Figure 76: Dielectric constant as a function ofωωT.. Calculating the expectation value of this operator in the ground state of the linear spin chain
E 0 =< 0 |Hˆexc| 0 >, where| 0 >=| 11 ... 1 > (136)one arrives at
E 0 =< 0 |Hˆexc| 0 >=− 2 JNS^2 (137)~was neglected here and also the fact that there are really N-1 interactions instead of N, the spin
expectation value was calculated classically so
√
S(S+ 1)is justS. Knowing the ground state energy
E 0 we can ask the question what does an elementary excited state look like and what is its energy.
For example we may be tempted to try a single spin flip like in figure 77 (b). Calculating its energy
we see that two terms of equation 135 are affected. Their new energy is+4JS^2 while their old energy
was− 4 JS^2. So we end up with a total energy increase of
E 1 =E 0 + 8JS^2 =− 2 JNS^2 + 8JS^2 (138). There are excitations which have a much lower energy like the one drawn in figure 77 (c). Such an
excited state may have an energy of the order of
E 1 ∝ E 0 + (nk+1
2
)·~(2JSa^2 )k^2 , wherek <≈
π
8 a(139)
<≈ E 0 +nk·JS^2
4
(140)
as we shall see later on.