Comprehensive Study Notes

The Radial Factor

Using (6.93), we have for the recursion relation (6.88)

\(
\begin{equation}
b{j+1}=\frac{2 Z}{n a} \frac{j+l+1-n}{(j+1)(j+2 l+2)} b{j} \tag{6.99}
\end{equation}
\)

The discussion preceding Eq. (6.91) shows that the highest power of $r$ in the polynomial $M(r)=\Sigma{j} b{j} r^{j}$ [Eq. (6.86)] is $k=n-l-1$. Hence use of $C=Z / n a$ [Eq. (6.93)] in $R(r)=e^{-C r} r l M(r)$ [Eq. (6.84)] gives the radial factor in the hydrogen-atom $\psi$ as

\(
\begin{equation}
R{n l}(r)=r^{l} e^{-Z r / n a} \sum{j=0}^{n-l-1} b_{j} r^{j} \tag{6.100}
\end{equation}
\)

where $a \equiv 4 \pi \varepsilon_{0} \hbar^{2} / \mu e^{2}$ [Eq. (6.63)]. The complete hydrogenlike bound-state wave functions are [Eq. (6.61)]

\(
\begin{equation}
\psi{n l m}=R{n l}(r) Y{l}^{m}(\theta, \phi)=R{n l}(r) S_{l m}(\theta) \frac{1}{\sqrt{2 \pi}} e^{i m \phi} \tag{6.101}
\end{equation}
\)

where the first few theta functions are given in Table 5.1.
How many nodes does $R(r)$ have? The radial function is zero at $r=\infty$, at $r=0$ for $l \neq 0$, and at values of $r$ that make $M(r)$ vanish. $M(r)$ is a polynomial of degree $n-l-1$, and it can be shown that the roots of $M(r)=0$ are all real and positive. Thus, aside from the origin and infinity, there are $n-l-1$ nodes in $R(r)$. The nodes of the spherical harmonics are discussed in Prob. 6.41.

Ground-State Wave Function and Energy

For the ground state of the hydrogenlike atom, we have $n=1, l=0$, and $m=0$. The radial factor (6.100) is

\(
\begin{equation}
R{10}(r)=b{0} e^{-Z r / a} \tag{6.102}
\end{equation}
\)

The constant $b_{0}$ is determined by normalization [Eq. (5.80)]:

\(
\left|b{0}\right|^{2} \int{0}^{\infty} e^{-2 Z r / a} r^{2} d r=1
\)

Using the Appendix integral (A.8), we find

\(
\begin{equation}
R_{10}(r)=2\left(\frac{Z}{a}\right)^{3 / 2} e^{-Z r / a} \tag{6.103}
\end{equation}
\)

Multiplying by $Y_{0}^{0}=1 /(4 \pi)^{1 / 2}$, we have as the ground-state wave function

\(
\begin{equation}
\psi_{100}=\frac{1}{\pi^{1 / 2}}\left(\frac{Z}{a}\right)^{3 / 2} e^{-Z r / a} \tag{6.104}
\end{equation}
\)

The hydrogen-atom energies and wave functions involve the reduced mass, given by (6.59) as

\(
\begin{equation}
\mu{\mathrm{H}}=\frac{m{e} m{p}}{m{e}+m{p}}=\frac{m{e}}{1+m{e} / m{p}}=\frac{m{e}}{1+0.000544617}=0.9994557 m{e} \tag{6.105}
\end{equation}
\)

where $m{p}$ is the proton mass and $m{e} / m_{p}$ was found from Table A.1. The reduced mass is very close to the electron mass. Because of this, some texts use the electron mass instead of the reduced mass in the H atom Schrödinger equation. This corresponds to assuming that the proton mass is infinite compared with the electron mass in (6.105) and that all the internal motion is motion of the electron. The error introduced by using the electron mass for the reduced mass is about 1 part in 2000 for the hydrogen atom. For heavier atoms, the error introduced by assuming an infinitely heavy nucleus is even less than this. Also, for many-electron atoms, the form of the correction for nuclear motion is quite complicated. For these reasons we shall assume in the future an infinitely heavy nucleus and simply use the electron mass in writing the Schrödinger equation for atoms.

