Quantum Mechanics¶

Normalized Schrodinger Equations for Hydrogen Atom¶

For each combination of quantum numbers ( $n, l, m$ ), the solution of the Schrodinger equation is a product of a radial solution $R(r)$ and an angular solution $Y(\theta, \phi)$ of the form $\Psi(r, \theta, \phi) = R(r) \cdot Y(\theta, \phi)$ .

The naming of the atomic orbitals depends on the angular momentum quantum number $l$ and the angular node(s) of the orbital.

To obtain the form required for real atomic orbital plots, linear combination of the complex angular wave functions has to be done such that:

$Y_{l, \ norm}^{+m} = \frac{1}{\sqrt{2}}\left(Y^{-|m|}_{1}+(-1)^{m}Y^{|m|}_{1}\right) \text{for } m > 0\\ Y_{l, \ norm}^{-m} = \frac{i}{\sqrt{2}}\left(Y^{-|m|}_{1}-(-1)^{m}Y^{|m|}_{1}\right) \text{for } m < 0$

n = 1, l = 0, m = 0¶

$\begin{align*} R^{1}_{0}(r) &= 2a^{-\frac{3}{2}} \ \text{exp} \left(-\frac{r}{a}\right) \\ \because m = 0, \therefore Y^{0}_{0}(\theta, \phi) &= \sqrt{\frac{1}{4\pi}}\\ \Rightarrow \Psi_{100}(r, \theta, \phi) &= R^{1}_{0}(r)\cdot Y^{0}_{0}(\theta, \phi) \\ &= \frac{1}{\sqrt{\pi}}a^{-\frac{3}{2}} \ \text{exp} \left(-\frac{r}{a}\right) \end{align*}$

$\because l = 0,\text{ and there are no angular nodes,} \therefore \Psi_{100}(r, \theta, \phi) \text{ describes the 1s orbital.}$

n = 2, l = 0, m = 0¶

$\begin{align*} R^{2}_{0}(r) &= \frac{1}{\sqrt{2}}a^{-\frac{3}{2}} \left(1-\frac{r}{2a}\right)\ \text{exp} \left(-\frac{r}{2a}\right) \\ \because m = 0, \therefore Y^{0}_{0}(\theta, \phi) &= \sqrt{\frac{1}{4\pi}} \\ \Rightarrow \Psi_{200}(r, \theta, \phi) &= R^{2}_{0}(r)\cdot Y^{0}_{0}(\theta, \phi) \\ &= \frac{1}{\sqrt{8\pi}}a^{-\frac{3}{2}} \left(1-\frac{r}{2a}\right)\ \text{exp} \left(-\frac{r}{a}\right) \end{align*}$

$\because l = 0, \text{and there are no angular nodes,} \therefore \Psi_{200}(r, \theta, \phi) \text{ describes the 2s orbital.}$

n = 2, l = 1, m = -1¶

$\begin{align*} R^{2}_{1}(r) &= \frac{1}{\sqrt{24}}a^{-\frac{3}{2}} \left(\frac{r}{a}\right)\ \text{exp} \left(-\frac{r}{2a}\right)\\ Y^{-1}_{1}(\theta, \phi) &= \sqrt{\frac{3}{8\pi}}\sin \theta \ \text{exp} \ (-i\phi)\\ \because m < 0, \therefore \text{Normalized angular solution, }Y^{-1}_{1, \ norm}(\theta, \phi)&=\frac{i}{\sqrt{2}}\left(Y^{-1}_{1}-(-1)^{-1}Y^{+1}_{1}\right)\\ &=\frac{i}{\sqrt{2}}\left(Y^{-1}_{1}+Y^{+1}_{1}\right)\\\\ &= \frac{i}{\sqrt{2}}\left(\sqrt{\frac{3}{8\pi}}\sin \theta\right)\ (\text{exp} \ (-i\phi) -\ \text{exp} \ (i\phi)) \\ &= -\frac{i}{\sqrt{2}}\left(\sqrt{\frac{3}{8\pi}}\sin \theta\sin\phi\right) \times 2i\\ &= \sqrt{2}\left(\sqrt{\frac{3}{8\pi}}\sin \theta\sin\phi\right)\\ &= \frac{1}{2}\sqrt{\frac{3}{\pi}}\sin \theta\sin\phi \end{align*}$

$\text{When } Y^{-1}_{1, \ norm} = 0, \sin \theta\sin\phi = 0$ .

