# Lecture 18: Orbital Angular Momentum, Spectroscopy and Multi-Electron Atoms

Recap of Lecture 16

Last lecture continues our discussion of the hydrogen atom. We started the lecture with the expression for the energy of electrons in the hydrogen atom and emphasize that while there are three quantum numbers in the solutions to the corresponding Schrodinger equation, that the energy only is a function of $$n$$. We continued our discussion of the radial component of the wavefunctions as a product of four terms that crudely results in an exponentially decaying amplitude as a function of distance from the nucleus scaled by a pair of polynomials (with the Leguerre polynomial as one). There is also a normalization constant to make the interpretation proper. We discussed the volume and shell element in spherical space and introduce the radial distribution function $$4\pi r^2 \psi^2$$ that quantifies the probability of finding the electron a specific radius (technically between two radii).

## The $$l$$ Quantum number characterizes Angular Momentum and Angular Component (Ignoring the Radial Component)

The angular momentum vector for a classical model of the atom. Image used with permission (CC BY-SA; OpenStax).

As $$n$$ increases the average value of $$r$$ increases, which agrees with the fact that the energy of the electron also increases as $$n$$ increases. The increased energy results in the electron being on the average pulled further away from the attractive force of the nucleus. As in the simple example of an electron moving on a line, nodes (values of $$r$$ for which the electron density is zero) appear in the probability distributions. The number of nodes increases with increasing energy and equals $$n - 1$$.

An electron possesses orbital angular momentum is the density distributions is no longer spherical.

The quantum number $$l$$ governs the magnitude of the angular momentum, just as the quantum number $$n$$ determines the energy. The magnitude of the angular momentum may assume only those values given by:

$|L| = \sqrt{l(l+1)} \hbar \label{4}$

with $$l = 0, 1, 2, 3, ... n-1$$.

When the electron possesses angular momentum the density distributions are no longer spherical. In fact for each value of $$\ell$$, the electron density distribution assumes a characteristic shape Figure $$\PageIndex{6}$$.

For any value of n, a value of $$\ell=0$$ places that electron in an s orbital. This orbital is spherical in shape:

When $$\ell=1$$ these are designated as p orbitals and have dumbbell shapes. Each of the p orbitals has a different orientation in three-dimensional space.

For $$\ell=2$$, the $$m_l$$ values can be -2, -1, 0, +1, +2 for a total of five d orbitals. Note that all five of the orbitals have specific three-dimensional orientations.

The most complex set of orbitals are the f orbitals. When $$\ell=3$$, the $$m_l$$ values can be -3, -2, -1, 0, +1, +2, +3 for a total of seven different orbital shapes. Again, note the specific orientations of the different f orbitals.

Exercise $$\PageIndex{1}$$

• For a Hydrogen atom what is the degeneracy for a specific $$n$$ and $$l$$?
• Does that appear familiar?
• For a Hydrogen atom what is the degeneracy for a specific $$n$$?

## The $$m_l$$ Quantum Number and Magnetic Fields

The magnetic quantum number, designated by the letter $$m_l$$, is the third quantum numbers which describe the unique quantum state of an electron. The magnetic quantum number distinguishes the orbitals available within a subshell, and is used to calculate the azimuthal component of the orientation of the orbital in space. As with our discussion of rigid rotors, the quantum number $$m_l$$ refers to the projection of the angular momentum in this arbitrarily chosen direction, conventionally called the $$z$$ direction or quantization axis. $$L_z$$, the magnitude of the angular momentum in the z direction, is given by the formula

$L_z = m_l \hbar$

The quantum number $$m_l$$ refers, loosely, to the direction of the angular momentum vector. The magnetic quantum number $$m_l$$ only affects the electron's energy if it is in a magnetic field because in the absence of one, all spherical harmonics corresponding to the different arbitrary values of $$m_l$$ are equivalent.

## Zeeman Effect

Magnetism results from the circular motion of charged particles. This property is demonstrated on a macroscopic scale by making an electromagnet from a coil of wire and a battery. Electrons moving through the coil produce a magnetic field, which can be thought of as originating from a magnetic dipole or a bar magnet.

