I have a linear chain of two atoms type connected by springs and a lattice's constant $a$. Let said that mass is $m_1=m_2=m$ and the spring's constant $C_1$ and $C_2$. I have found the dispersion relation $$ \omega_{\pm}^2= \frac{C_1+C_2}{m} \pm \frac{1}{m} \left[ C_1^2 + C_2^2 + 2 C_1 C_2 \, \cos\left( ka \right) \right]^{1/2} $$ If I set $C_1=C_2=C$ we get $$ \omega_{\pm}^2= \frac{2C}{m} \pm \frac{2C}{m} \cdot \left| \cos\left( \frac{ka}{2} \right) \right| $$ but now the chain is made of the same type of particles (due to the mass and interaction between particles is the same for both types), so the lattice constant change to $a/2$.

When $C_1 \neq C_2$ I have a gap between acoustic and optic branch. After set $C_1=C_2$ I should find just the acoustic one beacuse I now that cains with all atoms with the same mass and same springs have only one branch, but I find two conneted curves (without the gap)

I wonder about the physical reason why the optic branch is deleted when it cross with the acoustic one (i.e. when there is no gap)

EDIT: Inspired for one of the comments I have set $C_1=R\cdot C_2=R \cdot C$ then

$$ \omega_{\pm} \cdot \sqrt{\frac{m}{C}}= \sqrt {(1+R) \pm \left[ 1 + R^2 + 2 R \, \cos\left( ka \right) \right]^{1/2}} $$

and draw $\omega(k) \cdot \sqrt{m/C}$ vs $k\cdot a$ for different values of $R$. My case is $R=1$, also I have drawn $R=0.5$ and $R=1.5$

enter image description here enter image description here enter image description here

Thanks to the comment of @Gilbert I have understand that when there is no gap the curves are continuous (black line) and then the size of the 1º Brillouin Zone change (from $[-\pi/a,\pi/a]$ to $[-2\pi/a,2\pi/a]$), and we have

enter image description here

But I cannot understand why we just remove the yellow curve. I can see that if we move everything to the first Brillouin's zone for the old lattice ($[-\pi/a,\pi/a]$) the yellow curve an black overlap, but I though that now the first Brillouin's zone is $[-2\pi/a,2\pi/a]$. I think that my confusion is about the truly range that I have to use to the zone in each case


1 Answer 1


You’re right! When you have only one mass and spring constant, there is only one longitudinal mode. So why do you still see two? It’s because the size of the Brillouin zone changed when you set $C_1 = C_2$! Under that condition, the Brillouin zone is now twice the size, and the “optical” branch you’re seeing is the second half of the acoustic branch, folded back. If you recalculate the dispersion with the full Brillouin zone, you’ll find only one mode, as expected.

  • $\begingroup$ Thank you very much, I have understand more or less the answer, but I have edited my question to explain a bit more what I cannot understand. "If you recalculate the dispersion with the full Brillouin zone, you’ll find only one mode, as expected" I cannot see what you mean, I think that the two curves are still there, what you mean about have just one mode? $\endgroup$
    – user239504
    Dec 11, 2021 at 10:14
  • $\begingroup$ @user239504 you need to start the calculation from scratch, with only one atom in the unit cell. As long as there are two atoms, there will be two modes; however when the atoms are identical, the two modes are actually just one. $\endgroup$
    – Gilbert
    Dec 11, 2021 at 16:00

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