I assume some of you may be familiar with this model but to be clear, I will write some of the assumptions:
We have a network of little magnets, which we identify with $\mathbb{Z}^2$, for this model we restrict ourselves to some $\Lambda\subset\mathbb{Z}^2$, where $\Lambda$ is some rectangle and each point $X_i$ from $\Lambda$ is a random variable, which values belong to $\{{-1,1}\}$, $\Omega_\Lambda$ is a set of functions from $\Lambda$ to $\{{-1,1}\}$ $$\sigma_i:X_i\rightarrow\{{-1,1}\}$$ There is Hamiltonian defined as follows: $$ H_{\Lambda}(\sigma)=-\sum_{\substack{i,j\in\Lambda \\ |i-j|=1}} \sigma_i \sigma_j -h \sum_{\substack {i \in \Lambda }} \sigma_i $$ Magnetism is defined follows:$$M_\Lambda=\sum_{\substack {i \in \Lambda }}\sigma_i$$ Measure:$$ q(T,h,\Lambda)=\frac{e^{-\frac{1}{T}H_\Lambda}}{Q_{T\space h\space \Lambda}}$$ $$ Q_{T\space h\space \Lambda}=\sum_{\substack {X \in \Omega_\Lambda }}e^{-\frac{1}{T}H_\Lambda(x)}$$ and the energy $$F_{T\space h\space \Lambda}=-T\ln Q_{T\space h\space \Lambda}$$ Finally here come my questions:
- I have read that $M$ is equal to $F$ differentiated with respect to $h$, but I am not sure how to perform this calculation, because of the sum over all $X \in \Omega_\Lambda$.
- Secondly supposing it is true, how do i use this result to calculate the mean of $M$ in point $h=0$? I mean, I can't see how this result is useful.
- Trying to calculate the mean at point $h=0$ from definition of $M$
$$E{M_\Lambda}=E{{\sum_{\substack{i\in\Lambda}} \sigma_i}}=|\Lambda|E{\sigma_i}=|\Lambda|*((P(\sigma_i=1)-P(\sigma_i=-1))=0$$Is this correct?
For part 1, just put the derivative inside the sum. The final answer should be $$M=\frac{1}{Q_{Th\Lambda}}\sum_{X\in\Omega_\Lambda}M_\Lambda e^{-\frac1T H_\Lambda}$$ This means that you are taking an average over all configuration, weighted by the probability to get that particular configuration. For part 2, let's call $-\Lambda$ the state where you flip all spins. If $h=0$, you get $H_\Lambda=H_{-\Lambda}$ and $M_\Lambda=-M_{-\Lambda}$. Since you sum over all configurations, you should get the same answer if you sum over $\Lambda$ or $-\Lambda$, so you will get $M=-M$ or $M=0$