The 2s atomic orbital is,

\[\begin{equation} \phi^Z_{2s}=\frac{1}{4} \sqrt{\frac{Z^3}{2\pi a_0^3}}\left(2-\frac{Zr}{a_0}\right)\exp\left(-\frac{Zr}{2a_0}\right). \end{equation}\]

$a_0^{3/2}\phi_{2s}$
	$\frac{Z_{\text{eff}}x}{a_0}$

To check that the 2s orbital solves the Schrödinger equation,

\begin{equation} - \frac{\hbar^2}{2m}\nabla^2\phi^Z_{2s}(\vec{r}) - \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}|}\phi^Z_{2s}(\vec{r}) = E\phi^Z_{2s}(\vec{r}). \end{equation}

we let the Laplacian operator act on the orbital,

\begin{equation} \nabla ^2\phi^Z_{2s}=\frac{1}{4}\sqrt{\frac{Z^5}{2\pi a^5_0}} \left(-\frac{Z^2r}{4a_0^2}+\frac{5Z}{2a_0}-\frac{4}{r}\right)\exp\left(-\frac{Zr}{2a_0}\right), \end{equation}

and then calculate the energy $E_1= - \frac{\hbar^2}{2m}\frac{\nabla^2\phi_{2s}(\vec{r})}{\phi_{2s}(\vec{r})} - \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}|}$. The code below defines the 2s orbital and its Laplacian in Cartesian coordinates centered at position $(x_i,y_i,z_i)$. It then chooses random numbers for $x$, $y$, $z$, $x_i$, $y_i$, and $y_i$ and calculates the energy. If the orbital has been programmed properly, the energy should be the same for every position.

<script> var a0 = 5.2917721E-11; // Bohr radius in meters var pi = 3.1415926535897932; var hbar = 1.05457266E-34; //h-bar var ec = 1.60217733E-19; //electron charge var eps0 = 8.854187817E-12; //permittivity constant var me = 9.1093897E-31; //electron mass var k1 = 0.5*hbar*hbar/me; var k2 = ec*ec/(4*pi*eps0); //Coulomb constant var Z_eff = 1.3; //effective nuclear charge for Li using Slater's rules function ao_2s(x,y,z,xi,yi,zi,Za) { //the 2s atomic orbital r = Math.sqrt(Math.pow(x-xi,2)+Math.pow(y-yi,2)+Math.pow(z-zi,2)); return 0.25*Math.sqrt(Math.pow(Za,3)/(2*pi*Math.pow(a0,3)))*(2-Za*r/a0)*Math.exp(-Za*r/(2*a0)); } function L_ao_2s(x,y,z,xi,yi,zi,Za) { //Lagrangian of the 2s atomic orbital r = Math.sqrt(Math.pow(x-xi,2)+Math.pow(y-yi,2)+Math.pow(z-zi,2)); return 0.25*Math.sqrt(Math.pow(Za,5)/(2*pi*Math.pow(a0,5)))*(-Za*Za*r/(4*a0*a0)+5*Za/(2*a0)-4/r)*Math.exp(-Za*r/(2*a0)); } for (i=0; i<10; i++) { //The energy is calculated for 10 random positions x = 2*a0*(0.5-Math.random())/Z_eff; xi = 2*a0*(0.5-Math.random())/Z_eff; y = 2*a0*(0.5-Math.random())/Z_eff; yi = 2*a0*(0.5-Math.random())/Z_eff; z = 2*a0*(0.5-Math.random())/Z_eff; zi = 2*a0*(0.5-Math.random())/Z_eff; E = -k1*L_ao_2s(x,y,z,xi,yi,zi,Z_eff)/ao_2s(x,y,z,xi,yi,zi,Z_eff)-k2*Z_eff/r; document.write("E₁ = "+E/ec+" eV<br>"); } </script>

The probability of finding a electron a distance $r$ from the nucleus is $P(r)=4\pi r^2|\phi_{2s}|^2$. The probability has a maximum at $a_0$ but by looking at the integral is clear that it is more probable to find the electrons further than $a_0$ from the nucleus than closer than $a_0$ from the nucleus.

	$\frac{Z_{\text{eff}}r}{a_0}$

Consider the overlap integral of two 2s orbitals located at positions $\vec{r}_1$ and $\vec{r}_2$,

\begin{equation} S_{12}=\langle \phi_{2s}(\vec{r}-\vec{r}_1)|\phi_{2s}(\vec{r}-\vec{r}_2)\rangle . \end{equation}

For a Li₂ molecule, $\vec{r}_1=-1.336\,\hat{x}$ Å and $\vec{r}_2=1.336\,\hat{x}$ Å. Below $\phi_{2s}(\vec{r}-\vec{r}_1)\phi_{2s}(\vec{r}-\vec{r}_2)$ is plotted along the $x$-axis and along the $y$-axis.

$a_0^3\phi_{2s}(\vec{r}-\vec{r}_1)\phi_{2s}(\vec{r}-\vec{r}_2)$
	$\frac{Z_{\text{eff}}x}{a_0}\,,\frac{Z_{\text{eff}}y}{a_0}$

The code below uses a Monte-Carlo method to integrate $\phi_{2s}(\vec{r}-\vec{r}_1)\phi_{2s}(\vec{r}-\vec{r}_2)$ and calculate $S_{12}$.

