University Physics IV · topic UPIV-12.06
Bloch's theorem and energy bands
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.
Bloch's theorem and energy bands: predict a finite change
Bloch's theorem says that in a potential with the period of the lattice, every stationary electron state is a plane wave multiplied by a function carrying that same period: ψ_k(x) = u_k(x)e^(ikx) with u_k(x + a) = u_k(x). Use the supplied topic model Y_out = 4.178*((k-1)^2+1)-sqrt(69.83*(k-1)^2+ug^2), where Y_out is E AT THE MARKER (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. ug is Potential strength |U_G|, measured in eV. Keep k (Marker position k) = 0.5 π/a fixed. For this exercise use only 0.2125 eV ≤ ug ≤ 1.7875 eV 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 ug = 0.559 eV and 0.7795 eV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δug across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
E AT THE MARKER against ug
Predictions from the supplied Bloch's theorem and energy bands 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): E AT THE MARKER / eV
Horizontal axis (x): ug: Potential strength |U_G| / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| ug: Potential strength |U_G| / eV | E AT THE MARKER / eV |
|---|---|
| 0.2125 | 1.0388823967766845 |
| 0.60625 | 1.0005290073829265 |
| 1 | 0.9262807318527146 |
| 1.3937499999999998 | 0.8179524565512981 |
| 1.7874999999999999 | 0.6779806359748015 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of ug separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At ug = 0.559 eV, substitute into Y_out = 4.178*((k-1)^2+1)-sqrt(69.83*(k-1)^2+ug^2): Y₀ = 1.00705 eV.
- At ug = 0.7795 eV, the same fixed-parameter model gives Y₁ = 0.972192 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0348627 eV; the input change is Δug = 0.2205 eV.
- Average sensitivity is ΔY_out/Δug = -0.158108 eV per eV. 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 = 1.00705 eV, Y₁ = 0.972192 eV; ΔY_out = -0.0348627 eV; average sensitivity = -0.158108 eV per eV.
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.
Bloch's theorem and energy bands: infer an input with uncertainty
Bloch's theorem says that in a potential with the period of the lattice, every stationary electron state is a plane wave multiplied by a function carrying that same period: ψ_k(x) = u_k(x)e^(ikx) with u_k(x + a) = u_k(x). Use the supplied topic model Y_out = 4.178*((k-1)^2+1)-sqrt(69.83*(k-1)^2+ug^2), where Y_out is E AT THE MARKER (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. ug is Potential strength |U_G|, measured in eV. Keep k (Marker position k) = 0.5 π/a fixed. For this exercise use only 0.2125 eV ≤ ug ≤ 1.7875 eV 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 = 0.88268254 eV, with a bounded reading uncertainty ±0.02343 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown ug by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001575 eV, 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.
E AT THE MARKER against ug
Predictions from the supplied Bloch's theorem and energy bands 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): E AT THE MARKER / eV
Horizontal axis (x): ug: Potential strength |U_G| / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| ug: Potential strength |U_G| / eV | E AT THE MARKER / eV |
|---|---|
| 0.2125 | 1.0388823967766845 |
| 0.60625 | 1.0005290073829265 |
| 1 | 0.9262807318527146 |
| 1.3937499999999998 | 0.8179524565512981 |
| 1.7874999999999999 | 0.6779806359748015 |
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(ug) = 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(0.2125) = 1.03888 eV and Y_out(1.7875) = 0.677981 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives ug_line = 0.2125 + (0.882683 − 1.03888)(1.7875 − 0.2125)/(0.677981 − 1.03888) = 0.894167 eV. 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 ug = 1.17325 eV; substituting back gives Y_out = 0.88268254 eV.
- The interpolation error in the input is |ug_line − ug| = 0.279083 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.85925254 eV and 0.90611254 eV. Sort the roots even if the response decreases: 1.08337 eV ≤ ug ≤ 1.25715 eV.
- The lower and upper input deviations are 0.0898753 eV and 0.0839009 eV. 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 ug ≈ 1.17325 eV; compatible interval [1.08337, 1.25715] eV. Endpoint interpolation gives 0.894167 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.