University Advanced Quantum Mechanics · topic UAQM-15.06
Holes in the Fermi sea: a non-relativistic Dirac sea
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.
Holes in the Fermi sea: a non-relativistic Dirac sea: predict a finite change
A hole in the Fermi sea is an empty state inside the filled Fermi sphere; measured from the neutral ground state it carries minus the missing electron's momentum, the opposite spin projection, charge +e, and the positive energy μ − εₖ. Use the supplied topic model Y_out = 1-kk^2, where Y_out is HOLE μ − ε(k) (ε_F). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. kk is Emptied state |k|, measured in k_F. Keep q (Pair momentum q) = 0.6 k_F; th (Angle θ between k and q) = 60 ° fixed. For this exercise use only 0.10625 k_F ≤ kk ≤ 0.89375 k_F 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 kk = 0.2795 k_F and 0.38975 k_F. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δkk across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
HOLE μ − ε(k) against kk
Predictions from the supplied Holes in the Fermi sea: a non-relativistic Dirac sea 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): HOLE μ − ε(k) / ε_F
Horizontal axis (x): kk: Emptied state |k| / k_F
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| kk: Emptied state |k| / k_F | HOLE μ − ε(k) / ε_F |
|---|---|
| 0.10625 | 0.9887109375 |
| 0.303125 | 0.908115234375 |
| 0.5 | 0.75 |
| 0.6968749999999999 | 0.5143652343750001 |
| 0.8937499999999999 | 0.20121093750000008 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of kk separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At kk = 0.2795 k_F, substitute into Y_out = 1-kk^2: Y₀ = 0.92188 ε_F.
- At kk = 0.38975 k_F, the same fixed-parameter model gives Y₁ = 0.848095 ε_F.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0737848 ε_F; the input change is Δkk = 0.11025 k_F.
- Average sensitivity is ΔY_out/Δkk = -0.66925 ε_F per k_F. 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 = 0.92188 ε_F, Y₁ = 0.848095 ε_F; ΔY_out = -0.0737848 ε_F; average sensitivity = -0.66925 ε_F per k_F.
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.
Holes in the Fermi sea: a non-relativistic Dirac sea: infer an input with uncertainty
A hole in the Fermi sea is an empty state inside the filled Fermi sphere; measured from the neutral ground state it carries minus the missing electron's momentum, the opposite spin projection, charge +e, and the positive energy μ − εₖ. Use the supplied topic model Y_out = 1-kk^2, where Y_out is HOLE μ − ε(k) (ε_F). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. kk is Emptied state |k|, measured in k_F. Keep q (Pair momentum q) = 0.6 k_F; th (Angle θ between k and q) = 60 ° fixed. For this exercise use only 0.10625 k_F ≤ kk ≤ 0.89375 k_F 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 = 0.65587111 ε_F, with a bounded reading uncertainty ±0.049926 ε_F. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown kk 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 k_F, 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.
HOLE μ − ε(k) against kk
Predictions from the supplied Holes in the Fermi sea: a non-relativistic Dirac sea 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): HOLE μ − ε(k) / ε_F
Horizontal axis (x): kk: Emptied state |k| / k_F
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| kk: Emptied state |k| / k_F | HOLE μ − ε(k) / ε_F |
|---|---|
| 0.10625 | 0.9887109375 |
| 0.303125 | 0.908115234375 |
| 0.5 | 0.75 |
| 0.6968749999999999 | 0.5143652343750001 |
| 0.8937499999999999 | 0.20121093750000008 |
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(kk) = 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) = 0.988711 ε_F and Y_out(0.89375) = 0.201211 ε_F. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives kk_line = 0.10625 + (0.655871 − 0.988711)(0.89375 − 0.10625)/(0.201211 − 0.988711) = 0.43909 k_F. 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 kk = 0.586625 k_F; substituting back gives Y_out = 0.65587111 ε_F.
- The interpolation error in the input is |kk_line − kk| = 0.147535 k_F. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.60594511 ε_F and 0.70579711 ε_F. Sort the roots even if the response decreases: 0.542405 k_F ≤ kk ≤ 0.627738 k_F.
- The lower and upper input deviations are 0.0442203 k_F and 0.0411129 k_F. 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 kk ≈ 0.586625 k_F; compatible interval [0.542405, 0.627738] k_F. Endpoint interpolation gives 0.43909 k_F.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.