<script> var a0 = 5.2917721E-11; // Bohr radius in meters var pi = Math.PI; var Z_eff = 1.3; //effective nuclear charge for Li using Slater's rules function ao_2s(x,y,z,xi,yi,zi,Za) { //the 2s atomic orbital r = Math.sqrt(Math.pow(x-xi,2)+Math.pow(y-yi,2)+Math.pow(z-zi,2)); return 0.25*Math.sqrt(Math.pow(Za,3)/(2*pi*Math.pow(a0,3)))*(2-Za*r/a0)*Math.exp(-Za*r/(2*a0)); } S12 = 0; N = 1000000; //number of random points x1 = -1.336E-10; x2 = 1.336E-10; L = 20*a0/Z_eff for (i=0; i<N; i++) { x = L*(0.5-Math.random()); y = L*(0.5-Math.random()); z = L*(0.5-Math.random()); S12 += ao_2s(x,y,z,x1,0,0,Z_eff)*ao_2s(x,y,z,x2,0,0,Z_eff); } document.write("S<sub>12<\/sub> = "+Math.pow(L,3)*S12/N); //multply by the volume and divide by the number of points </script>

Press the 'Execute' button a few times and notice that the answer keeps changing. By doing this you can estimate the error in the calculation. The error should decrease like $1/\sqrt{N}$ where $N$ is the number of random numbers chosen. By setting $x_1=x_2=0$ in the code and pressing 'Execute', you calculate $\langle \phi_{2s}(\vec{r})|\phi_{2s}(\vec{r})\rangle$ which should equal 1 if the wave function is properly normalized.

The diagonal Hamiltonian matrix element of a homonuclear diatomic molecule (H₂, O₂, N₂, etc.) with two 2s orbitals located at position $\vec{r}_1$ is,

\begin{equation} H_{11}=\Big \langle \phi_{2s}^Z(\vec{r}-\vec{r}_1)\left|- \frac{\hbar^2}{2m}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_1|}- \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_2|} \right|\phi_{2s}^Z(\vec{r}-\vec{r}_1) \Big \rangle . \end{equation}

This can be broken into two terms,

\begin{equation} H_{11}=\Big \langle \phi_{2s}^Z(\vec{r}-\vec{r}_1)\left|- \frac{\hbar^2}{2m}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_1|}\right|\phi_{2s}^Z(\vec{r}-\vec{r}_1) \Big \rangle + \Big \langle \phi_{2s}^Z(\vec{r}-\vec{r}_1)\left|- \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_2|} \right|\phi_{2s}^Z(\vec{r}-\vec{r}_1) \Big \rangle . \end{equation}

The wave function $\phi_{2s}^Z(\vec{r}-\vec{r}_1)$ is an eigenfunction of the atomic orbital Hamiltonian in the first term $H\phi_{2s}^Z(\vec{r}-\vec{r}_1) = E_2 \phi_{2s}^Z(\vec{r}-\vec{r}_1)$ where $E_2=\frac{13.6Z^2_{\text{eff}}}{4}=3.4Z^2_{\text{eff}}$ eV. The first term is easily evaluated,

\begin{equation} H_{11}=E_2 - \Big \langle \phi_{2s}^Z(\vec{r}-\vec{r}_1)\left| \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_2|} \right|\phi_{2s}^Z(\vec{r}-\vec{r}_1) \Big \rangle . \end{equation}

The second term has a singularity at $\vec{r}_2$ which makes it difficult to evaluate numerically. We break the second term into an integral over a spherical volume of radius $\delta$ centered around $\vec{r}_2$ and a second integral outside that volume.

\begin{equation} H_{11}=E_2 - \int\limits_{|\vec{r}-\vec{r}_2| < \delta} \phi_{2s}^Z(\vec{r}-\vec{r}_1) \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_2|} \phi_{2s}^Z(\vec{r}-\vec{r}_1) d^3r - \int\limits_{|\vec{r}-\vec{r}_2| > \delta} \phi_{2s}^Z(\vec{r}-\vec{r}_1) \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_2|} \phi_{2s}^Z(\vec{r}-\vec{r}_1) d^3r. \end{equation}

Close to $\vec{r}_2$, $\phi_{2s}^Z(\vec{r}-\vec{r}_1)\approx \phi_{2s}^Z(\vec{r}_2-\vec{r}_1)$. Using this approximation, the first integral which includes the singularity can be performed analytically for small $\delta$.

\begin{equation} H_{11}=E_2 -\frac{Z^4e^2\delta^2}{256\pi\epsilon_0a_0^3}\left(2-\frac{Z|\vec{r}_2-\vec{r}_1|}{a_0}\right)^2\exp\left(-\frac{Z|\vec{r}_2-\vec{r}_1|}{a_0}\right) - \int H(|\vec{r}-\vec{r}_2|-\delta ) \phi_{2s}^Z(\vec{r}-\vec{r}_1) \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_2|} \phi_{2s}^Z(\vec{r}-\vec{r}_1) d^3r, \end{equation}

The second integral integrates over all space but a Heaviside step function has been introduced. $H(|\vec{r}-\vec{r}_2|-\delta ) = 0$ for $|\vec{r}-\vec{r}_2| < \delta$ and is 1 otherwise. The second integral contains no singularity and can be evaluated numerically.

$H(|\vec{r}-\vec{r}_2|-\delta )\frac{a_0^3e\phi_{2s}(\vec{r}-\vec{r}_1)\phi_{2s}(\vec{r}-\vec{r}_1)}{4\pi\epsilon_0 |\vec{r}-\vec{r}_2|}$
	$\frac{x}{a_0}$
The integrand of the matrix element plotted along the $x$-axis for $\delta = a_0/10$.