I have the following relationship \begin{equation} \int_{-\infty}^\infty W_1(x,y)\Gamma(x,y)dxdy=\int_{-\infty}^\infty W_2(x,y)\Gamma(x,y)dxdy=\frac1{2\pi}\int_{-\infty}^\infty\Gamma(x,y)dxdy \end{equation} where $x$ and $y$ are real numbers, $W_1$ and $W_2$ are real valued functions, but $\Gamma$ is generally complex valued. Everything is known except $\Gamma$. Is $\Gamma$ uniquely defined by this relationship (barring $\Gamma=0$)? If so, how might one solve for it? And if not, how can I see that? I also know that \begin{align} 0&\leq\left|\int_{-\infty}^\infty\Gamma(x,y)dxdy\right|<1 & &\text{and} & \int_{-\infty}^\infty W_1(x,y)dxdy&=\int_{-\infty}^\infty W_2(x,y)dxdy=1 \end{align}
Thanks in advance and please let me know if there's anything I can clarify!
Edit: I also should note that $W_1\neq W_2$, and further that \begin{equation} 2\pi\int_{-\infty}^\infty \left(W_1(x,y)\right)^2dxdy=2\pi\int_{-\infty}^\infty (W_2(x,y))^2dxdy=1 \end{equation} This last relation comes from the fact that $W_1$ and $W_2$ are Wigner functions of pure quantum states. The norm of $\int_{-\infty}^\infty \Gamma dxdy$ is also known, but not the phase.