Figure $$\PageIndex{1}$$: Faraday’s apparatus for demonstrating that a magnetic field can produce a current. A change in the field produced by the top coil induces an emf and, hence, a current in the bottom coil. When the switch is opened and closed, the galvanometer registers currents in opposite directions. No current flows through the galvanometer when the switch remains closed or open. Image used with permission (CC BT 3.0; OpenStax).

Magnets are acted on by forces and torques when placed within an external applied magnetic field. In a uniform external field, a magnet experiences no net force, but a net torque. The torque tries to align the magnetic moment ($$\vec{\mu _m}$$ of the magnet with the external field $$\vec{B}$$. The magnetic moment of a magnet points from its south pole to its north pole.

A magnet will feel a force to realign in an external field, i.e. go from a higher energy to a lower energy. The energy of this system is determined by Equation $$ref{8.4.1}$$ and classical can vary since the angle between $$\vec{\mu _m}$$and $$\vec{B}$$ can vary continuously from 0 (low energy) to 180° (high energy).

The magnetic quantum number determines the energy shift of an atomic orbital due to an external magnetic field (this is called the Zeeman effect) — hence the name magnetic quantum number. However, the actual magnetic dipole moment of an electron in an atomic orbital arrives not only from the electron angular momentum, but also from the electron spin, expressed in the spin quantum number, which is the fourth quantum number. $$m_s$$ and discussed in the next chapter.

The orbiting electron with a non-zero $$l$$ value acts like a magnetic field with is no energetic difference for any particular orientation (only one energy state, on the left),. However, in external magnetic field there is a high-energy state and a low-energy state depending on the relative orientations of the magnet to the external field. Image used with permission (CC SA-BY 3.0; Darekk2).

Example $$\PageIndex{1}$$:

Electrons with non-zero angular momenta exhibit magnetic moments (depending on the wavefuntion the electron occupies). In the absence of an eternal magnetic field, different orientations of this magnetic field with respect to the will have the same energy, but will split. Assume the external magnetic field is oriented in the z-direction, predict the splitting patterns for an electron in the following orbitals:

• 1s orbital
• 2p orbitals (consider each orbital)
• 2s orbital
• 3d orbitals (consider each orbital)
• How do you think the splitting would be for a helium atom?

## Which $$m_l$$ number corresponds to which $$p$$-orbital?

The answer is complicated; while $$m_l=0$$ corresponds to the $$p_z$$, the orbitals for $$m_l=+1$$ and $$m_l=−1$$ lie in the xy-plane (see Spherical Harmonics), but not on the axes. The reason for this outcome is that the wavefunctions are usually formulated in spherical coordinates to make the math easier, but graphs in the Cartesian coordinates make more intuitive sense for humans. The $$p_x$$ and $$p_y$$ orbitals are constructed via a linear combination approach from radial and angular wavefunctions and converted into $$xy$$ (this was discussed previously). Thus, it is not possible to directly correlate the values of $$m_l=±1$$ with specific orbitals.

Note: We sometimes lie to students

The notion that we can directly correlate the values of $$m_l=±1$$ with specific orbitals is sometimes presented in introductory courses (not the Libretexts) to make a complex mathematical model just a little bit simpler and more intuitive, but it is incorrect.

The three wavefunctions for $$n=2$$ and $$l=1$$ are as follows (the spherical harmonics).

$|Ψ_{2,1,0} \rangle =r \cos θR(r)$

$|Ψ_{2,1,+1} \rangle =−\dfrac{r}{2} \sinθ e^{iϕ} R(r)$

$|Ψ_{2,1,-1} \rangle =+\dfrac{r}{2} \sinθ e^{-iϕ} R(r)$

The notation is $$|Ψ_{n,l,m_l} \rangle$$ with $$R(r)$$ is the radial component of this wavefuction, $$θ$$ is the angle with respect to the z-axis and $$ϕ$$ is the angle with respect to the $$xz$$-plane.

$R(r)=\sqrt{\dfrac{Z^5}{32\pi a_0^5}}\mathrm{e}^{-Zr/2a_0}$

in which $$Z$$ is the atomic number (or probably better nuclear charge) and $$a_0$$ is the Bohr radius.

