for loop not working????

fidelities value does not change with different values of r.

i'm trying to get the different values of fidelity with different r values,and plot(fidelity,r).

[Atsushi Ueno]

No, for loop is not working because the number sequence is not recognized as a column vector.

for r=0.01;0.02;0.03;0.04;0.05;0.06;0.07;0.08;0.09;0.10;
        r
end

r = 0.0100

But this fixing does not work as you expected:

for r = [0.01;0.02;0.03;0.04;0.05;0.06;0.07;0.08;0.09;0.10;]
    r
end

r = 10×1

0.0100 0.0200 0.0300 0.0400 0.0500 0.0600 0.0700 0.0800 0.0900 0.1000

It can be fixed as following:

for r = 0.01:0.01:0.1 % or r = [0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10]
    r
end

r = 0.0100

r = 0.0200

r = 0.0300

r = 0.0400

r = 0.0500

r = 0.0600

r = 0.0700

r = 0.0800

r = 0.0900

r = 0.1000

Answer 1

[Bhanu Prakash]

Hi Abu,

As per my understanding, you are facing some errors with the "for" loop.

Consider the line below:

plot(fidelity,r);

For the "plot" function to work, "r" must be of the same size as "fidelity", i.e., 1x15. If not, it throws an error.

To make "r" to be of size 1x15, the function "linspace" can be used.

I have made some changes in the code and the edited part of the code is attached below:

% Define the commutator of E and F
R_1 = E-F ;
r=linspace(0.01,0.1,15)
% Compute the exponential of R_1
for r_m=r
U = exp(0.01*(R_1));

With the help of "linspace", 15 equally spaced values are assigned to the matrix "r" and the "for" loop runs for 15 times, which is the size of "r".

You can refer to the documentation of "linspace" and "plot" functions, for more info:

Generate linearly spaced vector - MATLAB linspace - MathWorks India

2-D line plot - MATLAB plot - MathWorks India

Hope this answer helps you.

Thanks,

Bhanu Prakash.