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Corelation function

 
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Jul25-12, 01:43 PM   #1
 

Corelation function


In Ising model with nearest neighbours interaction
[tex]\langle S_iS_{i+j}\rangle[/tex] is different of 0, and this is corelation function between spin on lattice site i, and spin on lattice site i+j. What means that corelation?
 
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Jul25-12, 05:59 PM   #2
 
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The correlator gives you some information about whether neighbouring spins are independent of each other or not.

Consider the case where for each position the spin state varies over time, so that [tex]\langle S_i \rangle=\langle S_{i+j}\rangle=0[/tex].

However, the absence of some preferred spin state on average does not mean that there is no order in the system. For example a system where every spin state at every position fluctuates randomly and they all are independent of each other will give [tex]\langle S_i \rangle=0[/tex] as well as a system where all spins at all positions have a common value at each given time, but this common value fluctuates randomly over time. In the former case, the spins are independent of each other and the correlator will vanish. In the latter case neighboring spins will always share the same value and the expectation value of the correlator will take a positive value.

Similarly you can also get a negative value for the expectation value of the correlator if neighbouring spins tend to have opposite values.
 
Jul26-12, 02:51 AM   #3
 
As a consequence of translational symmetry the expectation value for all spins is the same. So
[tex]\langle S_i \rangle=\langle S_{i+j} \rangle, \forall i,j[/tex]
I'm not satisfied with answer.
If I calculate corelator
[tex]\langle S_i S_{i+j} \rangle[/tex]
and get result different then zero spins are in some corelation. Can you explain me with more details. For example, what the difference between cases of feromagnet and paramagnet.
[tex]\langle S_i S_{i+j} \rangle=\frac{1}{Z}\displaystyle\sum_{\{S\}}S_i S_{i+j}exp(\sum_k^{N-1}\frac{J_k}{k_B T}S_k S_{k+1})[/tex]
 
Jul26-12, 04:55 AM   #4
 
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Corelation function


Quote by LagrangeEuler View Post
As a consequence of translational symmetry the expectation value for all spins is the same. So
[tex]\langle S_i \rangle=\langle S_{i+j} \rangle, \forall i,j[/tex]
Correct. This is what I told you earlier already.

Quote by LagrangeEuler View Post
Can you explain me with more details. For example, what the difference between cases of feromagnet and paramagnet.
[tex]\langle S_i S_{i+j} \rangle=\frac{1}{Z}\displaystyle\sum_{\{S\}}S_i S_{i+j}exp(\sum_k^{N-1}\frac{J_k}{k_B T}S_k S_{k+1})[/tex]
The difference between ferromagnets and paramagnets is already visible without having a look at the correlator. Ferromagnets have a net magnetic moment whether or not there is some external magnetic field is present, while paramagnets have a net magnetic moment of zero exactly when no external field is present. Now one can have a look at correlations. Naively, one would always expect that nearby spins have some tendency to share the same alignment in both cases. In the case of a paramagnet, spins that have a huge separation are supposed to be independent of each other, while in the case of a ferromagnet there is net magnetization and spins are supposed to have the same alignment even over huge distances. The distance over which the tendency to share the same alignment decays is the correlation length. One can determine it by calculating the correlator for several spin-spin distances.

The correlation length as detemined by the correlator can now be used as a tool to identify phase transitions from a paramagnet to a ferromagnet. For example the correlation length may jump from some finite value towards infinity at the transition which is a sign for a first-order phase transition. It could also diverge continuously which would be an indicator of a second-order phase transition.
 
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