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I'm trying to solve the Schrödinger equation for the hydrogen atom in the following form numerically:

$$\left[-\frac{\hbar^2}{2m}\frac{d^2}{dr^2}+V(r)+\frac{\hbar^2l(l+1)}{2mr^2}\right]R(r)=ER(r).$$

With the Coulomb-Potential $$V(r) = -\frac{e^2}{4\pi\epsilon_0r^2}.$$ By introducing the dimensionless variables $x = r/a_0$ and $\epsilon = E/E_0$, $E_0=1\text{Ry}$, the Schrödinger equation can be written as follows:

$$\left[-\frac{d^2}{dx^2}-\frac{2}{x}+\frac{l(l+1)}{x^2}\right]R(x)=\epsilon R(x)$$

I used the method of Leandro M. described in this post to solve the Schrödinger equation by discretizing the derivative of $R$ and transforming it into a Matrix equation.

I implemented this solution in Python and I think it worked, the only problem I have is that I don't know how to get from the Eigenvalues and -vectors of the resulting Matrix to the Energies and the corresponding Wavefunctions of the Hydrogen Atom for $n=1,2,3,...$ and $l = 0$.

Edit: To clarify, in the last line of my code i do get some values for the Eigenenergies, they are however in the range of a few 10000 instead of 1 for $n=1$ for example. The problem is that I don't know how to get from the values that I calculated to the correct values.

Here is my code:

import numpy as np
from numpy.linalg import eig
from matplotlib import pyplot as plt

d = 0.001
# set values for r
N = 3000
rmax = 10
r = np.linspace(1e-20, rmax, N)

# create first matrix of schrodinger eq corresponding to the derivative with the following shape:
# (-2  1                   )
# ( 1 -2  1                )
# (    1 -2  1             )  * (-1/(d**2))
# (         ...            )
# (                1  -2  1)
# (                    1 -2)


m = np.empty((N, N))

x = -1 / d ** 2

m[0, 0] = -2 * x
m[0, 1] = 1 * x
m[N - 1, N - 1] = -2 * x
m[N - 1, N - 2] = 1 * x

for i in range(1, N - 1):
    m[i, i - 1] = 1 * x
    m[i, i] = -2 * x
    m[i, i + 1] = 1 * x

for i in range(2, 10):
    m[0, i] = 0

# create matrix corresponding to the potential with the following shape:
# (V1                 )
# (   V2              )
# (      V3           )
# (       ...         )
# (            VN-1   )
# (                 VN)

vdiag = []
l = 0

for i in range(0, N - 1):
    vdiag.append(-2 / r[i] + l * (l + 1) / (r[i] ** 2))

v = np.empty((N, N))
np.fill_diagonal(v, vdiag)

# add matrices to get eigenvalue equation: H*R(x) = E*R(x)
H = np.empty((N, N))

for i in range(0, len(v[0] - 1)):
    for j in range(0, len(v[1] - 1)):
        H[i, j] = m[i, j] + v[i, j]

# setting boundary conditions R_1 = R_N = 0
H[:, 0] = H[:, N - 1] = 0

# determine eigenvalues and eigenvectors
energies, wavefcts = eig(H)
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  • 2
    $\begingroup$ Well, actually eigenvalues are indeed energies and eigenvectors are the wavefunctions or orbitals. $\endgroup$ Nov 18 '21 at 18:52
  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Nov 18 '21 at 18:52
  • $\begingroup$ My Problem is that i don't really know how to get the Eigenenergy for the Quantum Numbers n=1, l=0 for example (which should be approximately 1), since the vector with the eigenvalues only contains values that are way too high. $\endgroup$ Nov 18 '21 at 19:01
  • $\begingroup$ Your lowest eigenvalue for l=0 should be 0.5. If you don't get close, there's an error in your implementation. Maybe you shouldnt start your r-grid from 1e-20 bur rather some reasonably larger number. Or, you let it start for r=0, but then impose the Dirichlet condition R(r=0)=0. $\endgroup$
    – davidhigh
    Nov 18 '21 at 20:05
  • $\begingroup$ And wait -- your Rmax, your N and your d are dependent on esch other. You can't simply choose all three separately as you do in your code, but rather calculate the third with the first two. $\endgroup$
    – davidhigh
    Nov 18 '21 at 23:36
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I finally found the solution to my problem, the Eigenenergies where inside the vector of the Eigenvalues, but not in the right order. With the following code I got the right result, if someone is interested:

import numpy as np
from numpy.linalg import eig
from matplotlib import pyplot as plt


# set values for x
xmax = 500
N = 3000
d = xmax/N
x = np.linspace(1e-20, xmax, N)

# create first matrix of schrodinger eq corresponding to the derivative with the following shape:
# (-2  1                   )
# ( 1 -2  1                )
# (    1 -2  1             )  * (-1/(d**2))
# (         ...            )
# (                1  -2  1)
# (                    1 -2)


m = np.zeros((N, N))

g = -1 / d ** 2

m[0, 0] = -2 * g
m[0, 1] = 1 * g
m[N - 1, N - 1] = -2 * g
m[N - 1, N - 2] = 1 * g

for i in range(1, N - 1):
    m[i, i - 1] = 1 * g
    m[i, i] = -2 * g
    m[i, i + 1] = 1 * g

for i in range(2, 10):
    m[0, i] = 0

# create matrix corresponding to the potential with the following shape:
# (V1                 )
# (   V2              )
# (      V3           )
# (       ...         )
# (            VN-1   )
# (                 VN)

vdiag = []
l = 0

for i in range(0, N - 1):
    vdiag.append(-2 / x[i] + l * (l + 1) / (x[i] ** 2))

v = np.zeros((N, N))
np.fill_diagonal(v, vdiag)

# add matrices to get eigenvalue equation: H*R(x) = E*R(x)
H = np.zeros((N, N))

H = m + v



# determine eigenvalues and eigenvectors
energies, wavefcts = eig(H)

energies[0] = energies[N-1] = 0

E = np.sort(energies)
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