Taylor expansion of an Ising-like Hamiltonian

AI Thread Summary
The discussion focuses on the Taylor expansion of the partition function Z for an Ising-like Hamiltonian under different magnetic field conditions. When the magnetic field B is zero, Z simplifies to a product form involving the sum of exponentials, yielding Z = [1 + e^(-βA)]^N. For non-zero B, the first-order approximation introduces a term that complicates the evaluation of sums, particularly concerning the contributions from neighboring spins. The participant expresses uncertainty about evaluating the sums and suggests that transforming the sums into products of exponentials could simplify the calculations. The conversation highlights the challenges in deriving Z and log(Z) in the presence of interactions between spins.
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Homework Statement
Compute ##Z## and ##\log{Z}## of a set of ##N## two-level systems with total energy specified by the given Hamiltonian by Taylor expanding in powers of ##B## to first and second order.
Relevant Equations
$$H(\{n_i\}) = A\sum_{i=1}^{N}n_i + B\sum_{i=1}^{N-1}n_i n_{i+1}$$
##n_i = 0, 1## for ##i = 1, ..., N##
For the case when ##B=0## I get: $$Z = \sum_{n_i = 0,1} e^{-\beta H(\{n_i\})} = \sum_{n_i = 0,1} e^{-\beta A \sum_i^N n_i} =\prod_i^N \sum_{n_i = 0,1} e^{-\beta A n_i} = [1+e^{-\beta A}]^N$$
For non-zero ##B## to first order the best I can get is:
$$Z = \sum_{n_i = 0,1} e^{-\beta(A\sum_{i=1}^{N}n_i + B\sum_{i=1}^{N-1}n_i n_{i+1})} \approx \sum_{n_i = 0,1} e^{-\beta A\sum_{i=1}^{N}n_i} \left[1-\beta B \sum_{i=1}^{N-1}n_i n_{i+1} \right]$$ $$=
[1+e^{-\beta A}]^N - \beta B \sum_{n_i = 0,1} \sum_{i=1}^{N-1}n_i n_{i+1} e^{-\beta A\sum_{i=1}^{N}n_i}
$$ At this point I'm not sure how to evaluate the sums. Obiously, the only case when the sums yield a non-zero contribution is when ##n_i = n_{i+1} = 1##, but I don't know what ##e^{-\beta A\sum_{i=1}^{N}n_i}## evaluates to in that case. Should this just be ##-\beta B (N-1) e^{-\beta A N}##? That doesn't seem like the right answer since I can't really evaluate ##\log{Z}## then.
 
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It seems like it would be easier to turn the sums in the arguments of the exponential into products of exponentials like you did in the ##B=## case. So you get
$$
Z = \prod_i \sum_{n_i = 0,1} \exp\left( - \beta A n_i - \beta B n_i n_{i+1} \right) \approx \prod_i \sum_{n_i = 0,1} \exp\left( - \beta A n_i \right) \left[ 1 - \beta B n_i n_{i+1} \right].
$$
Now the sum over ##n_i = 0,1## is straightforward, and the same idea applies to second-order in ##B##.
 
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