2
$\begingroup$

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.

$\endgroup$
3
  • 2
    $\begingroup$ "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
  • 2
    $\begingroup$ 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

1
+50
$\begingroup$

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

$\alpha ^2 = \frac{2mE}{\hbar^2}$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.