## Wave function in momentum space

The wave function often calculated in spatial coordinate. However, in experimental point of view, the momentum distribution can be extracted directly from the experimental data.

The conversion between momentum space and position space is the Fourier transform

$\displaystyle \phi(\vec{k}) = \frac{1}{\sqrt{2\pi}^3} \int Exp\left(-i \vec{k}\cdot \vec{r} \right) \phi(\vec{r}) d\vec{r}$

Using the plane wave expansion

$\displaystyle \exp(i k\cdot r) = \sum_{l=0}^\infty (2l+1) i^l j_l(kr) P_l(\hat{k}\cdot\hat{r})$

or

$\displaystyle \exp(i k\cdot r) = 4\pi \sum_{l=0}^\infty \sum_{m=-l}^{l} i^l j_l(kr) Y_{lm}(\Omega_k) Y_{lm}^{*}(\Omega_r)$

Thus,

$\displaystyle \phi(\vec{k}) = \frac{4\pi}{\sqrt{2\pi}^3} \sum_{l=0}^\infty (-i)^l \sum_{m=-l}^{l} \int j_l(k r) Y_{lm}(\Omega_k) Y_{lm}^*(\Omega) \phi(\vec{r}) r^2 dr d\Omega$

where $j_l (x)$ is spherical Bessel function. Usually, due to conservation of angular momentum, the angular part can be separated from the spatial part. Let assume the wave function in position space is

$\phi(\vec{r}) = \psi(r) Y_{l_r m_r}(\Omega)$

$\phi(\vec{k}) = \psi(k) Y_{l_k m_k}(\Omega_k)$

Then we have

$\displaystyle \psi(k) = \frac{4\pi}{\sqrt{(2\pi)^3}} \int j_l(k r) \psi(r) r^2 dr$

$\displaystyle \phi(\vec{k}) = \psi(k) (-i)^l Y_{lm}(\Omega_k)$

where $l = l_r = l_k, m=m_r = m_k$, due to the orthogonality of spherical harmonics.

For s, p, d, f-state, the spherical Bessel function is

$\displaystyle j_0(x) = \frac{\sin(x)}{x}$

$\displaystyle j_1(x) = \frac{\sin(x)}{x^2} - \frac{\cos(x)}{x}$

$\displaystyle j_2(x) = \sin(x)\frac{2-x^2}{x^3} - \cos(x)\frac{3}{x^2}$

$\displaystyle j_3(x) = \sin(x)\frac{15-6x^2}{x^4} - \cos(x)\frac{15-x^2}{x^3}$

For Hydrogen-like wave function, the non-normalized momentum distribution is

$\displaystyle \psi_{10}(k) = \frac{4 Z^{5/2}}{(k^2 + Z^2)^2}$

$\displaystyle \psi_{20}(k) = \frac{16 \sqrt{2}Z^{5/2}(4k^2-Z^2)}{(4k^2 + Z^2)^3}$

$\displaystyle \psi_{21}(k) =\sqrt{\frac{2}{3}}\frac{64 k Z^{7/2}}{(4k^2 + Z^2)^3}$

$\displaystyle \psi_{30}(k) = \frac{36 \sqrt{3} Z^{5/2} (81k^3-30k^2Z^2+Z^4)}{(9k^2 + Z^2)^4}$

$\displaystyle \psi_{31}(k) = \frac{144 \sqrt{6} Z^{7/2} (9k^3-kZ^2)}{(9k^2 + Z^2)^4}$

Here is the plot for momentum distribution $(\psi_{nl}(k))^2 k^2$.

It is interesting that, the number of node decrease with higher angular momentum. But be-aware that it is only in atomic case, not a universal.

The higher the principle quantum number, the smaller of the spread of momentum. This is because, the spread of position wave function getting larger, and the uncertainty in momentum space will be smaller. This is a universal principle.

We also plot the Hydrogen radial function in here $\psi(r)^2 x^2$, for reference,

## Complete derivation from Schrodinger equation to Laguerre equation for Coulomb potential

The Hamiltonian is

$\displaystyle H = -\frac{\hbar^2}{2m}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0r}$

$\displaystyle \left( -\frac{\hbar^2}{2m}\frac{1}{r^2}\left(\frac{d}{dr} r^2 \frac{d}{dr} \right) - \frac{Ze^2}{4\pi\epsilon_0 r} + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2} \right) R(r) = E R(r)$

rearrange, using Atomic unit

$\displaystyle \left( -\frac{1}{2}\frac{1}{r^2}\left(\frac{d}{dr} r^2 \frac{d}{dr} \right) - E -\frac{Z}{ r} + \frac{1}{2} \frac{l(l+1)}{r^2} \right) R(r) = 0$

Since the normalization condition of $R(r)$ is $\int_0^\infty R(r)R(r) r^2 dr = 1$, it is natural to define $u(r) = r R(r)$

