Chapter 7: The Hydrogen Atom

7.1Application of the Schrödinger equation to the hydrogen atom

We met the hydrogen line spectra in Chapter 3, the Bohr model in Chapter 4, and the Schrödinger equation in Chapter 6. Now we apply the Schrödinger equation to hydrogen. To a good approximation the potential energy of the electron–proton system is electrostatic,

\[ V(r) = -\frac{e^2}{4\pi\epsilon_0r}, \]

and the three-dimensional time-independent Schrödinger equation is

\[ -\frac{\hbar^2}{2\mu}\frac{1}{\psi(x,y,z)}\Big[\frac{\partial^2\psi}{\partial x^2}+\frac{\partial^2\psi}{\partial y^2}+\frac{\partial^2\psi}{\partial z^2}\Big] = E - V(r), \]

with the reduced mass of the proton–electron system, \(\mu_e = m_eM/(m_e+M) = m_e/(1+m_e/M)\approx0.999456\,m_e\). Other hydrogen-like (single-electron) atoms such as He\(^+\) and Li\(^{2+}\) are handled by inserting the appropriate reduced mass and replacing \(e^2\) with \(Ze^2\). Because \(V(r)\) depends only on the distance \(r\) between proton and electron, the radial symmetry suggests spherical polar coordinates \((r,\theta,\phi)\), with \(x = r\sin\theta\cos\phi\), \(y = r\sin\theta\sin\phi\), \(z = r\cos\theta\), the polar angle \(\theta = \cos^{-1}(z/r)\), and the azimuthal angle \(\phi = \tan^{-1}(y/x)\). The Laplacian becomes

\[ \nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\Big(r^2\frac{\partial}{\partial r}\Big) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\Big(\sin\theta\frac{\partial}{\partial\theta}\Big) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\phi^2}, \]

\[ \begin{aligned} &\frac{1}{r^2}\frac{\partial}{\partial r}\Big(r^2\frac{\partial\psi}{\partial r}\Big) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\Big(\sin\theta\frac{\partial\psi}{\partial\theta}\Big)\\ &\qquad + \frac{1}{r^2\sin^2\theta}\frac{\partial^2\psi}{\partial\phi^2} + \frac{2\mu}{\hbar^2}(E-V)\psi = 0. \end{aligned} \]

7.2Solution of the Schrödinger equation for hydrogen

Try a separable solution \(\psi(r,\theta,\phi) = R(r)f(\theta)g(\phi)\). Then \(\partial\psi/\partial r = fg\,dR/dr\), \(\partial\psi/\partial\theta = Rg\,df/d\theta\), \(\partial\psi/\partial\phi = Rf\,dg/d\phi\). Substituting and multiplying by \(r^2\sin^2\theta/Rfg\),

\[ \begin{aligned} -\frac{\sin^2\theta}{R}\frac{d}{dr}\Big(r^2\frac{dR}{dr}\Big) &- \frac{2\mu}{\hbar^2}r^2\sin^2\theta\,(E-V)\\ &- \frac{\sin\theta}{f}\frac{d}{d\theta}\Big(\sin\theta\frac{df}{d\theta}\Big) = \frac1g\frac{d^2g}{d\phi^2}. \end{aligned} \]

The left side depends only on \(r\) and \(\theta\), the right only on \(\phi\), so both equal a constant, set to \(-m_\ell^2\). The azimuthal equation \(d^2g/d\phi^2 = -m_\ell^2g\) has solutions \(g(\phi)\propto e^{\pm im_\ell\phi}\). A physically valid solution must be periodic, \(g(\phi+2\pi) = g(\phi)\), so \(m_\ell = 0,\pm1,\pm2,\dots\). (If \(d^2g/d\phi^2>0\), \(g\) could not be periodic in \(\phi\).) Dividing the rest by \(\sin^2\theta\) and rearranging,

\[ \frac1R\frac{d}{dr}\Big(r^2\frac{dR}{dr}\Big) + \frac{2\mu r^2}{\hbar^2}(E-V) = \frac{m_\ell^2}{\sin^2\theta} - \frac{1}{f\sin\theta}\frac{d}{d\theta}\Big(\sin\theta\frac{df}{d\theta}\Big). \]