If we replace the reduced mass of the hydrogen atom by the electron mass, the quantity $a$ defined by (6.63) becomes

\(
\begin{equation}
a{0}=\frac{4 \pi \varepsilon{0} \hbar^{2}}{m_{e} e^{2}}=0.529177 \AA \tag{6.106}
\end{equation}
\)

where the subscript zero indicates use of the electron mass instead of the reduced mass. $a_{0}$ is called the Bohr radius, since it was the radius of the circle in which the electron moved in the ground state of the hydrogen atom in the Bohr theory. Of course, since the ground-state wave function (6.104) is nonzero for all finite values of $r$, there is some probability of finding the electron at any distance from the nucleus. The electron is certainly not confined to a circle.

A convenient unit for electronic energies is the electronvolt (eV), defined as the kinetic energy acquired by an electron accelerated through a potential difference of 1 volt (V). Potential difference is defined as energy per unit charge. Since $e=1.6021766 \times 10^{-19} \mathrm{C}$ and $1 \mathrm{VC}=1 \mathrm{~J}$, we have

\(
\begin{equation}
1 \mathrm{eV}=1.6021766 \times 10^{-19} \mathrm{~J} \tag{6.107}
\end{equation}
\)

EXAMPLE

Calculate the ground-state energy of the hydrogen atom using SI units and convert the result to electronvolts.

The H atom ground-state energy is given by (6.94) with $n=1$ and $Z=1$ as $E=-\mu e^{4} / 8 h^{2} \varepsilon_{0}^{2}$. Use of (6.105) for $\mu$ gives

\(
\begin{aligned}
E & =-\frac{0.9994557\left(9.109383 \times 10^{-31} \mathrm{~kg}\right)\left(1.6021766 \times 10^{-19} \mathrm{C}\right)^{4}}{8\left(6.626070 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)^{2}\left(8.8541878 \times 10^{-12} \mathrm{C}^{2} / \mathrm{N}-\mathrm{m}^{2}\right)^{2}} \frac{n^{2}}{n^{2}} \
E & =-\left(2.178686 \times 10^{-18} \mathrm{~J}\right)\left(Z^{2} / n^{2}\right)\left[(1 \mathrm{eV}) /\left(1.6021766 \times 10^{-19} \mathrm{~J}\right)\right]
\end{aligned}
\)

\(
\begin{equation}
E=-(13.598 \mathrm{eV})\left(Z^{2} / n^{2}\right)=-13.598 \mathrm{eV} \tag{6.108}
\end{equation}
\)

a number worth remembering. The minimum energy needed to ionize a ground-state hydrogen atom is 13.598 eV .
EXERCISE Find the $n=2$ energy of $\mathrm{Li}^{2+}$ in eV ; do the minimum amount of calculation needed. (Answer: -30.60 eV .)

EXAMPLE

Find $\langle T\rangle$ for the hydrogen-atom ground state.
Equations (3.89) for $\langle T\rangle$ and (6.7) for $\nabla^{2} \psi$ give

\(
\begin{gathered}
\langle T\rangle=\int \psi^{} \hat{T} \psi d \tau=-\frac{\hbar^{2}}{2 \mu} \int \psi^{} \nabla^{2} \psi d \tau \
\nabla^{2} \psi=\frac{\partial^{2} \psi}{\partial r^{2}}+\frac{2}{r} \frac{\partial \psi}{\partial r}-\frac{1}{r^{2} \hbar^{2}} \hat{L}^{2} \psi=\frac{\partial^{2} \psi}{\partial r^{2}}+\frac{2}{r} \frac{\partial \psi}{\partial r}
\end{gathered}
\)

since $\hat{L}^{2} \psi=l(l+1) \hbar^{2} \psi$ and $l=0$ for an $s$ state. From (6.104) with $Z=1$, we have $\psi=\pi^{-1 / 2} a^{-3 / 2} e^{-r / a}$, so $\partial \psi / \partial r=-\pi^{-1 / 2} a^{-5 / 2} e^{-r / a}$ and $\partial^{2} \psi / \partial r^{2}=\pi^{-1 / 2} a^{-7 / 2} e^{-r / a}$.
Using $d \tau=r^{2} \sin \theta d r d \theta d \phi$ [Eq. (5.78)], we have

