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If I consider the 1D (nearest-neighbour) Potts model consisting of $N$ spins that can each take on $q$-states, I consider the Hamiltonian: $$ H = - J \sum_{j=1}^{N-1} \delta_{n_{j}, n_{j+1}} $$

Where $n_{j} \in \{ 1, \ldots, q \}$ for each $j$, and $\delta_{n_{j}, n_{j+1}}$ is the Kronecker delta.

In what literature I've found, this system is solved assuming Periodic Boundary Conditions (PBC) (so that $n_{N+1} = n_{1}$). This is usually done using the Transfer matrix method, and yields the partition function $Z_{\mathrm{PBC}} = \left( e^{\beta J} + 1 \right)^{N} + \left( e^{\beta J} - 1 \right)^{N}$.

If I consider this system $without$ boundary conditions; my calculation for the partition function goes like: $$ Z = \sum_{n_{1}=1}^{q} \cdots \sum_{n_{N-1}=1}^{q} \sum_{n_{N}=1}^{q} \exp\left( J \beta \sum_{j=1}^{N} \delta_{n_{j},n_{j+1}} \right) \\ = \sum_{n_{1}=1}^{q} \cdots \sum_{n_{N-1}=1}^{q} \sum_{n_{N}=1}^{q} \exp\left( J \beta \delta_{n_{1},n_{2}} \right) \cdots \exp\left( J \beta \delta_{n_{N-2},n_{N-1}} \right) \exp\left( J \beta \delta_{n_{N-1},n_{N}} \right) \\ = \sum_{n_{1}=1}^{q} \cdots \sum_{n_{N-1}=1}^{q} \exp\left( J \beta \delta_{n_{1},n_{2}} \right) \cdots \exp\left( J \beta \delta_{n_{N-2},n_{N-1}} \right) \exp\left( J \beta \delta_{n_{N-1},n_{N-1}} \right) \\ = \sum_{n_{1}=1}^{q} \cdots \sum_{n_{N-2}=1}^{q} \exp\left( J \beta \delta_{n_{1},n_{2}} \right) \cdots \exp\left( J \beta \delta_{n_{N-3},n_{N-2}} \right) \exp\left( J \beta \delta_{n_{N-2},n_{N-2}} \right) \exp\left( J \beta \delta_{n_{N-2},n_{N-2}} \right) \\ = \sum_{n_{1}=1}^{q} \cdots \sum_{n_{N-1}=1}^{q} \exp\left( J \beta \delta_{n_{1},n_{2}} \right) \cdots \exp\left( J \beta \delta_{n_{N-2},n_{N-1}} \right) \exp\left( J \beta \delta_{n_{N-1},n_{N-1}} \right) \\ = \sum_{n_{1}=1}^{q} \left[ \exp\left( J \beta \delta_{n_{1},n_{1}} \right) \right]^{N-1} \\ = q \left[ \exp\left( J \beta \right) \right]^{N-1} \\ $$

So my partition function $Z = q e^{(N-1) J \beta}$ without boundary conditions.

I am confused because this partition function is far simpler that the PBC partition functions. Isn't it true that PBC should be simplifying the partition function? This partition function seems to yield almost trivial thermodynamics...

Question: Why is this partition so simple as compared to the PBC version? Or have I calculated this incorrectly?

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I think that there is a mistake when going from the second line to the third one. Consider $$\sum_{n_N} e^{\beta J\delta_{n_{N-1},n_N}} =\sum_{n_N} \big[e^{\beta J}\delta_{n_{N-1},n_N}+e^0(1-\delta_{n_{N-1},n_N})\big]=\big(e^{\beta J}-1\big)+q$$ Your partition function should finally read $${\cal Z}=q\big[e^{\beta J}+q-1\big]^{N-1}$$

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