Skip to main content

matrix - Why is this Mathematica code unable to solve for the real parameter set?


I tried using Bill's code for obtaining the parameters θ,ϕ and λ for the unitary


Z=(1√2i√2i√21√2)

.


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 (θ,ϕ,λ)=(π/2,π/2,3π/2)

and clearly ZZ†=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.


Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more