# Helium atom

## Perturbative method

The hamiltonian describing the helium atom ( $Z=2$ : two electrons and a nuclear charge of $2e_{0}$ ) is :

${\hat {H}}=\underbrace {{\frac {{\hat {\vec {p_{1}}}}^{2}}{2m}}-{\frac {Ze_{0}^{2}}{4\pi \epsilon _{0}{\vec {r}}_{1}}}} _{{\hat {H}}_{1}}+\underbrace {{\frac {{\hat {\vec {p_{2}}}}^{2}}{2m}}-{\frac {Ze_{0}^{2}}{4\pi \epsilon _{0}{\vec {r}}_{2}}}} _{{\hat {H}}_{2}}+\underbrace {\frac {e_{0}^{2}}{4\pi \epsilon _{0}\left|{\vec {r}}_{1}-{\vec {r}}_{2}\right|}} _{{\hat {V}}_{ee}}$ Let us consider as unperturbed hamiltonian
${\hat {H}}^{\left(0\right)}={\hat {H}}_{1}+{\hat {H}}_{2}$ The solutions of each $H_{i},i=1,2$ are
{\begin{aligned}&\left\langle r,\theta ,\phi |n,l,m_{l},m_{s}\right\rangle =R_{n,l}\left(r\right)Y_{l,m_{l}}\left(\theta ,\phi \right)\chi _{m_{s}}\\&\epsilon _{n}={\frac {-Z^{2}R_{y}}{n^{2}}}\end{aligned}} Since ${\hat {H}}^{\left(0\right)}$ is the sum of ${\hat {H}}_{1}$ and ${\hat {H}}_{2}$ its solutions are tensor products of single- particle states, properly antisymmetrized according to the symmetrization postulate. Let us discuss the ground state: two electrons on the $1s$ level, with opposite spins because of Pauli's exclusion principle. We can obtain the appropriate antisymmetric wavefunction using Slater determinant

{\begin{aligned}\varphi \left({\vec {r}}_{1},{\vec {r}}_{2}\right)&={\frac {1}{\sqrt {2}}}{\begin{vmatrix}\psi _{1s}\left({\vec {r}}_{1}\right)\alpha \left(1\right)&\psi _{1s}\left({\vec {r}}_{1}\right)\beta \left(1\right)\\\psi _{1s}\left({\vec {r}}_{2}\right)\alpha \left(2\right)&\psi _{1s}\left({\vec {r}}_{2}\right)\beta \left(2\right)\end{vmatrix}}\\&={\frac {1}{\sqrt {2}}}\left(\psi _{1s}\left({\vec {r}}_{1}\right)\alpha \left(1\right)\cdot \psi _{1s}\left({\vec {r}}_{2}\right)\beta \left(2\right)-\psi _{1s}\left({\vec {r}}_{1}\right)\beta \left(1\right)\cdot \psi _{1s}\left({\vec {r}}_{2}\right)\alpha \left(2\right)\right)\\&={\frac {1}{\sqrt {2}}}\psi _{1s}\left({\vec {r}}_{1}\right)\psi _{1s}\left({\vec {r}}_{2}\right)\underbrace {\left(\alpha \left(1\right)\beta \left(2\right)-\alpha \left(2\right)\beta \left(1\right)\right)} _{{\mbox{spin singlet }}S=0}\end{aligned}} The $0^{th}$ order energy value is
$E^{\left(0\right)}=\epsilon _{1}+\epsilon _{2}\approx 108{\mbox{eV}}$ Now we consider the electron-electron repulsion ${\hat {V}}_{ee}$ as a perturbation and evalue the first order correction to the ground-state energy:
{\begin{aligned}E^{\left(1\right)}&=\left\langle G.S\left|{\hat {V}}_{ee}\right|G.S\right\rangle \\&=\underbrace {\left({\frac {1}{2}}\cdot {\frac {e_{0}^{2}}{4\pi \epsilon _{0}}}\cdot \int _{0}^{+\infty }d^{3}{\vec {r}}_{1}d^{3}{\vec {r}}_{2}\left|\psi _{1s}\left({\vec {r}}_{1}\right)\right|^{2}\left|\psi _{1s}\left({\vec {r}}_{2}\right)\right|^{2}{\frac {1}{\left|{\vec {r}}_{1}-{\vec {r}}_{2}\right|}}\right)} _{\mbox{direct integral}}\underbrace {\sum \left|{\mbox{spin part}}\right|^{2}} _{=1}\end{aligned}} Before solving the direct integral it's important to give it a physical meaning:

• $-e_{0}\left|\psi _{1s}\left({\vec {r}}_{1}\right)\right|^{2}$ can be seen as a charge density associated with electron 1 since it's the product of a charge and a spatial probability distribution. Similarly $-e_{0}\left|\psi _{1s}\left({\vec {r}}_{2}\right)\right|^{2}$ can be seen as the charge density associated with the other electron
• In this perspective it's easy to see that the direct integral represents the energy associated with coulomb interaction between electrons

The result of such an integral is ${\frac {5}{4}}ZR_{y}$ , which is positive and thus increases the ground state energy. In the case of helium, with $Z=2$ , we obtain:

{\begin{aligned}&E^{\left(1\right)}={\frac {10}{4}}R_{y}\approx 34eV\\&E_{\mbox{G.S}}=E^{\left(0\right)}+E^{\left(1\right)}\approx -74.8eV\end{aligned}} This result has to be compared with the experimental value $-79eV$ .

