I have a lattice model consisting of $N$ spins $s_{j}$ which can take the values $s_{j}=\pm1$. The spins are considered to be non interacting. The probability for a spin spin to be 1 is p and the probability for a spin taking the value -1 is 1-p. The value p is bounded by $0\le p\le1 $. The magnetization per volume is defined as $m_{N}=\sum_{i=1}^{N}s_{j}$.

a) what values can $m_{N}$ take

b) What is the average value of $m_{N}$

c) evaluate the fluctuations $\langle m_{N}^{2}\rangle-\langle m_{N}\rangle ^{2}$

d) For which values of p those fluctuations are minimal

e) compute the probability that $m_{N}=\frac{j}{N}$ with $j=-N,-N+2,..,N$

The magnetization per volume can take any value between all spins pointing down to all spins pointing up divided by N: a) $m_{N}=\left \{ \frac{-N}{N},\frac{-N+1}{N},...,\frac{N-1}{N},\frac{N}{N} \right \}$

b) Since the spins are independent I can compute the average magnetization by using the equation for averages:

$$ \langle x_{i} \rangle = \sum_{k=1}^{N}p_{k}x_{k} $$

Out of this I can compute the average spin value for a single lattice site: $$ \langle m_{i}\rangle=1*p+(-1)(1-p)=2p-1 $$

And therefore the average magnetization per volume is:

$$ \langle m_{N} \rangle = \frac{1}{N}\sum_{i}^{N}\langle m_{i} \rangle = 2p-1 $$

c) in the same manner I would compute the fluctuations:

$$ \langle m_{N}^{2}\rangle-\langle m_{N}\rangle^{2} = 1^{2}*p+(-1)^{2}(1-p)-(2p-1)^{2}=\\ 1-4p^{2}-2p-1=-4p^{2}+2p $$ Now my problem is that the fluctuations are negative for certain values of p what does not make sense.

  • $\begingroup$ $p$ lies between $0$ and $1$, right? $\endgroup$
    – user197851
    Oct 3, 2018 at 19:49
  • $\begingroup$ @ LonelyProf : Correct I forgot. I will add this imidiately $\endgroup$
    – zodiac
    Oct 3, 2018 at 19:51
  • $\begingroup$ No problem. Is it allowed to apply such a block averaging scheme like I do? To compute the average only for a single spin and archive out of this the average magnetization per volume? $\endgroup$
    – zodiac
    Oct 3, 2018 at 20:01
  • $\begingroup$ I've deleted my earlier misleading comments. My advice is to look again at your formula for the averaged total magnetization, and the average total magnetization squared, and relate them carefully to the analogous single spin quantities, remembering that different spins are not correlated. There will be some factors of $N$ in your formulae. That's probably all I can say for now. $\endgroup$
    – user197851
    Oct 3, 2018 at 20:07
  • $\begingroup$ I forgot to write the $\frac{1}{N}$ factor in front of $\langle m_{N} \rangle$ equation therefore the N should factor out I fear. My apologises! $\endgroup$
    – zodiac
    Oct 3, 2018 at 20:12

1 Answer 1


So i figured it out now. There was as stupid sign error and the expansion of the squared bracket was wrong

$$ \langle m_{N}^{2}\rangle-\langle m_{N}\rangle^{2}= \frac{1}{N}\sum_{i=1}^{N}\langle s_{i}^{2} \rangle - \left ( \frac{1}{N}\sum_{i=1}^{N}\langle s_{i} \rangle \right )^{2} =\\ \frac{1}{N}\sum_{i=1}^{N}(1)^{2}p+(-1)^{2}(1-p)-\left ( \frac{1}{N}\sum_{i=1}^{N} (1)p+(-1)(1-p) \right )^{2}=\\ \frac{1}{N}\sum_{i=1}^{N}1-\left ( \frac{1}{N}\sum_{i=1}^{N} (2p-1) \right )^{2}=\\ \frac{1}{N}N*1-\left ( \frac{1}{N}(2p-1)*N \right )^{2}=\\ 1-(2p-1)^{2}=1-4p^{2}+4p-1=-4p^{2}+4p>0 $$ for the considered interval $0 \le p \le 1$


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