The left side depends only on \(r\) (\(V\) is the Coulomb potential), the right only on \(\theta\), so both equal a constant, set to \(\ell(\ell+1)\). This gives two ordinary differential equations:

\[ \begin{gathered} \text{radial:}\quad\frac{1}{r^2}\frac{d}{dr}\Big(r^2\frac{dR}{dr}\Big) + \frac{2\mu}{\hbar^2}\Big[E - V - \frac{\hbar^2}{2\mu}\frac{\ell(\ell+1)}{r^2}\Big]R = 0,\\ \text{angular:}\quad\frac{1}{\sin\theta}\frac{d}{d\theta}\Big(\sin\theta\frac{df}{d\theta}\Big) + \Big[\ell(\ell+1) - \frac{m_\ell^2}{\sin^2\theta}\Big]f = 0. \end{gathered} \]

The radial equation is the associated Laguerre equation; its solutions satisfying the boundary conditions are the associated Laguerre functions. For the ground state, \(\ell = 0\), it reads \(\frac{d^2R}{dr^2} + \frac2r\frac{dR}{dr} + \frac{2\mu}{\hbar^2}\big(E + \frac{e^2}{4\pi\epsilon_0r}\big)R = 0\). Trying \(R = Ae^{-r/a_0}\), with \(dR/dr = -R/a_0\) and \(d^2R/dr^2 = R/a_0^2\),

\[ \Big(\frac{1}{a_0^2} + \frac{2\mu}{\hbar^2}E\Big) + \Big(\frac{2\mu e^2}{4\pi\epsilon_0\hbar^2} - \frac{2}{a_0}\Big)\frac1r = 0. \]

For this to hold at every \(r\), both brackets must vanish: \(a_0 = 4\pi\epsilon_0\hbar^2/\mu e^2\), the Bohr radius, and \(E = -\hbar^2/2\mu a_0^2 = -E_0 = -13.6\) eV, the ground-state energy. In general the energy levels are

\[ E_n = -\frac{\mu}{2}\Big(\frac{e^2}{4\pi\epsilon_0\hbar}\Big)^2\frac{1}{n^2} = -\frac{E_0}{n^2}, \]

in agreement with Bohr. The angular equation is the associated Legendre equation, which requires \(\ell = 0,1,2,\dots\) and \(m_\ell = -\ell,-\ell+1,\dots,0,\dots,\ell-1,\ell\), i.e. \(|m_\ell|\le\ell\). The angular parts combine into the spherical harmonics \(Y_{\ell m_\ell}(\theta,\phi) = f(\theta)g(\phi)\), for example \(Y_{00} = 1/2\sqrt\pi\), \(Y_{10} = \tfrac12\sqrt{3/\pi}\cos\theta\), \(Y_{1\pm1} = \mp\tfrac12\sqrt{3/2\pi}\sin\theta\,e^{\pm i\phi}\). Example 7.1 verifies that \(Y_{11}\propto\sin\theta\) satisfies the angular equation with \(\ell = m_\ell = 1\). The probability density is \(\psi^*\psi\), and the full wave function is

\[ \psi_{n\ell m_\ell}(r,\theta,\phi) = R_{n\ell}(r)\,Y_{\ell m_\ell}(\theta,\phi), \]

where \(R_{n\ell}\) depends only on \(n\) and \(\ell\), and \(Y_{\ell m_\ell}\) only on \(\ell\) and \(m_\ell\). For example \(R_{10} = \frac{2}{a_0^{3/2}}e^{-r/a_0}\) and \(R_{21} = \frac{r}{a_0}\frac{e^{-r/2a_0}}{\sqrt3(2a_0)^{3/2}}\). Example 7.2: \(\psi_{211}\) is normalized: with the volume element \(d\tau = r^2\sin\theta\,dr\,d\theta\,d\phi\), \(\int\psi^*_{211}\psi_{211}\,d\tau = \frac{1}{64\pi a_0^5}[24a_0^5][\tfrac43][2\pi] = 1\).

