PHY.K02UF Molecular and Solid State Physics

## 1s Atomic Orbital

The 1s atomic orbital is,

$$$\phi^Z_{1s}=\sqrt{\frac{Z^3}{\pi a^3_0}}\exp\left(-\frac{Zr}{a_0}\right).$$$
 $a_0^{3/2}\phi_{1s}$ $\frac{x}{a_0}$

To check that the 1s orbital solves the Schrödinger equation,

$$- \frac{\hbar^2}{2m}\nabla^2\phi(\vec{r}) - \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}|}\phi(\vec{r}) = E\phi(\vec{r}),$$

we let the Laplacian operator act on the orbital,

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

and then calculate the energy $E_1= - \frac{\hbar^2}{2m}\frac{\nabla^2\phi_{1s}(\vec{r})}{\phi_{1s}(\vec{r})} - \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}|}$. The code below defines the 1s 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 -13.6 eV for every position.

The probability of finding a electron a distance $r$ from the nucleus is $P(r)=4\pi r^2|\phi_{1s}|^2$. 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.

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

Consider the overlap integral of two 1s orbitals located at positions $\vec{r}_1$ and $\vec{r}_2$,

$$S_{12}=\langle \phi_{1s}(\vec{r}-\vec{r}_1)|\phi_{1s}(\vec{r}-\vec{r}_2)\rangle .$$

For a hydrogen molecule, $\vec{r}_1=-0.38\,\hat{x}$ Å and $\vec{r}_2=0.38\,\hat{x}$ Å. Below $\phi_{1s}(\vec{r}-\vec{r}_1)\phi_{1s}(\vec{r}-\vec{r}_2)$ is plotted along the $x$-axis and along the $y$-axis.

 $a_0^3\phi_{1s}(\vec{r}-\vec{r}_1)\phi_{1s}(\vec{r}-\vec{r}_2)$ $\frac{x}{a_0}\,,\frac{y}{a_0}$

The code below uses a Monte-Carlo method to integrate $\phi_{1s}(\vec{r}-\vec{r}_1)\phi_{1s}(\vec{r}-\vec{r}_2)$ and calculate $S_{12}$.



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 $\langle \phi_{1s}(\vec{r})|\phi_{1s}(\vec{r})\rangle$ 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 1s orbitals located at position $\vec{r}_1$ is,

$$H_{11}=\Big \langle \phi_{1s}^Z(\vec{r}-\vec{r}_1)\left|- \frac{\hbar^2}{2m}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_1|}- \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_2|} \right|\phi_{1s}^Z(\vec{r}-\vec{r}_1) \Big \rangle .$$

This can be broken into two terms,

$$H_{11}=\Big \langle \phi_{1s}^Z(\vec{r}-\vec{r}_1)\left|- \frac{\hbar^2}{2m}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_1|}\right|\phi_{1s}^Z(\vec{r}-\vec{r}_1) \Big \rangle + \Big \langle \phi_{1s}^Z(\vec{r}-\vec{r}_1)\left|- \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_2|} \right|\phi_{1s}^Z(\vec{r}-\vec{r}_1) \Big \rangle .$$

The wave function $\phi_{1s}^Z(\vec{r}-\vec{r}_1)$ is an eigenfunction of the atomic orbital Hamiltonian in the first term $H\phi_{1s}^Z(\vec{r}-\vec{r}_1) = E_1 \phi_{1s}^Z(\vec{r}-\vec{r}_1)$, so the first term is easily evaluated,

$$H_{11}=E_1 - \Big \langle \phi_{1s}^Z(\vec{r}-\vec{r}_1)\left| \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_2|} \right|\phi_{1s}^Z(\vec{r}-\vec{r}_1) \Big \rangle .$$

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.

$$H_{11}=E_1 - \int\limits_{|\vec{r}-\vec{r}_2| < \delta} \phi_{1s}^Z(\vec{r}-\vec{r}_1) \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_2|} \phi_{1s}^Z(\vec{r}-\vec{r}_1) d^3r - \int\limits_{|\vec{r}-\vec{r}_2| > \delta} \phi_{1s}^Z(\vec{r}-\vec{r}_1) \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_2|} \phi_{1s}^Z(\vec{r}-\vec{r}_1) d^3r.$$

Close to $\vec{r}_2$, $\exp\left(\frac{ -Z|\vec{r}-\vec{r}_1|}{a_0}\right)\approx \exp\left(\frac{ -Z|\vec{r}_2-\vec{r}_1|}{a_0}\right)$. Using this approximation, the first integral which includes the singularity can be performed analytically for small $\delta$.

$$H_{11}=E_1 - \frac{Z^4e^2\delta^2}{2\pi\epsilon_0a_0^3} \exp(-2Z|\vec{r}_1-\vec{r}_2|/a_0) - \int H(|\vec{r}-\vec{r}_2|-\delta ) \phi_{1s}^Z(\vec{r}-\vec{r}_1) \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_2|} \phi_{1s}^Z(\vec{r}-\vec{r}_1) d^3r,$$

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.

