# 9: Electron spin and multi-electron atoms


## Electron spin

The quantum numbers $$n, \ l, \ m$$ are not sufficient to fully characterize the physical state of the electrons in an atom. In 1926, Otto Stern and Walther Gerlach carried out an experiment that could not be explained in terms of the three quantum numbers $$n, \ l, \ m$$ and showed that there is, in fact, another quantum-mechanical degree of freedom that needs to be included in the theory. The experiment is illustrated in the figure 9.1:

A beam of atoms (e.g. hydrogen or silver atoms) is sent through a spatially inhomogeneous magnetic field with a definite field gradient toward one of the poles. It is observed that the beam splits into two beams as it passes through the field region.

It is known that a current loop in a nonuniform magnetic field experiences a net force. This is illustrated in Figure 9.1:

This force arises from an energy $$E$$ given by

$E= -M\cdot B \tag{9.1}$

where $$M$$ is called the magnetic moment of the current loop. A current loop, as the figure suggests, is caused by circulating charges. A circulating particle has an associated angular moment $$L$$. If the charge of the particle is $$q$$ and its mass is $$m$$, then the magnetic moment is given by

$M=\dfrac{q}{2m} L\tag{9.2}$

The fact that the beam splits into 2 beams suggests that the electrons in the atoms have a degree of freedom capable of coupling to the magnetic field. That is, an electron has an intrinsic magnetic moment $$M$$ arising from a degree of freedom that has no classical analog. The magnetic moment must take on only 2 values according to the Stern-Gerlach experiment. The intrinsic property that gives rise to the magnetic moment must have some analog to an angular momentum and hence, must be a property that, unlike charge and mass, which are simple numbers, is a vector property. This property is called the spin, $$S$$, of the electron. As the expression above suggests, the intrinsic magnetic moment $$M$$ of the electron must be propertional to the spin

$M=\gamma S\tag{9.3}$

In quantum mechanics, spin share numerous features in common with angular momentum, which is why we represent it as a vector. In particular, spin is quantized, i.e. we have certain allowed values of spin. Like angular momentum, the value of the magnitude squared $$S^2$$ of spin is fixed, and one of its components $$S_z$$ is as well. For an electron, the allowed values of $$S_z$$ are

$S_z \rightarrow m_s \hbar \ ; \ m_s=-\dfrac{1}{2} ,\dfrac{1}{2}\tag{9.4}$

while $$S^2$$ has just one value $$3(\hbar)^2 /4$$, corresponding to a general formula $$s(s+1)(\hbar)^2$$, where $$s=1/2$$. For this reason, the electron is called a spin-1/2 particle. The formula $$s(s+1)(\hbar)^2$$ is generally valid for any spin-s particle. The constant $$\gamma$$ is called the spin gyromagnetic ratio

$\gamma =-\dfrac{ge}{2m_e} \tag{9.5}$

where $$g=2$$ for electrons. Note that $$\gamma <0$$.

For an electron in a uniform magnetic field $$B$$, the energy is determined by the spin $$S$$:

$E=-\gamma S\cdot B \tag{9.6}$

We choose the uniform field $$B$$ to be along the z-direction $$B=(0,0,B)$$. Since the field lines flow from the north pole to the south pole, this choice of the $$B$$ field means that the north pole lies below the south pole on the z-axis. In this case,

$E=-\gamma BS_z \tag{9.7}$

Since $$\gamma <0$$, we see that the lowest energy configuration uas $$S$$ antiparallel to $$B$$. Given the two values of $$S_z$$, we have two allowed energies or two energy levels corresponding to the two values of $$m_s$$:

$E_{1/2}=-\dfrac{\gamma B\hbar}{2} \ ; \ E_{-1/2}=\dfrac{\gamma B\hbar}{2} \tag{9.8}$

Using the given expression for $$\gamma$$, which is negative, we obtain the two energy levels

$E_{1/2}=\dfrac{eB\hbar}{2m_e} \ ; \ E_{-1/2}=-\dfrac{eB\hbar}{2m_e} \\tag{9.9}$

Unlike position and momentum, which have clear classical analogs, spin does not. But if we think of spin in pseudoclassical terms, we can think of a spinning charged particle, which is similar to a loop of current. Thus, if the particle spins about the z-axis, then $$S$$ points along the z-axis. Since the spinning charge is negative, the left-hand rule can be applied. When the fingers of the left hand curl in the direction of the spin, the thumb points in the direction of the spin. A spinning charge produces a magnetic field similar to that of a tiny bar magnet. In this case, the spin vector $$S$$ points toward the south pole of the bar magnet. Now if the spinning particle is placed in a magnetic field, it tends to align the spin vector in the opposite to the magnetic field lines, as the figure above suggests.

Now, when the electron is placed in a nonuniform magnetic field, with the field increasing in strength toward the north pole of the field source, the spin-down $$(m_s =-1/2)$$ electrons have their bar-magnet poles oriented such that the south pole points toward the north pole of the field source, and these electrons will be attracted toward the region of stronger field. The spin-up $$(m_s =1/2)$$ electrons have their bar-magnet north poles oriented toward the north pole of the field source and will be repelled to the region of weaker field, thus causing the beam to split as observed in the Stern-Gerlach experiment.