$\Rightarrow y = 0$

$\therefore \text{There is one angular node where y = 0, which is the xz nodal plane.}$

$\begin{align*} \Rightarrow \Psi_{21-1}(r, \theta, \phi) &= R^{2}_{1}(r)\cdot Y^{-1}_{1, \ norm}(\theta, \phi) \\ &= \frac{1}{4\sqrt{2\pi}}a^{-\frac{3}{2}} \left(\frac{r}{a}\right)\ \text{exp} \left(-\frac{r}{2a}\right)\sin \theta\sin\phi \end{align*}$

$\because l = 1, \text{and there is an angular node at y = 0} \therefore \Psi_{21-1}(r, \theta, \phi) \ \text{describes the } 2p_{y} \ \text{orbital.}$

n = 2, l = 1, m = 0¶

$\begin{align*} R^{2}_{1}(r) &= \frac{1}{\sqrt{24}}a^{-\frac{3}{2}} \left(\frac{r}{a}\right)\ \text{exp} \left(-\frac{r}{2a}\right) \\ \because m = 0, \therefore Y^{0}_{1}(\theta, \phi) &= \sqrt{\frac{3}{4\pi}}cos \theta \\ \end{align*}$

$\text{When }Y^{0}_{1} = 0, \cos \theta = 0$ .

$\Rightarrow z = 0$

$\therefore \text{There is one angular node where z = 0, which is the xy nodal plane.}$

$\begin{align*} \Rightarrow \Psi_{210}(r, \theta, \phi) &= R^{2}_{1}(r)\cdot Y^{1}_{0}(\theta, \phi) \\ &= \frac{1}{4}\sqrt{\frac{1}{2\pi}}a^{-\frac{3}{2}} \left(\frac{r}{a}\right)\ \text{exp} \left(-\frac{r}{2a}\right) \cos \theta \end{align*}$

$\because l = 1, \text{and there is an angular node at z = 0} \therefore \Psi_{210}(r, \theta, \phi) \ \text{describes the } 2p_{z} \ \text{orbital.}$

n = 2, l = 1, m = +1¶

$\begin{align*} R^{2}_{1}(r) &= \frac{1}{\sqrt{24}}a^{-\frac{3}{2}} \left(\frac{r}{a}\right)\ \text{exp} \left(-\frac{r}{2a}\right)\\ \because m > 0, \therefore \text{Normalized angular solution, }Y^{+1}_{1, \ norm}(\theta, \phi)&=\frac{1}{\sqrt{2}}\left(Y^{-1}_{1}+(-1)^{1}Y^{+1}_{1}\right)\\ &=\frac{1}{\sqrt{2}}\left(Y^{-1}_{1}-Y^{+1}_{1}\right)\\ &= \frac{1}{\sqrt{2}}\left(\sqrt{\frac{3}{8\pi}}\sin \theta\right)\ (\text{exp} \ (-i\phi) +\ \text{exp} \ (i\phi)) \\ &= \frac{1}{\sqrt{2}}\left(\sqrt{\frac{3}{8\pi}}\sin \theta\cos\phi\right) \times 2\\ &= \sqrt{2}\left(\sqrt{\frac{3}{8\pi}}\sin \theta\cos\phi\right)\\ &= \frac{1}{2}\sqrt{\frac{3}{\pi}}\sin \theta\cos\phi \end{align*}$

$\text{When } Y^{+1}_{1, \ norm} = 0, \sin \theta\cos\phi = 0$ .

$\Rightarrow x = 0$

$\therefore \text{There is one angular node where x = 0, which is the yz nodal plane.}$

$\begin{align*} \Rightarrow \Psi_{21+1}(r, \theta, \phi) &= R^{2}_{1}(r)\cdot Y^{+1}_{1, \ norm}(\theta, \phi) \\ &= \frac{1}{4\sqrt{2\pi}}a^{-\frac{3}{2}} \left(\frac{r}{a}\right)\ \text{exp} \left(-\frac{r}{2a}\right)\sin \theta\cos\phi \end{align*}$

