# One-dimensional $q$-state Potts Model Partition Function - without Boundary Conditions

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?

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 $${\cal Z}=q\big[e^{\beta J}+q-1\big]^{N-1}$$
We start again with the Hamiltonian $$H(\vec s)=- J \sum_{i=1}^{N-1} \delta(s_i - s_{i+1})$$ where $$s_i \in \mathbb S$$ are entries of the global configuration $$\vec s \in \mathbb S^N$$. The partition function is $$Z = \sum_{\vec s \in \mathbb S^N} \exp(- \beta H(\vec s)) = \sum_{\vec s \in \mathbb S^N} \exp(\beta J \sum_{i=1}^{N-1} \delta(s_i - s_{i+1})) = \sum_{\vec s \in \mathbb S^N} \prod_{i=1}^{N-1} \exp(\beta J \delta(s_i - s_{i+1}))$$ We set $$q := |\mathbb S|$$ to be the number of spins possible. Now we make the substitution $$\theta_j = s_j - s_{j+1}$$, then $$Z$$ becomes $$Z = \sum_{\vec s \in \mathbb S^N} \prod_{j=1}^{N-1} \exp(\beta J \delta(\theta_j)) = \sum_{s_1 \in \mathbb S} \dots \sum_{s_N \in \mathbb S} e^{\beta J \delta(\theta_1)} \dots e^{\beta J \delta(\theta_{N-1})}$$ Now because of the substitution we effectively reduced the dimension of the global spin configuration by 1, so it follows $$Z = \sum_{\theta_1 \in \mathbb S} \dots \sum_{\theta_{N-1} \in \mathbb S} e^{\beta J \delta(\theta_1)} \dots e^{\beta J \delta(\theta_{N-1})} \sum_{s_N \in \mathbb S}$$ $$= \left(\sum_{\theta \in \mathbb S} e^{\beta J \delta(\theta)}\right)^{N-1} \sum_{s_N \in \mathbb S} = \left(\sum_{\theta \in \mathbb S} e^{\beta J \delta(\theta)}\right)^{N-1} | \mathbb S |$$ There is only one way of having $$\theta = 0$$ in the sum and exactly $$q - 1$$ ways of having $$\theta \neq 1$$. This leads to the final result $$Z = q \left(e^{\beta J} + q - 1 \right)^{N-1}$$