The implication of the Stern-Gerlach experiment is that we need to include a fourth quantum number, $$m_s$$ in our description of the physical state of the electron. That is, in addition to give its principle, angular, and magnetic quantum numbers, we also need to say if it is a spin-up electron or a spin-down electron.

Note that we have added spin into the our quantum theory as a kind of a posteriori consideration, which seems a little contrived. In fact, the existence of the spin degree of freedom can be derived in a very natural way in the relativistic version of quantum mechanics, where it simply pops out of the relativistic analog of the Schrödinger equation, known as the Dirac equation.

#### Spin Wavefunctions

So far, we have not explicitly considered the spin of the electrons. For the next type of approximation we will consider, the so-called valene bond approximation, we will need to consider spin explicitly. As we have already seen in the case of the hydrogen atom, the wavefunctions of an electron depend on the coordinates $$r=(x,y,z)$$ or $$(r,\phi ,\theta)$$ and on the z-component of spin spin $$S_z$$ (the total spin $$|S|$$ is fixed). While the coordinates can be anything, the spin $$S_z$$ can only take on two values $$\hbar /2$$ and $$-\hbar /2$$. Recall that these lead to the two values of the spin quantum number $$m_s=\pm 1/2$$. The value $$m_s=1/2$$ is what we call the spin-up'' state and $$m_s =-1/2$$ is what we call spin-down''.

We now define the spin wavefunctions. The spin-up wavefunction is denoted

$\psi_{1/2}(S_z)=\psi_{\uparrow}(S_z)$

Since $$S_z$$ can take on only two values $$\pm \hbar/2$$, the spin wavefunction only have two values:

$\psi_{\uparrow}(\hbar/2)=1 \ ; \ \psi_{\uparrow}(-\hbar/2)=0$

The meaning of this wavefunction is that when the electron is in the spin-up state, the probability that a measurement of $$S_z$$ will yield the value $$\hbar/2$$ is $$P(S_z =\hbar/2)=|\psi_{\uparrow}(\hbar/2)|^2 =1$$ and the probability that is value will be $$-\hbar/2$$ is $$0$$. Similarly, the spin-down wavefunction is
$\psi_{-1/2}(S_z)=\psi_{\downarrow}(S_z)$

where

$\psi_{\downarrow}(\hbar/2)=0 \ ; \ \psi_{\downarrow}(-\hbar/2)=1$

so that the probability that a measurement of $$S_z$$ yields the value $$-\hbar/2$$ is $$1$$ and that its value is $$\hbar/2$$ is $$0$$. Note that the spin wavefunctions are normalized, meaning that they satisfy

$\sum_{S_z =-\hbar/2}^{\hbar/2}|\psi_{m_s}(S_z)|^2 =1$

and they are orthogonal, meaning that they satisfy

$\sum_{S_z =-\hbar/2}^{\hbar/2}\psi_{\uparrow}(S_z)\psi_{\downarrow}(S_z)=0$
Now, for a hydrogen atom, there are four quantum numbers, $$n, \ l, \ m, \ m_s$$, and the wavefunction depends on four coordinates, $$r, \ \theta , \ \phi , \ s_z$$. While $$r, \ \theta , \ \phi$$ are continuous, $$s_z$$ takes on only two values. The wavefunction can be expressed as a simple product

$\psi_{nlmm_s}(r,\theta ,\phi ,s_z)=\psi_{nlm}(r,\theta ,\phi)\psi_{m_s}(s_z)$

As a shorthand notation, we can generally represent the complete set of spatial and spin coordinates with a vector $$x$$. The spatial coordinates can be $$r,\theta ,\phi$$ or $$x,y,z$$ or any other set of spatial coordinates useful for a given problem. For the hydrogen atom, using this notation, we would write

$\psi_{nlmm_s}(x)=\psi_{nlm}(r,\theta ,\phi)\psi_{m_s}(s_z)$

## Many-electron atoms

The hydrogen atom is the only atom for which exact solutions of the Schrödinger equation exist. For any atom that contains two or more electrons, no solution has yet been discovered (so no solution for the helium atom exists!) and we need to introduce approximation schemes.

Let us consider the helium atom. The nucleus has a charge of $$+2e$$, and if we place the nucleus at the origin, there will be an electron at a position $$r_1$$ with spin $$s_{z,1}$$ and an electron at position $$r_2$$ and spin $$s_{z,2}$$. As usual, we consider the nucleus to be fixed. The classical energy is then

$\dfrac{|p_1|^2}{2m_e}+\dfrac{|p_2|^2}{2m_e}+\dfrac{e^2}{4\pi\epsilon_0}\left [ \dfrac{1}{|r_1 -r_2|}-\dfrac{2}{|r_1|}-\dfrac{2}{|r_2|}\right ] =E$

Note that $$E$$ is not simply a sum of terms for electron 1 and electron 2, $$E\neq \varepsilon_1 +\varepsilon_2$$. Therefore, it is not possible to write the wavefunction $$\Psi(x_1 ,x_2)$$ as a simple product of the form $$\psi_1 (x_1) \psi_2(x_2)$$, nor is it even possible to use the special form we introduced for identical particles

$\dfrac{1}{\sqrt{2}}[\psi_1(x_1)\psi_2(x_2)-\psi_1(x_2)\psi_2(x_1)]$

because these simple products are not correct solutions to the Schrödinger equation.