$\displaystyle R = \frac{u}{r}, \frac{d}{dr}\left(r^2\frac{dR}{dr}\right) = r \frac{d^2u}{dr^2}$

$\displaystyle -\frac{1}{2}\frac{d^2u}{dr^2} - \left( E + \frac{Z}{ r} - \frac{l(l+1)}{2r^2} \right) u(r)= 0$

Substitute the eigen energy $E_n = - \frac{Z^2}{2n^2}$

$\displaystyle \frac{d^2u}{dr^2} + \left( - \frac{Z^2}{n^2} + \frac{2Z}{ r} - \frac{l(l+1)}{r^2} \right) u(r)= 0$

Set

$\displaystyle x = \frac{2Z}{n} r, \frac{d}{dr} = \frac{d}{dx} \frac{2Z}{n}$

$\displaystyle \left(\frac{2Z}{n}\right)^2\frac{d^2u(x)}{dx^2} - \left( - \frac{Z^2}{n^2} + \frac{4Z^2}{n x} - \frac{4Z^2}{n^2}\frac{l(l+1)}{x^2} \right) u(x)= 0$

Take out the $4Z^2/n^2$

$\displaystyle \frac{d^2u(x)}{dx^2} - \left(\frac{n}{ x} - \frac{1}{4} - \frac{l(l+1)}{x^2} \right) u(x)= 0$

Now, we know the short and long-range behaviour of $u(x)$, assume it to be

$u(x) = \exp\left(-\frac{x}{2} \right) x^{l+1} y(x)$

Then the equation of $y(x)$ is

$\displaystyle x \frac{d^2y}{dx^2} + \left( 2(l+1) - x\right) \frac{dy}{dx} +\left( n - l- 1 \right) y(x) = 0$

This is the Laguerre equation

$\displaystyle x \frac{d^2y}{dx^2} + \left( \alpha+1 - x\right) \frac{dy}{dx} + m y(x) = 0$

where $\alpha = 2l+1$ and $m = n-l -1$. Therefore, the solution of the radial equation is,

$R(r) = \frac{1}{r} A \exp\left(-\frac{x}{2} \right) x^{l+1} L_{n-l-1}^{2l+1}(x)$

where $A$ is normalization factor.

Notice that the Laguerre polynomial is only defined for $m \geq 0$, thus, $n > l$.

## Hartree method for Helium ground state

After long preparation, I am ready to do this problem.

The two electron in the helium ground state occupy same spacial orbital but difference spin. Thus, the total wavefunction is

$\displaystyle \Psi(x,y) = \frac{1}{\sqrt{2}}(\uparrow \downarrow - \downarrow \uparrow) \psi(x) \psi(y)$

Since the Coulomb potential is spin-independent, the Hartree-Fock method reduce to Hartree method. The Hartree operator is

$F(x) = H(x) + \langle \psi(y)|G(x,y) |\psi(y) \rangle$

where the single-particle Hamiltonian and mutual interaction are

$\displaystyle H(x) = -\frac{\hbar^2}{2m} \nabla^2 - \frac{Ze^2}{4\pi\epsilon_0 x} = -\frac{1}{2}\nabla^2 - \frac{Z}{x}$

$\displaystyle G(x,y) = \frac{e^2}{4\pi\epsilon_0|x-y|} = \frac{1}{|x-y|}$

In the last step, we use atomic unit, such that $\hbar = 1, m=1, e^2 = 4\pi\epsilon_0$. And the energy is in unit of Hartree, $1 \textrm{H} = 27.2114 \textrm{eV}$.

We are going to use Hydrogen-like orbital as a basis set.

$\displaystyle b_i(r) = R_{nl}(r)Y_{lm}(\Omega) \\= \sqrt{\frac{(n-l-1)!Z}{n^2(n+l)!}}e^{-\frac{Z}{n}r} \left( \frac{2Z}{n}r \right)^{l+1} L_{n-l-1}^{2l+1}\left( \frac{2Z}{n} r \right) \frac{1}{r} Y_{lm}(\Omega)$

I like the left the $1/r$, because in the integration $\int b^2 r^2 dr$, the $r^2$ can be cancelled. Also, the $i = nlm$ is a compact index of the orbital.

Using basis set expansion, we need to calculate the matrix elements of

$\displaystyle H_{ij}=\langle b_i(x) |H(x)|b_j(x)\rangle = -\delta \frac{Z^2}{2n^2}$

$\displaystyle G_{ij}^{hk} = \langle b_i(x) b_h(y) |G(x,y) |b_j(x) b_k(y) \rangle$

Now, we will concentrate on evaluate the mutual interaction integral.

