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Given the following two equations with three unknowns $\eta_1, \eta_2$ and $\tau$:

$\frac{i e^{-i \phi/2}}{\sqrt{2}} = \cos(2 \tau - \eta_1 -\eta_2)\cos(\eta_1-\eta_2) - i\sin(2 \tau - \eta_1-\eta_2)\sin(\eta_1+\eta_2)$

$\frac{i e^{i \phi/2}}{\sqrt{2}} = \cos(2 \tau - \eta_1 -\eta_2)\sin(\eta_1-\eta_2) + i\sin(2 \tau - \eta_1-\eta_2)\cos(\eta_1+\eta_2)$

I'm trying to write code that numerically searches for $\eta_1, \eta_2$ and $\tau$, each taken from $[0, 2\pi]$ but am getting errors (freezes basically). Please advise if there is a simpler way of doing this type of search as compared to what I have. My prototype code is as follows:

(*Define the system of equations for each equation independently*) equation1[eta1_, eta2_, tau_] := I Exp[-I phi/2]/ Sqrt[2] - (Cos[2 tau - eta1 - eta2] Cos[eta1 - eta2] + I Sin[2 tau - eta1 - eta2] Sin[eta1 + eta2]) equation2[eta1_, eta2_, tau_] := I Exp[I phi/2]/Sqrt[2] - (Cos[2 tau - eta1 - eta2] Sin[eta1 - eta2] - I Sin[2 tau - eta1 - eta2] Cos[eta1 + eta2]) (*Set the value of phi*) phi = 2.; (*This parameter is fixed*) (*Range for eta1,eta2,and tau*) eta1Values = Subdivide[0, 2 Pi, 1000]; eta2Values = Subdivide[0, 2 Pi, 1000]; tauValues = Subdivide[0, 2 Pi, 1000]; (*Iterate over all combinations of eta1,eta2,and tau*) Do[Do[Do[(*Use NSolve to find the roots for each equation*) solution1 = NSolve[{equation1[eta1, eta2, tau] == 0, 0 <= eta1 <= 2 Pi, 0 <= eta2 <= 2 Pi, 0 <= tau <= 2 Pi}, {eta1, eta2, tau}, Reals]; solution2 = NSolve[{equation2[eta1, eta2, tau] == 0, 0 <= eta1 <= 2 Pi, 0 <= eta2 <= 2 Pi, 0 <= tau <= 2 Pi}, {eta1, eta2, tau}, Reals]; (*Print the solutions*) If[Length[solution1] > 0, Print["Solution for equation 1: ", solution1]]; If[Length[solution2] > 0, Print["Solution for equation 2: ", solution2]];, {tau, tauValues}], {eta2, eta2Values}], {eta1, eta1Values}]

The solution should exist based on the Solovay-Kitaev decomposition. More specifically based on Eqs. (13)-(15) of the following paper, where I assume $u:= \frac{i e^{-i \phi/2}}{\sqrt{2}}$ and $w:= \frac{i e^{i \phi/2}}{\sqrt{2}}$.

I think part of the problem is that because I'm are choosing discrete values from [0, 2 Pi, 1000], unlikely we will find perfect roots. Hence the equations will never be satisfied. I might have to add some kind of tolerance like NSolve[{equation1[eta1, eta2, tau] <= 0.001 instead of NSolve[{equation1[eta1, eta2, tau] == 0. I'm not sure if there is anything else dodgy but the code simply freezes when I run it. Any advice on how to improve the code would be most appreciated.

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2 Answers 2

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A simple minimization of the square error ought to do it.

phi = 2; (*Define the system of equations for each equation independently*) equation1[eta1_, eta2_, tau_] := I Exp[-I phi/2]/ Sqrt[2] - (Cos[2 tau - eta1 - eta2] Cos[eta1 - eta2] + I Sin[2 tau - eta1 - eta2] Sin[eta1 + eta2]) equation2[eta1_, eta2_, tau_] := I Exp[I phi/2]/ Sqrt[2] - (Cos[2 tau - eta1 - eta2] Sin[eta1 - eta2] - I Sin[2 tau - eta1 - eta2] Cos[eta1 + eta2]) {err, sol} = NMinimize[{ Abs[equation1[eta1, eta2, tau]]^2 + Abs[equation2[eta1, eta2, tau]]^2, 0 <= eta1 <= 2 Pi, 0 <= eta2 <= 2 Pi, 0 <= tau <= 2 Pi }, {eta1, eta2, tau}] 
{3.42769*10^-18, {eta1 -> 5.49779, eta2 -> 3.14159, tau -> 2.4635}}