7.3Quantum numbers

The three quantum numbers are \(n\), the principal quantum number; \(\ell\), the orbital angular momentum quantum number; and \(m_\ell\), the magnetic quantum number. The boundary conditions require

\[ \begin{gathered} n = 1,2,3,\dots;\\ \ell = 0,1,2,\dots,n-1;\\ m_\ell = -\ell,-\ell+1,\dots,0,\dots,\ell-1,\ell, \end{gathered} \]

all integers, i.e. \(n>0\), \(\ell\lt n\), \(|m_\ell|\le\ell\). It is customary to name the \(\ell\) values by letters, s, p, d, f, g, h, … for \(\ell = 0,1,2,3,4,5,\dots\) (the first four from early spectroscopy: sharp, principal, diffuse, fundamental). Thus \(n = 2\), \(\ell = 1\) is a 2p state; 1s, 2s, 4d, and 6g are other examples, but 2d is impossible. Conceptual Example 7.3: for \(n = 4\), \(\ell = 0,1,2,3\), and for each \(\ell\), \(m_\ell\) runs from \(-\ell\) to \(+\ell\).

The role of \(n\). The quantized energy levels depend only on \(n\), \(E_n = -E_0/n^2\). The role of \(\ell\). It is associated with \(R(r)\) and \(f(\theta)\) and with the orbital angular momentum \(\vec L = \vec r\times\vec p\), of magnitude \(L = mv_{\rm orbital}r\) classically, quantized as

\[ L = \sqrt{\ell(\ell+1)}\,\hbar, \]

not \(n\hbar\) as in Bohr's model. For \(n = 3\), \(\ell = 0,1,2\): the \(n = 3\) level is degenerate with respect to \(\ell\), since \(E_n\) does not depend on \(\ell\). The role of \(m_\ell\). The angle \(\phi\) measures rotation about the \(z\) axis, and

\[ L_z = m_\ell\hbar. \]

The choice of the \(z\) axis is arbitrary unless an external magnetic field defines a preferred direction, which is why \(m_\ell\) is called the magnetic quantum number. For \(\ell = 2\), \(m_\ell = -2,-1,0,1,2\): only certain orientations of \(\vec L\) are allowed, space quantization. Quantum mechanics allows \(\vec L\) to be quantized along only one direction. If all three components were known, the direction of \(\vec L\) would be known, the electron's motion would be confined to the plane perpendicular to it, and its position and momentum components along \(\vec L\) would both be exactly zero, forbidden by the uncertainty principle. Only \(|\vec L|\) and \(L_z\) may be specified simultaneously ("sharp"); \(L_x\) and \(L_y\) are "fuzzy". Since \(L = \sqrt{\ell(\ell+1)}\hbar>|L_z|_{\max} = \ell\hbar\), \(\vec L\) never points exactly along \(z\).

The \(2\ell+1\) values of \(m_\ell\) are an experimental fact. Since nothing distinguishes \(x\), \(y\), and \(z\), \(\langle L_x^2\rangle = \langle L_y^2\rangle = \langle L_z^2\rangle\), so (an argument due to Feynman, simplified by Milonni)

\[ \langle L^2\rangle = 3\langle L_z^2\rangle = \frac{3}{2\ell+1}\sum_{m_\ell=-\ell}^{\ell}m_\ell^2\hbar^2 = \ell(\ell+1)\hbar^2, \]

using \(\sum_{m_\ell=-\ell}^{\ell}m_\ell^2 = 2\sum_{m_\ell=1}^{\ell}m_\ell^2 = \ell(\ell+1)(2\ell+1)/3\).