Using the well-known expansion,

$\displaystyle G(x,y) = \frac{1}{|x-y|}=\frac{1}{r_{12}} = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} \frac{4\pi}{2l+1} \frac{r_<^l}{r_>^{l+1}} Y_{lm}^{*}(\Omega_1)Y_{lm}(\Omega_2)$

The angular integral

$\displaystyle \langle Y_i(x) Y_h(y)| Y_{lm}^{*}(x) Y_{lm}(y) | Y_j(x) Y_k(y) \rangle \\ = \big( \int Y_i^{*}(x) Y_{lm}^{*}(x) Y_j(x) dx \big) \big( \int Y_h^{*}(y) Y_{lm}(y) Y_k(y) dy \big)$

where the integral $\int dx = \int_{0}^{\pi} \int_{0}^{2\pi} \sin(\theta_x) d\theta_x d\phi_x$.

From this post, the triplet integral of spherical harmonic is easy to compute.

$\displaystyle \int Y_h^{*}(y) Y_{lm}(y) Y_k(y) dy = \sqrt{\frac{(2l+1)(2l_k+1)}{4\pi (2l_h+1)}} C_{l0l_k0}^{l_h0} C_{lm l_km_k}^{l_hm_h}$

The Clebsch-Gordon coefficient imposed a restriction on $l,m$.

$\displaystyle \langle R_i(x) R_h(y)| \frac{r_<^l}{r_>^{l+1}} | R_j(x) R_k(y) \rangle \\ = \int_0^{\infty} \int_{0}^{\infty} R_i(x) R_h(y) \frac{r_<^l}{r_>^{l+1}} R_j(x) R_k(y) y^2 x^2 dy dx \\ = \int_0^{\infty} R_i(x) R_j(x) \\ \left( \int_{0}^{x} R_h(y) R_k(y) \frac{y^l}{x^{l+1}} y^2dy + \int_{x}^{\infty} R_h(x)R_k(x) \frac{x^l}{y^{l+1}} y^2 dy \right) x^2 dx$

The algebraic calculation of the integral is complicated, but after the restriction of $l$ from the Clebsch-Gordon coefficient, only few terms need to be calculated.

The general consideration is done. now, we use the first 2 even states as a basis set.

$\displaystyle b_{1s}(r) = R_{10}(r)Y_{00}(\Omega) = 2Z^{3/2}e^{-Zr}Y_{00}(\Omega)$

$\displaystyle b_{2s}(r) = R_{20}(r)Y_{00}(\Omega) = \frac{1}{\sqrt{8}}Z^{3/2}(2-Zr)e^{-Zr/2}Y_{00}(\Omega)$

These are both s-state orbital. Thus, the Clebsch-Gordon coefficient

$\displaystyle C_{lm l_k m_k}^{l_h m_h} = C_{lm00}^{00}$

The radial sum only has 1 term. And the mutual interaction becomes

$\displaystyle G(x,y) = \frac{1}{|x-y|}=\frac{1}{r_{12}} = 4\pi \frac{1}{r_>} Y_{00}^{*}(\Omega_1)Y_{00}(\Omega_2)$

The angular part

$\displaystyle \langle Y_i(x) Y_h(y)| Y_{lm}^{*}(x) Y_{lm}(y) | Y_j(x) Y_k(y) \rangle = \frac{1}{4\pi}$

Thus, the mutual interaction energy is

$G_{ij}^{hk} = \displaystyle \langle b_i(x) b_h(y) |G(x,y) |b_j(x) b_k(y) \rangle = \langle R_i(x) R_h(y)| \frac{1}{r_>} | R_j(x) R_k(y) \rangle$

$G_{ij}^{hk} = \displaystyle \langle R_i(x) R_h(y)| \frac{1}{r_>} | R_j(x) R_k(y) \rangle \\ \begin{pmatrix} G_{11}^{hk} & G_{12}^{hk} \\ G_{21}^{hk} & G_{22}^{hk} \end{pmatrix} = \begin{pmatrix} \begin{pmatrix} G_{11}^{11} & G_{11}^{12} \\ G_{11}^{21} & G_{11}^{22} \end{pmatrix} & \begin{pmatrix} G_{12}^{11} & G_{12}^{12} \\ G_{12}^{21} & G_{12}^{22} \end{pmatrix} \\ \begin{pmatrix} G_{21}^{11} & G_{21}^{12} \\ G_{21}^{21} & G_{21}^{22} \end{pmatrix} & \begin{pmatrix} G_{22}^{11} & G_{22}^{12} \\ G_{22}^{21} & G_{22}^{22} \end{pmatrix} \end{pmatrix} \\= \begin{pmatrix} \begin{pmatrix} 1.25 & 0.17871 \\ 0.17871 & 0.419753 \end{pmatrix} & \begin{pmatrix} 0.17871 & 0.0438957 \\ 0.0439857 & 0.0171633 \end{pmatrix} \\ \begin{pmatrix} 0.17871 & 0.0438957 \\ 0.0438957 & 0.0171633 \end{pmatrix} & \begin{pmatrix} 0.419753 & 0.0171633 \\ 0.0171633 & 0.300781 \end{pmatrix} \end{pmatrix}$

We can easy to see that $G_{ij}^{hk} = G_{ji}^{hk} = G_{ij}^{kh} = G_{hk}^{ij} = G_{ji}^{kh}$. Thus, if we flatten the matrix of matrix, it is Hermitian, or symmetric.

