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The next step is to define the 2D potential, `V`. We define `V` for the FD matrix one row at a time, first row, and then the next row. For this case we will just have alternating positive and negative ions as potential maximima and minima sites. A positive ion means we have a point(`x,y`) of positive potential and a negative ion means we have a point(`x,y`) of negative potential. And these potential points alternate with equally spaced rows and columns. That means that the potential is still periodic but is a very sparse triangular wave instead of a sine or cosine wave. This approach reduces the number of points needed to describe the potential. So what is inserted into the diagonal of the FD matrix is all of the column ions of the first potential row, then all the columns of the second potential row, etc. The periodic values of the potential are shown in canvas 1 where the black circle indicates a negative potential and the red circle indicates a positive potential. The border of canvas 1 is blue which indicates a much larger potential than the periodic ions and is what is required to keep the Schrodinger wave function particle mode contained. This border potential also has to be part of the row and column data used in the finite difference matrix below.

For a good lecture on the form of the finite difference matrix for 2D see Laplace Equation 2D at time 08:20. To express the Schrodinger equation (SE) the potential, V, is added onto the second derivative part of the diagonal elements of the matrix just as in a 1D FD matrix. And instead of the usual (1,-2+V,1) tridiagonal matrix we obtain the multipliers (1,-4+V,1) where the 4 expresses the fact that we have the sum `(del^2psi)/(delx^2)+(del^2psi)/(dely^2)` as the operator for the second derivative instead of just `(d^2psi)/(dx^2)` derivative. If the 2D potential consists of 2 rows and 4 columns then a 2D finite difference (FD) matrix diagonal that is appropriate for the first two rows of the potential, `V`, is:

`D=[[-4+V_(1,1),1,0,0,0,0,0,0],[1,-4+V_(1,2),1,0,0,0,0,0],[0,1,-4+V_(1,3),1,0,0,0,0], [0,0,1,-4+V_(1,4),1,0,0,0],[0,0,0,1,-4+V_(2,1),1,0,0],[0,0,0,0,1,-4+V_(2,2),1,0], [0,0,0,0,0,1,-4+V_(2,3),1], [0,0,0,0,0,0,1,-4+V_(2,4)]]`

where the first index, i, of `V_(i,j)` corresponds to the potential row number (`y` coordinate) and the second index, j, corresponds to the potential column number (`x` cooridinate). To simplify expressing this matrix we can rename the 4 quarters of this matrix as`D=[[D_1,0],[0,D_2]]`

where each of these elements are a 4x4 matrix and ssubscripts 1 and 2 correspond to the first and second row of the potential. In addiition to the present matrix we also need a unit matrix`I=[[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]]`

that is spaced 4 elements away from the diagonal so we have:`D=[[D_1,I],[I,D_2]]`

And similarly for all `D_i)` For a moderate number of potential points, this matrix becomes enormous, with `(N_x*N_y)^2` total elements. where `N_x` is the number of potential columns and `N_y` is the total number of potential rows. This gives rise to a `8x8=64` element matrix as seen in this example. That is the number of eigenmodes that need to be solved for and the time and accuracy needed for these rises very quickly.We saw in the 1D examples that for a well that extends from -1 to 1 the lowest energy level

`psi(x)=cospix/2`

and this means that its lowest energy is:`E_1=1/2 (pi/2)^2`

The second lowest energy level with mode number, `n_x = 2` is 4 times larger than this`E_2=n_x^2/2 (pi/2)^2=1/2pi^2`

However, in 2D, it gets more complicated because we can have,for example, `n_x=2` and `n_y=1` for a total energy of`E_21=(n_x^2+n_y^2)/2 (pi/2)^2=5/2pi^2`

Considering mode integers larger than 2, we get more and more possible energies. The total number of different mode energies is a sum of allowed states of the well for a given sum, `N`, of mode number indices`N_("Modes")=N-1`

and their `(n_x,n_y)` values are:`(n_(ix),n_(iy))=(1,N-1),(2,N-2),.....(N-2,2),(N-1,1)`

Also, there will be both even and odd versions (cosine and sine versions) of each of these modes and all 4 of these versions will have the same energy. The values of the energies out to `N_(total) are then:`E_(Modes)=(pi^2/4)[(1+(N-1)^2),(4+(-2)^2),.....((N-2)^2+4),((1+(N-1)^2)]`

So it can be seen that the mode energies ramp up as approximately the square of `N` and the energies of the intermediate index modes have energies slightly smaller than the end index modes.
For a truly **infinite** well depth, the cosine*cosine modes are just

`"eigenvector"_(n_x,n_y)(x,y)=cos((n_xpix)/(2L_x))*cos((n_ypiy)/(2L_y))\ \ \ \ (1)`

with simple modifications for odd (sine*sine) modes and hybrid (cosine*sine) modes. If the well depth is not infinite, then the waves will decay exponentially at the boundaries of the well. Then we must determine the value of the wave at the boundary as well as match the slope of the wave with the exponential slope. This involves computing the wave vector from the FD matrix discussed above.The plot on Canvas 2 shows the eigenvalues (energies) computed using the FD matrix discussed above. Canvas 2 also points out the energy band gap which is approximately twice the adjsutable amplitude of the potential. The 2D color block graph on Canvas 3 shows the eigenvectors (wave functions) one at a time. The wave functions can be stepped through by pressing the buttons just above Canvas 3.

In conclusion, I have shown here that the 2D semiconductor modes and band gap can be calculated by solving for modes of a finite difference equation that is similar to (but much larger than) the analogous 1D equation.

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Expand Mode Energies Around Band Gap

Plot All Mode Energies