$[(a+k_i)^2+(b+k_j)^2]X_{i,j}-\sum_{m,n}V_{m,n}X_{i-m,j-n}=\mu X_{i,j}$. where $-N\le i,j\le N$ Here we can set $ N=10,a =1, b=1$ and $V_{m,n}$ is the matrix element of $V$. Once I write the coefficient matrix of $X$, I can solve the eigenvalue $\mu$.


closed as off-topic by Daniel Lichtblau, MarcoB, Alex Trounev, José Antonio Díaz Navas, bbgodfrey Apr 6 at 2:18

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  • 1
    $\begingroup$ A matrix is a list of lists. For example if $A=\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, we would write this in Mathematica as A={{1,2},{3,4}}. Is this what you are asking? $\endgroup$ – mjw Mar 25 at 6:15
  • $\begingroup$ What are the bounds of the summation \sum_{m,n}? $\endgroup$ – Henrik Schumacher Mar 25 at 7:19

$X$ is a $N\times N$ matrix, which I'll Flatten into a vector of length $N^2$. The coefficient array is a $N\times N\times N\times N$ tensor, which I'll ArrayFlatten into a $N^2\times N^2$ matrix.

I'll use M=$N$ because the symbol N is already in use in Mathematica. The coefficient array is W. Here I'll use M=3 for brevity.

M = 3;
X = Flatten[Array[Subscript[x, ##] &, {M, M}]];
W = ArrayFlatten[SparseArray[{i_,i_,j_,j_} ->
                   (a+Subscript[k,i])^2 + (b+Subscript[k,j])^2, {M,M,M,M}] - 
                 SparseArray[{i1_,i2_,j1_,j2_} ->
                   Subscript[V,i1-i2,j1-j2], {M,M,M,M}]];

Note that the second component of W is not sparse, and can be written as

Array[Subscript[V, #1-#2, #3-#4] &, {M, M, M, M}]

which may be more efficient, but harder on the eye.

Check that this gives the desired eigenvalue equations:

Thread[W.X == μ X]

$$ \{((a+k_1){}^2+(b+k_1){}^2-V_{0,0}) x_{1,1}-V_{0,-1} x_{1,2}-V_{0,-2} x_{1,3}-V_{-1,0} x_{2,1}-V_{-1,-1} x_{2,2}-V_{-1,-2} x_{2,3}-V_{-2,0} x_{3,1}-V_{-2,-1} x_{3,2}-V_{-2,-2} x_{3,3}=\mu x_{1,1},\\ -V_{0,1} x_{1,1}+((a+k_1){}^2+(b+k_2){}^2-V_{0,0}) x_{1,2}-V_{0,-1} x_{1,3}-V_{-1,1} x_{2,1}-V_{-1,0} x_{2,2}-V_{-1,-1} x_{2,3}-V_{-2,1} x_{3,1}-V_{-2,0} x_{3,2}-V_{-2,-1} x_{3,3}=\mu x_{1,2},\\ -V_{0,2} x_{1,1}-V_{0,1} x_{1,2}+((a+k_1){}^2+(b+k_3){}^2-V_{0,0}) x_{1,3}-V_{-1,2} x_{2,1}-V_{-1,1} x_{2,2}-V_{-1,0} x_{2,3}-V_{-2,2} x_{3,1}-V_{-2,1} x_{3,2}-V_{-2,0} x_{3,3}=\mu x_{1,3},\\ -V_{1,0} x_{1,1}-V_{1,-1} x_{1,2}-V_{1,-2} x_{1,3}+((b+k_1){}^2+(a+k_2){}^2-V_{0,0}) x_{2,1}-V_{0,-1} x_{2,2}-V_{0,-2} x_{2,3}-V_{-1,0} x_{3,1}-V_{-1,-1} x_{3,2}-V_{-1,-2} x_{3,3}=\mu x_{2,1},\\ -V_{1,1} x_{1,1}-V_{1,0} x_{1,2}-V_{1,-1} x_{1,3}-V_{0,1} x_{2,1}+((a+k_2){}^2+(b+k_2){}^2-V_{0,0}) x_{2,2}-V_{0,-1} x_{2,3}-V_{-1,1} x_{3,1}-V_{-1,0} x_{3,2}-V_{-1,-1} x_{3,3}=\mu x_{2,2},\\ -V_{1,2} x_{1,1}-V_{1,1} x_{1,2}-V_{1,0} x_{1,3}-V_{0,2} x_{2,1}-V_{0,1} x_{2,2}+((a+k_2){}^2+(b+k_3){}^2-V_{0,0}) x_{2,3}-V_{-1,2} x_{3,1}-V_{-1,1} x_{3,2}-V_{-1,0} x_{3,3}=\mu x_{2,3},\\ -V_{2,0} x_{1,1}-V_{2,-1} x_{1,2}-V_{2,-2} x_{1,3}-V_{1,0} x_{2,1}-V_{1,-1} x_{2,2}-V_{1,-2} x_{2,3}+((b+k_1){}^2+(a+k_3){}^2-V_{0,0}) x_{3,1}-V_{0,-1} x_{3,2}-V_{0,-2} x_{3,3}=\mu x_{3,1},\\ -V_{2,1} x_{1,1}-V_{2,0} x_{1,2}-V_{2,-1} x_{1,3}-V_{1,1} x_{2,1}-V_{1,0} x_{2,2}-V_{1,-1} x_{2,3}-V_{0,1} x_{3,1}+((b+k_2){}^2+(a+k_3){}^2-V_{0,0}) x_{3,2}-V_{0,-1} x_{3,3}=\mu x_{3,2},\\ -V_{2,2} x_{1,1}-V_{2,1} x_{1,2}-V_{2,0} x_{1,3}-V_{1,2} x_{2,1}-V_{1,1} x_{2,2}-V_{1,0} x_{2,3}-V_{0,2} x_{3,1}-V_{0,1} x_{3,2}+((a+k_3){}^2+(b+k_3){}^2-V_{0,0}) x_{3,3}=\mu x_{3,3}\} $$

Find the eigenvalues:


You'll probably have to insert numerical values into W before calculating the eigenvalues, otherwise the code will be too slow.

  • $\begingroup$ It is a little perverse to use SparseArray[] to generate what is actually a dense matrix. In any event: ArrayFlatten[Table[If[i1 == i2 && j1 == j2, ((a + Subscript[k, i1])^2 + (b + Subscript[k, j1])^2), 0] - Subscript[V, i1 - i2, j1 - j2], {i1, M}, {i2, M}, {j1, M}, {j2, M}]] $\endgroup$ – J. M. will be back soon Mar 25 at 9:14
  • $\begingroup$ Yes I just commented on that. I tend to prefer SparseArray because it's easy to debug, and once it works I try to find more efficient forms. $\endgroup$ – Roman Mar 25 at 9:16
  • $\begingroup$ thank, I will understand it now $\endgroup$ – yun shi Mar 26 at 2:39
  • $\begingroup$ This code looks perfect, but it may be a little abstract. Maybe I can't understand and imagine the mathematical form. Because for one subscript, I can write the matrix by hand and to observe the form. But for two subscript, the matrix is very large even for $N=3$, the matrix is $9\times 9$. My question is : Can you give me some detailed examples which involved the later case. Thanks a lot! $\endgroup$ – yun shi Mar 26 at 4:35
  • 1
    $\begingroup$ I don't know what you mean by "detailed examples", other than the 9 equations I already gave explicitly for the case M=3. Can you please formulate your question more concretely? $\endgroup$ – Roman Mar 26 at 6:56

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