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$[(a+k_i)^2+(b+k_j)^2]X_{i,j}-\sum_{m,n}V_{m,n}X_{i-m,j-n}=\mu X_{i,j}$. 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 $\mu$.

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

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  • "This question cannot be answered without additional information. Questions on problems in code must describe the specific problem and include valid code to reproduce it. Any data used for programming examples should be embedded in the question or code to generate the (fake) data must be included." – Daniel Lichtblau, MarcoB, Alex Trounev, José Antonio Díaz Navas, bbgodfrey
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ 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
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$X$ 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:

Eigenvalues[W]

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

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  • $\begingroup$ 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
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    $\begingroup$ 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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