Now, we can start doing the Hartree method.

The general solution of the wave function is

$\psi(x) = a_1 b_{1s}(x) + a_2 b_{2s}(x)$

The Hartree matrix is

$F_{ij} = H_{ij} + \sum_{h,k} a_h a_k G_{ij}^{hk}$

The first trial wave function are the Hydrogen-like orbital,

$\psi^{(0)}(x) = b_{1s}(r)$

$F_{ij}^{(0)} = \begin{pmatrix} -2 & 0 \\ 0 & -0.5 \end{pmatrix} + \begin{pmatrix} 1.25 & 0.17871 \\ 0.17817 & 0.419753 \end{pmatrix}$

Solve for eigen system, we have the energy after 1st trial,

$\epsilon^{(1)} = -0.794702 , (a_1^{(1)}, a_2^{(1)}) = (-0.970112, 0.242659)$

After 13th trial,

$\epsilon^{(13)} = -0.880049 , (a_1^{(13)}, a_2^{(13)}) = (-0.981015, 0.193931)$

$F_{ij}^{(13)} = \begin{pmatrix} -2 & 0 \\ 0 & -0.5 \end{pmatrix} + \begin{pmatrix} 1.15078 & 0.155932 \\ 0.155932 & 0.408748 \end{pmatrix}$

Thus, the mixing of the 2s state is only 3.7%.

Since the eigen energy contains the 1-body energy and 2-body energy. So, the total energy for 2 electrons is

$E_2 = 2 * \epsilon^{(13)} - G = -2.82364 \textrm{H} = -76.835 \textrm{eV}$

In which ,

$G = \langle \psi(x) \psi(y) |G(x,y) |\psi(x) \psi(y) \rangle = 1.06354 \textrm{H} = 28.9403 \textrm{eV}$

So the energies for

From He to He++.  $E_2 = -2.82364 \textrm{H} = -76.835 \textrm{eV}$
From He+ to He++, $E_1^+ = -Z^2/2 = 2 \textrm{H} = -54.422 \textrm{eV}$.
From He to He+, is $E_1 = E_2 - E_1^+ = -0.823635 \textrm{H} = -22.4123 \textrm{eV}$

The experimental 1 electron ionization energy for Helium atom is

$E_1(exp) = -0.90357 \textrm{H} = -24.587 \textrm{eV}$
$E_1^+(exp) = -1.99982 \textrm{H} = -54.418 \textrm{eV}$
$E_2(exp) = -2.90339 \textrm{H} = -79.005 \textrm{eV}$

The difference with experimental value is 2.175 eV. The following plot shows the Coulomb potential, the screening due to the existence of the other electron, the resultant mean field, the energy, and $r \psi(x)$

Usually, the Hartree method will under estimate the energy, because it neglected the correlation, for example, pairing and spin dependence. In our calculation, the $E_2$ energy is under estimated.

From the $(a_1^{(13)}, a_2^{(13)}) = (-0.981015, 0.193931)$, we can see, the mutual interaction between 1s and 2s state is attractive. While the interaction between 1s-1s and 2s-2s states are repulsive. The repulsive can be easily understood. But I am not sure how to explain the attractive between 1s-2s state.

Since the mass correction and the fine structure correction is in order of $10^{-3} \textrm{eV}$, so the missing 0.2 eV should be due to something else, for example, the incomplete basis set.

If the basis set only contain the 1s orbit, the mutual interaction is 1.25 Hartree = 34.014 eV. Thus, the mixing reduce the interaction by 5.07 eV, just for 3.7% mixing

I included the 3s state,

$\epsilon^{(13)} = -0.888475 , (a_1^{(13)}, a_2^{(13)}, a_3^{(13)}) = (0.981096, -0.181995, -0.06579)$

The mutual energy is further reduced to 1.05415 Hartree = 28.6848 eV. The $E_2 = -77.038 \textrm{eV}$. If 4s orbital included, the $E_2 = -77.1058 \textrm{eV}$. We can expect, if more orbital in included, the $E_2$ will approach to $E_2(exp)$.

