Two-electron atom

The two-electron atom is a physical system modeling an atom with two electron around and arbitrary nucleus. It is a generalization of the hydrogenic atom. The key challenge of the system is the indistinguishability of electrons and their effective interaction due to the Pauli exclusion principle. The Hamiltonian of the system is

Spin states#

The solved eigenfunction (in position representation) will be of the form $\Psi\equiv \Psi(q_{1},q_{2})$, where $q_{1}$ and $q_{2}$ are the generalized coordinates of the electrons (containing both position and spin state). Electrons are Spin $1/2$ fermions and so $\Psi$ must be an antisymmetric state: $\Psi(q_{1},q_{2})=-\Psi(q_{2},q_{1})$. Since $\hat{H}$ does not directly depend on spin, we can factor out the spin and leave a coordinate-only and spin-only wavefunction:

The spin wavefunction $\chi(1,2)$ can be constructed from the individual spin wavefunctions $\chi_{1/2,m_{S}}(1)$ and $\chi_{1/2,m_{S}}(2)$. We denote $\mathbf{S}_{1}$ and $\mathbf{S}_{2}$ the spin operators of the electrons and by $(S_{1})_{z}$ and $(S_{2})_{z}$ their $z$ components. Each spin state is described by an eigenfunction, but since we are working with spin $1/2$, there's only two possible functions per electron, one for spin up and one for spin down. We'll call these $\alpha$ and $\beta$ respectively and denote which electron they refer to with a number. $\mathbf{S}_{1}$ only acts on $\alpha(1)$ and $\beta(1)$, whereas $\mathbf{S}_{2}$ only acts on $\alpha(2)$ and $\beta(2)$. In turn, this means that $\mathbf{S}_{1}$ and $\mathbf{S}_{2}$ commute.

The total spin is $\mathbf{S}=\mathbf{S}_{1}+\mathbf{S}_{2}$ and the total $z$ component is $S_{z}=(S_{1})_{z}+(S_{2})_{z}$. The square of the total is

since $\mathbf{S}_{1}^{2}=\mathbf{S}_{2}^{2}=3/4$. Since the interaction between electrons is purely electromagnetic, it is independent of spin. As such, each electron's spin is simply either up $(\uparrow)$ or down $(\downarrow)$ without reference to the other, which leads to four distinct spin states:

Applying the $z$ spin operator to $\chi_{1}$ leads to

Clearly, the total spin over $z$ in the $\chi_{1}$ spin state is $1$ (in $\hbar$ units). More generally, we have

By doing the math for all four $\chi_{i}$ functions, we find that $m_{S}=1,0,0,-1$, in that order. There's evidently some degeneracy on $m_{S}=0$. Moreover, while $\chi_{1}$ and $\chi_{4}$ are symmetric under label exchange, $\chi_{2}$ and $\chi_{3}$ are not, and are also not antisymmetric. As such, the actual $m_{S}=0$ states must be normalized linear combinations of the two that are (anti)symmetric under label exchange:

The $+$ is symmetric, the $-$ is antisymmetric. These two functions are eigenstates of $\mathbf{S}^{2}$ and $S_{z}$, with eigenvalues $S_{z}=1$ and $m_{S}=0$ for $+$ and $S_{z}=0$ and $m_{S}=0$ for $-$. We now have four spin states, divided in one state with $S_{z}=0$ and three states with $S_{z}=1$. The former is called the singlet state

and is antisymmetric. The latter three are known as the triplet states

and are all symmetric. At this point, since the wave functions that describe a two-electron atom must be antisymmetric (because fermions must obey the Pauli exclusion principle), we only have two options:

• Parastates are wavefunctions where the spin component is antisymmetric (spin singlet) and the spatial component is symmetric.
• Orthostates are wavefunctions where the spin component is symmetric (spin triplet) and the spatial component is antisymmetric.

Independent particle model#

The electrons repel each other electrically due to having identical electric charge. However, as a first, somewhat qualitative approximation, we can solve the system by neglecting the interaction term. This is a so-called independent particle model[^1] This makes for two independent electrons around the same nucleus. Following perturbation theory, we rewrite the Hamiltonian as

where the electron-electron interaction

is the perturbation. We will choose to ignore this and deal with the consequences later.

Now that we no longer have coupling due to interaction, we can safely claim that the spatial wavefunction is the product of two independent, hydrogen atom wavefunctions

where $m\equiv m_{l}$ for clearer notation.

$H_{0}$ must be invariant under particle exchange (to respect the indistinguishability of electrons. Since the spatial wavefunction can be either even or odd (since the spin state can also be either even or odd, so long it's the other kind), we can write the wavefunctions as one of two different linear combinations:

As before, $+$ is symmetric, $-$ is antisymmetric. A note: if $n_{1},l_{1},m_{1}=n_{2},l_{2},m_{2}$, the antisymmetric combination vanishes. This is correct and what we'd hoped for. After all, this is really just the Pauli exclusion principle at play: two electrons cannot possibly have the same quantum numbers at the same time. In fact, if two electrons have the same spatial state, their spins must differ (i.e., be opposite), but if the spins are opposite then the total spin is zero ($S_{z}=0$), which always ends up being a singlet state, a parastate.

We can now write an approximation of the parastates and orthostates. The parastates have singlet spin state and symmetric spatial state:

Orthostates have triplet spin state (a linear combination of them at least) and antisymmetric spatial state:

where $d_{m_{S}}$ is the degeneracy of that spin state. The $A_{\pm}$ values are Normalization constants.

In this approximation, symmetric and antisymmetric states have the same energy, which is the eigenvalue of $H_{0}\Psi^{0}_{\pm}(\mathbf{r}_{1},\mathbf{r}_{2})=E^{0}_{n_{1},n_{2}}\Psi^{0}_{\pm}(\mathbf{r}_{1},\mathbf{r}_{2})$ and comes out to be

For parastates and orthostates to not have the same energy we must surrender the approximation and consider the actual interaction between electrons.

\phi(r_{1},r_{2})=\frac{Z_{e}^{3}}{\pi a^{3}}e^{-Z_{e}(r_{1}+r_{2})/a}=\Psi_{1s}^{Z_{e}}(r_{1})\Psi_{1s}^{Z_{e}}(r_{2})

\psi_{1s}^{Z_{e}}(r)=\left( \frac{Z_{e}^{3}}{\pi a^{3}} \right)^{1/2}e^{Z_{e}r/a}

E(\phi)=\frac{\braket{ \phi | H|\phi }}{\braket{ \phi | \phi } } =\braket{ \phi | H | \phi }

H=\nabla_{1}^{2}+\nabla_{2}^{2}-\left( \frac{Z_{e}}{r_{1}}+ \frac{Z_{e}}{r_{2}}- \frac{1}{r_{12}} \right)

\langle H \rangle {\phi}=2Z^{2}E{1}

\langle H \rangle {\phi}= 2(Z-2) \frac{e^{2}}{4\pi \varepsilon{0}} \left\langle \frac{1}{r_{12}} \right\rangle_{\phi} +\langle V_{ee} \rangle_{\phi}

\braket{ \phi | \frac{1}{r_{12}} | \phi } =\frac{5}{8}Z_{e}

Z_{e}=Z- \frac{5}{16}