 $H(|\vec{r}-\vec{r}_2|-\delta )\frac{a_0^3e\phi_{1s}(\vec{r}-\vec{r}_1)\phi_{1s}(\vec{r}-\vec{r}_1)}{4\pi\epsilon_0 |\vec{r}-\vec{r}_2|}$ $\frac{x}{a_0}$ The integrand of the matrix element plotted along the $x$-axis for $\delta = a_0/10$.

The code below uses a Monte-Carlo method to and calculate $H_{11}$.



Similarly, the off-diagonal Hamiltonian matrix element of a homonuclear diatomic molecule (H2, O2, N2, etc.) with two 1s orbitals located at positions $\vec{r}_1$ and $\vec{r}_2$ is,

$$H_{12}=\Big \langle \phi_{1s}^Z(\vec{r}-\vec{r}_1)\left|- \frac{\hbar^2}{2m}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_1|}- \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_2|} \right|\phi_{1s}^Z(\vec{r}-\vec{r}_2) \Big \rangle .$$

This can be broken into two terms,

$$H_{12}=\Big \langle \phi_{1s}^Z(\vec{r}-\vec{r}_1)\left|- \frac{\hbar^2}{2m}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_2|}\right|\phi_{1s}^Z(\vec{r}-\vec{r}_2) \Big \rangle + \Big \langle \phi_{1s}^Z(\vec{r}-\vec{r}_1)\left|- \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_1|} \right|\phi_{1s}^Z(\vec{r}-\vec{r}_2) \Big \rangle .$$

The wave function $\phi_{1s}^Z(\vec{r}-\vec{r}_2)$ is an eigenfunction of the atomic orbital Hamiltonian in the first term $H\phi_{1s}^Z(\vec{r}-\vec{r}_2) = E_1 \phi_{1s}^Z(\vec{r}-\vec{r}_2)$, so the first term is easily evaluated,

$$H_{12}=E_1S_{12} - \Big \langle \phi_{1s}^Z(\vec{r}-\vec{r}_1)\left| \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_1|} \right|\phi_{1s}^Z(\vec{r}-\vec{r}_2) \Big \rangle .$$

The second term has a singularity at $\vec{r}_1$ which makes it difficult to evaluate numerically. We break the second term into an integral over a spherical volume of radius $\delta$ centered around $\vec{r}_1$ and a second integral outside that volume.

$$H_{12}=E_1S_{12} - \int\limits_{|\vec{r}-\vec{r}_1| < \delta} \phi_{1s}^Z(\vec{r}-\vec{r}_1) \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_1|} \phi_{1s}^Z(\vec{r}-\vec{r}_2) d^3r - \int\limits_{|\vec{r}-\vec{r}_1| > \delta} \phi_{1s}^Z(\vec{r}-\vec{r}_1) \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_1|} \phi_{1s}^Z(\vec{r}-\vec{r}_2) d^3r.$$

Close to $\vec{r}_1$, $\exp\left(\frac{ -Z|\vec{r}-\vec{r}_2|}{a_0}\right)\approx \exp\left(\frac{ -Z|\vec{r}_1-\vec{r}_2|}{a_0}\right)$. Using this approximation, the first integral which includes the singularity can be performed analytically for small $\delta$.

$$H_{12}=E_1S_{12} - \frac{Z^4e^2\delta^2}{2\pi a_0^3\epsilon_0} \exp(-Z|\vec{r}_1-\vec{r}_2|/a_0) - \int H(|\vec{r}-\vec{r}_1|-\delta ) \phi_{1s}^Z(\vec{r}-\vec{r}_1) \frac{Ze^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_1|} \phi_{1s}^Z(\vec{r}-\vec{r}_2) d^3r,$$

The second integral integrates over all space but a Heaviside step function has been introduced. $H(|\vec{r}-\vec{r}_1|-\delta ) = 0$ for $|\vec{r}-\vec{r}_1| < \delta$ and is 1 otherwise. The second integral contains no singularity and can be evaluated numerically.

 $H(|\vec{r}-\vec{r}_1|-\delta )\frac{a_0^3e\phi_{1s}(\vec{r}-\vec{r}_1)\phi_{1s}(\vec{r}-\vec{r}_2)}{4\pi\epsilon_0 |\vec{r}-\vec{r}_1|}$ $\frac{x}{a_0}$ The integrand of the matrix element plotted along the $x$-axis for $\delta = a_0/10$.

The code below uses a Monte-Carlo method to and calculate $H_{12}$.