## Integrations of Laguerre polynomial

Since the Laguerre polynomial is deeply connected to the hydrogen-like electron orbital. The radial solution is

$\displaystyle R_{nl}(r) = A \frac{1}{r} \exp(- \frac{x}{2}) x^{l+1}L_{n-l-1}^{2l+1}(x)$

$\displaystyle x = \frac{2Z}{na_0} r$

From the normalization condition, we can get the coefficient A

$\displaystyle \int_0^{\infty} R_{nl}^2(r) r^2 dr$

Beside of calculating the normalization factor, the general expectation value also need to evaluation of the integration of the Laguerre polynomial.

In here, we will show the calculation for 3 integrals:

$\displaystyle \int_0^{\infty} x^{\alpha+m} e^{-x} L_n^\alpha(x) dx, n,\alpha\geq0$

$\displaystyle \int_0^{\infty} x^{\alpha+k} e^{-x} (L_n^\alpha(x))^2 dx$

$\displaystyle \int_0^{\infty} x^{\alpha +k} e^{-x} L_m^{\beta}(x) L_n^\alpha(x) dx$

Using the Rodrigues formula

$\displaystyle L_n^\alpha(x) = \frac{1}{n!} x^{-\alpha} e^x \frac{d^n}{dx^n}(e^{-x}x^{n+\alpha})$

$\displaystyle \int_0^{\infty} x^{\alpha+m} e^{-x} L_n^\alpha(x) dx = \frac{1}{n!} \int_0^{\infty} x^{m} \frac{d^n}{dx^n}(e^{-x}x^{n+\alpha})dx$

For $m=0$,

$\displaystyle \int_0^{\infty} x^{\alpha+m} e^{-x} L_n^\alpha(x) dx \\ = \frac{1}{n!} \int_0^{\infty} \frac{d^n}{dx^n}(e^{-x}x^{n+\alpha})dx \\ = \frac{1}{n!} (F_{n-1}(\infty)-F_{n-1}(0) )$

$\displaystyle F_{n-1}(x) = \int_0^{\infty} \frac{d^n}{dx^n}(e^{-x}x^{n+\alpha})dx \\ = \sum_{i=0}^{n-1} (-1)^i \begin{pmatrix} n-1 \\ i \end{pmatrix} \frac{(n+\alpha)!}{(\alpha+1+i)!} x^{\alpha+1+i} e^{-x}$

It is obvious that $F(\infty) = 0$ due to the $e^{-x}$. Since $\alpha>0$, then $x^{\alpha+1+i} = 0$ at $x = 0$. Thus,

$\displaystyle \int_0^{\infty} x^{\alpha+m} e^{-x} L_n^\alpha(x) dx = 0 , m=0$

Using integration by path,

$\displaystyle \int_0^{\infty} x^{m} d \left(\frac{d^{n-1}}{dx^{n-1}}(e^{-x}x^{n+\alpha}) \right)dx \\ = [ x^{m} \frac{d^{n-1}}{dx^{n-1}}(e^{-x}x^{n+\alpha}) ]_0^{\infty} - (m) \int_0^{\infty} x^{m-1} \frac{d^{n-1}}{dx^{n-1}}(e^{-x}x^{n+\alpha})dx$

For the same reason, the first term is zero for $\alpha + m \geq 0$. Repeat the integration by path $k$ times,

$\displaystyle \int_0^{\infty} x^{m} d \left(\frac{d^{n-1}}{dx^{n-1}}(e^{-x}x^{n+\alpha}) \right)dx \\ = (-1)^k \frac{(m)!}{(m-k)!} \int_0^{\infty} x^{m-k} \frac{d^{n-k}}{dx^{n-k}}(e^{-x}x^{n+\alpha})dx$

When $m \geq n || 0 > m$, $k$ can go to $n$, then,

$\displaystyle \int_0^{\infty} x^{m} d \left(\frac{d^{n-1}}{dx^{n-1}}(e^{-x}x^{n+\alpha}) \right)dx \\ = (-1)^{n} \frac{(m)!}{(m-n)!} \int_0^{\infty} x^{m-n}(e^{-x}x^{n+\alpha})dx \\ = (-1)^{n} \frac{(m)!}{(m-n)!} (\alpha+m)!$

When $n > m > 0$, $k$ can only go to $m$. then,

$\displaystyle \int_0^{\infty} x^{m} d \left(\frac{d^{n-1}}{dx^{n-1}}(e^{-x}x^{n+\alpha}) \right)dx \\ = (-1)^{m} (m)! \int_0^{\infty} \frac{d^{n-k}}{dx^{n-k}}(e^{-x}x^{n+\alpha})dx \\ = (-1)^{m} (m)! \frac{d^{n-m-1}}{dx^{n-m-1}}(e^{-x}x^{n+\alpha}) = 0$

For the similar result as $F_{n-1}(x)$, the result is zero.