\(
\begin{gathered}
\langle T\rangle=-\frac{\hbar^{2}}{2 \mu} \frac{1}{\pi a^{4}} \int{0}^{2 \pi} \int{0}^{\pi} \int{0}^{\infty}\left(\frac{1}{a} e^{-2 r / a}-\frac{2}{r} e^{-2 r / a}\right) r^{2} \sin \theta d r d \theta d \phi \
=-\frac{\hbar^{2}}{2 \mu \pi a^{4}} \int{0}^{2 \pi} d \phi \int{0}^{\pi} \sin \theta d \theta \int{0}^{\infty}\left(\frac{r^{2}}{a} e^{-2 r / a}-2 r e^{-2 r / a}\right) d r=\frac{\hbar^{2}}{2 \mu a^{2}}=\frac{e^{2}}{8 \pi \varepsilon_{0} a}
\end{gathered}
\)

where Appendix integral A. 8 and $a=4 \pi \varepsilon{0} \hbar^{2} / \mu e^{2}$ were used. From (6.94), $e^{2} / 8 \pi \varepsilon{0} a$ is minus the ground-state H -atom energy, and (6.108) gives $\langle T\rangle=13.598 \mathrm{eV}$. (See also Sec. 14.4.)

EXERCISE Find $\langle T\rangle$ for the hydrogen-atom $2 p{0}$ state using (6.113).
(Answer: $e^{2} / 32 \pi \varepsilon{0} a=(13.598 \mathrm{eV}) / 4=3.40 \mathrm{eV}$.)
Let us examine a significant property of the ground-state wave function (6.104). We have $r=\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}$. For points on the $x$ axis, where $y=0$ and $z=0$, we have $r=\left(x^{2}\right)^{1 / 2}=|x|$, and

\(
\begin{equation}
\psi_{100}(x, 0,0)=\pi^{-1 / 2}(Z / a)^{3 / 2} e^{-Z|x| / a} \tag{6.109}
\end{equation}
\)

Figure 6.7 shows how (6.109) varies along the $x$ axis. Although $\psi_{100}$ is continuous at the origin, the slope of the tangent to the curve is positive at the left of the origin but negative

FIGURE 6.7 Cusp in the hydrogen-atom groundstate wave function.
at its right. Thus $\partial \psi / \partial x$ is discontinuous at the origin. We say that the wave function has a cusp at the origin. The cusp is present because the potential energy $V=-Z e^{2} / 4 \pi \varepsilon_{0} r$ becomes infinite at the origin. Recall the discontinuous slope of the particle-in-a-box wave functions at the walls of the box.

We denoted the hydrogen-atom bound-state wave functions by three subscripts that give the values of $n, l$, and $m$. In an alternative notation, the value of $l$ is indicated by a letter:

The letters $s, p, d, f$ are of spectroscopic origin, standing for sharp, principal, diffuse, and fundamental. After these we go alphabetically, except that $j$ is omitted. Preceding the code letter for $l$, we write the value of $n$. Thus the ground-state wave function $\psi{100}$ is called $\psi{1 s}$ or, more simply, $1 s$.

Wave Functions for $\boldsymbol{n}=\mathbf{2}$

For $n=2$, we have the states $\psi{200}, \psi{21-1}, \psi{210}$, and $\psi{211}$. We denote $\psi{200}$ as $\psi{2 s}$ or simply as $2 s$. To distinguish the three $2 p$ functions, we use a subscript giving the $m$ value and denote them as $2 p{1}, 2 p{0}$, and $2 p{-1}$. The radial factor in the wave function depends on $n$ and $l$, but not on $m$, as can be seen from (6.100). Each of the three $2 p$ wave functions thus has the same radial factor. The $2 s$ and $2 p$ radial factors may be found in the usual way from (6.100) and (6.99), followed by normalization. The results are given in Table 6.1. Note that the exponential factor in the $n=2$ radial functions is not the same as in the $R{1 s}$ function. The complete wave function is found by multiplying the radial factor by the appropriate spherical harmonic. Using (6.101), Table 6.1, and Table 5.1, we have

\(
\begin{align}
2 s & =\frac{1}{\pi^{1 / 2}}\left(\frac{Z}{2 a}\right)^{3 / 2}\left(1-\frac{Z r}{2 a}\right) e^{-Z r / 2 a} \tag{6.111}\
2 p_{-1} & =\frac{1}{8 \pi^{1 / 2}}\left(\frac{Z}{a}\right)^{5 / 2} r e^{-Z r / 2 a} \sin \theta e^{-i \phi} \tag{6.112}
\end{align}
\)

