Matrix form for a square lattice arrangement.

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Question: I want to have a matrix form for hopping between nearest points on a square lattice.
My Attempt: To start and explain the idea I do that on a 1D lattice like shown in the figure
The matrixForm for this case comes from this.
I create a 1D chain using the following code (I use a fucntion SquareGrid defined below)
N = 3; % Number of points
pts = squareGrid([0 0 (N-1) 0], [1 1]);
scatter(pts(:,1),pts(:,2),50,'o', 'filled')
Now I want a matrixform such that I can see hopping between the sites which are neigbors. First and last sites are linked. For this i do the following.
N =5; % Number of sites
t = 1; % value for hopping
t_vec = repmat(t, [1,N-1]); % making a vector of size N-1 with t
H_tb = diag(t_vec, 1) + diag(t_vec, -1);
H_tb(1,5) = t; H_tb(5,1) = t; % linking first and last points
H_tb
H_tb = 5×5
0 1 0 0 1 1 0 1 0 0 0 1 0 1 0 0 0 1 0 1 1 0 0 1 0
Now I have the matricform as shown above which does what I wanted for a 1D chain. Now i want to do the same for 2D, square lattice. This is where I am stuck. How do make a matrixform such as above which takes into account all the hoppings between nearest sites? Any help would be appreciated. Thank You.
pts = squareGrid([0 0 1 1], [1 1]);
scatter(pts(:,1),pts(:,2),50,'o', 'filled')
For this I use a function SquareGrid as below.
function varargout = squareGrid(bounds, dim)
%bounds(xmin,ymin,xmax,ymax)
% dim(xsep, ysep) eg. dim(4,2) makes separation b/w points xpoints as 4 and
% ypoints as 2
x1 = bounds(1);
x2 = bounds(3);
lx = (x1: dim(1):x2)';
y1 = bounds(2);
y2 = bounds(4);
ly = (y1: dim(2):y2)';
%number of points in each coordinate
nx = length(lx);
ny = length(ly);
np = nx * ny;
%Creating Points
pts = zeros(np,2);
for i = 1:ny
pts((1:nx)' + (i-1) * nx, 1) =lx;
pts( (1:nx)'+(i-1)*nx, 2) = ly(i);
end
if nargout > 0 % if number of function arguments is > zero
varargout{1} = pts;
end
end
  2 Comments
Vira Roy
Vira Roy on 6 Oct 2021
Thanks for your reply Matt. I do want to genralise it for arbitrary n but for this sake I am looking at a 2 *2 grid of square lattice. In the end i would like to do higher n like 3 or 4. I merely want to undersatnd the technique to do that.

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Accepted Answer

Matt J
Matt J on 6 Oct 2021
For an arbitray lattice, you can always just use pdist2,
D=pdist2(pts,pts,'cityblock');
H_tp=(D==1);
This is without your circulant end conditions. The circulancy doesn't generalize in any obvious way to hexagonal or other arbitrary lattices, so I don't know in general what you would want to do. For a rectangular lattice, you would do,
e11=pts(:,1)==1; e1N=pts(:,1).'==N;
e21=pts(:,2)==1; e2N=pts(:,2).'==N;
E1=(e11&e1N); E1=E1|E1.';
E2=(e21&e2N); E2=E2|E2.';
H_tp=(D==1) | E1 | E2;
  1 Comment
Vira Roy
Vira Roy on 7 Oct 2021
Matt, thank you for the solution. It has helped me and so I am accepting the answer.

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More Answers (1)

Matt J
Matt J on 6 Oct 2021
Edited: Matt J on 6 Oct 2021
If N^2 isn't too large,
N=4;
h_tb=sparse( toeplitz([0,1,zeros(1,N-3),1]) ); %1D
A=kron(ones(N),h_tb);
B=kron(h_tb,ones(N));
H_tb=xor(A,B); %2D
H_tb = 16×16 sparse logical array
(2,1) 1 (4,1) 1 (5,1) 1 (7,1) 1 (10,1) 1 (12,1) 1 (13,1) 1 (15,1) 1 (1,2) 1 (3,2) 1 (6,2) 1 (8,2) 1 (9,2) 1 (11,2) 1 (14,2) 1 (16,2) 1 (2,3) 1 (4,3) 1 (5,3) 1 (7,3) 1 (10,3) 1 (12,3) 1 (13,3) 1 (15,3) 1 (1,4) 1 (3,4) 1 (6,4) 1 (8,4) 1 (9,4) 1 (11,4) 1 (14,4) 1 (16,4) 1 (1,5) 1 (3,5) 1 (6,5) 1 (8,5) 1 (9,5) 1 (11,5) 1 (14,5) 1 (16,5) 1 (2,6) 1 (4,6) 1 (5,6) 1 (7,6) 1 (10,6) 1 (12,6) 1 (13,6) 1 (15,6) 1 (1,7) 1 (3,7) 1 (6,7) 1 (8,7) 1 (9,7) 1 (11,7) 1 (14,7) 1 (16,7) 1 (2,8) 1 (4,8) 1 (5,8) 1 (7,8) 1 (10,8) 1 (12,8) 1 (13,8) 1 (15,8) 1 (2,9) 1 (4,9) 1 (5,9) 1 (7,9) 1 (10,9) 1 (12,9) 1 (13,9) 1 (15,9) 1 (1,10) 1 (3,10) 1 (6,10) 1 (8,10) 1 (9,10) 1 (11,10) 1 (14,10) 1 (16,10) 1 (2,11) 1 (4,11) 1 (5,11) 1 (7,11) 1 (10,11) 1 (12,11) 1 (13,11) 1 (15,11) 1 (1,12) 1 (3,12) 1 (6,12) 1 (8,12) 1 (9,12) 1 (11,12) 1 (14,12) 1 (16,12) 1 (1,13) 1 (3,13) 1 (6,13) 1 (8,13) 1 (9,13) 1 (11,13) 1 (14,13) 1 (16,13) 1 (2,14) 1 (4,14) 1 (5,14) 1 (7,14) 1 (10,14) 1 (12,14) 1 (13,14) 1 (15,14) 1 (1,15) 1 (3,15) 1 (6,15) 1 (8,15) 1 (9,15) 1 (11,15) 1 (14,15) 1 (16,15) 1 (2,16) 1 (4,16) 1 (5,16) 1 (7,16) 1 (10,16) 1 (12,16) 1 (13,16) 1 (15,16) 1
  1 Comment
Vira Roy
Vira Roy on 6 Oct 2021
This seems to work Matt and I am thankful for your solution as well but unfortunately this is not what i was hoping for. I wanted to use the function SquareGrid to make the points and then loop around them to find the matrix elements of the matrix. The reason being that I wanted to learn the technqiue of doing that so later i can try the idea for different grid system like say hexagonal lattice for which I have function as well similar to SquareGrid.
Thanks for the help though and if you have some idea of how should i approach like i mentioned it would be helpful.

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