1-dimensional Ising model: Difference between revisions

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The energy of the system will be given by
The energy of the system will be given by


<math>  U = -J \sum_{i=1}^{N-1} S_{i} S_{i+1} </math>,  
:<math>  U = -J \sum_{i=1}^{N-1} S_{i} S_{i+1} </math>,  


where each variable <math> S_j </math> can be either -1 or +1.
where each variable <math> S_j </math> can be either -1 or +1.
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The partition function of the system will be:
The partition function of the system will be:


<math> Q_N = \sum_{\Omega^N }  \exp \left[ K \sum_{i=1}^{N-1} S_i S_{i+1}  \right]</math>,  
:<math> Q_N = \sum_{\Omega^N }  \exp \left[ K \sum_{i=1}^{N-1} S_i S_{i+1}  \right]</math>,  




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and <math> K = J/k_B T </math>
and <math> K = J/k_B T </math>


<math> Q_{N} = \sum_{S_1} \sum_{S_2} e^{K S_1S_2} \sum_{S_3} e^{K S_2 S_3} \cdots \sum_{S_{N-1}} e^{K S_{N-2} S_{N-1}}\sum_{S_{N}} e^{K S_{N-1} S_{N} }
:<math> Q_{N} = \sum_{S_1} \sum_{S_2} e^{K S_1S_2} \sum_{S_3} e^{K S_2 S_3} \cdots \sum_{S_{N-1}} e^{K S_{N-2} S_{N-1}}\sum_{S_{N}} e^{K S_{N-1} S_{N} }
</math>
</math>


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


<math> Q_{N} = \sum_{S_1} \sum_{S_2} e^{K S_1S_2} \sum_{S_3} e^{K S_2 S_3} \cdots \sum_{S_{N-2}} e^{K S_{N-2} S_{N-1}} \left[ 2 \cosh ( K S_{N-1} ) \right]
:<math> Q_{N} = \sum_{S_1} \sum_{S_2} e^{K S_1S_2} \sum_{S_3} e^{K S_2 S_3} \cdots \sum_{S_{N-2}} e^{K S_{N-2} S_{N-1}} \left[ 2 \cosh ( K S_{N-1} ) \right]
</math>
</math>


Taking into account that <math> \cosh(K) = \cosh(-K) </math>
Taking into account that <math> \cosh(K) = \cosh(-K) </math>


<math> Q_{N} = \sum_{S_1} \sum_{S_2} e^{K S_1S_2} \sum_{S_3} e^{K S_2 S_3} \cdots \sum_{S_{N-1}} e^{K S_{N-2} S_{N-1}} \left[ 2 \cosh ( K ) \right]
:<math> Q_{N} = \sum_{S_1} \sum_{S_2} e^{K S_1S_2} \sum_{S_3} e^{K S_2 S_3} \cdots \sum_{S_{N-1}} e^{K S_{N-2} S_{N-1}} \left[ 2 \cosh ( K ) \right]
</math>
</math>


Therefore:
Therefore:


<math> Q_{N} = \left( 2 \cosh K \right) Q_{N-1} </math>
:<math> Q_{N} = \left( 2 \cosh K \right) Q_{N-1} </math>


<math> Q_N = 2^{N} \left( \cosh K \right)^{N-1} \approx ( 2 \cosh K )^N </math>
:<math> Q_N = 2^{N} \left( \cosh K \right)^{N-1} \approx ( 2 \cosh K )^N </math>


The Helmholtz free energy in the thermodynamic limit will be
The [[Helmholtz energy function]] in the thermodynamic limit will be


<math> A = - N k_B T \log \left( 2 \cosh K \right) </math>
:<math> A = - N k_B T \log \left( 2 \cosh K \right) </math>
 
[[Category: Models]]

Revision as of 18:23, 27 February 2007

Model: Consider a system with spins in a row.

The energy of the system will be given by

,

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

The partition function of the system will be:

,


where represents the possible configuration of the N spins of the system, and

Performing the sum of the possible values of we get:

Taking into account that

Therefore:

The Helmholtz energy function in the thermodynamic limit will be