We assume that the chains of matrix B can introduce a numerical statistical solution for ψ(r)^2 (total quantum energy) in the same way that they present a numerical statistical solution for the heat diffusion energy density without go through the PDE heat itself. .
The question arises whether this solution exists, how important is it, and whether it is as accurate as the SE solution?
it is possible to solve the Schrödinger equation via a statistical transition matrix and with numerical analysis without needing the Schrödinger equation itself
Another way to describe quantum particle dynamics is to use statistical transition matrices that completely ignore the Schrödinger equation as if it never existed in the same way that one solves the heat diffusion equation without going through thermal PDE itself.
today we only know one physical transition matrix which is the transition matrix B resulting from the so-called Cairo technique.
Step 1
Construct the 2D statistical matrix B corresponding to Figure 1 which represents a quantum particle in a 2D infinite potential well.
📷
Fig 1. A quantum particle in a 2D infinite potential well
The basis for generating an eigen or proper matrix is the 2D matrix B with 9 equidistant free nodes as shown in Figure 2, nodes 1-9.
Note that RO=0 because PE is zero.
It is expressed by the same matrix M1 explained previously in the example of 2D thermal conduction.
step 2
Compose the proper or eigen matrix M2 as given by,
M2=M1+S(x,y)
Where S(x,y ) is a diagonal matrix and S=C1*V(x,y)
The resulting eigenmatrix M2 will be given by,
M2=
1/14 1/4 0 1/4 0 0 0 0 0
1/4 4/14 1/4 0 1/4 0 0 0 0
0 1/4 1/14 0 0 1/4 0 0 0
1/4 0 0 4/14 1/4 0 1/4 0 0
0 1/4 0 1/4 9/14 1/4 0 1/4 0
0 0 1/4 0 1/4 4/14 0 0 1/4
0 0 0 1/4 0 0 1/14 1/4 0
0 0 0 0 1/4 0 1/4 4/14 1/4
0 0 0 0 0 1/4 0 1/4 1/14
Where C1 is substituted for by the factor 1/14.
step 3
The energy eigenvector E(x,y) is equal to the principal diagonal of the matrix A which gives the following eigenvector equation,
2/14 1/4 0 1/4 0 0 0 0 0
1/4 4/14 1/4 0 1/4 0 0 0 0
0 1/4 2/14 0 0 1/4 0 0 0
1/4 0 0 4/14 1/4 0 1/4 0 0
0 1/4 0 1/4 9/14 1/4 0 1/4 0
0 0 1/4 0 1/4 4/14 0 0 1/4
0 0 0 1/4 0 0 2/14 1/4 0
0 0 0 0 1/4 0 1/4 4/14 1/4
0 0 0 0 0 1/4 0 1/4 2/14
*
[2/14 4/14 2/14 4/14 9/14 4/14 2/14 4/14 2/14]T
is equal to,
[8/49 123/392 8/49 123/392 137/196 123/392 8/49 123/392 8/49] T
Showing that the energy eigenvector is=
[2/14 4/14 2/14 4/14 9/14 4/14 2/14 4/14 2/14 ] T
with a dominant eigenvalue almost equal to 1.
The reason why we multiply the nodes 1,3,6 and 9 by the factor 2 is that these nodes are located at the four intersections of the two axes x and y where the rule E=Ex+Ey applies.
The x-oriented eigenvectors and the y-oriented eigenvectors are shown in Figure 2 in black and red lines.
Please take a look at,
A 3D numerical statistical solution for the time-independent Schrödinger equation
Researchgate, IJISRT journal, December 2023.
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