In switching from spherical to Cartesian coordinates, we make the substitution $$z=r \cosθ$$, so:

$|Ψ_{2,1,0} \rangle =z R(r)$

This is $$Ψ_{2p_z}$$ since the value of $$Ψ$$ is dependent on $$z$$: when $$z=0$$; $$Ψ=0$$, which is expected since $$z=0$$ describes the $$xy$$-plane.

The other two wavefunctions are degenerate in the $$xy$$-plane. An equivalent statement is that these two orbitals do not lie on the x- and y-axes, but rather bisect them. Thus it is typical to take linear combinations of them to make the equation look prettier. If any set of wavefunctions is a solution to the Schrödinger equation, then any set of linear combinations of these wavefunctions must also be a solution. We can do this because of the linearity of the Schrödinger equation.

In the equations below, we're going to make use of some trigonometry, notably Euler's formula:

$\mathrm{e}^{\mathrm{i}\phi}=\cos{\phi}+\mathrm{i}\sin{\phi}$

$\sin{\phi} = \frac{\mathrm{e}^{\mathrm{i}\phi}-\mathrm{e}^{-\mathrm{i}\phi}}{2\mathrm{i}}$

$\cos{\phi} = \frac{\mathrm{e}^{\mathrm{i}\phi}+\mathrm{e}^{-\mathrm{i}\phi}}{2}$

We're also going to use $$x=\sin θ\cos ϕ$$ and $$y=\sin θ \sinϕ$$.

\begin{align} \psi_{2p_x}=\frac{1}{\sqrt{2}}\left(\psi_{2,1,+1}-\psi_{2,1,-1}\right)=\frac{1}{2}\left(\mathrm{e}^{\mathrm{i}\phi}+\mathrm{e}^{-\mathrm{i}\phi} \right)r\sin{\theta}f(r)=r\sin{\theta}\cos{\phi}f(r)=xf(r) \\ \psi_{2p_y}=\frac{\mathrm{i}}{\sqrt{2}}\left(\psi_{2,1,+1}+\psi_{2,1,-1}\right)=\frac{1}{2\mathrm{i}}\left(\mathrm{e}^{\mathrm{i}\phi}-\mathrm{e}^{-\mathrm{i}\phi} \right)r\sin{\theta}f(r)=r\sin{\theta}\sin{\phi}f(r)=yf(r)\\ \end{align}

So, while $$m_l=0$$ corresponds to $$|p_z \rangle$$, $$m_l=+1$$ and $$m_l=−1$$ cannot be directly assigned to $$|p_x \rangle$$ and $$|p_y \rangle$$.

Rather $$m_l=\pm 1$$ corresponds to {$$p_x \rangle$$, $$p_y \rangle$$}. An alternative discription is that $$m_l=+1$$ might correspond to $$(|p_x \rangle\ + |p_y \rangle )$$ and $$m_l=−1$$ might correspond to $$(|p_x \rangle\ - |p_y \rangle)$$.

## The Math is the same for Solving any Hydrogen-like (One electron) ion

The Hamiltonian for a one electron atom is given by is given by

$\hat{H}_{electron} = -\dfrac {\hbar^2}{2\mu} \bigtriangledown^2 - \dfrac {Ze^2}{4\pi \epsilon_0 r}\label{7}$

This only differs by the $$Z^2$$ component from the hydrogen atom.

Solving $$\hat{H}\psi_n = E_n \psi_n$$ gives $$E=-RZ^2/n^2$$.

For example with He+, the 2s orbital

$$E_{He_{2s}}$$ has $$z=2, n=2$$, thus $$E_{He_{2s}} = -R$$

## Spectroscopy of Hydrogen atoms

The solution of the Schrödinger equation for the hydrogen atom predicts that energy levels with $$n > 1$$ can have several orbitals with the same energy. In fact, as the energy and n increase, the degeneracy of the orbital energy level increases as well. The number of orbitals with a particular energy and value for $$n$$ is given by $$n_2$$. Thus, each orbital energy level is predicted to be $$n_2$$-degenerate. This high degree of orbital degeneracy is predicted only for one-electron systems. For multi-electron atoms, the electron-electron repulsion removes the $$l$$ degeneracy so only orbitals with the same $$m_l$$ quantum numbers are degenerate.

