University Advanced Quantum Mechanics · course UAQM
University Advanced Quantum Mechanics
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.
The spin-dependent amplitude g(θ) + i h(θ) σ⋅n: infer an input with uncertainty
For a spin-half particle scattering from a spin-zero target, the amplitude is the 2 × 2 spin matrix M = g(θ) + i h(θ) σ⋅n, where g leaves the spin alone, h acts through the spin component along the normal n to the scattering plane, and an unpolarised beam scatters with dσ/dΩ = |g|² + |h|². Use the supplied topic model Y_out = -2*deg(atan(w*kb^2*sind(th)/4)), where Y_out is SPIN ROTATION ABOUT n (°). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees; deg converts radians to degrees; unmarked trigonometric functions use radians. th is Marker angle θ, measured in °. Keep kb (Momentum times range kb) = 2.5; w (Spin-orbit depth ratio W/U) = 0.4 fixed. For this exercise use only 94.725 ° ≤ 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 = -47.628653 °, with a bounded reading uncertainty ±2.431 °. 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.
SPIN ROTATION ABOUT n against th
Predictions from the supplied The spin-dependent amplitude g(θ) + i h(θ) σ⋅n 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): SPIN ROTATION ABOUT n / °
Horizontal axis (x): th: Marker angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Marker angle θ / ° | SPIN ROTATION ABOUT n / ° |
|---|---|
| 94.725 | -63.835572394075136 |
| 111.26249999999999 | -60.43779102071531 |
| 127.8 | -52.56463894734074 |
| 144.3375 | -40.04155729129355 |
| 160.875 | -23.144833057771717 |
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(94.725) = -63.8356 ° and Y_out(160.875) = -23.1448 °. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives th_line = 94.725 + (-47.6287 − -63.8356)(160.875 − 94.725)/(-23.1448 − -63.8356) = 121.072 °. 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 = 135.077 °; substituting back gives Y_out = -47.628653 °.
- The interpolation error in the input is |th_line − th| = 14.0043 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -50.059653 ° and -45.197653 °. Sort the roots even if the response decreases: 131.66 ° ≤ th ≤ 138.243 °.
- The lower and upper input deviations are 3.41685 ° and 3.16616 °. 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 ≈ 135.077 °; compatible interval [131.66, 138.243] °. Endpoint interpolation gives 121.072 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Covariant notation and the Klein-Gordon equation: infer an input with uncertainty
The Klein-Gordon equation, (1/c²)∂²φ/∂t² − ∇²φ + (mc/ħ)²φ = 0, is the wave equation obtained by making E²/c² − p² = m²c² an operator identity with E = iħ∂/∂t and p = −iħ∇; written as (□ + (mc/ħ)²)φ = 0 it has the same form in every inertial frame. Use the supplied topic model Y_out = 197.327/m, where Y_out is COMPTON LENGTH ħ/mc (fm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. m is Rest energy mc², measured in MeV. Keep p (Momentum pc) = 100 MeV fixed. For this exercise use only 30.1875 MeV ≤ m ≤ 179.813 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 = 1.6246421 fm, with a bounded reading uncertainty ±0.079086 fm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown m by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0149625 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.
COMPTON LENGTH ħ/mc against m
Predictions from the supplied Covariant notation and the Klein-Gordon equation 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): COMPTON LENGTH ħ/mc / fm
Horizontal axis (x): m: Rest energy mc² / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Rest energy mc² / MeV | COMPTON LENGTH ħ/mc / fm |
|---|---|
| 30.1875 | 6.536712215320911 |
| 67.59375 | 2.919308368007397 |
| 105 | 1.8793047619047618 |
| 142.40625 | 1.3856624972569673 |
| 179.8125 | 1.0974042405283282 |
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(m) = 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(30.1875) = 6.53671 fm and Y_out(179.813) = 1.0974 fm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives m_line = 30.1875 + (1.62464 − 6.53671)(179.813 − 30.1875)/(1.0974 − 6.53671) = 165.309 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 m = 121.459 MeV; substituting back gives Y_out = 1.6246421 fm.
- The interpolation error in the input is |m_line − m| = 43.8504 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.5455561 fm and 1.7037281 fm. Sort the roots even if the response decreases: 115.821 MeV ≤ m ≤ 127.674 MeV.
- The lower and upper input deviations are 5.63804 MeV and 6.21504 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 m ≈ 121.459 MeV; compatible interval [115.821, 127.674] MeV. Endpoint interpolation gives 165.309 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Dirac's requirements and the algebra of α and β: infer an input with uncertainty
Dirac's requirements are that the Hamiltonian H = cα⋅p + βmc² be Hermitian and square to p²c² + m²c⁴; they hold exactly when α₁, α₂, α₃ and β are Hermitian, square to 1 and anticommute in pairs, which no ordinary numbers can do. Use the supplied topic model Y_out = 1.022*m*sqrt(max(1-s^2,0)), where Y_out is SOLID GAP AT ITS MINIMUM (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. s is Overlap s, where αβ + βα = 2s, dimensionless. Keep m (Mass m in electron masses) = 1 mₑ; p0 (Probe momentum p₀) = 1 MeV/c fixed. For this exercise use only 0.10625 ≤ s ≤ 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 = 0.82767559 MeV, with a bounded reading uncertainty ±0.028281 MeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown s 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.
SOLID GAP AT ITS MINIMUM against s
Predictions from the supplied Dirac's requirements and the algebra of α and β 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): SOLID GAP AT ITS MINIMUM / MeV
Horizontal axis (x): s: Overlap s, where αβ + βα = 2s
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Overlap s, where αβ + βα = 2s | SOLID GAP AT ITS MINIMUM / MeV |
|---|---|
| 0.10625 | 1.0162149156766742 |
| 0.303125 | 0.9739157214363764 |
| 0.5 | 0.8850779626676962 |
| 0.6968749999999999 | 0.7329708435271745 |
| 0.8937499999999999 | 0.45843386092625193 |
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(s) = 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) = 1.01621 MeV and Y_out(0.89375) = 0.458434 MeV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives s_line = 0.10625 + (0.827676 − 1.01621)(0.89375 − 0.10625)/(0.458434 − 1.01621) = 0.372438. 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 s = 0.586625; substituting back gives Y_out = 0.82767559 MeV.
- The interpolation error in the input is |s_line − s| = 0.214187. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.79939459 MeV and 0.85595659 MeV. Sort the roots even if the response decreases: 0.54639 ≤ s ≤ 0.623044.
- The lower and upper input deviations are 0.0402351 and 0.0364194. 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 s ≈ 0.586625; compatible interval [0.54639, 0.623044]. Endpoint interpolation gives 0.372438.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.