TABLE 6.1 Radial Factors in the Hydrogenlike-Atom

Wave Functions\(
\begin{aligned}
R{1 s} & =2\left(\frac{Z}{a}\right)^{3 / 2} e^{-Z r / a} \
R{2 s} & =\frac{1}{\sqrt{2}}\left(\frac{Z}{a}\right)^{3 / 2}\left(1-\frac{Z r}{2 a}\right) e^{-Z r / 2 a} \
R{2 p} & =\frac{1}{2 \sqrt{6}}\left(\frac{Z}{a}\right)^{5 / 2} r e^{-Z r / 2 a} \
R{3 s} & =\frac{2}{3 \sqrt{3}}\left(\frac{Z}{a}\right)^{3 / 2}\left(1-\frac{2 Z r}{3 a}+\frac{2 Z^{2} r^{2}}{27 a^{2}}\right) e^{-Z r / 3 a} \
R{3 p} & =\frac{8}{27 \sqrt{6}}\left(\frac{Z}{a}\right)^{3 / 2}\left(\frac{Z r}{a}-\frac{Z^{2} r^{2}}{6 a^{2}}\right) e^{-Z r / 3 a} \
R{3 d} & =\frac{4}{81 \sqrt{30}}\left(\frac{Z}{a}\right)^{7 / 2} r^{2} e^{-Z r / 3 a}
\end{aligned}
\)

\(
\begin{align}
& 2 p{0}=\frac{1}{\pi^{1 / 2}}\left(\frac{Z}{2 a}\right)^{5 / 2} r e^{-Z r / 2 a} \cos \theta \tag{6.113}\
& 2 p{1}=\frac{1}{8 \pi^{1 / 2}}\left(\frac{Z}{a}\right)^{5 / 2} r e^{-Z r / 2 a} \sin \theta e^{i \phi} \tag{6.114}
\end{align}
\)

Table 6.1 lists some of the normalized radial factors in the hydrogenlike wave functions. Figure 6.8 graphs some of the radial functions. The $r^{l}$ factor makes the radial functions zero at $r=0$, except for $s$ states.

The Radial Distribution Function

The probability of finding the electron in the region of space where its coordinates lie in the ranges $r$ to $r+d r, \theta$ to $\theta+d \theta$, and $\phi$ to $\phi+d \phi$ is [Eq. (5.78)]

\(
\begin{equation}
|\psi|^{2} d \tau=\left[R{n l}(r)\right]^{2}\left|Y{l}^{m}(\theta, \phi)\right|^{2} r^{2} \sin \theta d r d \theta d \phi \tag{6.115}
\end{equation}
\)

We now ask: What is the probability of the electron having its radial coordinate between $r$ and $r+d r$ with no restriction on the values of $\theta$ and $\phi$ ? We are asking for the probability of finding the electron in a thin spherical shell centered at the origin, of inner radius $r$ and outer radius $r+d r$. We must thus add up the infinitesimal probabilities (6.115) for all

FIGURE 6.8 Graphs of the radial factor $R_{n l}(r)$ in the hydrogen-atom ( $Z=1$ ) wave functions. The same scale is used in all graphs. (In some texts, these functions are not properly drawn to scale.)
possible values of $\theta$ and $\phi$, keeping $r$ fixed. This amounts to integrating (6.115) over $\theta$ and $\phi$. Hence the probability of finding the electron between $r$ and $r+d r$ is

\(
\begin{equation}
\left[R{n l}(r)\right]^{2} r^{2} d r \int{0}^{2 \pi} \int{0}^{\pi}\left|Y{l}^{m}(\theta, \phi)\right|^{2} \sin \theta d \theta d \phi=\left[R_{n l}(r)\right]^{2} r^{2} d r \tag{6.116}
\end{equation}
\)

since the spherical harmonics are normalized:

\(
\begin{equation}
\int{0}^{2 \pi} \int{0}^{\pi}\left|Y_{l}^{m}(\theta, \phi)\right|^{2} \sin \theta d \theta d \phi=1 \tag{6.117}
\end{equation}
\)

as can be seen from (5.72) and (5.80). The function $R^{2}(r) r^{2}$, which determines the probability of finding the electron at a distance $r$ from the nucleus, is called the radial distribution function; see Fig. 6.9.

