University Advanced Quantum Mechanics · unit UAQM-07
The Dirac Equation in a Central Field and Hydrogen Solved Exactly
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.
Conserved operators in a central field: J and parity, not L: infer an input with uncertainty
In a central electrostatic potential the Dirac Hamiltonian commutes with the total angular momentum J = L + (ħ/2)Σ and with the parity operator γ⁰P₀, but not with L² or S⋅L, so states are labelled by j, m_j and parity η instead of by l. Use the supplied topic model Y_out = (2*j+1)*sqrt((wl/100)*(1-wl/100)), where Y_out is SPREAD ΔL² (ħ²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. wl is Lower-component weight w, measured in %. Keep j (Total angular momentum j) = 1.5; par (Parity η) = -1 fixed. For this exercise use only 3.1875 % ≤ wl ≤ 26.8125 % 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 = 1.5232381 ħ², with a bounded reading uncertainty ±0.037304 ħ². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown wl by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0023625 %, 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 ΔL² against wl
Predictions from the supplied Conserved operators in a central field: J and parity, not L 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 ΔL² / ħ²
Horizontal axis (x): wl: Lower-component weight w / %
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| wl: Lower-component weight w / % | SPREAD ΔL² / ħ² |
|---|---|
| 3.1875 | 0.7026690188132675 |
| 9.09375 | 1.1500808395499857 |
| 14.999999999999998 | 1.42828568570857 |
| 20.906249999999996 | 1.6265564661271368 |
| 26.812499999999996 | 1.7719322080711777 |
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(wl) = 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(3.1875) = 0.702669 ħ² and Y_out(26.8125) = 1.77193 ħ². The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives wl_line = 3.1875 + (1.52324 − 0.702669)(26.8125 − 3.1875)/(1.77193 − 0.702669) = 21.3177 %. 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 wl = 17.5987 %; substituting back gives Y_out = 1.5232381 ħ².
- The interpolation error in the input is |wl_line − wl| = 3.71894 %. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.4859341 ħ² and 1.5605421 ħ². Sort the roots even if the response decreases: 16.5336 % ≤ wl ≤ 18.7279 %.
- The lower and upper input deviations are 1.06515 % and 1.12918 %. 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 wl ≈ 17.5987 %; compatible interval [16.5336, 18.7279] %. Endpoint interpolation gives 21.3177 %.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The operator K = β(Σ⋅L + ħ) and the quantum number κ: infer an input with uncertainty
κ is the non-zero integer defined by Kψ = −κħψ, where K = β(Σ⋅L + ħ) is Dirac's conserved operator: |κ| = j + ½, and its sign says whether the upper component has l = j − ½ (κ < 0) or l = j + ½ (κ > 0). Use the supplied topic model Y_out = w*(j+0.5), where Y_out is κ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. j is Total angular momentum j, dimensionless. Keep w (Sign of κ) = -1 fixed. For this exercise use only 0.5 ≤ j ≤ 4.5, at the specified test settings separated by 1. 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, with a bounded reading uncertainty ±0.32. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of j consistent with that bound. Compare with a straight-line interpolation between the two endpoint predictions, and explain why a fractional setting may be inadmissible. Do not assume that the supplied calibration points prove monotonicity outside this branch.
κ against j
Predictions from the supplied The operator K = β(Σ⋅L + ħ) and the quantum number κ 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): κ
Horizontal axis (x): j: Total angular momentum j
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| j: Total angular momentum j | κ |
|---|---|
| 0.5 | -1 |
| 1.5 | -2 |
| 2.5 | -3 |
| 3.5 | -4 |
| 4.5 | -5 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Enumerate the permitted settings in the stated range, evaluate Y_out at each, and retain every setting satisfying |Y_out − Y_measured| ≤ u. The plotted predictions are supplied calibration evidence.
Reveal the worked solution
- Endpoint predictions are Y_out(0.5) = -1 and Y_out(4.5) = -5. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives j_line = 0.5 + (-3 − -1)(4.5 − 0.5)/(-5 − -1) = 2.5. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Evaluate the model at each permitted setting. The central reading corresponds to j = 2.5; only the allowed settings 2.5 lie inside the stated reading bound.
- The interpolation error in the input is |j_line − j| = 0.
- The compatible allowed values span 2.5 to 2.5. Values between specified test settings are outside this exercise's input domain even if an interpolation passes through them.
- The quoted output bound restricts which specified settings remain compatible. It does not change physical restrictions on quantum numbers or counts, and the chosen grid alone does not establish physical quantisation.
