Kronig-Penney model does not give us an equation of how the electron energy $E$ depends on its wavenumber $k$. Then how do they draw the $E-k$ curves in different zone schemes such as extended, periodic and reduced? If one can explain the method of drawing the extended zone scheme that will solve for my purpose.

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    $\begingroup$ "Kronig-Penney model does not give us an equation of how the electron energy 𝐸 depends on its wavenumber 𝑘." -- Why not? Isn't that the point - that it is a nice simple solvable model? $\endgroup$ Commented Feb 29, 2020 at 22:40
  • $\begingroup$ Here is a reference to the KP model en.wikipedia.org/wiki/… Could you tell me what is $E$ as a function of $k$? $\endgroup$ Commented Feb 29, 2020 at 22:46
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    $\begingroup$ It seems to be all there, just after "For energy values inside the well (E < 0), we get: " $\endgroup$ Commented Feb 29, 2020 at 22:51

1 Answer 1


$$\cos(ka) = \cos(\alpha (a-b))\cos(\beta b) – \frac{\alpha ^2 + \beta^2}{2\alpha \beta}\sin(\alpha (a-b))\sin(βb) $$

$\alpha ^2 = \frac{2mE}{\hbar^2}$ and $\beta^2 = \frac{2m(E + V_0 )}{\hbar^2}$

When the electron energy increases, values of $\alpha$ e $\beta$ come close. When $\alpha \approx \beta$ the expression: $\frac{\alpha ^2 + \beta^2}{2\alpha \beta} \approx 1$. And the formula tends to: $$\cos(ka) = \cos(\alpha (a-b))\cos(\beta b)–\sin(\alpha (a-b))\sin(\beta b)$$ But the right side of that equation is $\cos(\alpha(a-b)+\beta b)$.

One simple solution is: $ka = \alpha (a-b) + \beta b$. When $\alpha \approx \beta$, $k \approx \alpha$, and how energy is proportional to $\alpha^2$, graphics $E$ x $k$ is approximately a parabola.

But according to Bloch theorem, $k$ is restricted to the Brillouin zone. This can be achieved by $$k = \frac{\cos^{-1}[\cos(\alpha (a - b)+\beta b)] }{a}$$ No matter how energy grows, the values of $k$ oscillates in the interval $\left[- \frac{\pi}{a+b} , \frac{\pi}{a+b}\right]$, what is the Brillouin zone of that periodic frame.


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