For the $1 s$ ground state of H , the probability density $|\psi|^{2}$ is from Eq. (6.104) equal to $e^{-2 r / a}$ times a constant, and so $\left|\psi{1 s}\right|^{2}$ is a maximum at $r=0$ (see Fig. 6.14). However, the radial distribution function $\left[R{1 s}(r)\right]^{2} r^{2}$ is zero at the origin and is a maximum at $r=a$ (Fig. 6.9). These two facts are not contradictory. The probability density $|\psi|^{2}$ is proportional to the probability of finding the electron in an infinitesimal box of volume $d x d y d z$, and this probability is a maximum at the nucleus. The radial distribution function is proportional to the probability of finding the electron in a thin spherical shell of inner and outer radii $r$ and $r+d r$, and this probability is a maximum at $r=a$. Since $\psi{1 s}$ depends only on $r$, the $1 s$ probability density is essentially constant in the thin spherical shell. If we imagine the thin shell divided up into a huge number of infinitesimal boxes each of volume $d x d y d z$, we can sum up the probabilities $\left|\psi{1 s}\right|^{2} d x d y d z$ of being in

FIGURE 6.9 Plots of the radial distribution function $\left[R_{n \prime}(r)\right]^{2} r^{2}$ for the hydrogen atom.

each tiny box in the thin shell to get the probability of finding the electron in the thin shell as being $\left|\psi{1 s}\right|^{2} V{\text {shell }}$. The volume $V_{\text {shell }}$ of the thin shell is

\(
\frac{4}{3} \pi(r+d r)^{3}-\frac{4}{3} \pi r^{3}=4 \pi r^{2} d r
\)

where terms in $(d r)^{2}$ and $(d r)^{3}$ are negligible compared with the $d r$ term. Therefore the probability of being in the thin shell is

\(
\left|\psi{1 s}\right|^{2} V{\text {shell }}=R{1 s}^{2}\left(Y{0}^{0}\right)^{2} 4 \pi r^{2} d r=R{1 s}^{2}\left[(4 \pi)^{-1 / 2}\right]^{2} 4 \pi r^{2} d r=R{1 s}^{2} r^{2} d r
\)

in agreement with (6.116). The $1 s$ radial distribution function is zero at $r=0$ because the volume $4 \pi r^{2} d r$ of the thin spherical shell becomes zero as $r$ goes to zero. As $r$ increases from zero, the probability density $\left|\psi{1 s}\right|^{2}$ decreases and the volume $4 \pi r^{2} d r$ of the thin shell increases. Their product $\left|\psi{1 s}\right|^{2} 4 \pi r^{2} d r$ is a maximum at $r=a$.

EXAMPLE

Find the probability that the electron in the ground-state H atom is less than a distance $a$ from the nucleus.

We want the probability that the radial coordinate lies between 0 and $a$. This is found by taking the infinitesimal probability (6.116) of being between $r$ and $r+d r$ and summing it over the range from 0 to $a$. This sum of infinitesimal quantities is the definite integral

\(
\begin{aligned}
\int{0}^{a} R{n l}^{2} r^{2} d r & =\frac{4}{a^{3}} \int{0}^{a} e^{-2 r / a} r^{2} d r=\left.\frac{4}{a^{3}} e^{-2 r / a}\left(-\frac{r^{2} a}{2}-\frac{2 r a^{2}}{4}-\frac{2 a^{3}}{8}\right)\right|{0} ^{a} \
& =4\left[e^{-2}(-5 / 4)-(-1 / 4)\right]=0.323
\end{aligned}
\)

where $R{10}$ was taken from Table 6.1 and the Appendix integral A. 7 was used.
EXERCISE Find the probability that the electron in a $2 p{1} \mathrm{H}$ atom is less than a distance $a$ from the nucleus. Use a table of integrals or the website integrals.wolfram.com.
(Answer: 0.00366.)

Real Hydrogenlike Functions

The factor $e^{i m \phi}$ makes the spherical harmonics complex, except when $m=0$. Instead of working with complex wave functions such as (6.112) and (6.114), chemists often use real hydrogenlike wave functions formed by taking linear combinations of the complex functions. The justification for this procedure is given by the theorem of Section 3.6: Any linear combination of eigenfunctions of a degenerate energy level is an eigenfunction of the Hamiltonian with the same eigenvalue. Since the energy of the hydrogen atom does not depend on $m$, the $2 p{1}$ and $2 p{-1}$ states belong to a degenerate energy level. Any linear combination of them is an eigenfunction of the Hamiltonian with the same energy eigenvalue.

