Lee–Yang zeros

Remarks

The identity holds since $\sum_v \sigma_v = N - 2k(\sigma)$. The coefficients of $P$ are positive, so $P$ has no zero on $[0, \infty)$, which is the set of values of $z$ for real $h$.

Example: two vertices joined by one edge with coupling $J$. Then $P(z) = e^{\beta J} (1 + 2 e^{-2\beta J} z + z^2)$, whose roots $z = -e^{-2\beta J} \pm i \sqrt{1 - e^{-4\beta J}}$ satisfy $\abs{z}^2 = e^{-4\beta J} + 1 - e^{-4\beta J} = 1$ when $J > 0$. This agrees with the Lee–Yang theorem, which says that all zeros lie on $\abs{z} = 1$ when every $J_e \ge 0$.

Non-example: the zeros of $Z$ as a function of a complex inverse temperature $\beta$ at fixed $h$ are not Lee–Yang zeros. They are called Fisher zeros and need not lie on a circle. For $J < 0$ in the example above, the roots are real and negative with product $1$, so they are off the unit circle.