This means that the wavefunction $$\Psi(x_1 ,x_2)$$ depends on the full set of 6 coordinates $$x_1 ,y_1 ,z_1 ,x_2 ,y_2 ,z_2$$ or $$r_1 ,\theta_1 ,\phi_1 ,r_2 ,\theta_2 ,\phi_2$$ if spherical coordinates are used, and 2 spin coordinates $$s_{z,1}$$, $$s_{z,2}$$, and that this dependence is not simple! In fact, as the number of electrons increases, the number of variables on which $$\Psi$$ depends increases as well. For an atom with $$M$$ electrons, the wavefunction $$\Psi$$ depends on $$3M$$ coordinates! Thus, it is clear that the wavefunction for a many-electron atom is a very unwieldy object!

As a side bar, we note that the 1998 Nobel prize in chemistry was awarded to Walter Kohn for the development of an extremely elegant theory of electronic structure known as density functional theory. In this theory, it is shown that the wavefunction $$\Psi$$, which depends on $$3M$$ coordinates, can be replaced by a much simpler object called the electron density denoted $$\rho (r)=\rho (x,y,z)$$. This object depends on only three variables for a system of any number of electrons. In density functional theory, it is shown that any physical quantity can be computed from this electron density $$\rho (r)$$.

Consider now an imaginary form of helium in which the two electrons do not interact. For this simplified case, the energy is simply

\begin{align*}E &= \dfrac{|p_1|^2}{2m_e}+\dfrac{|p_2|^2}{2m_e}-\dfrac{e^2}{4\pi \epsilon_0}\left [ \dfrac{2}{|r_1|}+\dfrac{2}{|r_2|}\right ] \\ &= \left [ \dfrac{|p_1|^2}{2m_e}-\dfrac{2e^2}{4\pi \epsilon_0 |r_1|}\right ]+\left [ \dfrac{|p_2|^2}{2m_e}-\dfrac{2e^2}{4\pi \epsilon_0 |r_2|}\right ] \\ &= \varepsilon_1 +\varepsilon_2 \end{align*}

For this imaginary helium atom, the wavefunction can be expressed as an antisymmetric product, and because we have included spin, in the ground state, both electrons can be in the 1s$$(Z=2)$$ spatial orbital without having the wavefunction vanish. Recall that, when $$Z=2$$, the 1s orbital is

$\psi_{100}(r,\theta ,\phi)=\dfrac{1}{\sqrt{\pi}}\left ( \dfrac{2}{a_0}\right )^{3/2}e^{-2r/a_0}\equiv \psi_{1s(Z=2)}(r)$

Multiplying this by a spin wavefunction, the wavefunction for one of the electrons is

$\psi (x)=\psi_{1s(Z=2)}(r)\psi_{m_s}(S_z)$

Given this form, the antisymmetrized two-electron wavefunction becomes

\begin{align*}\Psi (x_1 ,x_2) &= \dfrac{1}{\sqrt{2}}[\psi_{1s(Z=2)}(r_1)\psi_{\uparrow}(S_{z1})\psi_{1s(Z=2)}(r_2)\psi_{\downarrow}(S_{z2})-\psi_{1s(Z=2)}(r_2)\psi_{\uparrow}(S_{z2})\psi_{1s(Z=2)}(r_1)\psi_{\downarrow}(S_{z1})]\\ &= \dfrac{1}{\sqrt{2}}\psi_{1s(Z=2)}(r_1)\psi_{1s(Z=2)}(r_2)[\psi_{\uparrow}(S_{z1})\psi_{\downarrow}(S_{z2})-\psi_{\uparrow}(S_{z2})\psi_{\downarrow}(S_{z1})]\end{align*}

which, by virtue of the spin wavefunctions, does not vanish.

Given this wavefunction, the energy would just be the sum of the energies of two electrons interacting with a nucleus of charge $$+2e$$. We would need two quantum numbers $$n_1$$ and $$n_2$$ for this, and from our study of hydrogen-like atoms, the energy would be

$E_{n_1 n_2}=-\dfrac{4}{n_{1}^{2}}-\dfrac{4}{n_{2}^{2}}$

in Rydbergs. This comes from the fact that the energy of one electron interacting with a nucleus of charge $$+Ze$$ is $$E_n =-Z^2 /n^2$$ in Rydbergs. So, the ground-state energy $$(n_1 =1,n_2=1)$$ would be -8 Ry. In real helium, the electron-electron Coulomb repulsion $$e^2 /(4\pi \epsilon_0 r_{12})$$ increases the ground state energy above this value (the experimentally measured ground-state energy is -5.8 Ry). The tendency is for the electrons to arrange themselves such that the Coulomb repulsion is as small as possible.