Energy levels predicted by the Bohr model. Electron transitions and their resulting wavelengths for hydrogen. Energy levels are not to scale. Image used with permission from Stephen Lower.

To understand the hydrogen atom spectrum, we also need to determine which transitions are allowed and which transitions are forbidden. This issue is addressed next by using selection rules that are obtained from the transition moment integral. In previous chapters we determined selection rules for the particle in a box, the harmonic oscillator, and the rigid rotor. Now we will apply those same principles to the hydrogen atom case by starting with the transition moment integral.

Transition requires a transfer from one state with its quantum numbers $$(n_i, l_i, m_i)$$ to another state $$(n_f, l_f, m_f)$$. The transition moment integral for a transition between an initial (i) state and a final (f) state of a hydrogen atom is given by

$\left \langle \mu _T \right \rangle = \int \psi ^* _{n_f, l_f, m_{l_f}} (r, \theta , \psi ) \hat {\mu} \psi _{n_i, l_i, m_{l_i}} (r, \theta , \psi ) d \tau \label {8.3.2a}$

or in bra ket notation

$\left \langle \mu _T \right \rangle = \langle \psi ^*_{n_f, l_f, m_{l_f}} | \hat {\mu} | \psi _{n_i, l_i, m_{l_i}} \rangle \label{8.3.2b}$

where the dipole moment operator is given by

$\hat {\mu} = - e \hat {r} \label {8.3.3}$

The dipole moment operator expressed in spherical coordinates is

$\hat {\mu} = -er (\bar {x} \sin \theta \cos \psi + \bar {y} \sin \theta \sin \psi + \bar {z} \cos \theta \label {8.3.4}$

The right hand side of Equation 8.3.4 shows that there are three components of $$\left \langle \mu _T \right \rangle$$ to evaluate in (8.3.2), where each component consists of three integrals: an $$r$$ integral, a $$\theta$$ integral, and a $$\psi$$ integral.

Evaluation reveals that the $$r$$ integral always differs from zero so

$\color{red} \Delta n = n_f - n_i = \text {not restricted} \label {8.3.5}$

There is no restriction on the change in the principal quantum number during a spectroscopic transition; $$\Delta n$$ can be anything. For absorption, $$\Delta n > 0$$, for emission $$\Delta n < 0$$, and $$\Delta n = 0$$ when the orbital degeneracy is removed by an external field or some other interaction.

The selection rules for $$\Delta l$$ and $$\Delta m_l$$ come from the transition moment integrals involving $$\theta$$ and $$\varphi$$ in Equation (8.3.2). These integrals are the same ones that were evaluated for the rotational selection rules, and the resulting selection rules are

$\color{red} \Delta l = \pm 1$

and

$\color{red} \Delta m_l = 0, \pm 1$

These rules demand conservation of angular momentum. Since a photon carries an intrinsic angular momentum of 1.

Figure 2: A Grotrian diagram of the hydrogen atom shows the allowed electronic transitions between the energy levels of atoms. Image used with permission from Tufts OCW (CC SA-BY; Gary Goldstein).

## One Electron ion (e.g., Helium+)

The Hamiltonian for a one electron atom is given by is given by

$\hat{H}_{electron} = -\dfrac {\hbar^2}{2\mu} \bigtriangledown^2 - \dfrac {Ze^2}{4\pi \epsilon_0 r}$

Solving

$\hat{H} |\psi_n \rangle = E_n | \psi_n \rangle$

gives energies of

$$E_n=-R\dfrac{Z^2}{n^2}$$.

Which as with the hydrogen atom are degenerate with repsect to the $$l$$ and $$m_l$$ quantum numbers.