Space quantization of \(\vec L\)

\(\ell\)

Left: the \(2\ell+1\) allowed values of \(L_z = m_\ell\hbar\) and the vectors of length \(\sqrt{\ell(\ell+1)}\hbar\) that reach them (book's Figure 7.3); none points straight up. Right: the same vectors drawn in perspective on cones about \(z\), because \(L_x\) and \(L_y\) are undetermined, only \(|\vec L|\) and \(L_z\) are sharp. The readout checks Feynman's \(3\langle L_z^2\rangle = \ell(\ell+1)\hbar^2\).

Example 7.4: the \(n = 3\) level contains \(\ell = 0,1,2\) with \(1+3+5 = 9\) states, all with the same energy (in the absence of a magnetic field). In general the degeneracy is \(\sum_{\ell=0}^{n-1}(2\ell+1) = n^2\) (Problem 16).

Counting the states of level \(n\)

\(n\)

7.4Magnetic effects on atomic spectra: the normal Zeeman effect

In 1896 Pieter Zeeman found that the spectral lines of atoms placed in a magnetic field broaden and appear to split: the Zeeman effect, the splitting of an energy level into multiple levels in an external magnetic field. A line split into three is the "normal" Zeeman effect; more than three lines is the anomalous Zeeman effect. Both can be understood by considering the atom as a small magnet. The orbiting electron is a current loop with magnetic moment \(\mu = IA\), where the current \(I = dq/dt = -e/T = -ev/2\pi r\) and the area is \(\pi r^2\):

\[ \mu = IA = -\frac{evr}{2} = -\frac{e}{2m}L,\qquad \vec\mu = -\frac{e}{2m}\vec L, \]

with \(L = mvr\); \(\vec\mu\) and \(\vec L\) point in opposite directions because of the negative charge. An external field \(\vec B\) exerts a torque \(\vec\tau = \vec\mu\times\vec B\), tending to align the dipole with the field, and the potential energy is \(V_B = -\vec\mu\cdot\vec B\). As with a spinning top in a gravitational field, \(\vec L\) precesses about \(\vec B\) (Example 7.5, the Larmor precession, \(\omega_L = eB/2m\)). With \(\vec B\) along \(z\),

\[ \begin{gathered} \mu_z = -\frac{e}{2m}L_z = -\frac{e\hbar}{2m}m_\ell = -\mu_Bm_\ell,\\ \mu_B = \frac{e\hbar}{2m},\\ V_B = -\mu_zB = +\mu_Bm_\ell B, \end{gathered} \]

where \(\mu_B\) is a unit of magnetic moment called the Bohr magneton (and \(L = \sqrt{\ell(\ell+1)}\hbar>m_\ell\hbar\)). The potential energy is quantized according to \(m_\ell\): each degenerate level of given \(\ell\) splits into \(2\ell+1\) energies. For \(\ell = 1\): \(E_0+\mu_BB\) (\(m_\ell = 1\)), \(E_0\) (\(m_\ell = 0\)), \(E_0-\mu_BB\) (\(m_\ell = -1\)). Conceptual Example 7.6: the lowest state with 5 \(m_\ell\) values is 3d (\(\ell = 2\) requires \(n\ge3\)). Example 7.7: \(\mu_B = (1.602\times10^{-19}\ \text{C})(1.055\times10^{-34}\ \text{J·s})/2(9.11\times10^{-31}\ \text{kg}) = 9.27\times10^{-24}\) J/T, so in a 2.00 T field the \(m_\ell = 0\) and \(+1\) components of the 2p state differ by \(\Delta E = \mu_BB\Delta m_\ell = 1.85\times10^{-23}\ \text{J} = 1.16\times10^{-4}\) eV, easily observed optically.

