I have a 2x2 complex-valued matrix, $U=\pmatrix{U_{1,1}e^{i\theta_{1,1}}&U_{1,2}e^{i\theta_{1,2}}\\ U_{2,1}e^{i\theta_{2,1}}&U_{2,2}e^{i\theta_{2,2}}}$ which I must impose two conditions: 1) it is unitary $UU^\dagger=\mathcal{I}$ and 2) the product with its transpose is a fixed value $UU^{T}=M$, where this matrix has zero at his diagonal. This gives me a set of equations:

From condition 1:





From condition 2:




So we have 8 variables and 7 equations of motion, so we get one free parameter.

From the first equation of both conditions, we can already see what a solution looks like,

$U_{\rm sol}=\frac{1}{\sqrt{2}}\pmatrix{1&-i\\ 1&i}$ and this fulfils the conditions I need. The problem is that I cannot rely on 'observations' like this because I have to scale it up to more dimensions like 3x3, which is much more involved. I tried to get the same solution using Mathematica, but it just doesn't work and I cannot figure out why. Here is my code:

$Assumptions = {{u11, u12, u21, 
     u22, \[Theta]11, \[Theta]21, \[Theta]12, \[Theta]22, 
     m} \[Element] Reals};
U = ({
    {u11*Exp[I*\[Theta]11], u12*Exp[I*\[Theta]12]},
    {u21*Exp[I*\[Theta]21], u22*Exp[I*\[Theta]22]}
UIden = U.U\[ConjugateTranspose] // FullSimplify;
UTrans = U.U\[Transpose] // FullSimplify;
EqIm1 = UIden[[1, 1]] == 1;
EqIm2 = UIden[[2, 2]] == 1;
EqIm12 = UIden[[1, 2]] == 0;
EqIm21 = UIden[[2, 1]] == 0;
Eqm1 = UTrans[[1, 1]] == 0;
Eqm2 = UTrans[[2, 2]] == 0;
Eqm12 = UTrans[[1, 2]] == 1;
Solve[{EqIm1, EqIm2, EqIm12, Eqm1, Eqm2, Eqm12}, {u11, u12, u21, 
  u22, \[Theta]11, \[Theta]21, \[Theta]12, \[Theta]22}]

Where the equation list is just the 7 equations I have just described and I don't get any result, just $\{\}$.

Does anybody know what I should do?

EDIT: In general, the matrix $M$ has 0 in all the diagonal, but it has non-zero elements in the off-diagonal terms. In this example, I choose $m_{1,2}=m_{2,1}=1$.

EDIT 2: I have written the code I use.

  • 3
    $\begingroup$ To clarify: $M$ is all $0$ in its diagonal, and $1$ everywhere else? $\endgroup$ Jun 20, 2022 at 15:32
  • $\begingroup$ The matrix M is 0 in the diagonal, but the off-diagonal terms are generally $m_{i,j}$. The case m=1 gives the solution I gave. I question I ask myself is if it is the one solution possible or if there are other matrices U for different m's. $\endgroup$
    – J.Agusti
    Jun 20, 2022 at 15:43
  • $\begingroup$ Please include complete code that we can run. It may be a simple syntax issue -- for example, EqIm1=u_{1,1}^{2}+u_{1,2}^{2}==0 is not legal Mathematica code (you cannot use underscore in this way). Also, you need to tell Solve to use your assumptions. $\endgroup$
    – bill s
    Jun 20, 2022 at 18:32
  • $\begingroup$ @bills I have written my code. Question: By defining the Assumption at the beginning of the code, do I need to define them again when using Solve? $\endgroup$
    – J.Agusti
    Jun 20, 2022 at 19:39
  • $\begingroup$ The only solutions are for M=={{0,p},{p,0}} with p!=0. There are 32 solutions, 2 each for the signs of u11,u12,u21,u22 and 2 for a sign associated with t11 which then determines the values of t12,t21,t22 This gives you the 32 solutions: ClearAll["Global`*"]; Assuming[Element[{u11,u12,u21,u22,t11,t21,t12,t22},Reals], U={{u11*Exp[I*t11],u12*Exp[I*t12]},{u21*Exp[I*t21],u22*Exp[I*t22]}}; LogicalExpand[Simplify[Reduce[Simplify[{U.ConjugateTranspose[U]=={{1,0},{0,1}},U.Transpose[U]=={{0,p},{p,0}}}],{u11,u12,u21,u22,t11,t21,t12,t22}]]]] $\endgroup$
    – Bill
    Jun 21, 2022 at 14:24


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