# Product Rule for Partition Sums $Z_N=(Z_1)^N$

For the 1D Ising model with the Hamiltonian

$$H=const.-\mu h' \sum_i S_i^z$$

we can write the canonical partition sum as

$$Z_N = \sum_{ \{ S_i^z \}_N } e^{-\beta \mu h \sum_i S^z_i} = \sum_{ \{ S_i^z \}_N } \prod_i e^{-\beta \mu h S^z_i}$$

for which we then later used

$$Z_N=(Z_1)^N$$with the single particle partition sum $$Z_1$$

We didn't go further into the proof during the lecture, and it's not immediately obvious to me.

My attempt:

For the classical ideal gas I could use the Multinomial Theorem

$$(x_1+x_2+\dots+x_n)^k=\sum_{k_1+\dots+k_n=k} \frac{k!}{k_1!\dots k_n!} \ x_1^{k_1}x_2^{k_2} \dots x_n^{k_n}$$

but for that the factor $$\frac{k!}{k_1!\dots k_n!}$$ would have to be $$1$$.

How do I best prove $$Z_N=(Z_1)^N$$?

In fact, $$Z_N = (Z_1)^N$$ is true for all non-interacting systems with identical components, where the Hamiltonian can be written as $$H = H_1 + ... + H_N$$ where $$H_k$$ only depends on the state of the $$k$$-th particle (/spin/...) and all $$H_k$$ are the same.
$$Z_N = \sum_{states \ s_1...s_N} e^{- \beta H} = \sum_{s_1} \sum_{s_2} ...\sum_{s_N} e^{- \beta (H_1+...+H_N)}= \sum_{s_1} e^{- \beta (H_1)} \sum_{s_2} e^{- \beta (H_2)}...\sum_{s_N} e^{- \beta (H_N)} = ( \sum_{s_1} e^{-\beta H_1})^N = Z_1^N$$
Note: For identical particles, a factor $$1/N!$$ might additionally be necessary.
First of all, the last manipulation of the partition sum is not correct. You cannot pull out a global factor $$e^{-\beta\mu h}$$.
To give you a hint: It might be helpful to write $$\sum_{\{S_i\}} = \prod_i\sum_{S_i=\pm 1} = \sum_{S_1=\pm 1}\sum_{S_2=\pm 1}\cdots\sum_{S_N=\pm 1}$$ and exploit that the Boltzmann factor factorizes.