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Argument on why spin correlation functions in Ising model decay exponentially with a correlation length?

Physics Asked by RicknJerry on January 15, 2021

I’m reading Quantum Field Theory in Strongly Correlated Electronic Systems, Nagaosa.

Consider 1D Ising model,
$$H=J_zsum_i S^z_iS^z_{i+1}.$$
on page 3, it says

The groud stae is 2-fold degenerate because the Hamiltonian is invariant
under the transformation $S^i_z rightarrow -S^i_z$, performed at all sites $i$.
Calling these two ground states $A$ and $B$ and assuming that the system at the right-hand side is in state $A$, and at the left-hand side in state $B$, then somewhere there
must exist a boundary between region $A$ and region $B$. This boundary is called a kink or soliton. Because at finite temperature this excitation occurs
with a finite density, the spin correlation function $F(r) =langle S^z_iS^z_{i+r}rangle$ will decay exponentially with a correlation length $xi$.

I know how to directly calculate the correlation function, but I wonder how the argument for exponential decay of correlation function is made here and how to understand it.

Any help would be highly appriciated!!

One Answer

Let me write the Hamiltonian $$ H = -J sum_i S_i^z S_{i+1}^z. $$ This choice will avoid some annoying (and irrelevant) signs.

One way to formulate the statement in the OP precisely is as follows.

Consider the variables $delta_i=S_i^zS_{i+1}^z$. Since $delta_i=1$ when the spins at $i$ and $i+1$ agree and $delta_i=-1$ when the spins at $i$ and $i+1$ disagree, you can identify them with the kinks in your question (that is, there is a kink between $i$ and $i+1$ when $delta_i=-1$).

Introducing the variables $delta_i=S_i^zS_{i+1}^z$, the Hamiltonian becomes $$ H = J^z sum_i delta_i. $$ It follows that the random variables $delta_i$ are independent and identically distributed. One can easily compute their expectation: since $$ P(delta_i = 1) = frac{e^{beta J^z}}{e^{beta J^z} + e^{-beta J^z}}, $$ one has $$ langle delta_i rangle = frac{e^{beta J^z} - e^{-beta J^z}}{e^{beta J^z} + e^{-beta J^z}} = tanh(beta J^z). $$ Finally, noting that $S_i^zS_{i+r}^z = delta_idelta_{i+1}cdotsdelta_{i+r-1}$, we obtain $$ langle{S_i^zS_{i+r}^z}rangle = langledelta_idelta_{i+1}cdotsdelta_{i+r-1}rangle = langle delta_i rangle^r = (tanh(beta J^z))^r. $$


In words, the fact that kinks proliferate in the system (at each $i$, there is a positive probability that a kink is present, so there will be a positive density of them in the system) prevents the ordering of the spins.

Correct answer by Yvan Velenik on January 15, 2021

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