A linear chain of diatomic molecules can be modeled by a chain of molecules with different spring constants $C_1$ and $C_2$ (See Figure)

enter image description here

The corresponding equations of motion are:

$$M\ddot{u}=-c_1[u_n-v_n]-c_2[u_n-v_{n-1}] \\ M \ddot{v}=-c_1[v_n-u_n]-c_2[v_n-v_{n+1}]$$

One can use the Ansatz $u=\epsilon_1 e^{i(kna-\omega t)}; v=\epsilon_2 e^{i(kna-\omega t)}$ and obtain the following system of equations:

$$\epsilon_1(M\omega^2-c_1-c_2)+\epsilon_2(c_2e^{-ika}+c_1)=0 \\ \epsilon_1(c_1+c_2 e^{ika})+\epsilon_2(M\omega^2-c_2-c_1)=0$$

Provided the determinant of the coefficients of $\epsilon_1, \epsilon_2$ vanishes, the system will have a solution. Calculating the determinant and solving for $\omega$ yields:

$$\omega^2=\frac{c_+c_2}{M} \pm \frac{1}{M}\sqrt{c_1^2+c_2^2+2c_1c_2 \cos ka}$$

(The identical derivation can be found in Ashcroft/Mermin, Solid state physics, p.433-435)

In Ashcroft/Mermin the dispersion relation is drawn like this:

enter image description here

The upper branch is the optical branch and lower branch is the acoustic branch.

My question:

  1. Why are there two branches and not four? If we look at the dispersion relation $$\omega^2=\frac{c_+c_2}{M} \pm \frac{1}{M}\sqrt{c_1^2+c_2^2+2c_1c_2 \cos ka}$$

then the "$\pm$" already gives two solutions. But the authors are graphing $\omega$ and not $\omega^2$. Wouldn't this lead to $$\omega=\pm \sqrt{\frac{c_+c_2}{M} \pm \frac{1}{M}\sqrt{c_1^2+c_2^2+2c_1c_2 \cos ka}}$$

which would mean four solutions?

  • $\begingroup$ It looks like two of the four solutions would be imaginary (in any case if $\cos ka<0$). $\endgroup$ Commented Jun 5, 2016 at 13:59
  • 1
    $\begingroup$ $\omega$ is a positive quantity $\endgroup$
    – jim
    Commented Jun 5, 2016 at 14:32

1 Answer 1


The result $\omega^2=\frac{c_1+c_2}{M} \pm \frac{1}{M}\sqrt{c_1^2+c_2^2+2c_1c_2 \cos ka}$ leads to two real solutions for $\omega^2$, since $ -1 \lt \cos ka \lt 1$ and the square root lies between $|c_1 - c_2|$ (for $\cos ka = -1$) and $c_1 + c_2$ (for $\cos ka = +1$), so that $\omega^2$ is always positive. The frequency $\omega$ is taken as a positive quantity, since a negative value is just taken to be the same motion in the opposite sense (and doesn't represent anything new).

  • $\begingroup$ Thank you for an answer. My original question has been answered but I have a follow up question. What exactly does this graph tell me "physically". In other words, what do these curves represent with respect to lattice vibrations? Could you say that the larger the wave number (higher energy) the higher the frequency of the lattice vibrations? $\endgroup$
    – qmd
    Commented Jun 5, 2016 at 15:23
  • $\begingroup$ @qmd I think that "larger" and "higher" in your last statement does not make sense as the optical and the acusic phonon brach are developing in the opposite direction. But you are right, you can see it as it is: A dispersion relation for phonons, i.e. the connection between frequency and wave number of the lattice vibrations. $\endgroup$
    – user_na
    Commented Jun 5, 2016 at 19:07
  • $\begingroup$ You can see for the top figure, that $\omega$ gets smaller as $k$ increases, while $\omega$ gets larger as $k$ increases in the bottom figure. $\endgroup$
    – jim
    Commented Jun 5, 2016 at 19:31

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