As shown in Problem 5.14 , the symmetry group for a finite 1D periodic lattice having cyclic

Question:

As shown in Problem 5.14 , the symmetry group for a finite 1D periodic lattice having cyclic boundary conditions with \(N\) periods between boundaries is the cyclic group of order \(N\), with irreps of the form

\[\Gamma^{(p)}=e^{2 \pi i p / N}(p=1,2,3, \ldots, N) .\]

Show that if the total length of the finite 1D lattice is \(L=a N\), wavefunctions defined on the lattice obey \(\psi_{k}(x+a)=e^{i k a} \psi_{k}(x)\), or equivalently \(\psi_{k}(x)=u_{k}(x)\), where \(u_{k}(x)\) is periodic, \(u_{k}(x)=u_{k}(x+a)\), and \(k \equiv 2 \pi p / L\) is called the crystal momentum. This result is a 1D version of Bloch's theorem, described in Section 5.4 .

Data from Problem  5.14

If cyclic boundary conditions are imposed on a periodic 1D lattice by identifying the two ends with \(N\) cycles between the boundaries, the translation group becomes a cyclic group of order \(N\). Show that the irreps of this group are of the form \[\Gamma^{(p)}(C)=e^{2 \pi i p / N}(p=1,2,3, \ldots, N),\]
where \(C\) denotes group elements. Hint: Elements of the cyclic group of order \(N\) are \(C_{1}=c, C_{2}=c^{2}, C_{3}=c^{3}, \ldots, C_{N}=c^{N}=1\), because of the closure condition \(C_{N}=1\), where 1 is the group identity. Thus the group is abelian and irreps are 1D and labeled by complex numbers.

Fantastic news! We've Found the answer you've been seeking!

Step by Step Answer:

Question Posted: