The partition function

The RIS model can be treated in considerable detail. As a first step we calculate the partition function 

$$Z=\sum_{\varphi _{2}\in \{t,g^{+},g^{-}\}}\cdots \sum_{\varph... ...^{-}\}}\exp \{-\beta E(\varphi _{2},\ldots ,\varphi _{N-1})\}.$$ (1.3)

Introducing the energy (1.2) and using a short hand notation for the summations we get

$$Z=\sum_{\varphi _{2}}\cdots \sum_{\varphi _{N-1}}e^{-\beta \e... ...rphi _{2},\varphi _{3})\cdots t(\varphi _{N-2},\varphi _{N-1})$$ (1.4)

$$t(\varphi _{i-1},\varphi _{i})=\exp \{-\beta \epsilon _{1}(\varphi _{i})-\beta \epsilon _{2}(\varphi _{i-1},\varphi _{i})\}.$$ (1.5)

We recognize a sequence of matrix products in Eq. (1.4), which makes it possible to write

$$Z=\sum_{\varphi _{2}}\sum_{\varphi _{N-1}}e^{-\beta \epsilon _{1}(\varphi _{2})}[T^{N-3}]_{\varphi _{2},\varphi _{N-1}}$$ (1.6)

T is a matrix with elements $t(\varphi _{i-1},\varphi _{i})$. In the case of polyethylene  it is given by

 

$$\begin{array}{ccc} t & g^{+} & g^{-} \end{array}$$

$$T= \begin{array}{c} t \\ g^{+} \\ g^{-} \end{array}\left( {\beg... ... & \sigma & \sigma \\ 1 & \sigma & 0 \\ 1 & 0 & \sigma \end{array}} \right) ,$$ (1.7)

where $\sigma =\exp \{-\beta \epsilon _{1}(g^{+})\}=\exp \{-\beta \epsilon _{1}(g^{-})\}$ and $\epsilon _{2}(\varphi _{i-1},\varphi _{i})=0$ except for $\epsilon _{2}(g^{+},g^{-})=\epsilon _{2}(g^{-},g^{+})=\infty$. The zero of energy has been chosen such that $\epsilon _{1}(t)=0$.

Eq. (1.6) may also be written as

$$Z = \left( {\begin{array}{ccc} 1 &\sigma &\sigma \end{array}... ...{\left( {\begin{array}{c} 1 \\ 1 \\ 1 \end{array}} \right)} .$$ (1.8)

Moreover, using

$$\left( {\begin{array}{ccc} 1 &\sigma &\sigma \end{array}} \right) = \left( {\begin{array}{ccc} 1 &0 &0 \end{array}} \right) T$$ (1.9)

we may write

Z = Y^T T^N-2 X (1.10)

with $Y^T = ( \begin{array}{ccc} 1 & 0 & 0 \end{array} )$ and $X^T = ( \begin{array}{ccc} 1 & 1 & 1 \end{array})$.

It is useful to decompose the matrix T in terms of its eigenvectors:

$$TA = A\Lambda$$ (1.11)

$$T = A\Lambda A^{-1} \equiv A\Lambda B$$ (1.12)

where $\Lambda$ is the diagonal matrix containing the eigenvalues of T. For notational convenience we have introduced B = A^-1.

Eq. (1.12) may be written like

$$T = \sum_i \lambda_i A_i B_i^T$$ (1.13)

where the A_i are the columns of A and the B_i^T the rows of B. Then

$$(A\Lambda A^{-1}) (A\Lambda A^{-1}) \cdots (A\Lambda A^{-1}) = A \Lambda^N A^{-1} = A \Lambda^N B$$ T^N =

$$\sum_i \lambda_i^N A_i B_i^T.$$ T^N = (1.14)

The partition function then reads

$$Z = \sum_i \lambda_i^{N-2} (Y^T A_i) (B_i^T X) = \sum_i c_i \lambda_i^{N-2}.$$ (1.15)

In practice we are usually interested in the free energy  per monomer

$$\frac{1}{N}\ln Z=\frac{N-2}{N}\ln \lambda _{\mathrm{max}}+\fr... ...lambda _{i}}{\lambda _{\mathrm{max}}} \right) ^{N-2}\right\} ,$$ (1.16)

where $\lambda _{\mathrm{max}}$ is the largest eigenvalue of T. In the limit of N going to infinity the second term goes to zero

$$\lim_{N\rightarrow \infty }\frac{1}{N}\ln Z=\ln \lambda _{\mathrm{max}}.$$ (1.17)

In the case of matrix (1.7) one easily calculates the eigenvectors. From $\det (T-\lambda I)=0$ one gets

$$(\lambda ^{2}-\lambda (1+\sigma )-\sigma )(\sigma -\lambda )=0$$

$$\lambda _{\mathrm{max}}=\frac{1}{2}\{(1+\sigma )+\sqrt{1+6\sigma +\sigma ^{2} }\}.$$ (1.18)