(Edited) For finding the ground state wave function of:
$ H\psi(x) = (-1/2)d^2\psi(x)/dx^2 + (1/2)x^2\psi(x) = E \psi(x)$
I have written:
mOneDSchEq[n_] :=
Table[Switch[i - j, -1, p[x[i]],
0, (10/(n + 1))^2 q[x[i]] - 2 p[x[i]], 1, p[x[i]], _, 0], {i,
n}, {j, n}];
q[x_] := -x^2; p[x_] := 1;
Xarray[n_] := Do[x[i] = -5 + i 10/(n + 1), {i, 0, n + 1}];
EigVec[n_] := Eigenvectors[mOneDSchEq[n]];
lisEigVec = EigVec[35];
OneEigVec[j_] := Part[Reverse[lisEigVec], j];
y[i_] := Part[OneEigVec[1], i];
listOfPoints =
Join[{{x[0], 0}}, Table[{x[i], y[i]}, {i, 1, 35}], {{x[36], 0}}];
ListPlot[listOfPoints, PlotJoined -> True, PlotRange -> All,
PlotLabel -> "Ground State Wave Function of Harmonic Oscillator",
AxesLabel -> {"x", "y"}]
Which I have obtained the Gaussian, correctly.
The question that came to my mind is that:
Is it possible by knowing the ground stat eigenvalue, i.e. 1/2, solving the Schrödinger equation numerically, and obtain the ground state wave function? in other words, to solve:
$ H\psi(x) = (-1/2)d^2\psi(x)/dx^2 + (1/2)x^2\psi(x) = (1/2) \psi(x)$
or
$ H\psi(x) = (-1/2)d^2\psi(x)/dx^2 + (1/2)x^2\psi(x) = (3/2) \psi(x)$
So, I wrote:
s = NDSolve[{-(1/2) \[Psi]''[x] + (1/2) x^2(\[Psi][x]) == (1/2) \[Psi][
x], \[Psi][-5] == 0, \[Psi][5] == 0}, \[Psi], {x, -5, 5}]
Plot[Evaluate[\[Psi][x] /. s], {x, -5, 5}, PlotRange -> All]
BUT, I got nothing. What is the problem?
The other question is that, I was traveling through the website and found an elegant approach to two dimensional Harmonic Oscillator here.
My question is, if again we want to solve the Schrödinger equation numercally and obtain wave functions, now two dimensional, by knowing the eigenvalues, what should we do? For example:
$ H\psi(x,y) = (-1/2)(d^2/x^2 + d^2/dy^2)\psi(x,y) + (1/2)(x^2 + y^2)\psi(x,y) = (1) \psi(x,y) $
and
$ H\psi(x,y) = (-1/2)(d^2/x^2 + d^2/dy^2)\psi(x,y) + (1/2)(x^2 + y^2+ x y)\psi(x,y) = (0.96) \psi(x,y) $
Thanks for your attention!
1/(2 a^2)
bit there to take into account the factor of 1/2 in front of the laplacian? Also, thePartition
is there because he is representing 2d space in a 1d vector (basically, he discretises space, then take the 2d matrix and set the rows one after the other to each other so as to form a 1d vector; thePartition
undoes this). I'll write an answer tomorrow if I remember, have time and nobody else does! $\endgroup$