The [Lee-Yang circle theorem](https://freedommathdance.blogspot.com/2018/05/a-theorem-of-lee-yang.html) 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?



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