To summarize, the following formula suitable for all $m > n || 0>m$

$\displaystyle \int_0^{\infty} x^{\alpha+m} e^{-x} L_n^\alpha(x) dx = (-1)^n (\alpha+m)! \begin{pmatrix} m \\ n \end{pmatrix}$

To evaluate the integral,

$\displaystyle \int_0^{\infty} x^{\alpha+k} e^{-x} (L_n^\alpha(x))^2 dx$

We need the know that the Laguerre polynomial can be expressed as,

$\displaystyle L_n^\alpha(x) = \sum_{i=0}^{n} (-1)^i \begin{pmatrix} n+\alpha \\ n-i \end{pmatrix} \frac{x^i}{i!}$

Thus, we can evaluate

$\displaystyle \int_0^{\infty} x^{\alpha+k} e^{-x} x^i L_n^\alpha(x) dx$

Then combine everything,

$\displaystyle \int_0^{\infty} x^{\alpha+k+i} e^{-x} L_n^\alpha(x) dx = (-1)^n (\alpha+i+k)! \begin{pmatrix} i+k \\ n \end{pmatrix}$

put inside the series expression,

$\displaystyle \int_0^{\infty} x^{\alpha+k} e^{-x} (L_n^\alpha(x))^2 dx = \sum_{i=0}^{n} (-1)^{i+n} \begin{pmatrix} n+\alpha \\ n-i \end{pmatrix} \begin{pmatrix} i+k \\ n \end{pmatrix} \frac{(\alpha+k+i)!}{i!}$

In general, $i + k \geq n || 0 > i+k$.

For $k = 0$, the sum only has 1 term for $i = n$

$\displaystyle \int_0^{\infty} x^{\alpha} e^{-x} (L_n^\alpha(x))^2 dx = \frac{(\alpha+n)!}{n!}$

Using the formula, for $k=1$, we have

$\displaystyle \int_0^{\infty} x^{\alpha+1} e^{-x} (L_n^\alpha(x))^2 dx = \frac{(\alpha+n)!}{n!} (2n+\alpha +1)$

To evaluate the expectation value of radial function

$\displaystyle \langle R_{nl} | \frac{1}{r^m} | R_{nl} \rangle$

We have to calculate the integral,

$\displaystyle \int_0^\infty e^{-x} x^{2l+2-m} \left( L_{n-l-1}^{2l+1} (x) \right)^2 dx$

For $m = 0$, we get the normalization constant,

$\displaystyle A^2 \frac{na_0}{2Z} \int_0^{\infty} e^{-x} x^{2l+2} \left( L_{n-l-1}^{2l+1}(x) \right)^2 dx = 1$

$\displaystyle \langle \frac{1}{r} \rangle = \frac{1}{n^2} \frac{Z}{a_0}$

$\displaystyle \langle \frac{1}{r^2} \rangle = \frac{2}{n^3(2l+1)} \frac{Z^2}{a_0^2}$

In general,

$\displaystyle \langle \frac{1}{r^m} \rangle = \left(\frac{Z}{a_0}\right)^m \frac{2^{m-1}(n-l-1)!}{n^{m+1}(n+l)!} G_{nlm}$

$\displaystyle G_{nlm} = \sum_{i=0}^{n-l-1} (-1)^{n-l-1+i} \begin{pmatrix} n+l \\ n-l-1+i \end{pmatrix} \begin{pmatrix} 1-m+i \\ n-l-1 \end{pmatrix} \frac{(2l+2-m+i)!}{i!}$

For the integral, for $m \neq n$

$\displaystyle \int_0^{\infty} x^{\alpha +k} e^{-x} L_m^{\beta}(x) L_n^\alpha(x) dx$

When $k = 0$, when $m < n$, the integral is zero.

When $k \neq 0, m

$\displaystyle \int_0^{\infty} x^{\alpha +k} e^{-x} L_m^{\beta}(x) L_n^\alpha(x) dx \\ = \sum_{i=0}^{m} \frac{(-1)^i}{i!}\begin{pmatrix} m+\beta \\ m-i \end{pmatrix} \int_0^{\infty} x^{\alpha+k+i} e^{-x} L_n^\alpha(x) dx \\ = \sum_{i=0}^{m} \frac{(-1)^{i+n}}{i!}\begin{pmatrix} m+\beta \\ m-i \end{pmatrix} \begin{pmatrix} k+i \\ n \end{pmatrix} (\alpha+k+i)!$

The sum only non-zero when $m\geq i > n-k$ or $-k > i \geq 0$

This shows that the Laguerre polynomial is orthogonal with weighting $e^{-x} x^{\alpha}$.

## Reason for n>l in atomic orbit

If we study atomic structure, we will find the principle quantum number must larger than azimuthal quantum number (the number of angular momentum), i.e.