One way to combine these two functions to obtain a real function is

\(
\begin{equation}
2 p{x} \equiv \frac{1}{\sqrt{2}}\left(2 p{-1}+2 p_{1}\right)=\frac{1}{4 \sqrt{2 \pi}}\left(\frac{Z}{a}\right)^{5 / 2} r e^{-Z r / 2 a} \sin \theta \cos \phi \tag{6.118}
\end{equation}
\)

where we used (6.112), (6.114), and $e^{ \pm i \phi}=\cos \phi \pm i \sin \phi$. The $1 / \sqrt{2}$ factor normalizes $2 p_{x}$ :

\(
\begin{aligned}
\int\left|2 p{x}\right|^{2} d \tau & =\frac{1}{2}\left(\int\left|2 p{-1}\right|^{2} d \tau+\int\left|2 p{1}\right|^{2} d \tau+\int\left(2 p{-1}\right) 2 p{1} d \tau+\int\left(2 p{1}\right) 2 p_{-1} d \tau\right) \
& =\frac{1}{2}(1+1+0+0)=1
\end{aligned}
\)

Here we used the fact that $2 p{1}$ and $2 p{-1}$ are normalized and are orthogonal to each other, since

\(
\int{0}^{2 \pi}\left(e^{-i \phi}\right) * e^{i \phi} d \phi=\int{0}^{2 \pi} e^{2 i \phi} d \phi=0
\)

The designation $2 p_{x}$ for (6.118) becomes clearer if we note that (5.51) gives

\(
\begin{equation}
2 p_{x}=\frac{1}{4 \sqrt{2 \pi}}\left(\frac{Z}{a}\right)^{5 / 2} x e^{-Z r / 2 a} \tag{6.119}
\end{equation}
\)

A second way of combining the functions is

\(
\begin{gather}
2 p{y} \equiv \frac{1}{i \sqrt{2}}\left(2 p{1}-2 p{-1}\right)=\frac{1}{4 \sqrt{2 \pi}}\left(\frac{Z}{a}\right)^{5 / 2} r \sin \theta \sin \phi e^{-Z r / 2 a} \tag{6.120}\
2 p{y}=\frac{1}{4 \sqrt{2 \pi}}\left(\frac{Z}{a}\right)^{5 / 2} y e^{-Z r / 2 a} \tag{6.121}
\end{gather}
\)

The function $2 p_{0}$ is real and is often denoted by

\(
\begin{equation}
2 p{0}=2 p{z}=\frac{1}{\sqrt{\pi}}\left(\frac{Z}{2 a}\right)^{5 / 2} z e^{-Z r / 2 a} \tag{6.122}
\end{equation}
\)

where capital $Z$ stands for the number of protons in the nucleus, and small $z$ is the $z$ coordinate of the electron. The functions $2 p{x}, 2 p{y}$, and $2 p{z}$ are mutually orthogonal (Prob. 6.42). Note that $2 p{z}$ is zero in the $x y$ plane, positive above this plane, and negative below it.

The functions $2 p{-1}$ and $2 p{1}$ are eigenfunctions of $\hat{L}^{2}$ with the same eigenvalue: $2 \hbar^{2}$. The reasoning of Section 3.6 shows that the linear combinations (6.118) and (6.120) are also eigenfunctions of $\hat{L}^{2}$ with eigenvalue $2 \hbar^{2}$. However, $2 p{-1}$ and $2 p{1}$ are eigenfunctions of $\hat{L}{z}$ with different eigenvalues: $-\hbar$ and $+\hbar$. Therefore, $2 p{x}$ and $2 p{y}$ are not eigenfunctions of $\hat{L}{z}$.

We can extend this procedure to construct real wave functions for higher states. Since $m$ ranges from $-l$ to $+l$, for each complex function containing the factor $e^{-i|m| \phi}$ there is a function with the same value of $n$ and $l$ but having the factor $e^{+i|m| \phi}$. Addition and subtraction of these functions gives two real functions, one with the factor $\cos (|m| \phi)$, the other with the factor $\sin (|m| \phi)$. Table 6.2 lists these real wave functions for the hydrogenlike atom. The subscripts on these functions come from similar considerations as for the $2 p{x}, 2 p{y}$, and $2 p{z}$ functions. For example, the $3 d{x y}$ function is proportional to $x y$ (Prob. 6.37).

