Course:Statistical Mechanics/Statistical mechanics of phase
transitions/Bulk free energy, thermodynamic limit and absence of phase
transitions
Let us therefore consider a one-dimensional Ising model with N spins and
free boundary conditions, i.e. the first and the last spin can assume any
value. Using the nearest neighbour interaction Hamiltonian we have:
-\mathcal{H} =HΣ_i S_i +J Σ_{\left \langle i,j \right \rangle }S_i S_j
and defining h=βH and K=βJ for the sake of simplicity, the partition
function of the system will be:
Z_N=\operatorname{Tr} e^{-β\mathcal{H} }=Σ_{\lbrace S_i = \pm 1 \rbrace
}e^{hΣ_i S_i +K Σ_i S_i S_{i+1}}
If we now set h=0, namely if the system is not subjected to any external
field, then:
Z_N = Σ_{\lbrace S_i = \pm 1 \rbrace }e^{K Σ_i S_i S_{i+1}}=Σ_{S_1=\pm
1}⋯Σ_{S_N=\pm 1}e^{KS_1 S_2 + ⋯KS_{N-1}S_N}
In order to compute Z_N we use the so called recursion method: from the
expression of Z_N we can deduce the expression of the partition function
Z_{N+1} of the same system with one additional spin added to the lattice:
Z_{N+1}=Σ_{S_1=\pm 1}⋯Σ_{S_N=\pm 1}Σ_{S_{N+1}=\pm 1} e^{K(S_1 S_2 +
⋯S_{N-1}S_N)}e^{KS_N S_{N+1}}
However, the sum over S_{N+1} gives:
Σ_{S_{N+1}=\pm 1}e^{KS_N S_{N+1}}=e^{KS_N}+e^{-KS_N}=2\cosh (KS_N)=2\cosh K
where we have used the evenness of the hyperbolic cosine, namely the fact
that \cosh (\pm x)=\cosh x. Therefore we have:
Z_{N+1}=Z_N ·2 \cosh K
and iterating the relation N times we get:
Z_{N+1} = Z_1 (2\cosh K)^N ⇒ Z_N=Z_1 (2\cosh K)^{N-1}
Since¹ :
Z_1 = Σ_{S_1=\pm 1}1 = 1+1 = 2
we get the final result:
Z_N = 2(2\cosh K)^{N-1}
Now, following all the prescriptions we know, we get to:
F(T)=-k_B T \left{\ln 2 +(N-1)\ln \left (2\cosh \frac{J}{k_B T} \right
)\right }
and in the thermodynamic limit:
f(T)=-k_B T \ln \left (2 \cosh \frac{J}{k_B T} \right )
Let us note that f does indeed respect the properties we have seen in
Analytic properties of the Ising model. Furthermore, since the logarithm
and the hyperbolic cosine are analytic functions we see that f(T) is itself
an analytic function of T: it is therefore impossible that the system will
exhibit any kind of phase transition (at least for T\neq 0² ) even in the
thermodynamic limit.
- This can be justified as follows: Z₁ is the partition function of a
single-spin Ising model and we are considering two-spin interactions, so
this single spin will not interact with anything. Therefore the
Hamiltonian of the system in this case is null, and so e^{-β\mathcal{H}
}=e⁰⁼¹.
- In this case, in fact, there can be problems. From this fact we can state
that the only "phase transition" that can happen in a one-dimensional
Ising model occurs at T=0 (which is of course an unphysical case), where
all the spins are aligned.