# Standing wave and optical mode in a crystal

I want to understand how when the wavelength $$\lambda$$ is equal to 2a, with a being the lattice constant, two neighboring building particles of the crystal move in opposite direction. The general formula for the displacement of a particle from its equilibrium is:

$$U_n=Ue^{i(k\cdot a\cdot n -\omega t) }$$.

Now if the wavelength $$\lambda = 2a$$ that means that the wave-number is $$k=\frac \pi a$$. At any moment in time the displacement between the particle in position n and the one in position n+1 should be opposite, meaning if for the particle in position n the displacement has a negative value, that means that it is in the left of the equilibrium position, while the particle in position n+1 should be on the right, meaning a positive value of the displacement, but the absolute values of the displacements for both particles should be equal. When i plug in the following values : $$k=\frac \pi a$$, t=0, I get something which has a complex number and I don't see the opposite movement. Can someone help me, by showing me, through calculation that indeed, for $$k=\frac \pi a$$ two neighboring particles move in the opposite direction.

And regarding the optical phonons. The first condition for their existence is that in the crystal lattice we must have at least 2 different atoms. The other one from wikipedia is :

"Optical phonons are out-of-phase movements of the atoms in the lattice, one atom moving to the left, and its neighbor to the right."

And that type of movement happens only for the shortest wavelength/ largest value of the wavenumber. Does this mean that optical phonons exist only when the wavelength of the propagating wave is minimal, meaning $$k=\frac \pi a$$ ?

1. About displacement, if you plug $$k=\pi/a$$ into $$e^{ikna}$$, you'll get $$e^{i\pi n}$$. Now, $$n$$ is an integer (number of an atom in the 1-d chain), so this equals $$1$$ if $$n$$ is even and $$-1$$ if $$n$$ is odd. I suppose this is what you meant by the opposite movement of the neighbouring atoms.
2. $$k=\pi/a$$ is the largest wavevector in the periodic lattice (or chain) of atoms (right end of the Brillouin zone). So the k-numbers can be from $$0$$ to $$\pi/a$$. The largest wavevector corresponds to the shortest wavelength, so excitations with larger wavelengths are possible too. Check with the dispersion in the wikipedia article.