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Manipulating a multidimensional array.

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Vira Roy
Vira Roy am 23 Apr. 2022
Kommentiert: Bruno Luong am 24 Apr. 2022
PROBLEM: I have managed to create a multidimensional array storing eigenvectors of a hamiltonian. Now I want to manipulate the array to find the expectation value of a matrix and make a quiver plot.
EXPLANATION : I have the following Hamiltonian with . I want to find the eigenvectors for a set of values on a meshgrid and then use it to find and and then to do a quiver plot.
It can be done by solving analyticaly as well and the results are,
The above can be easily coded as follows,
kx = linspace(-0.5,0.5,15);
ky = linspace(-0.5,0.5,15);
[KX,KY] = meshgrid(kx,kx);
X = -sin(atan2(KY,KX));
Y = cos(atan2(KY,KX));
quiver(KX,KY,X,Y)
My Attempt
kx = linspace(-0.5,0.5,15);
ky = linspace(-0.5,0.5,15);
[KX,KY] = meshgrid(kx,kx);
alpha_R = 0.18851;
M = length(kx);
% SPIN MATRICES
sigma_x = [0,1;1,0]; % sigma_x
sigma_y = [0,-1j;1j,0]; %sigma_y
I = [1,0;0,1]; % Identity
%Pre-allocation
E = nan(M,M,2);
Psi = nan(M,M,2);
Psi_prime = nan(M,M,2);
for i = 1: length(kx)
for j = 1: length(ky)
kx_t = kx(i);
ky_t = ky(j);
eps_k = kx_t^2 + ky_t^2; % epsilon_k
%hamiltonian
H = eps_k * I + alpha_R * (sigma_x .* ky_t - sigma_y .* kx_t);
E(i,j,:) = eig(H);
[V,D] = eig(H);
Psi(i,j,:) = V(:,1);
end
end
Psi_prime = pagectranspose(Psi);
%%%ERROR HERE
expx = Psi_prime .* sigma_x .* Psi ;
% expy
% quiver(KX,KY,expx,expy)
I want to find the expectation value of x and y and make a quiver plot of that to get the same texture as above.
Any help or guidance would be helpful. I just am unable to visualize a multidimensional array.
Thank You.

Akzeptierte Antwort

Bruno Luong
Bruno Luong am 23 Apr. 2022
Bearbeitet: Bruno Luong am 23 Apr. 2022
Try this:
kx = linspace(-0.5,0.5,15);
ky = linspace(-0.5,0.5,15);
[KX,KY] = meshgrid(kx,kx);
alpha_R = 0.18851;
M = length(kx);
% SPIN MATRICES
sigma_x = [0,1;1,0]; % sigma_x
sigma_y = [0,-1j;1j,0]; %sigma_y
I = [1,0;0,1]; % Identity
%Pre-allocation
E = nan(M,M,2);
Psi = nan(2,1,M,M); % BL's change here
for i = 1: length(kx)
for j = 1: length(ky)
kx_t = kx(i);
ky_t = ky(j);
eps_k = kx_t^2 + ky_t^2; % epsilon_k
%hamiltonian
H = eps_k * I + alpha_R * (sigma_x .* ky_t - sigma_y .* kx_t);
E(i,j,:) = eig(H);
[V,D] = eig(H);
Psi(:,:,i,j) = V(:,1); % BL's change here
end
end
% Psi_prime = pagectranspose(Psi); % BL's change here
%%%NO LONGER ERROR HERE
expx = pagemtimes(pagemtimes(Psi,'ctranspose',sigma_x,'none'),Psi); % BL's change here
expx = reshape(expx,[M M]); % BL's change here
expy = pagemtimes(pagemtimes(Psi,'ctranspose',sigma_y,'none'),Psi); % BL's change here
expy = reshape(expy,[M M]); % BL's change here
% expy
quiver(KX,KY,expx,expy)
  5 Kommentare
SHUBHAM PATEL
SHUBHAM PATEL am 24 Apr. 2022
Could you please tell how can we add a third feature (sigma_z, and expz) which is represented by the background color in the quiver plot?
In the code above if we want to add another two line.
% before the 'for' loop
sigma_z = [1,0;0,-1]; %sigma_z
%After the for loop
expz = pagemtimes(pagemtimes(Psi,'ctranspose',sigma_z,'none'),Psi); % BL's change here
expz = reshape(expz,[M M]); % BL's change here
And if we want to reflect the magnitude of espectation value of z by background color of the quiver plot, is it possible in MATLAB?
PS: The magnitude of expz is vecry small (zero), still I want to know the way.
Bruno Luong
Bruno Luong am 24 Apr. 2022
It sounds lie a graphical/plotting related question. I suggest you to open a new question so other participants can give you answers.

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