I tried using Bill's code for obtaining the parameters $\theta, \phi$ and $\lambda$ for the unitary
$$Z=\begin{pmatrix}\frac{1}{\sqrt 2} & \frac{i}{\sqrt 2} \\ \frac{i}{\sqrt 2} & \frac{1}{\sqrt 2} \end{pmatrix}$$.
Here's my code:
Z = {{1/Sqrt[2], I/Sqrt[2]}, {I/Sqrt[2], 1/Sqrt[2]}};
U = {{Cos[\[Theta]/
2], -E^(I \[Lambda]) Sin[\[Theta]/
2]}, {E^(I \[CurlyPhi]) Sin[\[Theta]/2],
E^(I (\[Lambda] + \[CurlyPhi])) Cos[\[Theta]/2]}};
N[Reduce[Z == U && 0 <= \[CurlyPhi] < 2 Pi && 0 <= \[Theta] <= Pi &&
0 <= \[Lambda] < 2 Pi, {\[Theta], \[CurlyPhi], \[Lambda]}, Reals]]
But it says:
"The system {1/Sqrt[2],I/Sqrt[2]}=={Cos[\[Theta]/2],-E^(I\\[Lambda])\ \
Sin[\[Theta]/2]}&&{I/Sqrt[2],1/Sqrt[2]}=={E^(I\\[CurlyPhi])\ Sin[\
\[Theta]/2],E^(I\(\[Lambda]+\[CurlyPhi]))\ Cos[\[Theta]/2]}&&0<=\
\[CurlyPhi]<2\ \[Pi]&&0<=\[Theta]<=\[Pi]&&0<=\[Lambda]<2\ \[Pi] \
contains a nonreal constant I/Sqrt[2]. With the domain Reals \
specified, all constants should be real. "
I don't quite understand this error. By inspection, one simple solution for the given $Z$ is $$(\theta, \phi,\lambda) = (\pi/2,\pi/2,3\pi/2)$$ and clearly $ZZ^{\dagger}=I$.
So why is Mathematica returning me complex values for the angle parameters instead of the simple real solution? Am I missing something?
Answer
Try this
Z = {{1/Sqrt[2], I/Sqrt[2]}, {I/Sqrt[2], 1/Sqrt[2]}};
U = {{Cos[θ/2], -E^(I λ) Sin[θ/2]}, {E^(I ϕ) Sin[θ/2],E^(I (λ + ϕ)) Cos[θ/2]}};
FullSimplify[Reduce[Z == U && 0<=ϕ<2 Pi && 0<=θ<=Pi && 0<=λ<2 Pi, {θ, ϕ, λ}]]
which quickly returns
2*θ == Pi && 2*ϕ == Pi && 2*λ == 3*Pi
There are a vast number of algorithms hiding inside of Reduce and it is often difficult to diagnose exactly what is happening behind the curtain. But it is still very impressive what it is able to accomplish with the tap of a key.
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