## Hartree-Fock theory

For the helium atom, what happens if we try to use a product form as a guess'' wavefunction? Ignoring spin for the moment, we take as a guess of the solution for the ground-state wavefunction

$\Psi_g (r_1 ,r_2)=\dfrac{1}{\sqrt{2}}[\psi_1 (r_1)\psi_2 (r_2)-\psi_1 (r_2)\psi_2 (r_1)]$

Then we know from problem set #4 that the corresponding guess of the ground-state energy

$E_g=\int \Psi_g \hat{H}\Psi_g dV_1 dV_2 >E_0$

where $$E_0$$ is the true ground-state energy. Here, $$dV_1$$ and $$dV_2$$ are the volume elements of electron 1 and electron 2, respectively, and $$\hat{H}$$ is the Hamiltonian

$\hat{H}=\hat{K_1}+\hat{K_2}+\dfrac{e^2}{4\pi \epsilon_0}\left [ \dfrac{1}{|r_1 -r_2|}-\dfrac{2}{|r_1|}-\dfrac{2}{|r_2|} \right ]$

That is, we know that this guess is not correct, however, we can try to optimize the form of the functions $$\psi_1 (r)$$ and $$\psi_2 (r)$$ so as to make $$E_g$$ as small as possible, thereby making it approach $$E_0$$.

If we do this, we find that $$\psi_1 (r)$$ and $$\psi_2 (r)$$ satisfy a set of 2 coupled Schrödinger-like equations and that the potential energy $$V$$ in each equation corresponds to what the electron would experience if we averaged over all possible positions of the other electron (think back to the classical shell model!).

If we solve these equations, we find that both $$\psi_1 (r)$$ and $$\psi_2 (r)$$ bear a strikingly resemblance to the hydrogenic function $$\psi_{100}(r,\phi ,\theta)=\psi_{1s}(r,\theta ,\phi)$$, the 1s wavefunction. The difference, however, is that it decays a little faster, as if its $$Z$$ value were between 1 and 2. In fact, the value one obtains from an actual calculation is $$Z\approx 1.69$$. We will denote this as $$\psi_{1s(Z)}(r, \theta ,\phi)$$. We can imagine one of the electrons as occupying'' the state $$\psi_1 (r)$$ and the other as occupying'' the state $$\psi_2 (r)$$, but of course, we cannot say which electron is in which state, which is why we need the above form of the guess wavefunction. We also obtain two energies $$\varepsilon_1$$ and $$\varepsilon_2$$ that are close in energy but that the energies are lower than the energy $$E_1$$ of the hydrogen atom.

Now, we see the problem with ignoring spin. If $$\psi_1$$ and $$\psi_2$$ are both $$\psi_{1s(Z)}$$, then the guess wavefunction becomes

$\Psi_g (r_1 ,r_2)=\dfrac{1}{\sqrt{2}}[\psi_{1s(Z)} (r_1)\psi_{1s(Z)} (r_2)-\psi_{1s(Z)}(r_2)\psi_{1s(Z)}(r_1)]=0$

However, if we multiply the spatial wavefunctions by spin wavefunctions, then the two orbitals we obtain are

$\psi_1 (x)=\psi_{1s(Z)} (r)\psi_{\uparrow}(s_z) \ ; \ \psi_2 (x)=\psi_{1s(Z)}(r)\psi_{\downarrow}(s_z)$

Thus, if we substitute this into the expression for $$\Psi_g(x_1 ,x_2)$$, we find that we can factor out $$\psi_{1s(Z)}(r_1)\psi_{1s(Z)}(r_2)$$ and we are left with the antisymmetric combination of spin wavefunctions. The only possibility for the combination of spin wavefunctions leads to

$\Psi_g (x_1 ,x_2)=\dfrac{1}{\sqrt{2}}\psi_{1s(Z)}(r_1)\psi_{1s(Z)}(r_2)[\psi_{\uparrow}(s_{z,1})\psi_{\downarrow}(s_{z,2})-\psi_{\uparrow}(s_{z,2})\psi_{\downarrow}(s_{z,1})]$

Note that another possible spin wavefunction would be $$[\psi_{\uparrow}(s_{z,2})\psi_{\downarrow}(s_{z,1})-\psi_{\uparrow}(s_{z,1})\psi_{\downarrow}(s_{z,2})]/\sqrt{2}$$ however, this is different from the above wavefunction by an overall minus sign and is, therefore, not actually different.

As another shorthand notation, we can use symbols "1" and "2" to represent any of the variables of electrons 1 and 2. Thus, the guess wavefunctions can be represented simply as

\begin{align*}\psi_g (1,2) &= \dfrac{1}{\sqrt{2}}[\psi_1 (1)\psi_2 (2)-\psi_1 (2)\psi_2 (1)]\\ &=\dfrac{1}{\sqrt{2}}\psi_{1s(Z)}(1)\psi_{1s(Z)}(2)[\psi_{\uparrow}(1)\psi_{\downarrow}(2)-\psi_{\uparrow}(2)\psi_{\downarrow}(1)]\end{align*}

Given that the spin wavefunctions are rather simple, we can focus on the spatial parts of the wavefunctions in the proceeding discussion. For short, we denote the electronic configuration as $$1s^2$$.

