In the prior exercises we used a unit cell that looked like (K1,K2,K3) as this is the smallest repeating pattern. Now in part 4 we get a symmetric matrix and we make a symmetry argument [1,1,1]^T [1,a,1]^T [1,a,-1]^T which I agree is valid. But what confuses me is that now we have a unit cell (K1,K1,K3) if we take K1=K2 and ka=0. Which is assymetric so no symmetry argument can be made for this specific unit cell. So the question is:
Am I allowed to shift the unit cell to (K1,K3,K1) as if you see this pattern stays the same and now the symmetry argument can still be made?
….K3)(K1,K1,K3)(K1,K1,K3)(K1,K1,K3)(K1,….=….K1)(K1,K3,K1)(K1,K3,K1)(K1,K3….
Or does this change the physics of the equations of motion? In my opinion it does change the physics but I just don’t understand why a symmetry argument can be made for a asymetric unit cell.
If anyone understands this I would like a response. Thank you!
Or am I looking at it wrong? should I look at system as K3–m–K1–m–K1–m–K3 for a (K1,K1,K3) unit cell. Because if I shift my unit cell to (K1,K3,K1) I get K1–m–K1–m–K3–m–K1 which is not symmetric. Since my eigenvectors can only describe the motion of 3 masses. The masses determine the symmetry.
Absolutely, you are allowed to shift the unit cell. That must be possible because it does not change the physics; it only changes how we label the same infinite chain.
However, you do not need to shift it here. The symmetry is already present if you include the springs to the neighboring unit cells. For ka=0, all unit cells move in phase. So the unit cell to the left and the unit cell to the right have the same displacements as the one we are looking at.
Now set \kappa_1=\kappa_2=\kappa. Then atom 1 and atom 3 have equivalent environments: atom 1 is connected to atom 2 by \kappa and to atom 3 in the previous cell by \kappa_3, while atom 3 is connected to atom 2 by \kappa and to atom 1 in the next cell by \kappa_3.
That is why the spring matrix is symmetric under exchanging atoms 1 and 3. So we can choose eigenvectors that are symmetric or antisymmetric under this exchange:
[1,a,1]^T
or
[1,0,-1]^T.
The middle entry is zero in the antisymmetric vector because atom 2 maps onto itself. If the whole vector changes sign under the exchange, then u_2=-u_2, so u_2=0.
So: shifting the unit cell is allowed, but the symmetry argument already works in the original unit cell once you include the neighboring cells and use ka=0.