$\because l = 1, \text{and there is an angular node at x = 0} \therefore \Psi_{21+1}(r, \theta, \phi) \ \text{describes the } 2p_{x} \ \text{orbital.}$

n = 3, l = 0, m = 0¶

$\begin{align*} R^{3}_{0}(r) &= \frac{2}{81\sqrt{3}}\left[27-18\left(\frac{r}{a}\right)+2\left(\frac{r}{a}\right)^{2}\right] \ \text{exp} \left(-\frac{r}{3a}\right)\\ Y^{0}_{0}(\theta, \phi) &= \sqrt{\frac{1}{4\pi}}\\ \Psi_{300}(r, \theta, \phi) &= \frac{1}{81\sqrt{3\pi}}a^{-\frac{3}{2}}\left[27-18\left(\frac{r}{a}\right)+2\left(\frac{r}{a}\right)^{2}\right] \ \text{exp} \left(-\frac{r}{3a}\right) \end{align*}$

$\therefore \Psi_{300} (r, \theta, \phi) \text{ describes the }3s\text{ orbital.}$

n = 3, l = 1, m = -1¶

$\begin{align*} R^{3}_{1}(r) &= \frac{8}{27\sqrt{6}}a^{-\frac{3}{2}}\left(1-\frac{r}{6a}\right)\left(\frac{r}{a}\right) \ \text{exp}\left(-\frac{r}{3a}\right)\\ Y^{-1}_{1} &= \sqrt{\frac{3}{8\pi}} \sin\theta \ \text{exp}\left(-i\phi\right)\\ \text{Normalized }Y^{-1}_{1} &= \frac{1}{2}\sqrt{\frac{3}{\pi}} \sin\theta \sin\phi\\ \Psi_{31-1} &= \frac{4}{27}\sqrt{\frac{1}{2\pi}}a^{-\frac{3}{2}}\left(1-\frac{r}{6a}\right)\left(\frac{r}{a}\right) \ \text{exp}\left(-\frac{r}{3a}\right)\sin\theta\sin\phi \end{align*}$

$\therefore \Psi_{31-1}(r, \theta, \phi)\text{ describes the }3p_{y}\text{ orbital.}$

n = 3, l = 1, m = 0¶

$\begin{align*} R^{3}_{1}(r) &= \frac{8}{27\sqrt{6}}a^{-\frac{3}{2}}\left(1-\frac{r}{6a}\right)\left(\frac{r}{a}\right) \ \text{exp}\left(-\frac{r}{3a}\right)\\ Y^{0}_{1} &= \sqrt{\frac{3}{4\pi}} \cos\theta\\ \Psi_{310} &= \frac{4}{27}\sqrt{\frac{1}{2\pi}}a^{-\frac{3}{2}}\left(1-\frac{r}{6a}\right)\left(\frac{r}{a}\right) \ \text{exp}\left(-\frac{r}{3a}\right)\cos\theta \end{align*}$

$\therefore \Psi_{310}(r, \theta, \phi)\text{ describes the }3p_{z}\text{ orbital.}$

n = 3, l = 1, m = +1¶

$\begin{align*} R^{3}_{1}(r) &= \frac{8}{27\sqrt{6}}a^{-\frac{3}{2}}\left(1-\frac{r}{6a}\right)\left(\frac{r}{a}\right) \ \text{exp}\left(-\frac{r}{3a}\right)\\ Y^{+1}_{1} &= \sqrt{\frac{3}{8\pi}} \sin\theta \ \text{exp}\left(i\phi\right)\\ \text{Normalized }Y^{+1}_{1} &= \frac{1}{2}\sqrt{\frac{3}{\pi}} \sin\theta \cos\phi\\ \Psi_{311} &= \frac{4}{27}\sqrt{\frac{1}{2\pi}}a^{-\frac{3}{2}}\left(1-\frac{r}{6a}\right)\left(\frac{r}{a}\right) \ \text{exp}\left(-\frac{r}{3a}\right)\sin\theta\cos\phi \end{align*}$

$\therefore \Psi_{311}(r, \theta, \phi)\text{ describes the }3p_{x}\text{ orbital.}$

n = 3, l = 2, m = -2¶

$\begin{align*} R^{3}_{2}(r) &= \frac{4}{81\sqrt{30}}a^{-\frac{3}{2}}\left(\frac{r}{a}\right)^{2} \ \text{exp}\left(-\frac{r}{3a}\right)\\ Y^{-2}_{2} &= \sqrt{\frac{15}{32\pi}}\sin^{2}\theta \ \text{exp} \ (-2i\phi)\\ \text{Normalized }Y^{-2}_{2} &= \frac{1}{4}\sqrt{\frac{15}{\pi}}\sin^{2}\theta\sin 2\phi\\ \Psi_{32-2} &= \frac{1}{81}\sqrt{\frac{1}{2\pi}}a^{-\frac{3}{2}}\left(\frac{r}{a}\right)^{2} \ \text{exp}\left(-\frac{r}{3a}\right)\sin^{2}\theta\sin 2\phi \end{align*}$

