University Quantum Mechanics II · unit UQMII-09
Hydrogen in External Fields and Hyperfine Structure
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.
The Zeeman Hamiltonian, and what counts as a weak field: infer an input with uncertainty
The Zeeman Hamiltonian H'_Z = (μ_B B/ħ)(L_z + 2S_z) is the energy of hydrogen's orbital and spin moments in a uniform field B along z; the field is weak or strong for a level as μ_B B is much smaller or much larger than the fine-structure splitting it competes with. Use the supplied topic model Y_out = 10^lb, where Y_out is FIELD B (T). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lb is Field, log₁₀(B / 1 T), dimensionless. Keep n (Principal quantum number n (l = 1)) = 2 fixed. For this exercise use only -2.25625 ≤ lb ≤ 3.25625 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 = 12.775415 T, with a bounded reading uncertainty ±1.9155 T. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lb by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00055125, 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 B against lb
Predictions from the supplied The Zeeman Hamiltonian, and what counts as a weak 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): FIELD B / T
Horizontal axis (x): lb: Field, log₁₀(B / 1 T)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lb: Field, log₁₀(B / 1 T) | FIELD B / T |
|---|---|
| -2.25625 | 0.0055430653660835045 |
| -0.8781250000000003 | 0.13239604139104352 |
| 0.49999999999999956 | 3.162277660168376 |
| 1.8781249999999994 | 75.53095919585759 |
| 3.256249999999999 | 1804.055940091783 |
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(lb) = 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(-2.25625) = 0.00554307 T and Y_out(3.25625) = 1804.06 T. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives lb_line = -2.25625 + (12.7754 − 0.00554307)(3.25625 − -2.25625)/(1804.06 − 0.00554307) = -2.21723. 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 lb = 1.10638; substituting back gives Y_out = 12.775415 T.
- The interpolation error in the input is |lb_line − lb| = 3.32361. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 10.859915 T and 14.690915 T. Sort the roots even if the response decreases: 1.03583 ≤ lb ≤ 1.16705.
- The lower and upper input deviations are 0.0705486 and 0.0606738. 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 lb ≈ 1.10638; compatible interval [1.03583, 1.16705]. Endpoint interpolation gives -2.21723.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The weak-field Zeeman effect and the Landé g-factor: infer an input with uncertainty
In a field weak beside the fine structure, a level of angular momentum j splits into 2j + 1 evenly spaced sublevels at E_Z = gⱼμ_B B mⱼ, where the Landé factor gⱼ measures how much of the moment L + 2S lies along J. Use the supplied topic model Y_out = 100*(4/9)*(gs-1)*(gs-1)*(13.996*b/10.969)/(1-(gs-1)/3), where Y_out is 2P₁/₂ SECOND-ORDER ERROR (%). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. gs is Spin g-factor gₛ in the Zeeman term, dimensionless. Keep b (Magnetic field B) = 0.1 T fixed. For this exercise use only 1.10625 ≤ gs ≤ 1.89375 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 = 2.4258948 %, with a bounded reading uncertainty ±0.35393 %. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown gs 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.
2P₁/₂ SECOND-ORDER ERROR against gs
Predictions from the supplied The weak-field Zeeman effect and the Landé g-factor 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): 2P₁/₂ SECOND-ORDER ERROR / %
Horizontal axis (x): gs: Spin g-factor gₛ in the Zeeman term
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| gs: Spin g-factor gₛ in the Zeeman term | 2P₁/₂ SECOND-ORDER ERROR / % |
|---|---|
| 1.10625 | 0.06637010474115115 |
| 1.3031249999999999 | 0.579639974180545 |
| 1.5 | 1.7012793630534537 |
| 1.696875 | 3.587301621662364 |
| 1.8937499999999998 | 6.452051501313068 |
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(gs) = 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) = 0.0663701 % and Y_out(1.89375) = 6.45205 %. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives gs_line = 1.10625 + (2.42589 − 0.0663701)(1.89375 − 1.10625)/(6.45205 − 0.0663701) = 1.39723. 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 gs = 1.58663; substituting back gives Y_out = 2.4258948 %.
- The interpolation error in the input is |gs_line − gs| = 0.189392. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.0719648 % and 2.7798248 %. Sort the roots even if the response decreases: 1.54662 ≤ gs ≤ 1.62319.
