With or without magnetic field.

  • #1
LagrangeEuler
708
19
1d Ising model without magnetic field has hamiltonian ##H=-J\sum^{N-1}_{i=1}S_iS_{i+1}## with no boundary condition has partition function ##Z_N=2^{N}\cosh^{N-1}(\beta J)##. In this model with magnetic field people often put into the game circular boundary condition because then they could solve problem using transfer matrix method. Is there any easy way to solve 1d Ising model without boundary condition in extern magnetic field. Hamiltonian of that model would be ##H=-J\sum^{N-1}_{i=1}S_iS_{i+1}-B\sum^{N}_{i=1}S_i##.
 

Answers and Replies

  • #2
DrDu
Science Advisor
6,258
906
There is no problem to apply the transfer matrix method in the case of other boundary conditions. Just try it out.
 
  • #3
LagrangeEuler
708
19
I tried, problem is with ##Z=\sum_{allposibilities}\langle S_1|\hat{T}|S_2 \rangle \langle S_2|\hat{T}|S_3 \rangle...\langle S_N|\hat{T}|S_1 \rangle=\sum_{S_1}\langle S_1 |\hat{T}^{N}|S_1 \rangle=Tr(\hat{T}^{N})##
I can't get trace in case without circular boundary condition.
 
  • #4
DrDu
Science Advisor
6,258
906
I can't get trace in case without circular boundary condition.
So what? The term <S_1|T|S_N> is missing. Ok.
You get [itex] \sum_{S_1} \sum_{S_N} \langle S_1| T^{N-1} |S_N\rangle [/itex].
What happens when you decompose |S_1/N> into eigenfunctions of T?
 
  • #5
LagrangeEuler
708
19
I'm not quite sure what you think? One eigenvector is just ##|+1\rangle##.
 
  • #6
LagrangeEuler
708
19
[tex]\hat{T}= \left( \begin{array}{cc}
\langle +|\hat{T}|+ \rangle & \langle +|\hat{T}|- \rangle \\
\langle -|\hat{T}|+ \rangle & \langle -|\hat{T}|- \rangle \end{array} \right)[/tex]
For ##|S_1/_N\rangle## there are two possibilities ##|+\rangle## or ##|-\rangle##.
But I don't know how to write that in some normal way and calculate.
 
  • #7
DrDu
Science Advisor
6,258
906
That's a 2x2 matrix in the basis of the two states |+> and |->. I suppose you know how to find the eigenvectors of a 2x2 matrix?
 
  • #8
LagrangeEuler
708
19
Yes. For matrix
[tex] \left( \begin{array}{ccc}
e^{j+h} & e^{-j} \\
e^{-j} & e^{j-h}
\end{array} \right) [/tex]
eigenvectors are
[tex] \left( \begin{array}{ccc}
\frac{1}{2}e^{-h}(-e^{2j}+e^{2h+2j}\pm \sqrt{4e^{2h}+e^{4j}-2e^{2h+4j}+e^{4h+4j}}) \\
1
\end{array} \right) [/tex]
 
  • #9
DrDu
Science Advisor
6,258
906
What I meant is that you can write
[itex] U^T T U=\Lambda[/itex] with [itex]\Lambda[/itex] being a diagonal matrix of the eigenvalues [itex]\lambda_1[/itex] and [itex] \lambda_2[/itex]. I assume [itex]\lambda_1>\lambda_2[/itex] and call V_0 the matrix V with j=0.
Hence [itex]Z=Tr(V^{N-1}V_0)=Tr(U^T\Lambda^{N-1}UV_0)=Tr(\Lambda^{N-1} UV_0U^T)\approx
\lambda_1^{N-1} (UV_0U^T)_{11}[/itex].
Hence Z differs from the Z for periodic boundary conditions only by the factor [itex] (UV_0U^T)_{11}/\lambda_1[/itex]. As we are only interested in Log Z, this makes a term which is 1/N times smaller than Z. This is of the same order as the error resulting from the replacement of [itex]\Lambda^{N-1}[/itex] by [itex]\lambda_1^{N-1}[/itex].
 
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