$\therefore \Psi_{32-2}(r, \theta, \phi)\text{ describes the }3d_{xy}\text{ orbital.}$

n = 3, l = 2, m = -1¶

$\begin{align*} R^{3}_{2}(r) &= \frac{4}{81\sqrt{30}}a^{-\frac{3}{2}}\left(\frac{r}{a}\right)^{2} \ \text{exp}\left(-\frac{r}{3a}\right)\\ Y^{-1}_{2} &= \sqrt{\frac{15}{8\pi}}\cos\theta\sin\theta \ \text{exp} \ (-i\phi)\\ \text{Normalized }Y^{-1}_{2} &= \frac{1}{2}\sqrt{\frac{15}{\pi}}\cos\theta\sin\theta\sin\phi\\ \Psi_{32-1} &= \frac{2}{81}\sqrt{\frac{1}{2\pi}}a^{-\frac{3}{2}}\left(\frac{r}{a}\right)^{2} \ \text{exp}\left(-\frac{r}{3a}\right)\cos\theta\sin\theta\sin\phi \end{align*}$

$\therefore \Psi_{32-1}(r, \theta, \phi)\text{ describes the }3d_{yz}\text{ orbital.}$

n = 3, l = 2, m = 0¶

$\begin{align*} R^{3}_{2}(r) &= \frac{4}{81\sqrt{30}}a^{-\frac{3}{2}}\left(\frac{r}{a}\right)^{2} \ \text{exp}\left(-\frac{r}{3a}\right)\\ Y^{0}_{2} &= \sqrt{\frac{5}{16\pi}}\left(3\cos^{2}\theta -1\right)\\ \Psi_{320} &= \frac{1}{81}\sqrt{\frac{1}{6\pi}}a^{-\frac{3}{2}}\left(\frac{r}{a}\right)^{2} \ \text{exp}\left(-\frac{r}{3a}\right)\left(3\cos^{2}\theta -1\right) \end{align*}$

$\therefore \Psi_{320}(r, \theta, \phi)\text{ describes the }3d_{z^{2}}\text{ orbital.}$

n = 3, l = 2, m = +1¶

$\begin{align*} R^{3}_{2}(r) &= \frac{4}{81\sqrt{30}}a^{-\frac{3}{2}}\left(\frac{r}{a}\right)^{2} \ \text{exp}\left(-\frac{r}{3a}\right)\\ Y^{+1}_{2} &= \sqrt{\frac{15}{8\pi}}\cos\theta\sin\theta \ \text{exp} \ (i\phi)\\ \text{Normalized }Y^{+1}_{2} &= \frac{1}{2}\sqrt{\frac{15}{\pi}}\cos\theta\sin\theta\cos\phi\\ \Psi_{321} &= \frac{2}{81}\sqrt{\frac{1}{2\pi}}a^{-\frac{3}{2}}\left(\frac{r}{a}\right)^{2} \ \text{exp}\left(-\frac{r}{3a}\right)\cos\theta\sin\theta\cos\phi \end{align*}$

$\therefore \Psi_{321}(r, \theta, \phi)\text{ describes the }3d_{xz}\text{ orbital.}$

n = 3, l = 2, m = +2¶

$\begin{align*} R^{3}_{2}(r) &= \frac{4}{81\sqrt{30}}a^{-\frac{3}{2}}\left(\frac{r}{a}\right)^{2} \ \text{exp}\left(-\frac{r}{3a}\right)\\ Y^{+2}_{2} &= \sqrt{\frac{15}{32\pi}}\sin^{2}\theta \ \text{exp} \ (2i\phi)\\ \text{Normalized }Y^{-2}_{2} &= \frac{1}{4}\sqrt{\frac{15}{\pi}}\sin^{2}\theta\cos 2\phi\\ \Psi_{32-2} &= \frac{1}{81}\sqrt{\frac{1}{2\pi}}a^{-\frac{3}{2}}\left(\frac{r}{a}\right)^{2} \ \text{exp}\left(-\frac{r}{3a}\right)\sin^{2}\theta\cos 2\phi \end{align*}$

$\therefore \Psi_{322}(r, \theta, \phi)\text{ describes the }3d_{x^{2}-y^{2}}\text{ orbital.}$