Studying those numerical figures, it looks like there is an exact solution for $\phi=2$ at $$ \eta_1=7 \pi/4 \\ \eta_2=\pi\\ \tau=(4 + 5\pi)/8 $$

FullSimplify[equation1[7 Pi/4, Pi, 1/8 (4 + 5 Pi)]] == FullSimplify[equation2[7 Pi/4, Pi, 1/8 (4 + 5 Pi)]] == 0 (* True *)
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This can be done symbolically in such a way.

phi = 2; TrigReduce[I Exp[-I phi/2]/ Sqrt[2] - (Cos[2 tau - eta1 - eta2] Cos[eta1 - eta2] + I Sin[2 tau - eta1 - eta2] Sin[eta1 + eta2])]

-(1/2) E^-I (-I Sqrt[2] + E^I Cos[2 eta1 - 2 tau] + E^I Cos[2 eta2 - 2 tau] + I Exp[I phi/2]/ Sqrt[2] - (Cos[2 tau - eta1 - eta2] Sin[eta1 - eta2] - I Sin[2 tau - eta1 - eta2] Cos[eta1 + eta2]) I E^ I Cos[2 eta1 + 2 eta2 - 2 tau] - I E^I Cos[2 tau])

TrigReduce[-I* Exp[I phi/2]/ Sqrt[2] - (Cos[2 tau - eta1 - eta2] Sin[eta1 - eta2] - I Sin[2 tau - eta1 - eta2] Cos[eta1 + eta2])]

1/2 (I Sqrt[2] E^I - Sin[2 eta1 - 2 tau] + Sin[2 eta2 - 2 tau] - I Sin[2 eta1 + 2 eta2 - 2 tau] + I Sin[2 tau])

We see the same arguments in the above. In view of it

Reduce[{a == 2 eta1 - 2 tau, b == eta1 - eta2, c == 2 tau - eta1 - eta2, d == 2 tau}, {eta1, eta2, tau}]

a == b - c && eta1 == b/2 - c/2 + d/2 && eta2 == -(b/2) - c/2 + d/2 && tau == d/2

Now

eq1 = Simplify[ TrigReduce[ I Exp[-I phi/2]/ Sqrt[2] - (Cos[2 tau - eta1 - eta2] Cos[eta1 - eta2] + I Sin[2 tau - eta1 - eta2] Sin[eta1 + eta2])] /. {eta1 -> b/2 - c/2 + d/2, eta2 -> -(b/2) - c/2 + d/2, tau -> d/2, a -> b - c}] == 0; eq2 = Simplify[TrigReduce[ I Exp[I phi/2]/ Sqrt[2] - (Cos[2 tau - eta1 - eta2] Sin[eta1 - eta2] - I Sin[2 tau - eta1 - eta2] Cos[eta1 + eta2])] /. {eta1 -> b/2 - c/2 + d/2, eta2 -> -(b/2) - c/2 + d/2, tau -> d/2, a -> b - c}] == 0; FindInstance[eq1 && eq2 && 0 < b/2 - c/2 + d/2 < 2*Pi && 0 < -(b/2) - c/2 + d/2 < 2*Pi && 0 < d/2 < 2*Pi, {b, c, d}]

The latest command produces a long output and a warning .

FullSimplify[%]

{{b -> -(\[Pi]/4), c -> 2 (\[Pi] + ArcSin[Sqrt[1/2 (1 - Sin[1])]]), d -> 2 (\[Pi] + ArcTan[(2 (1 + Sqrt[2]) Cos[ 1] (Cos[1] - Sin[1]) (-198 + 140 Sqrt[2] + (-140 + 99 Sqrt[2]) Cos[1] + 140 Sin[1] - 99 Sqrt[2] Sin[1]) Sqrt[-1 + 2/( 1 + Sin[1])])/((-2 + Sqrt[2]) (-4 + 3 Sqrt[2] + (-6 + 4 Sqrt[2]) Cos[1])^2 (-1 + Sin[1]))])}}

FullSimplify[b/2 - c/2 + d/2 /. %]

{-(\[Pi]/8) - ArcSin[Sqrt[1/2 (1 - Sin[1])]] + ArcTan[(-140 + 99 Sqrt[2] + 2 (99 - 70 Sqrt[2]) Sin[1])/(-140 + 99 Sqrt[2] + 2 (-99 + 70 Sqrt[2]) Cos[1])]}

Similarly with eta2 and tau.

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