Answer Compatible j: 2.5. Endpoint interpolation predicts 2.5; check admissibility against the discrete settings.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Spinor spherical harmonics and the action of σ⋅r̂: infer an input with uncertainty
A spinor spherical harmonic χ_κᵐ is Quantum Mechanics II's coupled state |l, ½, j, m⟩ written as a two-component function of direction; a central-field Dirac state carries χ_κᵐ in its upper pair and χ_(−κ)ᵐ in its lower, and σ⋅r̂ turns each into minus the other. Use the supplied topic model Y_out = max(0,1-abs(n-1))+max(0,1-abs(n-2))*(1+3*cosd(th)^2)/2+max(0,1-abs(n-3))*(15*cosd(th)^4-6*cosd(th)^2+3)/4, where Y_out is 4π|χ|² AT θ, BOTH (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; abs(x) means |x|; max(a,b) selects the larger value; sind/cosd/tand use degrees. n is |κ| = j + ½, dimensionless. Keep s (Sign of κ) = -1; th (Polar angle θ at φ = 0) = 60 ° fixed. For this exercise use only 1 ≤ n ≤ 3, at the specified test settings separated by 1. 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 = 0.875, with a bounded reading uncertainty ±0.03125. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of n consistent with that bound. Compare with a straight-line interpolation between the two endpoint predictions, and explain why a fractional setting may be inadmissible. Do not assume that the supplied calibration points prove monotonicity outside this branch.
4π|χ|² AT θ, BOTH against n
Predictions from the supplied Spinor spherical harmonics and the action of σ⋅r̂ model, with all other inputs fixed as stated in the question. 3 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): 4π|χ|² AT θ, BOTH
Horizontal axis (x): n: |κ| = j + ½
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| n: |κ| = j + ½ | 4π|χ|² AT θ, BOTH |
|---|---|
| 1 | 1 |
| 2 | 0.8750000000000002 |
| 3 | 0.609375 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Enumerate the permitted settings in the stated range, evaluate Y_out at each, and retain every setting satisfying |Y_out − Y_measured| ≤ u. The plotted predictions are supplied calibration evidence.
Reveal the worked solution
- Endpoint predictions are Y_out(1) = 1 and Y_out(3) = 0.609375. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives n_line = 1 + (0.875 − 1)(3 − 1)/(0.609375 − 1) = 1.64. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Evaluate the model at each permitted setting. The central reading corresponds to n = 2; only the allowed settings 2 lie inside the stated reading bound.
- The interpolation error in the input is |n_line − n| = 0.36.
- The compatible allowed values span 2 to 2. Values between specified test settings are outside this exercise's input domain even if an interpolation passes through them.
- The quoted output bound restricts which specified settings remain compatible. It does not change physical restrictions on quantum numbers or counts, and the chosen grid alone does not establish physical quantisation.
Answer Compatible n: 2. Endpoint interpolation predicts 1.64; check admissibility against the discrete settings.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The coupled radial equations for G(r) and F(r): infer an input with uncertainty
The coupled radial equations are the two real first-order equations, one from each block row of the Dirac equation, obeyed by the radial functions G(r) and F(r) of an electron's upper and lower components in any central potential. Use the supplied topic model Y_out = (p/(1+sqrt(1+p^2))), where Y_out is F/G AMPLITUDE cp/(E + mc²) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. p is Momentum p, measured in mc. Keep st (State: κ = −2, −1, +1, +2) = 2 fixed. For this exercise use only 0.469687 mc ≤ p ≤ 3.58031 mc 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 = 0.66312351, with a bounded reading uncertainty ±0.014377. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown p 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 mc, 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.
F/G AMPLITUDE cp/(E + mc²) against p
Predictions from the supplied The coupled radial equations for G(r) and F(r) 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): F/G AMPLITUDE cp/(E + mc²)
Horizontal axis (x): p: Momentum p / mc
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Momentum p / mc | F/G AMPLITUDE cp/(E + mc²) |
|---|---|
| 0.4696875 | 0.22314953731926265 |
| 1.24734375 | 0.4799861607722101 |
| 2.025 | 0.6214599166488232 |
| 2.80265625 | 0.7049438851187437 |
| 3.5803125 | 0.758968054915741 |
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(p) = 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) = 0.22315 and Y_out(3.58031) = 0.758968. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives p_line = 0.469687 + (0.663124 − 0.22315)(3.58031 − 0.469687)/(0.758968 − 0.22315) = 3.0239 mc. 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 p = 2.36717 mc; substituting back gives Y_out = 0.66312351.
- The interpolation error in the input is |p_line − p| = 0.656731 mc. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.64874651 and 0.67750051. Sort the roots even if the response decreases: 2.24043 mc ≤ p ≤ 2.50466 mc.
- The lower and upper input deviations are 0.126743 mc and 0.137486 mc. 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 p ≈ 2.36717 mc; compatible interval [2.24043, 2.50466] mc. Endpoint interpolation gives 3.0239 mc.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.