The normal Zeeman effect on the 2p → 1s line

Magnetic field \(B\) (T)

Levels (splitting exaggerated)

Lyman-α line, shift from 121.567 nm (pm)

In a field, the 2p level splits into three equally spaced sublevels, \(\Delta E = \mu_BB\), while 1s (\(\ell = 0\)) does not. The selection rule \(\Delta m_\ell = 0,\pm1\) (Section 7.6) allows three transitions, so the Lyman-α line becomes three lines separated by \(\Delta\lambda = \lambda^2\Delta E/hc\), only about 0.7 pm per tesla. (Spin, ignored here, makes real hydrogen lines show the anomalous Zeeman effect.)

In 1922 O. Stern and W. Gerlach looked for space quantization directly. In an inhomogeneous magnetic field, stronger near one pole than the other, a magnet feels a net force as well as a torque. For atoms in an \(\ell = 1\) state with the field along \(z\), \(V_B = -\mu_zB\) gives \(F_z = -dV_B/dz = \mu_z\,dB/dz\): different forces for the three \(m_\ell\) states, so a beam should split into three. Stern and Gerlach sent silver atoms through such a field and observed two distinct lines, not three. That was clear evidence of space quantization, but the number of \(m_\ell\) states, \(2\ell+1\), is always odd. Why two?

The Stern–Gerlach experiment

Atoms from an oven pass through a magnet whose field grows toward the upper pole, and land on the screen at right. Classical magnets with randomly oriented moments would smear into a continuous band; orbital angular momentum \(\ell = 1\) would give three spots (\(m_\ell = +1, 0, -1\)); silver atoms give two, because their magnetic moment comes from a single electron's spin, \(m_s = \pm\tfrac12\).

7.5Intrinsic spin

Space quantization and the two lines of the Stern–Gerlach experiment posed a problem. Wolfgang Pauli suggested that a fourth quantum number, after \(n\), \(\ell\), and \(m_\ell\), might explain the anomalous optical spectra. In 1925 Goudsmit and Uhlenbeck proposed that the electron has an intrinsic angular momentum and therefore, being charged, a magnetic moment. A classical picture of a rotating electron fails: if the electron's electrostatic energy equals its rest energy, \(e^2/4\pi\epsilon_0R = mc^2\), its radius is the classical electron radius \(R = 2.8\times10^{-15}\) m, and giving a spherical shell of that radius angular momentum \(\hbar/2\) requires an equatorial speed of about 100\(c\) (Problem 44). Spin must be a purely quantum-mechanical property.

The electron has an intrinsic spin quantum number \(s = \tfrac12\). By analogy with \(\ell\), there are \(2s+1 = 2\) components of the spin angular momentum \(\vec S\), so the magnetic spin quantum number has only two values, \(m_s = \pm\tfrac12\): the electron's spin is "up" or "down" in a magnetic field, never exactly along \(z\). The spin angular momentum has magnitude \(|\vec S| = \sqrt{s(s+1)}\hbar = \sqrt{3/4}\,\hbar\), with \(S_z = m_s\hbar = \pm\hbar/2\), and the magnetic moment is \(\vec\mu_s = -(e/m)\vec S\), or \(-2\mu_B\vec S/\hbar\). The coefficient of \(\vec S/\hbar\) is \(-2\mu_B\) rather than \(-\mu_B\) as for \(\vec L\), a consequence of relativity (the Dirac equation). This factor is the gyromagnetic ratio, \(g_\ell = 1\) and \(g_s = 2\):

\[ \vec\mu_\ell = -\frac{g_\ell\mu_B\vec L}{\hbar} = -\frac{\mu_B\vec L}{\hbar},\qquad \vec\mu_s = -\frac{g_s\mu_B\vec S}{\hbar} = -2\frac{\mu_B\vec S}{\hbar}. \]