The lowest HF energy obtained for helium, $$\varepsilon_1 \approx -2.38 \ Ry$$. Note how close this is to the formula $$-Z^2 /n^2$$ for $$n=1$$ with $$Z=1.69$$! The next energy $$\varepsilon_2$$ is only slightly large than this value, so we see that $$\varepsilon_1 \approx \varepsilon_2$$, but strictly $$\varepsilon_1 <\varepsilon_2$$. However, the energies are so close that they can be considered as degenerate. In fact, there is a class of Hartree-Fock calculations (called closed-shell'' calculations), in which this degeneracy is assumed at the outset, and the pair of electrons is treated as a single unit or single quantum state.

Now, if we followed the same thing for lithium, we would need three functions to construct the guess wavefunction, which we could take as

$\psi_1 (x)=\phi_1 (r)\psi_{\uparrow}(S_z) \ ; \ \psi_2 (x)=\phi_2 (r)\psi_{\downarrow}(S_z) \ ; \ \psi_3 =\phi_3 (r)\psi_{\uparrow}(S_z)$

where $$\phi_1 (r)$$, $$\phi_2 (r)$$, and $$\phi_3 (r)$$ are arbitrary spatial wavefunctions whose shape we optimize so as to make $$E_g$$ as close to the true ground state energy $$E_0$$ as possible. For lithium, the Hamiltonian is

$\hat{H}=\hat{K_1}+\hat{K_2}+\hat{K_3}+\dfrac{e^2}{4\pi \epsilon_0}\left [ \dfrac{1}{|r_1 -r_2|}+\dfrac{1}{|r_1 -r_3|}+\dfrac{1}{|r_2 -r_3|}-\dfrac{3}{|r_1|}-\dfrac{3}{|r_2|}-\dfrac{3}{|r_3|}\right ]$

The guess wavefunction we would need is

\begin{align*}\Psi_g (x_1 ,x_2 ,x_3 ) &= \dfrac{1}{\sqrt{6}}[\psi_1 (x_1)\psi_2 (x_2)\psi_3 (x_3)+\psi_1 (x_3)\psi_2 (x_1)\psi_3 (x_2)+\psi_1 (x_2)\psi_2 (x_3)\psi_3 (x_1)\\ &- \psi_1 (x_1)\psi_2 (x_3)\psi_3 (x_2)-\psi_1 (x_3)\psi_2 (x_2)\psi_3 (x_1) -\psi_1 (x_2)\psi_2 (x_1)\psi_3 (x_3)]\end{align*}

and the guess energy is computed using

$E_g = \int \Psi_{g}^{*}\hat{H}\Psi_g dx_1 dx_2 dx_3$

where the notation $$\int dx$$ means

$\int dx = \sum_{S_z=-\hbar/2}^{\hbar/2}\int dV$

That is, we sum over the two spin values and integrate over all space using either Cartesian $$(dV=dxdydz)$$ or spherical $$(dV=r^2sin\theta drd\theta d\phi)$$ coordinates or any other coordinates we wish to use. Then, minimizing the guess energy by optimizing the shapes of the orbitals $$\phi_1$$, $$\phi_2$$, and $$\phi_3$$, we obtain two orbitals $$\phi_1 (r)$$ and $$\phi_2 (r)$$ that resemble 100 or 1s hydrogen-like orbitals $$\psi_{100(Z)}(r,\phi ,\theta)$$ and $$\phi_3$$ resembles $$\psi_{200(Z)}(r, \phi ,\theta)$$ but again, all of these orbitals decay faster than their counterparts for the hydrogen atom. The calculation gives an effective value of $$Z\approx 2.64$$. Thus, multiplying these by spin wavefunctions, we would obtain the three HF orbitals as

\begin{align*}\psi_1 (x) &= \psi_{1s(Z)} (r)\psi_{\uparrow}(s_z)\\ \psi_2 (x) &= \psi_{1s(Z)}(r)\psi_{\downarrow}(s_z)\\ \psi_3 (x) &= \psi_{2s(Z)}(r)\psi_{\uparrow}(s_z)\end{align*}

We would denote the electronic configuration as $$1s^2 2s^1$$. Here, we find $$\varepsilon_1 \approx \varepsilon_2$$, and $$\varepsilon_3 >\varepsilon_1 ,\varepsilon_2$$. Strictly, $$\varepsilon_1 <\varepsilon_2 <\varepsilon_3$$. The total HF energy for the ground state is -14.86 Ry, while the experimental value is -14.96 Ry. Note that if we neglected the electron-electron repulsion, the energy would be

$E_{n_1 n_2 n_3}=-\dfrac{9}{n_{1}^{2}}-\dfrac{9}{n_{2}^{2}}-\dfrac{9}{n_{3}^{2}}$

with $$n_1 =n_2 =n_3 =1$$, which gives -27 Ry.

