Application of triple integrals in quantum chemistry

Multiple integrals: Applications of multiple integrals

Application of triple integrals in quantum chemistry

The table below gives the wave functions for different orbitals of a hydrogen-like atom with nuclear charge $Z$ in atomic units (for hydrogen $Z=1$ ) via spherical coordinates $(r,\phi,\theta)$ , where $a_0$ is the Bohr radius and $\rho=\frac{Z\,r}{a_0}$ and the following norms used are $N_1=\sqrt{\dfrac{Z^3}{\pi a_0^3}},\quad N_2= \frac{1}{4}\sqrt{\dfrac{Z^3}{2\pi a_0^3}}\quad N_3=\frac{1}{81}\sqrt{\dfrac{Z^3}{2\pi a_0^3}}\text.$ $\begin{array}{|c|c|} \hline \mathit{orbital} &\mathit{wave function} \\ \hline 1s & \psi_{1s} = N_{1}\,e^{-\rho}\\[0.25cm] \hline 2s & \psi_{2s} =N_{2}\,(2-\rho)\,e^{-\rho/2}\\[0.25cm] 2p_x & \psi_{2p_x} = N_{2}\,\rho\, e^{-\rho/2}\cos\phi\sin\theta\\[0.25cm] 2p_y & \psi_{2p_y} =N_{2}\,\rho\, e^{-\rho/2}\sin\phi\sin\theta\\[0.25cm] 2p_z & \psi_{2p_z} =N_{2}\,\rho\, e^{-\rho/2}\cos\theta\\[0.25cm]\hline 3s & \psi_{3s} =\sqrt{\frac{2}{3}}\,N_{3}\,(27-18\rho+2\rho^2)\,e^{-\rho/3}\\[0.25cm] 3p_x & \psi_{3p_x} = 2N_{3}\,\rho\, (6-\rho)\, e^{-\rho/3}\cos\phi\sin\theta\\[0.25cm] 3p_y & \psi_{3p_y} = 2N_{3}\,\rho\,(6-\rho)\,\e^{-\rho/3}\sin\phi\sin\theta\\[0.25cm] 3p_z & \psi_{3p_z} = 2N_{3}\,\rho\, (6-\rho)\,e^{-\rho/3}\cos\theta\\[0.25cm] 3d_{z^2} & \psi_{3d_{z^2}} = \sqrt{\frac{1}{3}}\,N_3\,\rho^2\, e^{-\rho/3}(3\cos^2\theta-1)\\[0.25cm] 3d_{xz} & \psi_{3d_{xz}} = N_3\, \rho^2\, e^{-\rho/3}\cos\varphi\sin2\theta\\[0.25cm] 3d_{yz} & \psi_{3d_{yz}} = N_3\,\rho^2\, e^{-\rho/3}\sin\varphi\sin2\theta\\[0.25cm] 3d_{x^2-y^2} & \psi_{3d_{x^2-y^2}} = N_3\, \rho^2\, e^{-\rho/3}\cos2\varphi\sin^2\theta\\[0.25cm] 3d_{xy} & \psi_{3d_{xy}} = N_3\,\rho^2\, e^{-\rho/3}\sin2\varphi\sin^2\theta \\ \hline \end{array}$ The probability of the electron being in a volume element $\dd(x,y,z)$ at position $(x,y,z)$ is equal to $\psi(x,y,z)^2\,\dd(x,y,z)$ for a given wave function $\psi(x,y,z)$ .

This means that integration over the entire space $\iiint_{\text{3-D space}}\psi_{1s}^2\,\dd(x,y,z)$ must be equal to $1$ . In other words, the wave function is normalized.