- The lower and upper input deviations are 0.0400047 and 0.036562. 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 gs ≈ 1.58663; compatible interval [1.54662, 1.62319]. Endpoint interpolation gives 1.39723.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The Paschen-Back limit and the normal Zeeman pattern: infer an input with uncertainty
The Paschen-Back limit is the regime in which an atom's Zeeman energy μ_B B far exceeds its fine-structure splitting, so the good states are |n l mₗ mₛ⟩ with energies Eₙ + μ_B B(mₗ + 2mₛ), and fine structure is a small diagonal correction to them. Use the supplied topic model Y_out = k*k*1.9035/b, where Y_out is NEGLECTED 2ND-ORDER SHIFT (GHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. k is Fine-structure dial k (1 = hydrogen), dimensionless. Keep b (Magnetic field B) = 6 T fixed. For this exercise use only 0.10625 ≤ k ≤ 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 = 0.10917489 GHz, with a bounded reading uncertainty ±0.015839 GHz. 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 0.00007875, 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.
NEGLECTED 2ND-ORDER SHIFT against k
Predictions from the supplied The Paschen-Back limit and the normal Zeeman pattern 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): NEGLECTED 2ND-ORDER SHIFT / GHz
Horizontal axis (x): k: Fine-structure dial k (1 = hydrogen)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: Fine-structure dial k (1 = hydrogen) | NEGLECTED 2ND-ORDER SHIFT / GHz |
|---|---|
| 0.10625 | 0.003581455078125 |
| 0.303125 | 0.029150441894531243 |
| 0.5 | 0.0793125 |
| 0.6968749999999999 | 0.1540676293945312 |
| 0.8937499999999999 | 0.25341583007812496 |
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(0.10625) = 0.00358146 GHz and Y_out(0.89375) = 0.253416 GHz. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives k_line = 0.10625 + (0.109175 − 0.00358146)(0.89375 − 0.10625)/(0.253416 − 0.00358146) = 0.43909. 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 = 0.586625; substituting back gives Y_out = 0.10917489 GHz.
- The interpolation error in the input is |k_line − k| = 0.147535. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.09333589 GHz and 0.12501389 GHz. Sort the roots even if the response decreases: 0.542405 ≤ k ≤ 0.627738.
- The lower and upper input deviations are 0.0442202 and 0.0411129. 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 ≈ 0.586625; compatible interval [0.542405, 0.627738]. Endpoint interpolation gives 0.43909.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The intermediate field: diagonalising the n = 2 matrix: infer an input with uncertainty
In the intermediate field μ_B B is comparable with the fine-structure splitting Δ, so the n = 2 energies come from diagonalising the fine structure and the Zeeman term together, exactly, inside each block of fixed l and mⱼ. Use the supplied topic model Y_out = (delta-(13.996*bmax))/2+sqrt(delta^2-2*delta*(13.996*bmax)/3+(13.996*bmax)^2)/2, where Y_out is UPPER mⱼ = −1/2, RIGHT EDGE (GHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. bmax is Field range, 0 to B, measured in T. Keep delta (Fine-structure splitting Δ/h) = 10.95 GHz; lamb (Lamb shift L/h of 2S₁/₂) = 0 GHz fixed. For this exercise use only 0.575938 T ≤ bmax ≤ 4.47406 T 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 = 4.3441434 GHz, with a bounded reading uncertainty ±0.036856 GHz. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown bmax by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000389812 T, 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.
UPPER mⱼ = −1/2, RIGHT EDGE against bmax
Predictions from the supplied The intermediate field: diagonalising the n = 2 matrix 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): UPPER mⱼ = −1/2, RIGHT EDGE / GHz
Horizontal axis (x): bmax: Field range, 0 to B / T
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| bmax: Field range, 0 to B / T | UPPER mⱼ = −1/2, RIGHT EDGE / GHz |
|---|---|
| 0.5759375000000001 | 7.057862575625717 |
| 1.5504687499999998 | 5.021880795002332 |
| 2.525 | 4.4696064148134464 |
| 3.4995312499999995 | 4.230376967772273 |
| 4.4740625 | 4.098437540096157 |
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(bmax) = 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(0.575938) = 7.05786 GHz and Y_out(4.47406) = 4.09844 GHz. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives bmax_line = 0.575938 + (4.34414 − 7.05786)(4.47406 − 0.575938)/(4.09844 − 7.05786) = 4.15042 T. 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 bmax = 2.95379 T; substituting back gives Y_out = 4.3441434 GHz.
- The interpolation error in the input is |bmax_line − bmax| = 1.19663 T. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 4.3072874 GHz and 4.3809994 GHz. Sort the roots even if the response decreases: 2.81288 T ≤ bmax ≤ 3.11021 T.
- The lower and upper input deviations are 0.140912 T and 0.156419 T. 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 bmax ≈ 2.95379 T; compatible interval [2.81288, 3.11021] T. Endpoint interpolation gives 4.15042 T.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.