## Variational method

Theorem

We can improve the previous result by using a different approach, based on the following theorem:

$\forall \varphi \left\langle \varphi \left|{\hat {H}}\right|\varphi \right\rangle \geq E_{\mbox{ground state}}$ Proof

Let $\left|u_{n}\right\rangle$ be the set of eigenstates of ${\hat {H}}$ . Since they constitute a complete orthonormal system we can expand each $\varphi$ on this set:

{\begin{aligned}&{\hat {H}}\left|u_{n}\right\rangle =\epsilon _{n}\left|u_{n}\right\rangle \\&\left|\varphi \right\rangle =\sum _{n}c_{n}\left|u_{n}\right\rangle \end{aligned}} Computing $\left\langle \varphi \left|{\hat {H}}\right|\varphi \right\rangle$ we have:
{\begin{aligned}\left\langle \varphi \left|{\hat {H}}\right|\varphi \right\rangle &=\sum _{n,n'}c_{n'}^{*}c_{n}\underbrace {\left\langle u_{n'}\left|{\hat {H}}\right|u_{n}\right\rangle } _{\delta _{n',n}\epsilon _{n}}\\&=\sum _{n}\epsilon _{n}\left|c_{n}\right|^{2}\geq \sum _{n}E_{\mbox{ground state}}\left|c_{n}\right|^{2}=E_{\mbox{ground state}}\\&\Rightarrow \left\langle \varphi \left|{\hat {H}}\right|\varphi \right\rangle \geq E_{\mbox{ground state}}\end{aligned}} We can take advantage of this fact choosing a ground-state single particle wavefunction depending on a variational parameter:

