Converse of the Lee-Yang circle theorem for polynomials with unitary roots

Question

The Lee-Yang circle theorem states that if $\left( a_{ij} \right)$ is a Hermitian square $n \times n$ matrix whose entries are in the closed unit disc, then the polynomial $$ P\left(Z \right) = \sum_{S\subseteq\left[n\right]}\left(\prod_{i\in S,j\notin S}a_{ij}\right)Z^{\left|S\right|} $$ has all of its roots on the unit circle. Here $\left[n\right] = \{1,2,\dots,n\}$.

Let $P\left( Z \right)$ be the polynomial above, and denote the coefficient of $Z^k$ by $\alpha_k$. Note that $\alpha_0 = \alpha_n = 1$. Then it follows from the Lee-Yang circle theorem that for every real $\lambda$ with $\left|\lambda \right| \le 1$, the polynomial $$P_\lambda \left( Z \right)= \sum_{k=0}^n{\alpha_k \lambda^{k\left(n - k\right)} Z^k}$$ has all of its roots on the unit circle (this follows from replacing the matrix $\left( a_{ij} \right)$ with the matrix $\left( \lambda a_{ij} \right)$).

My question: suppose $$Q\left( Z \right) = \sum_{k=0}^n{\beta_k Z^k}$$ is a polynomial with complex coefficients, such that $\beta_0 = \beta_n = 1$, and such that for every $\lambda$ with $\left| \lambda \right| \le 1$ the polynomial $$ Q_\lambda\left(Z\right) = \sum_{k=0}^n{\beta_k \lambda^{k \left(n - k\right)} Z^k}$$ has all of its roots on the unit circle. Then is it possible to write $Q\left(Z\right)$ as $P\left(Z\right)$ above for some Hermitian matrix $\left(a_{ij}\right)$ with entries in the closed unit disc? Is there some algorithm to find such matrix?

Answer 1

It looks to be enough to require this for $\lambda=1$.

Indeed, assume that the polynomial $Q(Z)=\sum_{k=0}^n \beta_k Z^k$, $\beta_0=\beta_n=1$, has all roots on the unit circle. Denote the roots by $-\theta_i^{-1}$, $i=1,\ldots,n$. Then $Q(Z)=\prod (1+\theta_i Z)$ and $\prod_i \theta_i=1$.

There exists an Hermitian matrix $(a_{ij})$ with $|a_{ij}|=1$ such that $\theta_i=\prod_{j\ne i} a_{ij}$. Indeed, we may put $a_{ij}=1$ if $\max(i,j)<n$, $a_{in}=\overline{a_{ni}}=\theta_i$ for $i=1,\ldots,n-1$, $a_{nn}=1$. Then $\theta_i=\prod_{j\ne i} a_{ij}$ for all $i=1,\ldots,n-1$ by construction and also for $i=n$ due to $\theta_n=\prod_{i=1}^{n-1} \overline{\theta_i}=\prod_{i=1}^{n-1} a_{ni}$.

Now $$\beta_k=\sum_{|S|=k}\prod_{i\in S} \theta_i=\sum_{|S|=k}\prod_{i\in S} \prod_{j\ne i} a_{ij}=\sum_{|S|=k}\prod_{i\in S} \prod_{j\notin S} a_{ij}$$ since for $j\in S$ the multiples $a_{ij}$ and $a_{ji}$ cancel out.