University Physics IV · topic UPIV-15.05
Cooper pairing, the BCS gap, Josephson effects, and high Tc
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.
Cooper pairing and the BCS gap: predict a finite change
A Cooper pair is the bound state formed by two electrons of opposite momentum and spin just outside a filled Fermi sea, and the BCS gap Δ is the minimum energy of each quasiparticle released when one breaks, so breaking a pair costs 2Δ. Use the supplied topic model Y_out = 0.30419*1.13*td*sqrt(200/max(m,1))*exp(-1/max(lam,0.01)), where Y_out is GAP 2Δ(0) = 3.53 kB Tc (meV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; exp(x) means eˣ; max(a,b) selects the larger value. lam is Coupling N(0)V, dimensionless. Keep td (Debye temperature Θ at 200 u) = 275 K; m (Ion mass M) = 200 u fixed. For this exercise use only 0.16375 ≤ lam ≤ 0.63625 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 lam = 0.2677 and 0.33385. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δlam across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
GAP 2Δ(0) = 3.53 kB Tc against lam
Predictions from the supplied Cooper pairing and the BCS gap 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): GAP 2Δ(0) = 3.53 kB Tc / meV
Horizontal axis (x): lam: Coupling N(0)V
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lam: Coupling N(0)V | GAP 2Δ(0) = 3.53 kB Tc / meV |
|---|---|
| 0.16375 | 0.2105600756797623 |
| 0.281875 | 2.7215841455545844 |
| 0.3999999999999999 | 7.75925215353372 |
| 0.518125 | 13.719933418476174 |
| 0.6362499999999999 | 19.632310891347434 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of lam separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At lam = 0.2677, substitute into Y_out = 0.30419*1.13*td*sqrt(200/max(m,1))*exp(-1/max(lam,0.01)): Y₀ = 2.25548 meV.
- At lam = 0.33385, the same fixed-parameter model gives Y₁ = 4.72813 meV.
- Signed change: ΔY_out = Y₁ − Y₀ = 2.47265 meV; the input change is Δlam = 0.06615.
- Average sensitivity is ΔY_out/Δlam = 37.3794 meV per unit input. 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 = 2.25548 meV, Y₁ = 4.72813 meV; ΔY_out = 2.47265 meV; average sensitivity = 37.3794 meV 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.
Cooper pairing and the BCS gap: infer an input with uncertainty
A Cooper pair is the bound state formed by two electrons of opposite momentum and spin just outside a filled Fermi sea, and the BCS gap Δ is the minimum energy of each quasiparticle released when one breaks, so breaking a pair costs 2Δ. Use the supplied topic model Y_out = 0.30419*1.13*td*sqrt(200/max(m,1))*exp(-1/max(lam,0.01)), where Y_out is GAP 2Δ(0) = 3.53 kB Tc (meV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; exp(x) means eˣ; max(a,b) selects the larger value. lam is Coupling N(0)V, dimensionless. Keep td (Debye temperature Θ at 200 u) = 275 K; m (Ion mass M) = 200 u fixed. For this exercise use only 0.16375 ≤ lam ≤ 0.63625 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 = 10.343664 meV, with a bounded reading uncertainty ±1.3933 meV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lam by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00004725, 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.
GAP 2Δ(0) = 3.53 kB Tc against lam
Predictions from the supplied Cooper pairing and the BCS gap 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): GAP 2Δ(0) = 3.53 kB Tc / meV
Horizontal axis (x): lam: Coupling N(0)V
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lam: Coupling N(0)V | GAP 2Δ(0) = 3.53 kB Tc / meV |
|---|---|
| 0.16375 | 0.2105600756797623 |
| 0.281875 | 2.7215841455545844 |
| 0.3999999999999999 | 7.75925215353372 |
| 0.518125 | 13.719933418476174 |
| 0.6362499999999999 | 19.632310891347434 |
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(lam) = 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(0.16375) = 0.21056 meV and Y_out(0.63625) = 19.6323 meV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives lam_line = 0.16375 + (10.3437 − 0.21056)(0.63625 − 0.16375)/(19.6323 − 0.21056) = 0.410272. 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 lam = 0.451975; substituting back gives Y_out = 10.343664 meV.
- The interpolation error in the input is |lam_line − lam| = 0.0417028. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 8.950364 meV and 11.736964 meV. Sort the roots even if the response decreases: 0.424234 ≤ lam ≤ 0.479354.
- The lower and upper input deviations are 0.0277414 and 0.0273786. 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 lam ≈ 0.451975; compatible interval [0.424234, 0.479354]. Endpoint interpolation gives 0.410272.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.