First, I got a real version of equation set
Solve[{d u^2 - d v^2 == 2 e u v, (u u + v v) == 1}, {u, v}]
This can be correctly solved in the blink of an eye.
Now I want to solve this complex version
Solve[{d u^2 - d\[Conjugate] v^2 - 2 e u v ==
0, (u u\[Conjugate] + v v\[Conjugate]) == 1}, {u, v}]
mathematica can not be solve the above code.
But I can show how this set of equation can be easily solved.
take
$$ - {v^2}{d^*} + d{u^2} - 2euv = 0$$
multiply both side by $\frac{{{d^*}}}{{{u^2}}}$, we got $$- {\left( {{d^*}\frac{v}{u}} \right)^2} - 2e\left( {{d^*}\frac{v}{u}} \right) + {\left| d \right|^2} = 0$$ we can see this is a simple quadratic equation relative to ${d^*}\frac{v}{u}$, and got $${d^*}\frac{v}{u} = e \pm \sqrt {{e^2} + {{\left| d \right|}^2}} $$ define $$\xi \equiv \sqrt {{e^2} + {{\left| d \right|}^2}} $$ then $$v = \frac{u}{{{d^*}}}\left( {e + \xi } \right)$$ and $${\left| v \right|^2} = \frac{{{{\left| u \right|}^2}}}{{{{\left| d \right|}^2}}}{\left( {e + \xi } \right)^2}$$ plug this back into $${\left| u \right|^2} + {\left| v \right|^2} = 1$$ we got $${\left| u \right|^2}\left[ {1 + \frac{1}{{{{\left| d \right|}^2}}}{{\left( {e + \xi } \right)}^2}} \right] = 1$$ at this step, we can say we solved the absolute value of u and v. their relative phase is determined by previous $v = \frac{u}{{{d^*}}}\left( {e + \xi } \right)$
After I solved the equation, I notice that I missed a condition in Solve
. Because there is phase uncertainty, if we restrict that u is real, then v should be also fixed.
So I thought this can be solved
Solve[{d u^2 - d\[Conjugate] v^2 - 2 e u v ==
0, (u u\[Conjugate] + v v\[Conjugate]) == 1, Im[u]==0}, {u, v}]
With Im[u]
condition added.
But mathematica still can not give an answer. Am I missing something here? How to solve this set of equation?