1-dimensional Ising model: Difference between revisions

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</math>
</math>


Performing the sum of the possible <math> S_{N+1} </math> values we get:
Performing the sum of the possible values of <math> S_{N+1} </math> we get:


<math> Q_{N+1} = \sum_{S_1} \sum_{S_2} e^{kS_1S_2} \sum_{S_3} e^{K S_2 S_3} \cdots \sum_{S_N} e^{K S_{N-1} S_N} \left[ 2 \cosh ( K S_n ) \right]
<math> Q_{N+1} = \sum_{S_1} \sum_{S_2} e^{kS_1S_2} \sum_{S_3} e^{K S_2 S_3} \cdots \sum_{S_N} e^{K S_{N-1} S_N} \left[ 2 \cosh ( K S_N ) \right]
</math>
</math>
Taking into account that <math> \cosh(K) = \cosh(-K) </math>


<math> Q_{N+1} = \sum_{S_1} \sum_{S_2} e^{kS_1S_2} \sum_{S_3} e^{K S_2 S_3} \cdots \sum_{S_N} e^{K S_{N-1} S_N} \left[ 2 \cosh ( K ) \right]
<math> Q_{N+1} = \sum_{S_1} \sum_{S_2} e^{kS_1S_2} \sum_{S_3} e^{K S_2 S_3} \cdots \sum_{S_N} e^{K S_{N-1} S_N} \left[ 2 \cosh ( K ) \right]

Revision as of 11:44, 23 February 2007

Model: Consider a system with N spins in a row.

The energy of the system will be given by

U=−J∑i=1N−1SiSi+1,

where each variable Sj can be either -1 or +1.

The partition function of the system will be:

QN=∑ΩNexp[K∑i=1N−1SiSi+1],


where ΩN represents the possible configuration of the N spins of the system, and K=J/kBT

QN+1=∑S1∑S2ekS1S2∑S3eKS2S3⋯∑SNeKSN−1SN∑SN+1eKSNSN+1

Performing the sum of the possible values of SN+1 we get:

QN+1=∑S1∑S2ekS1S2∑S3eKS2S3⋯∑SNeKSN−1SN[2cosh(KSN)]

Taking into account that cosh(K)=cosh(−K)

QN+1=∑S1∑S2ekS1S2∑S3eKS2S3⋯∑SNeKSN−1SN[2cosh(K)]

Therefore:

QN+1=(2coshK)QN+1

QN=2N(coshK)N−1≈(2coshK)N

The Helmholtz free energy in the thermodynamic limit will be

A=−NkBTlog(2coshK)