c The effect of a virtual transition is to lower the ground state energy of the

# C the effect of a virtual transition is to lower the

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(c) The effect of a virtual transition is to lower the ground state energy of the system(the electrons have a bigger space to move in), with the maximum effect being to arrange the M 2+ n anti-ferromagnetally. 3.7 Ferromagnetic Ground-State and Excitations of the Heisenberg Hamiltonian In this section we will consider only a 1D chain of ions. The Heisenberg Hamiltonian is ˆ H = - | J | ~ 2 X i ˆ S i · ˆ S i +1 (3.46) Suppose all ions have spin 1/2, and the chain forms a closed loop so that ˆ S N +1 = ˆ S 1 . If the spins are aligned, the energy takes the maximum negative value. Ground States Consider χ 0 = 1 2 3 ... N (3.47) 45 PHYS40352 Spring 2016 3 MAGNETISM We will focus on 1 2 . We will start by applying ˆ S 1 · ˆ S 2 on 1 2 , where ˆ S 1 · ˆ S 2 = 1 2 ˆ S 1 + ˆ S 2 2 - ˆ S 2 1 - ˆ S 2 2 (3.48) Note that 1 2 is one of the triplet states with total spin of ions as S = 1 . Therefore ˆ S 1 · ˆ S 2 1 2 = 1 2 ˆ S 1 + ˆ S 2 2 - 3 2 ~ 2 1 2 = 1 2 1(1 + 1) ~ 2 - 3 2 ~ 2 1 2 = 1 4 ~ 2 1 2 Note that as ˆ S 1 , ˆ S 2 only apply to 1 , 2 , we could multiply this by 3 , ..., N without changing anything. Hence ˆ S 1 · ˆ S 2 χ 0 = 1 4 ~ 2 χ 0 (3.49) So in general for neighbouring spins ˆ S m · ˆ S m +1 χ 0 = 1 4 ~ 2 χ 0 (3.50) Therefore ˆ 0 = - | J | ~ 2 X m ˆ S m · ˆ S m +1 χ 0 = - | J | ~ 2 N ~ 2 4 χ 0 = - 1 4 | J | N χ 0 hence χ 0 is an energy eigenstate with energy eigenvalue E (0) = - 1 4 | J | N (3.51) where E (0) is the ground state energy. Now, in this state S z = N 2 ~ , the maximum possible, with S = N 2 . The degeneracy is 2 S + 1 , giving a degeneracy of N + 1 which is huge (the ground state is not unique - there are many different possibilities). This high degeneracy is closely connected with the broken symmetry of χ 0 . ˆ H has full rotational symmetry (in spin space), but χ 0 has much lower symmetry - only rotation about the z-axis. Excited States Consider a state with 1 spin flipped: χ n = 1 2 ... n - 1 n n +1 Note that ˆ S m · ˆ S m +1 χ n = - 1 4 ~ 2 χ n provided that m 6 = n or m 6 = n - 1 . 46 PHYS40352 Spring 2016 3 MAGNETISM Let’s consider the effect of ˆ S n · ˆ S n +1 on χ n ˆ S n · ˆ S n +1 n n +1 = ˆ S n · ˆ S n +1 1 2 ( n n +1 - ↑ n n +1 ) + 1 2 ( n n +1 + n n +1 ) = 1 2 ˆ S n + ˆ S n +1 2 - 3 2 ~ 2 1 2 ( n n +1 - ↑ n n +1 ) + 1 2 ( n n +1 + n n +1 ) = 1 2 1 2 0(0 + 1) ~ 2 - 3 2 ~ 2 ( n n +1 - ↑ n n +1 ) + 1 2 1(1 + 1) ~ 2 - 3 2 ~ 2 ( n n +1 + n n +1 ) = 1 4 - 3 2 + 1 2 ( n n +1 ) + 3 2 + 1 2 n n +1 ~ 2 = 1 4 ~ 2 ( - ↓ n n +1 +2 n n +1 ) ˆ S n · ˆ S n +1 χ n = 1 4 ~ 2 ( - χ n + 2 χ n +1 ) Hence the effect is to either leave the flipped spin where it is, or to transfer it to the adjacent ion. So far we have the spin-flip on site n , ˆ S n · ˆ S n +1 χ n = 1 4 ~ 2 ( - χ n + 2 χ n +1 ) (3.52) or for m 6 = n , with N - 2 states, ˆ S m · ˆ S m +1 χ n = - ~ 2 4 χ n (3.53) and lastly we can apply ˆ S n - 1 · ˆ S n χ n = 1 4 ~ 2 ( - χ n + 2 χ n - 1 ) . (3.54) We can combine these three equations to find the total spins, ˆ n = - | J | ~ 2 N X m =1 ˆ S m · ˆ S m +1 χ n = - | J | ~ 2 ( N - 2) 1 4 ~ 2 χ n + ~ 2 4 ( - χ n + 2 χ n +1 ) + ~ 2 4 ( - χ n + 2 χ n - 1 ) = - 1 4 N | J | χ n - | J | - 2 4 - 1 4 - 1 4 χ n + 2 4 χ n +1 + 2 4 χ n - 1  #### You've reached the end of your free preview.

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