In the experiment regarding modelling of 1D diatomic lattice via LC circuits,I was able to plot the dispersion relation of frequency vs wave-vector. As should be expected, I get a jump from the acoustic to the optic branch. We use harmonic approximation to model the lattice.

My question is how to calculate the energy band gap of this lattice from its Dispersion relation. $\theta$ is the phase(=wave-vector x lattice parameter 'a') and $\omega$ is 2$\pi$ x frequency. $$ \omega^2=K\Big(\frac{1}{M_1}+\frac{1}{M_2}\Big)\pm K\sqrt{\Big(\frac{1}{M_1}+\frac{1}{M_2}\Big)^2-\frac{4sin^2\theta}{M_1M_2}} $$ enter image description here

  • $\begingroup$ M1 and M2 are the masses of the two different atoms. $\endgroup$ Feb 3, 2017 at 8:52
  • $\begingroup$ $\omega = 2\pi\nu$, and $E=h\nu$, aren't they? $\endgroup$
    – nicoguaro
    Feb 3, 2017 at 16:38
  • $\begingroup$ Since harmonic approximation is used, and energy of a harmonic oscillator is quantised=$E_n=\hbar\omega(n+\frac{1}{2})$, I was not sure whether to use this formula. $\endgroup$ Feb 5, 2017 at 10:42

1 Answer 1


The band gap occurs at $\Theta=90^\circ$ , i.e. $\sin\Theta = 1$. This yields \begin{align} \omega_\pm^2 &= K \left( \frac{1}{M_1}+\frac{1}{M_2} \pm \sqrt{\left(\frac{1}{M_1}-\frac{1}{M_2}\right)^2} \right)\\ \end{align} If we w.l.o.g. assume that $M_1<M_2$ it follows \begin{align} \omega_+ &= \sqrt{\frac{2K}{M_1}}\\ \omega_- & = \sqrt{\frac{2K}{M_2}}\\ \Delta \omega &= \sqrt{2K}\left( \sqrt{\frac{1}{M_1}}- \sqrt{\frac{1}{M_2}}\right) \end{align}

  • $\begingroup$ Thanks,but how will I find the energy associated with it? $\endgroup$ Feb 3, 2017 at 13:35
  • $\begingroup$ Depends on what you mean by energy. The energy of one phonon with frequency $\omega$ is $E=\hbar \omega$. $\endgroup$
    – Jannick
    Feb 3, 2017 at 16:50

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