Given $A,B\in\mathbb{C}^{n\times n}$. Lets define the following set (Davis-Wielandt shell):
$$DW(A):=\{(x^*Ax,x^*A*Ax) \mid x\in\mathbb{C^n},\|x\|_2=1\}\subseteq\mathbb{C}\times\mathbb{R}\sim\mathbb{R}^3.$$
Assume that $DW(A)\cap DW(-B)$ is nonempty, then there are orthogonal pairs $(x_1,x_2)\in\mathbb{C^n}$ and $(y_1,y_2)\in\mathbb{C^n}$ such that $$Ax_1=\alpha x_1+\beta x_2\qquad\text{and}\qquad -By_1=\alpha y_1+\beta y_2.$$
with $(\alpha, \alpha^2+\beta^2)\in DW(A)\cap DW(-B)$. Suppose $U$ is unitary with $x_1,x_2$ as its first two columns, and $V$ is unitary with $y_1,y_2$ as its first two columns. Then $U^*AU+V^*BV$ has zero first column.
I want to construct these $U$ and $V$. I know $A,B,\alpha,\beta$. And also I know that $DW(A)$ is convex set.
My attempt:
From $Ax_1=\alpha x_1+\beta x_2$ we can get $\frac{A-\alpha I}{\beta}x_1=Cx_1=x_2$. Then we need to find $x_1$ such that the following three equations hold: $\|x_1\|=1$, $\|Cx_1\|=1$, $x_1^*x_2=0$.
If $C=U\Sigma V^*$ has singular values $\sigma_{max}>1$ and $\sigma_{max}<1$, then one possible way to construct $x_1$ might be $x_1=V[\sqrt{\frac{1-\sigma_{min}^2}{\sigma_{max^2}-\sigma_{min}^2}}\quad0\quad\cdots\quad0\quad \sqrt{1-\frac{1-\sigma_{min}^2}{\sigma_{max^2}-\sigma_{min}^2}}]^T.$
Then both $\|x\|=1$ and $\|Cx\|=1$ but I don't think $x_1^*x_2=0$ is satisfied. And also this procedure won't work when $C$ has all singular values larger than 1.
If I can get correct $x_1$ and $x_2$, then by using the following Matlab code I can get $U$:
U = [x1,x2,null([x1,x2]')]
It does not seem solvable. Consider e.g., the case $A=2I_n$ ($I_n$ is the identity matrix) andd $\alpha=0, \beta=1$. Clearly, $A,\alpha$ and $\beta$ fulfill your assumptions. However, for any $x_1$ with $\|x_1\|=1$ we have $2x_1=Ax_1\stackrel{!}{=} \alpha x_1 + \beta x_2 = \beta x_2$, thus $\|x_2\|=2$ and it is impossible to find an orthonormal pair $(x_1,x_2)$.