PARITY WEIGHT / % |
|---|---|
| 19.5625 | 7.379725441629182 |
| 37.28125 | 2.404802325049932 |
| 55 | 1.1492119295003878 |
| 72.71875 | 0.6671887047177805 |
| 90.4375 | 0.434426655324921 |
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(gap) = 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(19.5625) = 7.37973 % and Y_out(90.4375) = 0.434427 %. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives gap_line = 19.5625 + (0.888659 − 7.37973)(90.4375 − 19.5625)/(0.434427 − 7.37973) = 85.8022 meV. 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 gap = 62.7963 meV; substituting back gives Y_out = 0.88865932 %.
- The interpolation error in the input is |gap_line − gap| = 23.0059 meV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.82052432 % and 0.95679432 %. Sort the roots even if the response decreases: 60.4559 meV ≤ gap ≤ 65.4197 meV.
- The lower and upper input deviations are 2.34038 meV and 2.62344 meV. 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 gap ≈ 62.7963 meV; compatible interval [60.4559, 65.4197] meV. Endpoint interpolation gives 85.8022 meV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Why time reversal must be antiunitary: infer an input with uncertainty
Time reversal Θ reverses motion, ΘxΘ⁻¹ = x and ΘpΘ⁻¹ = −p, while evolution keeps running forward, and it must be antiunitary: a unitary Θ obeying ΘU(t) = U(−t)Θ would send every energy E to −E. Use the supplied topic model Y_out = (ell-f)^2, where Y_out is ENERGY OF STATE ℓ (ε). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. f is Flux Φ/Φ₀, with Φ₀ = h/q, dimensionless. Keep ell (Angular momentum ℓ = L_z/ħ) = 2 fixed. For this exercise use only -0.354375 ≤ f ≤ 0.354375 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 = 3.6942282 ε, with a bounded reading uncertainty ±0.14792 ε. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown f by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000070875, 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.
ENERGY OF STATE ℓ against f
Predictions from the supplied Why time reversal must be antiunitary 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): ENERGY OF STATE ℓ / ε
Horizontal axis (x): f: Flux Φ/Φ₀, with Φ₀ = h/q
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Flux Φ/Φ₀, with Φ₀ = h/q | ENERGY OF STATE ℓ / ε |
|---|---|
| -0.354375 | 5.543081640625 |
| -0.1771875 | 4.74014541015625 |
| 0 | 4 |
| 0.17718749999999994 | 3.32264541015625 |
| 0.354375 | 2.7080816406249997 |
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(f) = 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(-0.354375) = 5.54308 ε and Y_out(0.354375) = 2.70808 ε. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives f_line = -0.354375 + (3.69423 − 5.54308)(0.354375 − -0.354375)/(2.70808 − 5.54308) = 0.107838. 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 f = 0.0779625; substituting back gives Y_out = 3.6942282 ε.
- The interpolation error in the input is |f_line − f| = 0.0298759. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.5463082 ε and 3.8421482 ε. Sort the roots even if the response decreases: 0.0398602 ≤ f ≤ 0.116836.
- The lower and upper input deviations are 0.0381023 and 0.0388731. 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 f ≈ 0.0779625; compatible interval [0.0398602, 0.116836]. Endpoint interpolation gives 0.107838.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.