For Beryllium, we would obtain four orbitals, two of which resemble 1s orbitals and two of which resemble 2s orbitals. Again, these decay as if they have an effective $$Z$$ value between 3 and 4. We can, therefore, represent these orbitals as

\begin{align*}\psi_1 (x) &= \psi_{1s(Z)}(r)\psi_{\uparrow}(s_z)\\ \psi_2 (x) &= \psi_{1s(Z)}(r)\psi_{\downarrow}(s_z)\\ \psi_3 (x) &= \psi_{2s(Z)}(r)\psi_{\uparrow}(s_z)\\ \psi_4 (x) &= \psi_{2s(Z)}(r)\psi_{\downarrow}(s_z)\end{align*}

We would denote the electronic configuration as $$1s^2 2s^2$$.

For Boron, we would obtain 5 orbitals, two of which would resemble $$1s$$, two would resemble $$2s$$, and the fifth would resemble a $$2p_x$$ orbital. Again, all of these would decay as if they have an effective $$Z$$ between $$4$$ and $$5$$. We can thus represent the 5 orbitals as

\begin{align*}\psi_1 (x) &= \psi_{1s(Z)}(r)\psi_{\uparrow}(s_z)\\ \psi_2 (x) &= \psi_{1s(Z)}(r)\psi_{\downarrow}(s_z)\\ \psi_3 (x) &= \psi_{2s(Z)}(r)\psi_{\uparrow}(s_z)\\ \psi_4 (x) &= \psi_{2s(Z)}(r)\psi_{\downarrow}(s_z)\\ \psi_5 (x) &= \psi_{2p_x (Z)}(r)\psi_{\uparrow}(s_z)\end{align*}

and denote the electronic configuration as $$1s^2 2s^2 2p_{x}^{1}$$. Here the HF energies would follow the pattern $$\varepsilon_1 \approx \varepsilon_2$$, $$\varepsilon_3 \approx \varepsilon_4$$, $$\varepsilon_3 ,\varepsilon_4 >\varepsilon_1 ,\varepsilon_2$$, and $$\varepsilon_5 >\varepsilon_3 ,\varepsilon_4$$. Strictly, $$\varepsilon_1 <\varepsilon_2 <\varepsilon_3 <\varepsilon_4 <\varepsilon_5$$.

For carbon, we obtain 6 orbitals, and this time, the 6th orbital does not resemble $$2p_x$$ but rather $$2p_y$$. Hence, we would represent the 6 orbitals as

\begin{align*}\psi_1 (x) &= \psi_{1s(Z)}(r)\psi_{\uparrow}(s_z)\\ \psi_2 (x) &= \psi_{1s(Z)}(r)\psi_{\downarrow}(s_z)\\ \psi_3 (x) &= \psi_{2s(Z)}(r)\psi_{\uparrow}(s_z)\\ \psi_4 (x) &= \psi_{2s(Z)}(r)\psi_{\downarrow}(s_z)\\ \psi_5 (x) &= \psi_{2p_x (Z)}(r)\psi_{\uparrow}(s_z)\\ \psi_6 (x) &= \psi_{2p_y (Z)}(r)\psi_{\uparrow}(s_z)\end{align*}

These all decay as if they had an effective $$Z$$ value between 5 and 6. The HF energies follow the pattern $$\varepsilon_1 \approx \varepsilon_2$$, $$\varepsilon_3 \approx \varepsilon_4$$, $$\varepsilon_5 \approx \varepsilon_6$$, $$\varepsilon_3 ,\varepsilon_4 >\varepsilon_1 ,\varepsilon_2$$, and $$\varepsilon_5 ,\varepsilon_6 >\varepsilon_3 ,\varepsilon_4$$. Note that for Boron and Carbon, the energies of the $$2p$$ orbitals are larger than for the $$2s$$ orbitals, which is not the case for hydrogen. The electronic configuration would be $$1s^2 2s^2 2p_{x}^{1} 2p_{y}^{1}$$.

For Nitrogen, the 7th orbital would resemble a $$2p_z$$ orbital, and we would just add $$\psi_7 (x)=\psi_{2p_z (Z)}\psi_{\uparrow}(s_z)$$. Now, for oxygen, the 8th orbital again resembles a $$2p_x$$ orbital, and we would add this to the list but with a down spin wavefunction$$\psi_8(x)=\psi_{2p_x (Z)}\psi_{\downarrow}(s_z)$$. For oxygen, we find that the 5-8 HF energies are approximately equal, $$\varepsilon_5 \approx \varepsilon_6 \approx \varepsilon_7 \approx \varepsilon_8$$, although strictly $$\varepsilon_5 <\varepsilon_6 <\varepsilon_7 < \varepsilon_8$$.

These examples illustrate a more general procedure known as Hartree-Fock theory or the Hartree-Fock approximation (named after Douglas Hartree and Vladimir Fock) developed in 1930. Hartree-Fock (HF) theory makes several important assumptions:

1. The guess wavefunction is always composed of combinations of products of single-electron functions $$\psi_1 (r)$$, $$\psi_2 (r)$$,... that properly account for the identical nature of the electrons.
2. When the shapes of these functions are optimized, each electron is subject to an effective potential $$V$$ that is generated by averaging over the positions of all the other electrons. For the $$i$$th electron, $$V$$ depends only on $$r_i$$.
3. For an atom, $$V$$ depends only on the magnitude $$|r_i|$$, which means that the effective potential is spherically symmetric.

