hw204a - Physics 512 Homework Set#4 Solutions Winter 2003 1...

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Physics 512 Winter 2003 Homework Set #4 – Solutions 1. A molecule is composed of six identical atoms which form a regular hexagon. Consider a single electron which can be localized on any one of the atoms. Let | n i denote the state in which it is localized on the n -th atom. (This is a six-dimensional Hilbert space, spanned by the states | 1 i through | 6 i .) 1 3 4 5 6 2 Let R be the discrete rotation operator, R | i i = | i +1 i (where we let | 7 i≡| 1 i and | 0 6 i ). We assume that the Hamiltonian has the form H = 6 X i =1 ± E 0 | i ih i |− λ ( | i ih i | + | i ih i 1 | ) ² ( λ represents a nearest-neighbor electron hopping interaction). a ) Find the eigenvalues and eigenstates of R , and show that they are non-degenerate. Since R is the discrete clockwise rotation operator, it is easy to see that if we act with R six times, we end up with the identity, R 6 | ψ i = | ψ i . This indicates that the eigenvalues of R may be the sixth roots of unity R | ψ k i = ω k | ψ i = e πi/ 3 (1) Strictly speaking, we do not yet know that all the distinct sixth roots of unity show up non-degenerately as eigenvalues. However, we can go ahead and construct the eigenstates and show that they do correspond to non-degenerate roots. To do so, we write | ψ k i = 6 X i =1 c i | i i and substitute it into the eigenvalue equation (1) . This results in a set of equations c i +1 = ω k c i (with periodicity implied on i ). The solution is then c i = ω ki (where i is an index). With standard normalization, we Fnd | ψ k i = 1 6 6 X i =1 ω | i i , eigenvalue = ω k (2) Since we have constructed six states all with distinct eigenvalues, they are ob- viously non-degenerate. Note that these states are a discrete version of Bloch wavefunctions. After all, this ring is essentially a one-dimensional lattice with periodic boundary conditions. 1
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Since the structure is a regular hexagon, the rotation operator R forms part of the symmetry group of the molecule. The complete symmetry group of the hexagon is the dihedral group D 6 , which may be generated by the rotation R along with a “reflection”. b ) Show that R commutes with the Hamiltonian, and fnd the eigenvalues and eigen- states oF H . Note that the rotation operator R may be written as R = 6 X i =1 | i ih i 1 | while its inverse is R 1 = R 5 = 6 X i =1 | i ih i +1 | Hence the Hamiltonian may be written as H = E 0 I λ ( R 1 + R ) (3) where I is the identity operator. Since R commutes with itself and its inverse (and clearly with the identity as well), we see that R commutes with H . Since [ R, H ]=0 , we may Fnd simultaneous eigenstates of both R and H . In particular, the non-degenerate states found in a ) above must also be eigenstates of H .I t should be obvious from (1) and (3) that the eigenenergies of H are given by E k = E 0 λ ( ω k + ω k )= E 0 2 λ cos( kπ/ 3) where the k th eigenstate is given in (2) . This problem is easily generalized to a regular n -gon, corresponding to a lattice with n sites and with periodic boundary conditions. In this case, k is essentially a Bloch momentum, and the eigenenergy
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This note was uploaded on 02/15/2011 for the course PHYS 512 taught by Professor Unknow during the Winter '03 term at Cornell College.

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hw204a - Physics 512 Homework Set#4 Solutions Winter 2003 1...

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