University Quantum Mechanics II · topic UQMII-12.01
Coefficient equations for H_0 + H'(t)
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.
Further practice
Work through a fresh case before tackling the extension.
Coefficient equations for H₀ + H′(t): predict a finite change
The coefficient equations are the Schrödinger equation for H₀ + H′(t) rewritten, with no approximation, as coupled first-order equations for the amplitudes cₙ(t) of the unperturbed stationary states, each state keeping its own phase e^(−iEₙt/ħ). Use the supplied topic model Y_out = (1-bz/sqrt(bz^2+bx^2))/2, where Y_out is UPPER STATE OF H₀ + H′ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. bz is Static field B₀ along z, measured in μT. Keep bx (Transverse field B₁ along x) = 30 μT; tau (Switch-off time τ) = 0.3 ms fixed. For this exercise use only 19.5625 μT ≤ bz ≤ 90.4375 μT 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. Calculate Y_out at bz = 35.155 μT and 45.0775 μT. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δbz across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
UPPER STATE OF H₀ + H′ against bz
Predictions from the supplied Coefficient equations for H₀ + H′(t) 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 STATE OF H₀ + H′
Horizontal axis (x): bz: Static field B₀ along z / μT
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| bz: Static field B₀ along z / μT | UPPER STATE OF H₀ + H′ |
|---|---|
| 19.5625 | 0.2268927264393928 |
| 37.28125 | 0.11045915002833195 |
| 55 | 0.06105221354280782 |
| 72.71875 | 0.037788577136887824 |
| 90.4375 | 0.025429272890197974 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of bz separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At bz = 35.155 μT, substitute into Y_out = (1-bz/sqrt(bz^2+bx^2))/2: Y₀ = 0.119662.
- At bz = 45.0775 μT, the same fixed-parameter model gives Y₁ = 0.0837548.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0359076; the input change is Δbz = 9.9225 μT.
- Average sensitivity is ΔY_out/Δbz = -0.0036188 1 per μT. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 0.119662, Y₁ = 0.0837548; ΔY_out = -0.0359076; average sensitivity = -0.0036188 1 per μT.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Coefficient equations for H₀ + H′(t): infer an input with uncertainty
The coefficient equations are the Schrödinger equation for H₀ + H′(t) rewritten, with no approximation, as coupled first-order equations for the amplitudes cₙ(t) of the unperturbed stationary states, each state keeping its own phase e^(−iEₙt/ħ). Use the supplied topic model Y_out = (1-bz/sqrt(bz^2+bx^2))/2, where Y_out is UPPER STATE OF H₀ + H′ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. bz is Static field B₀ along z, measured in μT. Keep bx (Transverse field B₁ along x) = 30 μT; tau (Switch-off time τ) = 0.3 ms fixed. For this exercise use only 19.5625 μT ≤ bz ≤ 90.4375 μT 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 = 0.048840679, with a bounded reading uncertainty ±0.0035117. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown bz 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 μT, 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 STATE OF H₀ + H′ against bz
Predictions from the supplied Coefficient equations for H₀ + H′(t) 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 STATE OF H₀ + H′
Horizontal axis (x): bz: Static field B₀ along z / μT
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| bz: Static field B₀ along z / μT | UPPER STATE OF H₀ + H′ |
|---|---|
| 19.5625 | 0.2268927264393928 |
| 37.28125 | 0.11045915002833195 |
| 55 | 0.06105221354280782 |
| 72.71875 | 0.037788577136887824 |
| 90.4375 | 0.025429272890197974 |
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(bz) = 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) = 0.226893 and Y_out(90.4375) = 0.0254293. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives bz_line = 19.5625 + (0.0488407 − 0.226893)(90.4375 − 19.5625)/(0.0254293 − 0.226893) = 82.2013 μT. 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 bz = 62.7962 μT; substituting back gives Y_out = 0.048840679.
- The interpolation error in the input is |bz_line − bz| = 19.4051 μT. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.045328979 and 0.052352379. Sort the roots even if the response decreases: 60.2929 μT ≤ bz ≤ 65.5698 μT.
- The lower and upper input deviations are 2.50339 μT and 2.77352 μT. 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 bz ≈ 62.7962 μT; compatible interval [60.2929, 65.5698] μT. Endpoint interpolation gives 82.2013 μT.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.