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Let’s tackle a problem like the 2D problem of the final exam, but first more generally. Instead of a square lattice consider the lattice with the geometry illustrated in fig. 1.1.
Fig 1.1: Oblique one atom basis
Here, and are the vector differences between the equilibrium positions separating the masses along the and interaction directions respectively. The equilibrium spacing for the cross coupling harmonic forces are
Based on previous calculations, we can write the equations of motion by inspection
Inserting the trial solution
and using the matrix form for the projection operators, we have
This fully specifies our eigenvalue problem. Writing
we wish to solve
Neglecting the specifics of the matrix at hand, consider a generic two by two matrix
for which the characteristic equation is
So our angular frequencies are given by
The square root can be simplified slightly
so that, finally, the dispersion relation is
Our eigenvectors will be given by
So, our eigenvectors, the vectoral components of our atomic displacements, are
There is not too much to gain by expanding out the projection operators explicitly in general. However, let’s do this for the specific case of a square lattice (as on the exam problem). In that case, our projection operators are
Our matrix is
where, specifically, the squared sines for this geometry are
Using eq. 1.0.6, the dispersion relation and eigenvectors are
This calculation is confirmed in oneAtomBasisPhononSquareLatticeEigensystem.nb. Mathematica calculates an alternate form (equivalent to using a zero dot product for the second row), of
Either way, we see that leads to only horizontal or vertical motion.
With the exam criteria
In the specific case that we had on the exam where and , these are
For horizontal and vertical motion we need , or for a difference in the absolute values of the sine arguments
That is, one of
In the first BZ, that is one of or .
System in rotated coordinates
On the exam, where we were asked to solve for motion along the cross directions explicitly, there was a strong hint to consider a rotated (by ) coordinate system.
The rotated the lattice basis vectors are , and the projection matrices. Writing and , where , or . In the basis the projection matrices are
The dot products that show up in the squared sines are
So that in this basis
With the rotated projection operators eq. 22.214.171.124 takes the form
This clearly differs from eq. 1.0.16d, and results in a different expression for the eigenvectors, but the same as eq. 126.96.36.199 for the angular frequencies.
For the and case of the exam, this is
Similar to the horizontal coordinate system, we see that we have motion along the diagonals when
or one of
The exam asked why the cross coupling is required for stability. Clearly we have more complex interaction. The constant surfaces will also be more complex. However, I still don’t have a good intuition what exactly was sought after for that part of the question.
A Manipulate allowing for choice of the spring constants and lattice orientation, as shown in fig. 1.2, is available in phy487/oneAtomBasisPhonon.nb. This interface also provides a numerical calculation of the distribution relation as shown in fig. 1.3, and provides an animation of the normal modes for any given selection of and (not shown).
Fig 1.2: 2D Single atom basis Manipulate interface
Fig 1.3: Sample distribution relation for 2D single atom basis.