If an atom has $$M$$ electrons, then HF theory yields $$M$$ functions $$\psi_1 (r),...,\psi_M (r)$$ and $$M$$ energies $$\varepsilon_1 ,...,\varepsilon_M$$. These energies are all negative, $$\varepsilon_i <0$$, and they are ordered such that

$\varepsilon_1 <\varepsilon_2 <...<\varepsilon_M$

We will give a physical interpretation of these energies shortly.

So how do we calculate probabilities in HF theory? What we are interested in is the probability of finding one of the $$M$$ electrons in a small volume $$dV$$ about the point $$r$$ independent of the positions of the remaining $$M-1$$ electrons. This probability is given in terms of the electron density introduced earlier. That is

$P(electron \ is \ in \ dV \ about \ r)=\rho (r)dV$

and the electron density can be computed from the squares of the HF orbitals:

$\rho (r)=\sum_{S_z =-\hbar/2}^{\hbar/2}\sum_{i=1}^{M}|\psi_i (r,S_z)|^2$

This is the same density that lead to the chemistry Nobel prize!

To give an idea of how well HF theory can predict the ground state energies of several atoms, consider the table below (all energies are in Ry):

Table 9.2: Hartree Fock Calculations of Ground Energies
Atom Hartree-Fock Energy Experiment
$$He$$ $$-5.72$$ $$-5.80$$
$$Li$$ $$-14.86$$ $$-14.96$$
$$Ne$$ $$-257.10$$ $$-257.88$$
$$Ar$$ $$-1053.64$$ $$-1055.20$$

The HF theory motivates the so-called aufbau'' principle for expressing/building up electron configurations in atoms. Aufbau'' is German for building up'' (Aufbauen = to build up''). We see that the HF orbitals are ordered according to their energies and that as the energy increases, their shapes follow the ladder'' of wavefunctions of the hydrogen atom. Thus, as we consider atoms with more and more electrons, we simply have to occupy'' more of the HF orbitals in order of increasing energy. Since groups of these orbitals are close in energy, there are two rules we need to introduce for this procedure of occupying orbitals:

1. Pauli exclusion principle: No two electrons in an atom can have the same set of quantum numbers.
2. Hund's rule: When electrons are added to a set of HF orbitals of approximately equal energy, a single electron enters each orbital before a second one can enter any orbital. The lowest energy configuration is always the one with parallel spins.

Now, let us apply these rules to fill the orbitals in the first row elements (which are all we care about for the next two chapters). The filling will refer to the figure below:

Figure: Table of electronic configurations

Starting with hydrogen $$(Z=1)$$, HF theory gives the exact answer because the H atom is exactly soluble. Thus, we generate the exact function $$\psi_{100}(r,\phi ,\theta)$$, and we put the one electron into this orbital for the ground state. We write this as

$1s^1$

For Helium $$(Z=2)$$, HF theory is no longer exact, however as we discussed, HF theory gives two orbitals $$\psi_1 (r)$$ and $$\psi_2 (r)$$, both of which resemble $$\psi_{100}$$ (but decays a little faster). The energies $$\epsilon_1$$ and $$\epsilon_2$$ are also very similar. Thus, we regard these orbitals as having roughly the same energy, and we place one electron into $$\psi_1$$ and one into $$\psi_2$$. We regard this as placing two electrons into the single orbital $$\psi_{100}$$ and assigning the electrons the same set of H-like quantum numbers $$n,l,m$$. To be consistent with Pauli's exclusion principle, they need to have one quantum number that is different, so we assign one electron $$m_s =1/2$$ and the other $$m_s =-1/2$$ (see figure). We write the electron configuration as

$1s^2$

signifying that two electrons occupy the $$\psi_{100}$$ orbital (although in reality, they occupy two different orbitals that happen to be very similar to each other).

For Lithium $$(Z=3)$$, HF theory yields three orbitals $$\psi_1$$, $$\psi_2$$, and $$\psi_3$$. The first two closely resemble $$\psi_{100}$$, while the third resembles $$\psi_{200}$$. HF theory also yields three energies $$\varepsilon_1$$, $$\varepsilon_2$$ and $$\varepsilon_3$$ with $$\varepsilon_1 \approx \varepsilon_2$$ and $$\varepsilon_3 >\varepsilon_2 >\varepsilon_1$$. Each electron occupies on HF orbital, but we regard this as two electrons in a $$\psi_{100}$$ orbital and the third in a $$\psi_{200}$$ orbital (see figure). We write this as

$1s^2 2s^1$

Note that it takes two electrons to fill the 1s orbital, after which we move to a 2s-like orbital. In this case, we say that the 1s electrons form a closed shell, since we have now moved on to a different value of $$n$$, i.e., $$n=2$$. The 2 1s electrons are called core electrons and the 2s electron is called a valence electron.

