0
$\begingroup$

I am facing a problem in plotting the Piecewise function. I simplified my problem to take the following form:

enter image description here

where \[Sigma][i] are Pauli matrices. My code is:

\[Sigma][1] = PauliMatrix[1]; \[Sigma][2] = 
 PauliMatrix[2];    \[Sigma][3] = PauliMatrix[3];
DM = {{2 t, 0, 0, 8}, {0, 8, 10, 0}, {0, 15, 5, 0}, {2, 0, 0, 3}};
x[i_] = Tr[DM.(KroneckerProduct[\[Sigma][i], \[Sigma][i]])];
X = {{x[1], x[2], x[3]}};
{\[Lambda]1, \[Lambda]2, \[Lambda]3, \[Lambda]4} = Eigenvalues[DM];
MR = Piecewise[{{Tr[DM.DM\[Transpose]] - Norm[X], 
    X != 0}, {Max[\[Lambda]1, \[Lambda]2, \[Lambda]3, \[Lambda]4], 
    X = 0}}]
Plot[MR, {t, 0, 100}] 
$\endgroup$
4
$\begingroup$

In your piecewise definition, you're comparing a vector with a number (zero). This won't work:

{1, 1} == 0
(* {1, 1} == 0 *)

Instead, you need to compare to the zero vector:

{1, 1} == {0, 0}
(* False *)

Also, you had a typo in the piecewise, it should be X == {0, 0} instead of X = {0, 0}, although MMA 12.1 auto-corrected it. Here's the final code that produces a plot:

σ[1] = PauliMatrix[1]; 
σ[2] = PauliMatrix[2]; 
σ[3] = PauliMatrix[3];
DM = {{2 t, 0, 0, 8}, {0, 8, 10, 0}, {0, 15, 5, 0}, {2, 0, 0, 3}};
x[i_] = Tr[DM.(KroneckerProduct[σ[i], σ[i]])];
X = {{x[1], x[2], x[3]}};
{λ1, λ2, λ3, λ4} = Eigenvalues[DM];
MR = Piecewise[{{Tr[DM.DM\[Transpose]] - Norm[X], 
    X != {0, 0}}, 
    {Max[λ1, λ2, λ3, λ4], 
    X == {0, 0}}}]
Plot[MR, {t, 0, 100}]

Plot

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you very much, I got the right results. $\endgroup$ – Ragab Zidan Apr 11 at 12:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.