We check normalization for the wave function of the $1s$ orbital: $\begin{aligned}\iiint_{\text{3D space}}\psi_{1s}^2\,\dd(x,y,z) &= N_1^2\cdot \lim_{N\to\infty}\int_{r=0}^{r=N}\left(\int_{\varphi=0}^{\varphi=2\pi}\left(\int_{\theta=0}^{\theta=\pi}e^{-2\rho}\,r^2\sin\theta\,\dd\theta\right)\dd\varphi\right)\dd r\\[0.25cm] &= \frac{1}{\pi}\cdot \lim_{N\to\infty}\int_{\rho=0}^{\rho=N}\left(\int_{\varphi=0}^{\varphi=2\pi}\left(\int_{\theta=0}^{\theta=\pi}e^{-2\rho}\,\rho^2\sin\theta\,\dd\theta\right)\dd\varphi\right)\dd \rho\\[0.25cm]&\phantom{abcuva}\blue{\text{substitution }r=\frac{a_0}{Z}\,\rho, \dd r = \frac{a_0}{Z}\,\dd\rho}\\[0.25cm] &= \frac{1}{\pi}\cdot\left(\lim_{N\to\infty}\int_{\rho=0}^{\rho=N}\rho\,e^{-2\rho}\,\dd\rho\right)\cdot \left(\int_{\varphi=0}^{\varphi=2\pi}\,\dd\varphi\right)\cdot \left(\int_{\theta=0}^{\theta=\pi}\sin\theta\right)\\[0.25cm] &= \frac{1}{\pi} \cdot\left(\lim_{N\to\infty}\int_{\rho=0}^{\rho=N}\rho\,e^{-2\rho}\,\dd\rho\right)\cdot \biggl[\varphi\biggr]_{\varphi=0}^{\varphi=2\pi}\cdot \bigg[-\cos\theta\biggr]_{\theta=0}^{\theta}-\cos\theta\biggr]_{\theta=0}^{\theta=\pi}\\[0.25cm] &= \frac{1}{\pi}\cdot 4\pi \cdot \lim_{N\to\infty}\int_{\rho=0}^{\rho=N}\rho^2\,e^{-\rho}\dd\rho\\[0.25cm] &\phantom{abcuva}\blue{\int_{0}^{\infty} x^n\,e^{-\alpha\,x}=\dfrac{n!}{\alpha^{n+1}}, \quad n\in\mathbb{N}, \alpha>0}\\[0.25cm] &= 4 \cdot \frac{2!}{2^3}\\[0.25cm] &= 1 \end{aligned}$

The average distance of an electron to the nucleus $\langle r\rangle$ in the $1s$ orbital of a hydrogen-like atom with nuclear charge $Z$ is the expectation value of $r$ for the wave function $N_1\,e^{-\rho}$ . This can be calculated as follows: $\langle r\rangle=\iiint_{\text{3-D ruimte}}r\,\psi_{1\,\!s}^2\,\dd(x,y,z)$ We do this in spherical coordinates: $\begin{aligned}\iiint_{\text{3-D ruimte}}&r\,\psi_{1\,\!s}^2\,\dd(x,y,z)\\[0.25cm] &= \frac{a_0}{Z}\cdot N_1^2 \int_{r=0}^{r=\infty}\left(\int_{\varphi=0}^{\varphi=2\pi}\left(\int_{\theta=0}^{\theta=\pi}r\,e^{-2\rho}\,r^2\,\sin\theta\,\dd\theta\right)\dd\varphi\right)\dd r\\[0.25cm] &=\frac{a_0}{Z}\cdot \frac{1}{\pi} \int_{\rho=0}^{\rho=\infty}\left(\int_{\varphi=0}^{\varphi=2\pi}\left(\int_{\theta=0}^{\theta=\pi}\rho^3\,e^{-2\rho}\,\sin\theta\,\dd\theta\right)\dd\varphi\right)\dd \rho\\[0.25cm] &=\frac{a_0}{Z}\cdot \frac{1}{\pi} \left(\int_{0}^{\infty} \rho^3\,e^{-2\rho}\,\dd\rho\right)\cdot \left(\int_{0}^{2\pi}\dd\varphi\right)\cdot \left(\int_{0}^{\pi}\,\sin\theta\,\dd\theta\right)\\[0.25cm] &= \frac{a_0}{Z}\cdot \frac{1}{\pi} \left(\int_{0}^{\infty} \rho^3\,e^{-2\rho}\,\dd\rho\right)\cdot \biggl[\varphi\biggr]_0^{2\pi}\cdot \biggl[-\cos\theta\biggr]_0^{\pi}\\[0.25cm] &= \frac{a_0}{Z}\cdot \frac{1}{\pi} \cdot\frac{3! }{2^4}\cdot 2\pi\cdot 2\\[0.25cm] &= \frac{a_0}{Z}\cdot \frac{1}{\pi} \cdot \frac{3}{8}\cdot 2\pi\cdot 2\\[0.25cm] &= \frac{3}{2}\,\frac{a_0}{Z}\end{aligned}$