Coming to Beryllium $$(Z=4)$$, we have 4 HF orbitals, two of which resemble $$\psi_{100}$$ and two of which resemble $$\psi_{200}$$, so we imagine that an electron in each HF orbital is equivalent to 2 in the $$\psi_{100}$$ and 2 in the $$\psi_{200}$$. In each pair, the electrons have opposite spins to be consistent with the Pauli exclusion principle. We write this as

$1s^2 2s^2$

Now, we have two core electrons and two valence electrons.

For Boron $$(Z=5)$$, there are 5 HF orbitals. $$\psi_1$$ and $$\psi_2$$ resemble $$\psi_{100}$$, $$\psi_3$$ and $$\psi_4$$ resemble $$\psi_{200}$$, and $$\psi_5$$ resembles $$\psi_{21_m}$$, i.e. a generic p-orbital (we do not specify $$m$$ because the direction of the orbital does not matter). The energies are ordered such that $$\varepsilon_1 \approx \varepsilon_2$$, $$\varepsilon_3 \approx \varepsilon_4$$. Interestingly, $$\varepsilon_5$$ is larger than all of these and not close to $$\varepsilon_4$$. Unlike in hydrogen, where orbitals with the same $$n$$ but different values of $$l$$ have the same energy, here orbitals with the same $$n$$ but different values of $$l$$ have different energies. Thus, $$\psi_5$$ is at a higher energy. Hence, when we fill, we start with $$1s$$, then $$2s$$ and finally $$2p$$. We can choose any of the $$2p$$ orbitals, so we just start with $$2p_x$$. We thus write the electronic configuration as

$1s^2 2s^2 2p_{x}^{1}$

For carbon $$(Z=6)$$, the situation is much the same. However, here we have $$\psi_5$$ and $$\psi_6$$ both with p-like character and similar in energy. Here, however, we need Hund's rule. We first put a spin-up electron in the $$2p_x$$ orbital and then another spin-up electron in the $$2p_y$$ orbital. The electronic configuration is

$1s^2 2s^2 2p_{x}^{1}2p_{y}^{1}$

For nitrogen $$(Z=7)$$, we would place a spin-up electron in the $$2p_z$$ orbital and write $$1s^2 2s^2 2p_{x}^{1}2p_{y}^{1}2p_{z}^{1}$$. When we get to oxygen $$Z=8$$, we have now placed one electron in each of the three p-like orbitals. We now have to return to $$p_x$$ and place another electron in it, which will now be spin-down to be consistent with the Pauli exclusion principle. Its electronic configuration is $$1s^2 2s^2 2p_{x}^{2}2p_{y}^{1}2p_{z}^{1}$$. For fluorine $$Z=9$$, it is $$1s^2 2s^2 2p_{x}^{2}2p_{y}^{2}2p_{z}^{1}$$. Finally, when we get to Neon $$(Z=10)$$, we have$$1s^2 2s^2 2p_{x}^{2}2p_{y}^{2}2p_{z}^{2}$$, which we can write simply as $$1s^2 2s^2 2p^{6}$$. With Neon, as with Helium, we have completed a shell, and the next HF orbital will be like $$\psi_{300}$$ and have a higher energy, thereby starting the next shell. Thus, we see that for Neon, 8 electrons are needed to complete a shell, the 2 2s electrons and the 6 2p electrons. The fact that 8 electrons close a shell for Neon lend this atom special stability and make it highly unreactive.

#### Probing energy levels and the Koopmans approximation

Quantum theory successfully predicts the shell structure of atoms and can explain the observed features of the periodic table (see end of chapter 5). The shell structure and the ionization energies of atoms can be probed by a technique known as photoelectron spectroscopy.

The technique works very much like the photoelectric effect in metals except that we use incident EM radiation (usually X-rays or UV rays) to knock electrons out of individuals atoms, thereby leading to measures of the ionization energies. In this method, the ionization energy IE plays the same role as the work function of a metal.

Suppose a photon of energy $$h\nu$$ strikes an atom or cation in order to probe one of the ionization energies $$IE_i$$. The electron will be ejected from the atom or ion and have a residual kinetic energy $$p^2 /2m_e$$. Thus, by energy conservation

$h\nu =IE_i +\dfrac{p^2}{2m_e}$

Given the experimental numbers, we can compare the measured ionization energies to the orbital energies generated in a HF calculation. We find a very interesting result. If the energies are ordered according to $$\varepsilon_1 <\varepsilon_2 <...<\varepsilon_M$$, then the first ionization energy $$IE_1$$ is approximately equal to the negative of the highest HF energy $$-\varepsilon_M$$:

$IE_1 \approx -\varepsilon_M$

Similarly, the second ionization energy has the property

$IE_2 \approx -\varepsilon_{M-1}$

Finally, the $$M$$th ionization energy $$IE_M$$ satisfies

$IE_M \approx -\varepsilon_1$

These relations are known as Koopmans's approximation.

The essential approximations made in these relations are, firstly, the approximations inherent in the HF theory. But beyond this, it is also assumed that the energies do not change much as the electrons are sequentially stripped out of the atom. Obviously, these energies should change, as the screening and repulsion effects of the other electrons is reduced, and the electrons are drawn closer to the positively charged nucleus, but the effects are small. This type of approximation is known as the frozen core approximation.