For example with an electron in the 2s orbital of $$He^+$$ with $$Z=2, \;n=2$$, has an energy:

$E_{He_{2s}} = -R$

but the electron in the lowest eigenstate with $$Z=2, \;n=1$$ is

$E_{He_{2s}} = -4R$

## The Eigenvalue Problem for the Helium Atom

Step 1: Define the potential for the problem

$V(r_1,r_2) = \underset{\text{nucleaus-electron 1 attraction}}{- \dfrac {Ze^2}{4\pi\epsilon_0 r_1}} - \underset{\text{nucleaus-electron 2 attraction}}{\dfrac {Ze^2}{4\pi\epsilon_0 r_2}} + \underset{\text{electron-electron repulsion}}{\dfrac {e^2}{4\pi \epsilon_0 r_{12}}}$

where

• $$r_1$$ and $$r_2$$ are the ditances of electron 1 and electron 2 from the nucleus
• $$r_{12}$$ is the distance between the two electrons
• $$Z$$ is the proton number (i.e., charge of the nucleas), which for Helium is $$Z=2$$

Step 2: Define the Schrödinger Equation for the problem

With every quantum eigenvalue problem, we define the Hamiltonian as such:

$\hat {H} = T + V$

The potential is defined above and the Kinetic energy of each electron is given by

$T = -\dfrac {\hbar^2}{2m_e} \bigtriangledown^2$

For a 2 electron atom, the Hamiltonian should consist of the Hamiltonian for two electrons, and an electron repulsion factor, given by

$\dfrac {e^2}{4\pi\epsilon_0 r_{12}}$

The $$r_{12}$$ term is the distance from electron 1 to electron 2, which is incredibly hard to solve for. The Hamiltonian for the Helium atom is:

$\hat{H} = -\dfrac{\hbar^2}{2m_e}\bigtriangledown_{el_{1}}^2 -\dfrac{\hbar^2}{2m_e}\bigtriangledown_{el_{2}}^2 - \dfrac {Ze^2}{4\pi\epsilon_0 r_1} - \dfrac {Ze^2}{4\pi\epsilon_0 r_2} + \dfrac {e^2}{4\pi \epsilon_0 r_{12}}\label{8}$

Since the potential $$V(r_1,r_2)$$ has no time-dependence, we can se the time independent Schrödinger Equation to "solve" the QM problem:

$\hat {H} | \psi (r_1,r_2) \rangle = E | \psi (r_1,r_2) \rangle$

Step 3: Solve the Schrödinger Equation for the problem

Multi-electron systems are "three (or more) body problems", which we cannot solve analytically. However, we can introduce approximations.

Step 4: Do something with the Eigenstates and associated energies

Let's talk about the solutions and approximations first.

## Approximations

### The "ignorance is bliss" approximation

Ignore the $$\dfrac {e^2}{4\pi\epsilon_0 r_{12}}$$ electron-electron repulsion term. This assume electron 1 is separable from electron 2, so that

$\psi_{total} = \psi_{el_{1}}\psi_{el_{2}}$

or in braket notation

$| \psi_{total} \rangle = \hat{H} | \psi_{el_1} \rangle | \psi_{el_2} \rangle$

With some operator algebra, something important arises - the one electron energies are additive:

$\hat{H} \psi_{total} = (\hat{H}_{el_1} + \hat{H}_{el_2}) \psi_{n\ {el_1}} \psi_{n\ {el_2}} = (E_{n_1} + E_{n_2}) \psi_{n\ {el_1}} \psi_{n\ {el_2}}$

or in bra-ket notation

$\hat{H} | \psi_{total} \rangle = \hat{H} | \psi_{el_1} \rangle | \psi_{el_2} \rangle = (E_{n_1} + E_{n_2}) | \psi_{1} \rangle | \psi_{1} \rangle$

The energy for a ground state Helium atom (both electrons in lowest state) is then

$E_{He_{1s}} = \underset{\text{energy of single electron in helium}}{E_{n_1}} + \underset{\text{energy of single electron in helium}}{E_{n_2}} = -R\left(\dfrac{Z^2}{1}\right) -R \left(\dfrac{Z^2}{1}\right) = -8R$

where $$R$$ is the Rydberg constant ($$13.6 \; eV$$ that also maps the lowest energy of the hydrogen atom and $$Z=2$$ for the helium nucleus. Experimentally, we find that the total energy for a ground state Helium atom $$E_{He_{1s}} = -5.8066\,R$$. The big difference in results is due to the approximation we used to get to the value of $$-8\;R$$. The "ignorance is bliss" approximation overestimates the energy of the helium atom greatly; this is a poor approximation and we need to address electron-electron repulsion properly (or better at least).