- #1
Dom Tesilbirth
- 4
- 1
- Homework Statement
- Consider a one-dimensional Ising model with ##N## spins at very low temperature. Let there be ##r## spin flips with each costing energy ##2J##. The total energy of the system is ##E=-NJ+2rJ## and the number of configurations is ##C(N, r)##, where ##r## varies from ##0## to ##N##. Find the partition function.
- Relevant Equations
- ##E=-NJ+2rJ## and
##Z=\sum ^{N}_{r=0}C\left( N,r\right) e^{-\beta \left[ -NJ+2rJ\right] }##
Attempt at a solution:
\begin{aligned}Z=\sum ^{N}_{r=0}C\left( N,r\right) e^{-\beta \left[ -NJ+2rJ\right] }\\
\Rightarrow Z=e^{\beta NJ}\sum ^{N}_{r=0}C\left( N,r\right) e^{-2\beta rJ}\end{aligned}
Let ##e^{-2\beta J}=x##. Then ##e^{-2\beta rJ}=x^{r}##.
\begin{aligned}\therefore Z=e^{\beta NJ}\sum ^{N}_{r=0}C(N, r)x^{r}\\
\Rightarrow Z=e^{\beta NJ}\left( 1+x\right) ^{N}=\left( e^{\beta J}+e^{\beta J}e^{-2\beta J}\right) ^{N}\\
\Rightarrow Z=\left( e^{\beta J}+e^{-\beta J}\right) ^{N}\\
\Rightarrow Z=\left( 2\cosh\beta J\right) ^{N}\end{aligned}
However, the answer provided is ##Z=\left(\cosh \beta J\right) ^{N}##. How do we remove the factor ##2##? Was the given answer wrong, or is there something that I still need to do?
\begin{aligned}Z=\sum ^{N}_{r=0}C\left( N,r\right) e^{-\beta \left[ -NJ+2rJ\right] }\\
\Rightarrow Z=e^{\beta NJ}\sum ^{N}_{r=0}C\left( N,r\right) e^{-2\beta rJ}\end{aligned}
Let ##e^{-2\beta J}=x##. Then ##e^{-2\beta rJ}=x^{r}##.
\begin{aligned}\therefore Z=e^{\beta NJ}\sum ^{N}_{r=0}C(N, r)x^{r}\\
\Rightarrow Z=e^{\beta NJ}\left( 1+x\right) ^{N}=\left( e^{\beta J}+e^{\beta J}e^{-2\beta J}\right) ^{N}\\
\Rightarrow Z=\left( e^{\beta J}+e^{-\beta J}\right) ^{N}\\
\Rightarrow Z=\left( 2\cosh\beta J\right) ^{N}\end{aligned}
However, the answer provided is ##Z=\left(\cosh \beta J\right) ^{N}##. How do we remove the factor ##2##? Was the given answer wrong, or is there something that I still need to do?
