I am not sure what exactly do you expect, but basically every answer to your question is just a reformulation of the required point sets. Suppose we are talking about vectors in $\mathbb{C}^n$. Let $U$ be a unitary matrix such that $UAU^H = \mathrm{diag}(d)$ for some $d\in\mathbb{R}^n$ ($d$ is a real vector because every eigenvalue of a Hermitian matrix is real). Define a polytope $P$ and a halfspace $H$ by
\begin{align}
P &= \left\{x=(x_1,\ldots,x_n)^T\in\mathbb{R}^n: x_i\ge 0\ \forall i,\ \sum_kx_k= 1\right\},\\
H &= \left\{x\in\mathbb{R}^n: x^Td\ge 0\right\}.
\end{align}
Note that $u=(u_1,\ldots,u_n)\in\mathbb{C}^n$ is a unit vector if and only if
$$
f(u)\equiv u\circ \overline{u}\equiv (|u_1|^2,\ldots,|u_n|^2)\in P.
$$
Then
$$
\left\{x\in\mathbb{C}^n: \|x\|=1,\ x^HAx\ge 0\right\}
= U^H\left(f^{-1}(H\cap P)\right).
$$
This is because
\begin{align}
x^HAx\ge 0
&\Leftrightarrow (Ux)^H\mathrm{diag}(d)(Ux) \ge 0\\
&\Leftrightarrow y^Td \ge 0,\ \textrm{ where } y=f(Ux)\\
&\Leftrightarrow y\in H\cap P\\
&\Leftrightarrow Ux\in f^{-1}(H\cap P)\\
&\Leftrightarrow x\in U^H\left(f^{-1}(H\cap P)\right).
\end{align}
So, the set $\{x\in\mathbb{C}^n: \|x\|=1,\ x^HA_1x, x^HA_2x\ge0\}$ would be something like $U_1^H\left(f^{-1}(H_1\cap P)\right) \cap U_2^H\left(f^{-1}(H_2\cap P)\right)$, but I wonder if this formulation could be useful.