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University Advanced Quantum Mechanics · topic UAQM-07.07

Reading the spectrum: degeneracy in n and j, and hydrogen-like uranium

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.

Further practice · 01

Reading the spectrum: n, j and hydrogen-like uranium: predict a finite change

In the Dirac–Coulomb spectrum a level is fixed by n and j alone: the two states of one shell with the same j, whose orbital labels are l = j ± ½ and whose parities are opposite, have exactly the same energy, and only the top level j = n − ½ is unpaired. Use the supplied topic model Y_out = 510.999*(sqrt(1-(z*z/18779)/4)-sqrt((1+sqrt(1-(z*z/18779)))/2)), where Y_out is 2P DOUBLET, EXACT (keV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. z is Nuclear charge Z, dimensionless. Keep terms (Series terms kept) = 1 fixed. For this exercise use only 1 ≤ z ≤ 81, 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. Calculate Y_out at z = 19 and 20. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δz across this interval. State whether the response increases or decreases; explain why a finite difference between these settings does not require interpolation between them.

Explore the question data

2P DOUBLET, EXACT against z

Predictions from the supplied Reading the spectrum: n, j and hydrogen-like uranium 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): 2P DOUBLET, EXACT / keV

2P DOUBLET, EXACT against zPredictions from the supplied Reading the spectrum: n, j and hydrogen-like uranium 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. Horizontal axis: z: Nuclear charge Z. Vertical axis: 2P DOUBLET, EXACT / keV. Exact values are available in the data table below.-0.10030.57661.2541.932.607-2.219.44162.684.2xy

Horizontal axis (x): z: Nuclear charge Z

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1z: Nuclear charge Z: 12P DOUBLET, EXACT / keV: 4.53e-8
Read the exact data as a table
2P DOUBLET, EXACT against z — supplied values
z: Nuclear charge Z2P DOUBLET, EXACT / keV
14.528342665344187e-8
210.008937780108471176
410.13556838145210978
610.7165043380142411
812.5070295532438527

These are the supplied values, not a worked answer. No fitted trend is drawn.

Show a hint

Substitute each value of z separately while keeping every other input fixed, then subtract in the same order.

Reveal the worked solution
  1. At z = 19, substitute into Y_out = 510.999*(sqrt(1-(z*z/18779)/4)-sqrt((1+sqrt(1-(z*z/18779)))/2)): Y₀ = 0.00597301 keV.
  2. At z = 20, the same fixed-parameter model gives Y₁ = 0.00734295 keV.
  3. Signed change: ΔY_out = Y₁ − Y₀ = 0.00136994 keV; the input change is Δz = 1.
  4. Average sensitivity is ΔY_out/Δz = 0.00136994 keV per unit input. Its positive sign means that this response increases over the specified interval.
  5. This finite difference compares two specified test settings without interpolating between them. A point derivative is not required. Any quantum numbers or particle counts must also obey their physical admissibility conditions.

Answer Y₀ = 0.00597301 keV, Y₁ = 0.00734295 keV; ΔY_out = 0.00136994 keV; average sensitivity = 0.00136994 keV per unit input.

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.

Extra challenging · 01

Reading the spectrum: n, j and hydrogen-like uranium: infer an input with uncertainty

In the Dirac–Coulomb spectrum a level is fixed by n and j alone: the two states of one shell with the same j, whose orbital labels are l = j ± ½ and whose parities are opposite, have exactly the same energy, and only the top level j = n − ½ is unpaired. Use the supplied topic model Y_out = 510.999*(sqrt(1-(z*z/18779)/4)-sqrt((1+sqrt(1-(z*z/18779)))/2)), where Y_out is 2P DOUBLET, EXACT (keV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. z is Nuclear charge Z, dimensionless. Keep terms (Series terms kept) = 1 fixed. For this exercise use only 1 ≤ z ≤ 81, 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 = 0.30885685 keV, with a bounded reading uncertainty ±0.20056 keV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of z 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.

Explore the question data

2P DOUBLET, EXACT against z

Predictions from the supplied Reading the spectrum: n, j and hydrogen-like uranium 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): 2P DOUBLET, EXACT / keV

2P DOUBLET, EXACT against zPredictions from the supplied Reading the spectrum: n, j and hydrogen-like uranium 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. Horizontal axis: z: Nuclear charge Z. Vertical axis: 2P DOUBLET, EXACT / keV. Exact values are available in the data table below.-0.10030.57661.2541.932.607-2.219.44162.684.2xy

Horizontal axis (x): z: Nuclear charge Z

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1z: Nuclear charge Z: 12P DOUBLET, EXACT / keV: 4.53e-8
Read the exact data as a table
2P DOUBLET, EXACT against z — supplied values
z: Nuclear charge Z2P DOUBLET, EXACT / keV
14.528342665344187e-8
210.008937780108471176
410.13556838145210978
610.7165043380142411
812.5070295532438527

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
  1. Endpoint predictions are Y_out(1) = 4.52834e-8 keV and Y_out(81) = 2.50703 keV. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives z_line = 1 + (0.308857 − 4.52834e-8)(81 − 1)/(2.50703 − 4.52834e-8) = 10.8557. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Evaluate the model at each permitted setting. The central reading corresponds to z = 50; only the allowed settings 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56 lie inside the stated reading bound.
  4. The interpolation error in the input is |z_line − z| = 39.1443.
  5. The compatible allowed values span 39 to 56. Values between specified test settings are outside this exercise's input domain even if an interpolation passes through them.
  6. 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 z: 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56. Endpoint interpolation predicts 10.8557; check admissibility against the discrete settings.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.