$n > l$

But most text book give this result using mathematics that there is no solution for the radial equation when $n \leq l$. Some text book solves the radial equation using power series, and argue that the series has to be terminated at power of $n-1$ because the number of node is $n-1$. The exactly solution is related to Laguerre polynomial, and the polynomial is only defined for $n \geq l+1$.

Also, in nuclear physics, there is no restriction and $n$ and $l$ can take any integers. Why these two cases are different? of course, the key point is the potential difference. Coulomb potential is used in atomic orbit. Wood-Saxon potential (or finite square well ) is used in nuclear orbit.

The Schrodinger equation for potential $V(r)$ is

$\displaystyle \left( \frac{1}{r^2} \frac{\partial}{\partial r}\left( r^2 \frac{\partial}{\partial r}\right) - \frac{1}{r^2} L^2 - \frac{2m}{\hbar^2} V(r) \right) \psi(r, \Omega) = -\frac{2mE}{\hbar^2} \psi(r,\Omega)$

Separate the radial and spherical part $\psi(r, \Omega) = R(r) Y(\Omega)$

$\displaystyle L^2 Y(\Omega) = l(l+1) Y(\Omega)$

$\displaystyle \left( \frac{1}{r^2} \frac{d}{d r}\left( r^2 \frac{d}{d r}\right) - \frac{l(l+1)}{r^2} - \frac{2m}{\hbar^2} V(r) \right) R(r) = -\frac{2mE}{\hbar^2} R(r)$

The radial equation further reduce by using $u(r) = r R(r)$

$\displaystyle \frac{d^2 u}{dr^2} - \frac{2m}{\hbar^2} U(r) u(r) = \frac{2mE}{\hbar^2} u(r)$

we can see, the effective potential is

$\displaystyle U(r) = V(r) + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2}$

For Coulomb potential,

$\displaystyle U(r) = -\frac{e^2}{4\pi \epsilon_0 r} + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2}$

Introduce dimensionless quantities, or Express everything using the fine structure constant. $a_0$ is Bohr radius,

$\displaystyle \rho = \frac{2r}{a_0}, \lambda = \frac{e^2}{4\pi \epsilon_0} \sqrt{ - \frac{m}{2\hbar^2 E}} = n$

The equation becomes

$\displaystyle \frac{d^2u}{d\rho^2} + \left(\frac{\lambda}{\rho} - \frac{l(l+1)}{\rho^2}\right) u = \frac{1}{4} u$

The simplified effective potential,

$\displaystyle U(\rho) = \frac{n}{\rho} - \frac{l(l+1)}{\rho^2}$

Now, we related the principle quantum number and azimuthal quantum number in one simply formula.

Since the repulsive part due to the rotation is getting larger and larger, the minimum point of the effective potential is

$\displaystyle \rho_{min} = \frac{2l(l+1)}{n}$

When $l = n$, $\rho_{min} = 2(n+1) , r_{min} = (n+1)a_0$.

I plot the case for $n=2$

The bottom line is $l = 0$. We can see, when $l = 2$, the minimize point of the potential is at $r = 3 a_0$. The potential is vary shallow, and the bound state is very diffuses. Since the 2nd derivative of the wave function $u$ around the minimum point is

$\displaystyle \frac{d^2u}{d\rho^2} = \left( \frac{l(l+1)}{\rho^2} - \frac{n}{\rho} + \frac{1}{4} \right) u \approx \frac{1}{4} \frac{l(l+1) - n^2}{l(l+1)} u = k u$

For $l = n-1$ , $k$ is negative, that give oscillating wave function.

For $l = n$, $k$ is positive, that gives explosive wave function that cannot be normalized.

Therefore, $l < n$.

In fact, if we plot the effective potential,

$\displaystyle U(r) = -\frac{e^2}{4\pi \epsilon_0 r} + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2}$

we can find that when $l \geq 1$, the potential is also very shallow. For such a picture, we can see the wave function must be oscillating that the number of node must be larger then 1, so the principle quantum number should be larger than 1.

In nuclear orbit, the effective potential using Wood-Saxon potential is

$\displaystyle U(r) = -\frac{1}{1-\exp(\frac{r-R}{a})} + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2}$

where $R \approx 1.25 A^{1/3} [fm]$ and $a = 0.6 [fm]$. For simplicity, we use

$\displaystyle U(r) = -\frac{1}{1-\exp(\frac{r-R}{a})} + \frac{l(l+1)}{r^2}$

The position of minimum potential is no simple formula. I plot R=10,  a=1

We can see, for any $l < l_{u}$ there is a “well” that can bound a nucleon.

I haven’t solve the finite spherical well, but the infinite spherical well is solved. The solution is the spherical Bessel function of first kind. that function only has 1 parameter, which is the angular momentum. To determine the energy is simply find the node position, and the number of node is related to the principle quantum number. Therefore, the principle quantum number and angular momentum is unrelated. A similar result should be applied well in the case of finite spherical well like Wood-Saxon type, that, for a fixed angular momentum, there can be many solution with different number of nodes.