Each state \((n,\ell,m_\ell)\) now becomes two distinct states, \(m_s = +\tfrac12\) and \(-\tfrac12\), degenerate in energy unless the atom is in a magnetic field, which splits them just as \(\Delta E = \mu_B\Delta m_\ell\) split the orbital levels. Now we understand Stern and Gerlach: their atoms were in a state with \(\ell = 0\), so there was no splitting due to \(m_\ell\), but the spin's space quantization still split the beam in two. The potential energy becomes \(V_B = -\vec\mu_s\cdot\vec B = +\frac em\vec S\cdot\vec B\). In the electron's own frame, the orbiting proton creates a magnetic field at the electron, so even without an external field the electron's energy depends on whether its spin is up or down: many levels are split into doublets. Dirac's relativistic quantum theory (1928) showed that the electron's intrinsic spin is required by relativity. Example 7.9: with spin, the degeneracy of level \(n\) is \(2n^2\); the 4d level (\(n = 4\), \(\ell = 2\)) has 5 values of \(m_\ell\), each with 2 values of \(m_s\), so 10 states.

The 21 cm line. The proton and electron in hydrogen both have spin, and when their magnetic moments are aligned the atom's energy is \(5.9\times10^{-6}\) eV higher than when they are opposed: a hyperfine structure. A spin flip from aligned to opposed emits a photon of 1420 MHz, wavelength 21.1 cm. Proposed by van de Hulst in 1944 and detected by Ewen and Purcell in 1951, this line is how radio astronomers map hydrogen across the galaxy.

7.6Energy levels and electron probabilities

In the absence of an external magnetic field, hydrogen's energy levels are degenerate with respect to \(\ell\), \(m_\ell\), and \(m_s\); a magnetic field removes the degeneracy. In heavier atoms with several electrons the degeneracy is removed even without a field, because the average potential energy from the nucleus plus electrons is non-Coulombic: for \(Z>1\), smaller \(\ell\) lie lower for a given \(n\), as in sodium and potassium, where \(E(4S)\lt E(4P)\lt E(4D)\lt E(4F)\) (Section 8.1). For hydrogen, the levels depend only on \(n\), as predicted by the Bohr theory. An atom in its ground state cannot emit radiation; it can absorb radiation or gain energy from inelastic collisions.

Selection rules. Transition probabilities computed from the wave functions show that electrons absorbing or emitting photons are much more likely to change states when \(\Delta\ell = \pm1\) (allowed transitions). Transitions with \(\Delta\ell\ne\pm1\) are possible but much less likely (forbidden transitions). There is no rule restricting \(\Delta n\); \(\Delta m_\ell = 0,\pm1\); and \(m_s\) can (but need not) change between \(\pm\tfrac12\):

\[ \Delta n = \text{anything},\qquad \Delta\ell = \pm1,\qquad \Delta m_\ell = 0,\pm1. \]

Thus 3P → 2P, 3D → 2S, and 3S → 1S do not occur, violating \(\Delta\ell = \pm1\). Conservation of angular momentum then implies that the photon carries one unit (\(\hbar\)) of angular momentum; quantizing Maxwell's equations confirms that electromagnetic radiation consists of photons with energy \(hf\) and intrinsic angular momentum \(\hbar\). Example 7.10: (a) \((2,0,0,\tfrac12)\to(3,1,1,\tfrac12)\): \(\Delta\ell = +1\), \(\Delta m_\ell = 1\), allowed, absorbing \(-13.6\ \text{eV}(\tfrac1{3^2}-\tfrac1{2^2}) = 1.89\) eV; (b) \((2,0,0,\tfrac12)\to(3,0,0,\tfrac12)\): \(\Delta\ell = 0\), not allowed; (c) \((4,2,-1,-\tfrac12)\to(2,1,0,\tfrac12)\): \(\Delta\ell = -1\), \(\Delta m_\ell = +1\), allowed (\(m_s\) does not matter), emitting 2.55 eV.

Allowed and forbidden transitions

From \(n\) \(\ell\) to \(n'\) \(\ell'\)

Hydrogen's levels arranged in columns by \(\ell\) (S, P, D, F), with energies \(-13.6\ \text{eV}/n^2\) (vertical scale compressed). Faint lines are all allowed transitions to the lower state you chose; the bold arrow is your transition, green if \(\Delta\ell = \pm1\) and red dashed if forbidden.

