abs(phi(x,y))=abs(phi(y,x)) implies phi(y,x)=exp(i•alpha)•phi(x,y) Making$$ \left| \phi(x,y) \right| = \left| \phi(y,x) \right| \implies \phi(y,x) = e^{i \alpha} \phi(x,y) $$
Making exchange twice, we we get phi(x,y)=exp(2i•alpha)•phi(x,y) Hence
$$\phi(x,y) = e^{2 i \alpha} \phi(x,y)$$
Hence,exp(2i•alpha)=1 so alpha=integer•pi If
$$e^{2 i \alpha} = 1$$
so $\alpha = n \pi$ with $n$ integer. If it's an odd integer then phi(x,y)=-phi(y,x) If then $\phi(x,y) = -\phi(y,x)$. If it's an even integer then phi(y,x)=+phi(x,y) then $\phi(y,x) = \phi(x,y)$ (assuming alpha$\alpha$ is a constant )