Wood-Saxon potential, the effective potential is always sloping, especially when outside of the well, that forces that wave function to be decay faster. However, the Coulomb potential is a long range force, the effective potential can be slowly changing for long distance, that allow the wave function to be oscillating for long range.

## Express Hydrogen orbit using fine structure constant

The fine structure constant is

$\displaystyle \alpha = \frac{e^2}{4\pi \epsilon_0} \frac{1}{\hbar c}$

$\displaystyle a_0 = \frac{4\pi \epsilon_0}{e^2} \frac{\hbar^2}{m} = \frac{\hbar}{m c \alpha}$

The energy

$\displaystyle E_n = -\frac{1}{2n^2} \left(\frac{e^2}{2\pi \epsilon_0}\right)^2 \frac{m}{\hbar^2} = - \frac{1}{2} m (c\alpha)^2 \frac{1}{n^2}$

We can remember that the electron is traveling at $\alpha c$ at ground state.

When solving the radial part of the schrodinger equation, we can use

$\displaystyle \rho = \frac{2r}{a_0} = r\sqrt{- \frac{8mE}{\hbar^2}}$

$\displaystyle \lambda = n = \sqrt{ -\frac{mc^2 \alpha^2}{2 E}}$

## Molecular orbital

Using 2 hydrogen atom, we started from 2 atomic orbital (AO),

$\phi_1(\vec{r}) = \phi(\vec{r} - \vec{R_1})$

$\phi_2(\vec{r}) = \phi(\vec{r} - \vec{R_2})$

The atomic orbital $\phi$ is the ground state of hydrogen atom. From these 2 AO, we can construct linear combination of atomic orbitals (LCAO),

$\psi_1(\vec{r}) = A_1( \phi_1(\vec{r}) + \phi_2(\vec{r}))$

$\psi_2(\vec{r}) = A_2( \phi_1(\vec{r}) - \phi_2(\vec{r}))$

Notice that an electron can be in either $\psi_1$ or $\psi_2$, that means the electron is not belong to any atom, but belong to the molecule.

The last step is adding the spin, there are 2 spins $\alpha, \beta$, The molecular orbital (MO) are,

$\chi_1(\vec{r}) = \psi_1(\vec{r}) \alpha$

$\chi_2(\vec{r}) = \psi_1(\vec{r}) \beta$

$\chi_3(\vec{r}) = \psi_2(\vec{r}) \alpha$

$\chi_4(\vec{r}) = \psi_2(\vec{r}) \beta$

The Slater determinant can be formed by any 2 of them.

$\Psi = \sqrt{2}^{-1} ( \chi_1(\vec{r_1})\chi_2(\vec{r_2}) - \chi_2(\vec{r_1})\chi_1(\vec{r_2}) ) = |\chi_1\chi_2 \rangle$

We started by 1 atomic orbital, and there are and 2 atoms,  then 2 LCAOs are formed, and then multiplied by 2 spin states, gives 4 molecular orbitals. From these 4 MO and 2 electrons, we can from 6 different Slater determinant.

If we have different number of AO for different atom, we have AO

$\vec{n} = (n_1, n_2,..., n_k)$

The LCAO are

$\displaystyle \psi_i = \sum_{r=1}^k C_{ir} \phi_r, r = 1,2,..., n_k$

The total number of LCAO are all permutation of the AOs. A simple construction is that only plus or minus sign are used. Then the number of LCAO ( not orthonormal ) is

$\displaystyle K = 2^{-1} \prod_{i=1}^{k} (2n_i) = 2^{k-1} \prod_{i=1}^{k} n_i$

The MO are

$\chi_{2i-1} = \psi_i \alpha \\ \chi_{2i} = \psi_i \beta , i = 1,2,..., K$

Thus the number of Slater determinant are

$\displaystyle N = {}_{2K}C_p = \frac{(2K)!}{(2K-p)!p!}$

For example, we have 3 atoms, there are $(1, 2, 2)$ AO, there are 16 LCAOs. Thus, we have 32 MO. With 2 electrons, there are 496 Slater determinant.

For C-H molecule, there are 6 electrons for carbon and 1 electron for hydrogen, use 6 AOs for carbon and 1 AO for hydrogen, the number of MO is 24 and 346,104 Slater determinant.

When using all Slater determinant, it is called full CI.

When only increase the AO, but limited number of Slater determinant, The energy is called Hartree limit.

A problem is to construct LCAOs for arbitrary AOs. For example, for $(1,1,1)$ AOs, The number of possible combination with pule or minus sign is 4. However. I cannot construct orthonormal 4 LCAOs, because the system are over constrained (16 constrains with 12 unknown of $C_{ij}$)