University Physics V · topic UPV-09.09
Tensor products and the addition of angular momenta
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.
Tensor products and adding angular momenta: predict a finite change
Adding two angular momenta means decomposing the tensor product of their state spaces into the total-J towers it contains: j takes every value from |j₁ − j₂| to j₁ + j₂ in integer steps, each tower carrying 2j+1 states. Use the supplied topic model Y_out = z*(l+0.5), where Y_out is SPLITTING (meV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. z is Spin–orbit constant ζ, measured in meV. Keep l (Orbital quantum number l) = 2 fixed. For this exercise use only 0.30625 meV ≤ z ≤ 1.09375 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. Calculate Y_out at z = 0.4795 meV and 0.58975 meV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δz across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
SPLITTING against z
Predictions from the supplied Tensor products and adding angular momenta 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): SPLITTING / meV
Horizontal axis (x): z: Spin–orbit constant ζ / meV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| z: Spin–orbit constant ζ / meV | SPLITTING / meV |
|---|---|
| 0.30625 | 0.765625 |
| 0.503125 | 1.2578125 |
| 0.7 | 1.75 |
| 0.896875 | 2.2421875 |
| 1.09375 | 2.734375 |
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
- At z = 0.4795 meV, substitute into Y_out = z*(l+0.5): Y₀ = 1.19875 meV.
- At z = 0.58975 meV, the same fixed-parameter model gives Y₁ = 1.47437 meV.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.275625 meV; the input change is Δz = 0.11025 meV.
- Average sensitivity is ΔY_out/Δz = 2.5 meV per meV. Its positive sign means that this response increases 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 = 1.19875 meV, Y₁ = 1.47437 meV; ΔY_out = 0.275625 meV; average sensitivity = 2.5 meV per meV.
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.
Tensor products and adding angular momenta: infer an input with uncertainty
Adding two angular momenta means decomposing the tensor product of their state spaces into the total-J towers it contains: j takes every value from |j₁ − j₂| to j₁ + j₂ in integer steps, each tower carrying 2j+1 states. Use the supplied topic model Y_out = z*(l+0.5), where Y_out is SPLITTING (meV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. z is Spin–orbit constant ζ, measured in meV. Keep l (Orbital quantum number l) = 2 fixed. For this exercise use only 0.30625 meV ≤ z ≤ 1.09375 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 = 1.9665625 meV, with a bounded reading uncertainty ±0.11517 meV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown z 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 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.
SPLITTING against z
Predictions from the supplied Tensor products and adding angular momenta 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): SPLITTING / meV
Horizontal axis (x): z: Spin–orbit constant ζ / meV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| z: Spin–orbit constant ζ / meV | SPLITTING / meV |
|---|---|
| 0.30625 | 0.765625 |
| 0.503125 | 1.2578125 |
| 0.7 | 1.75 |
| 0.896875 | 2.2421875 |
| 1.09375 | 2.734375 |
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(z) = 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(0.30625) = 0.765625 meV and Y_out(1.09375) = 2.73438 meV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives z_line = 0.30625 + (1.96656 − 0.765625)(1.09375 − 0.30625)/(2.73438 − 0.765625) = 0.786625 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 z = 0.786625 meV; substituting back gives Y_out = 1.9665625 meV.
- The interpolation error in the input is |z_line − z| = 0 meV. This is within the requested numerical tolerance.
- Invert both limits Y_out = 1.8513925 meV and 2.0817325 meV. Sort the roots even if the response decreases: 0.740557 meV ≤ z ≤ 0.832693 meV.
- The lower and upper input deviations are 0.046068 meV and 0.046068 meV. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer z ≈ 0.786625 meV; compatible interval [0.740557, 0.832693] meV. Endpoint interpolation gives 0.786625 meV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.