Result of A(:,:,1,1) coming as result of A(:,:,1,2)

4 Ansichten (letzte 30 Tage)
Gözde Üstün
Gözde Üstün am 6 Jul. 2020
Kommentiert: Christine Tobler am 10 Jul. 2020
Hello,
Ihave this code:
function [A,B] = trying(d)
A=zeros(d,d,2,d);
sigma_x = [0,1;1,0];
sigma_z = [1,0;0,-1];
if d==2
x = sigma_x;
z = sigma_z;
end
if d==4
x = kron(sigma_x,sigma_x);
z = kron(sigma_z,sigma_z);
end
[xv1,xv2] = eig(x);
%xv1 = [xv1(:,2),xv1(:,1)];
[zv1,zv2] = eig(z);
%zv1 = [zv1(:,2),zv1(:,1)];
for i=1:d
% for k = 1:2
A(:,:,1,i)= xv1(:,i)*transpose(xv1(:,i));
A(:,:,2,i)= zv1(:,i)*transpose(zv1(:,i));
% end
end
end
However, I have inverse result For instance the result should be like that:
A(:,:,1,1) =
0.5000 0.5000
0.5000 0.5000
A(:,:,2,1) =
1 0
0 1
A(:,:,1,2) =
0.5000 -0.5000
-0.5000 0.5000
But I have this result:
A(:,:,1,1) =
0.5000 -0.5000
-0.5000 0.5000
A(:,:,2,1) =
0 0
0 1
A(:,:,1,2) =
0.5000 0.5000
0.5000 0.5000
So I tried the that But it did not work
xv1 = [xv1(:,2),xv1(:,1)];
zv1 = [zv1(:,2),zv1(:,1)];
for i=1:d
for k = 1:2
A(:,:,1,i)= xv1(:,k)*transpose(xv1(:,k));
A(:,:,2,i)= zv1(:,k)*transpose(zv1(:,k));
end
end
Any idea?
  2 Kommentare
Geoff Hayes
Geoff Hayes am 6 Jul. 2020
Gözde - why should the output look like the first block of output that you have shown? What algorithm or equations are you basing your code on? Also, note the following
for i=1:d
% for k = 1:2
A(:,:,1,i)= xv1(:,i)*transpose(xv1(:,i));
A(:,:,2,i)= zv1(:,i)*transpose(zv1(:,i));
% end
end
Since i is either 1 or 2, then this means that you also have A(:,:,2,2) set with a value which in this case is
A(:,:,2,2) =
1 0
0 0
Is this expected or should you just have populated A(:,:1,1), A(:,:,1,2), and A(:,:,2,1)?
Gözde Üstün
Gözde Üstün am 6 Jul. 2020
Hello,
Because I am using "eig" function and in the eig function, first eigenvector is coming as second eigen vector.
You are right I (should) have also A(:,:,2,2) but for showing you I just put the other values.

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Antworten (1)

Christine Tobler
Christine Tobler am 7 Jul. 2020
The eigenvectors returned by EIG are returned in the same order as the eigenvalues, but the eigenvalues are not necessarily sorted. Could this be the issue here?
  2 Kommentare
Gözde Üstün
Gözde Üstün am 7 Jul. 2020
Yeah How can I solve this prolem? I used that:
xv1 = [xv1(:,2),xv1(:,1)];
zv1 = [zv1(:,2),zv1(:,1)];
for i=1:d
for k = 1:2
A(:,:,1,i)= xv1(:,k)*transpose(xv1(:,k));
A(:,:,2,i)= zv1(:,k)*transpose(zv1(:,k));
end
end
But it did not work for me
Christine Tobler
Christine Tobler am 10 Jul. 2020
You can sort the eigenvalues and eigenvectors before using them further, see this example for how to do that.

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