The real hydrogenlike functions are derived from the complex functions by replacing $e^{i m \phi} /(2 \pi)^{1 / 2}$ with $\pi^{-1 / 2} \sin (|m| \phi)$ or $\pi^{-1 / 2} \cos (|m| \phi)$ for $m \neq 0$; for $m=0$ the $\phi$ factor is $1 /(2 \pi)^{1 / 2}$ for both real and complex functions.

In dealing with molecules, the real hydrogenlike orbitals are more useful than the complex ones. For example, we shall see in Section 15.5 that the real atomic orbitals $2 p{x}, 2 p{y}$, and $2 p{z}$ of the oxygen atom have the proper symmetry to be used in constructing a wave function for the $\mathrm{H}{2} \mathrm{O}$ molecule, whereas the complex $2 p$ orbitals do not.

TABLE 6.2 Real Hydrogenlike Wave Functions

$1 s=\frac{1}{\pi^{1 / 2}}\left(\frac{Z}{a}\right)^{3 / 2} e^{-Z r / a}$
$2 s=\frac{1}{4(2 \pi)^{1 / 2}}\left(\frac{Z}{a}\right)^{3 / 2}\left(2-\frac{Z r}{a}\right) e^{-Z r / 2 a}$
$2 p{z}=\frac{1}{4(2 \pi)^{1 / 2}}\left(\frac{Z}{a}\right)^{5 / 2} r e^{-Z r / 2 a} \cos \theta$
$2 p{x}=\frac{1}{4(2 \pi)^{1 / 2}}\left(\frac{Z}{a}\right)^{5 / 2} r e^{-Z r / 2 a} \sin \theta \cos \phi$
$2 p{y}=\frac{1}{4(2 \pi)^{1 / 2}}\left(\frac{Z}{a}\right)^{5 / 2} r e^{-Z r / 2 a} \sin \theta \sin \phi$
$3 s=\frac{1}{81(3 \pi)^{1 / 2}}\left(\frac{Z}{a}\right)^{3 / 2}\left(27-18 \frac{Z r}{a}+2 \frac{Z^{2} r^{2}}{a^{2}}\right) e^{-Z r / 3 a}$
$3 p{z}=\frac{2^{1 / 2}}{81 \pi^{1 / 2}}\left(\frac{Z}{a}\right)^{5 / 2}\left(6-\frac{Z r}{a}\right) r e^{-Z r / 3 a} \cos \theta$
$3 p{x}=\frac{2^{1 / 2}}{81 \pi^{1 / 2}}\left(\frac{Z}{a}\right)^{5 / 2}\left(6-\frac{Z r}{a}\right) r e^{-Z r / 3 a} \sin \theta \cos \phi$
$3 p{y}=\frac{2^{1 / 2}}{81 \pi^{1 / 2}}\left(\frac{Z}{a}\right)^{5 / 2}\left(6-\frac{Z r}{a}\right) r e^{-Z r / 3 a} \sin \theta \sin \phi$
$3 d{z^{2}}=\frac{1}{81(6 \pi)^{1 / 2}}\left(\frac{Z}{a}\right)^{7 / 2} r^{2} e^{-Z r / 3 a}\left(3 \cos ^{2} \theta-1\right)$
$3 d{x z}=\frac{2^{1 / 2}}{81 \pi^{1 / 2}}\left(\frac{Z}{a}\right)^{7 / 2} r^{2} e^{-Z r / 3 a} \sin \theta \cos \theta \cos \phi$
$3 d{y z}=\frac{2^{1 / 2}}{81 \pi^{1 / 2}}\left(\frac{Z}{a}\right)^{7 / 2} r^{2} e^{-Z r / 3 a} \sin \theta \cos \theta \sin \phi$
$3 d{x^{2}-y^{2}}=\frac{1}{81(2 \pi)^{1 / 2}}\left(\frac{Z}{a}\right)^{7 / 2} r^{2} e^{-Z r / 3 a} \sin ^{2} \theta \cos 2 \phi$
$3 d_{x y}=\frac{1}{81(2 \pi)^{1 / 2}}\left(\frac{Z}{a}\right)^{7 / 2} r^{2} e^{-Z r / 3 a} \sin ^{2} \theta \sin 2 \phi$