Probability distribution functions. The probability of finding the electron in a volume element is \(dP = \psi^*\psi\,d\tau\), with \(d\tau = r^2\sin\theta\,dr\,d\theta\,d\phi\). The \(\phi\) part contributes \((e^{im_\ell\phi})^*e^{im_\ell\phi} = 1\). If we are interested only in the radial part, and \(f\) and \(g\) are already normalized,

\[ \begin{gathered} P(r)\,dr = r^2|R(r)|^2\,dr\int_0^\pi|f(\theta)|^2\sin\theta\,d\theta\int_0^{2\pi}|g(\phi)|^2\,d\phi\\ \Longrightarrow\quad P_{n\ell}(r) = r^2|R_{n\ell}(r)|^2, \end{gathered} \]

the radial probability density. Example 7.11: setting \(dP/dr = 0\) gives the most probable radius \(r = a_0\) for 1s and \(r = 4a_0\) for 2p, the Bohr radii; this holds only for the largest possible \(\ell\) for each \(n\) (Problem 36). Example 7.12: the average radius of a 1s electron is \(\langle r\rangle = \int_0^\infty rP(r)\,dr = \tfrac32a_0\), larger than the most probable value because of the tail of the distribution. Example 7.13: the probability that a 1s electron is farther than \(a_0\) from the nucleus is \(5e^{-2} = 0.68\). Example 7.14: a 3d electron has \(\langle r\rangle = 10.5a_0\), more than the Bohr radius \(9a_0\), and a probability 0.606 of being beyond \(9a_0\).

The \(\ell = 0\) states are spherically symmetric, because their wave functions have no \(\theta\) or \(\phi\) dependence. For \(\ell>0\) the distributions depend on \(f(\theta)\). For \(\ell = 1\), \(m_\ell = 0\), \(Y_{10}\propto\cos\theta\), so \(\psi^*\psi\propto\cos^2\theta\) peaks along the \(\pm z\) axis; for \(m_\ell = \pm1\), \(\psi^*\psi\propto\sin^2\theta\), a flattened ring in the \(xy\) plane. For \(\ell = 2\), \(|Y_{20}|^2\) peaks at \(\theta = 0°\) and \(180°\), \(|Y_{2\pm2}|^2\propto\sin^4\theta\) in the \(xy\) plane, and \(|Y_{2\pm1}|^2\propto\sin^2\theta\cos^2\theta\) at 45° and 135°. Although \(\langle r\rangle\) is smaller for 2p than 2s, the 2s distribution has a secondary maximum close to the nucleus, so a 2s electron spends more time very close to it. There are still small corrections, such as spin-orbit coupling (Section 8.2).

Orbital explorer

\(n\) \(\ell\) \(m_\ell\)

\(|\psi|^2\) in the \(xz\) plane (\(z\) up)

Radial wave function \(R_{n\ell}(r)\)

Radial probability \(P_{n\ell}(r) = r^2R_{n\ell}^2\)

Computed from the associated Laguerre functions and spherical harmonics, normalized numerically, for any \(n\le6\). The image is a slice through the nucleus in the \(xz\) plane (brightness \(\propto\sqrt{|\psi|^2}\) to show the faint lobes); because \(|e^{im_\ell\phi}|^2 = 1\), every density is symmetric about \(z\), and the 3D shape is this slice rotated about the vertical axis. The readout gives the most probable radius, \(\langle r\rangle\) (exact: \(\tfrac{a_0}2[3n^2-\ell(\ell+1)]\)), and the probabilities of Examples 7.13 and 7.14.

This completes Chapter 7. Chapter 8 continues with spin-orbit coupling and many-electron atoms.

Chapter 7: The hydrogen atomquantum numbers, spin, and the shapes of orbitals