$\psi _{1s}\propto e^{\frac {-Z_{eff}r}{a_{0}}}$ The wavefunction of helium ground state is thus:
$\varphi \left(Z_{eff},{\vec {r}}_{1},{\vec {r}}_{2}\right)\propto e^{\frac {-Z_{eff}r_{1}+r_{2}}{a_{0}}}$ We have neglected both the spin part and the angular part of the wave function since they will not contribute to the following calculation. Now we apply the previous theorem:
{\begin{aligned}&\left\langle \varphi \left(Z_{\mbox{eff}}\right)\left|{\hat {H}}\right|\varphi \left(Z_{\mbox{eff}}\right)\right\rangle =E\left(Z_{\mbox{eff}}\right)\\&E\left(Z_{\mbox{eff}}\right)\geq E_{\mbox{ground state}},\\&{\hat {H}}={\hat {H}}_{1}+{\hat {H}}_{2}+{\frac {e_{0}^{2}}{4\pi \epsilon _{0}}}{\frac {1}{\left|{\vec {r}}_{1}-{\vec {r}}_{2}\right|}}\end{aligned}} That means we have to find the minimum of $E\left(Z_{\mbox{eff}}\right)$ in order to get the best approximation, that is computing ${\frac {\partial E\left(Z_{\mbox{eff}}\right)}{\partial Z_{\mbox{eff}}}}$ {\begin{aligned}\left\langle \varphi \left(Z_{\mbox{eff}}\right)\left|{\hat {H}}_{\mbox{tot}}\right|\varphi \left(Z_{\mbox{eff}}\right)\right\rangle &=\left\langle \varphi \left(Z_{\mbox{eff}}\right)\left|{\hat {H}}_{1}+{\hat {H}}_{2}+{\frac {e_{0}^{2}}{4\pi \epsilon _{0}}}{\frac {1}{\left|{\vec {r}}_{1}-{\vec {r}}_{2}\right|}}\right|\varphi \left(Z_{\mbox{eff}}\right)\right\rangle \\&={\frac {5Z_{\mbox{eff}}R_{y}}{4}}+2\left\langle \psi _{1s}\left(Z_{\mbox{eff}}\right)\left|{\hat {H}}_{1}\right|\psi _{1s}\left(Z_{\mbox{eff}}\right)\right\rangle \end{aligned}} Now
$2\left\langle \psi _{1s}\left(Z_{\mbox{eff}}\right)\left|{\hat {H}}_{1}\right|\psi _{1s}\left(Z_{\mbox{eff}}\right)\right\rangle =2\left\langle \psi _{1s}\left(Z_{\mbox{eff}}\right)\left|{\frac {{\hat {\vec {p_{1}}}}^{2}}{2m}}-{\frac {Ze_{0}^{2}}{4\pi \epsilon _{0}{\vec {r}}_{1}}}\right|\psi _{1s}\left(Z_{\mbox{eff}}\right)\right\rangle$ We need to evaluate two terms:
{\begin{aligned}&\left\langle \psi _{1s}\left(Z_{\mbox{eff}}\right)\left|{\frac {{\hat {\vec {p}}}_{1}^{2}}{2m}}\right|\psi _{1s}\left(Z_{\mbox{eff}}\right)\right\rangle \\&\left\langle \psi _{1s}\left(Z_{\mbox{eff}}\right)\left|-{\frac {Ze_{0}^{2}}{4\pi \epsilon _{0}{\vec {r}}_{1}}}\right|\psi _{1s}\left(Z_{\mbox{eff}}\right)\right\rangle \end{aligned}} The latter is simple to evaluate since
$\left\langle \psi _{1s}\left(Z_{\mbox{eff}}\right)\left|-{\frac {1}{{\vec {r}}_{1}}}\right|\psi _{1s}\left(Z_{\mbox{eff}}\right)\right\rangle ={\frac {Z_{\mbox{eff}}}{a_{0}}}$ so that this term gives
{\begin{aligned}\left\langle \psi _{1s}\left(Z_{\mbox{eff}}\right)\left|-{\frac {Ze_{0}^{2}}{4\pi \epsilon _{0}{\vec {r}}_{1}}}\right|\psi _{1s}\left(Z_{\mbox{eff}}\right)\right\rangle &=-\underbrace {\frac {e_{0}^{2}}{4\pi \epsilon _{0}a_{0}}} _{={\frac {R_{y}}{2}}}ZZ_{\mbox{eff}}\\&=-2R_{y}ZZ_{\mbox{eff}}\end{aligned}} The former can be easily computed writing
${\frac {{\hat {\vec {p}}}_{1}^{2}}{2m}}={\hat {H}}_{1}\left(Z_{\mbox{eff}}\right)+{\frac {Z_{\mbox{eff}}e_{0}^{2}}{4\pi \epsilon _{0}r_{1}}}$ so that
{\begin{aligned}\left\langle \psi _{1s}\left(Z_{\mbox{eff}}\right)\left|{\frac {{\hat {\vec {p}}}_{1}^{2}}{2m}}\right|\psi _{1s}\left(Z_{\mbox{eff}}\right)\right\rangle &=\left\langle \psi _{1s}\left(Z_{\mbox{eff}}\right)\left|{\hat {H}}_{1}+{\frac {Z_{\mbox{eff}}e_{0}^{2}}{4\pi \epsilon _{0}r_{1}}}\right|\psi _{1s}\left(Z_{\mbox{eff}}\right)\right\rangle \\&=-Z_{\mbox{eff}}^{2}R_{y}+\underbrace {\frac {e_{0}^{2}}{4\pi \epsilon _{0}a_{0}}} _{={\frac {R_{y}}{2}}}Z_{\mbox{eff}}^{2}\\&=-Z_{\mbox{eff}}^{2}R_{y}+2Z_{\mbox{eff}}^{2}R_{y}\\&=Z_{\mbox{eff}}^{2}R_{y}\end{aligned}} With these result we can compute
{\begin{aligned}E\left(Z_{\mbox{eff}}\right)&=\left\langle \varphi \left(Z_{\mbox{eff}}\right)\left|{\hat {H}}_{\mbox{tot}}\right|\varphi \left(Z_{\mbox{eff}}\right)\right\rangle \\&={\frac {5Z_{\mbox{eff}}R_{y}}{4}}+2Z_{\mbox{eff}}^{2}R_{y}-4ZZ_{\mbox{eff}}R_{y}\end{aligned}} Differentiation with respect to $Z_{\mbox{eff}}$ gives
{\begin{aligned}{\frac {\partial E\left(Z_{\mbox{eff}}\right)}{\partial Z_{\mbox{eff}}}}={\frac {5}{4}}+4Z_{\mbox{eff}}-4Z\end{aligned}} this result must equal 0 if we are looking for the minimum:
{\begin{aligned}&{\frac {5}{4}}+4Z_{\mbox{eff}}-4Z=0\\&\Rightarrow Z_{\mbox{eff}}=Z-{\frac {5}{16}}\end{aligned}} In case of helium we find $Z_{\mbox{eff}}={\frac {27}{16}}$ , which is greater than $Z=2$ : with this method electron screening has been taken into account. If we put this value of $Z_{\mbox{eff}}$ in the function $\varphi \left(Z_{eff},{\vec {r}}_{1},{\vec {r}}_{2}\right)\propto e^{\frac {-Z_{eff}r_{1}+r_{2}}{a_{0}}}$ we get the wavefunction of the ground state, and if we compute the energy we get
{\begin{aligned}E_{\mbox{ground state}}\left(Z_{\mbox{eff}}={\frac {27}{16}}\right)&={\frac {5Z_{\mbox{eff}}R_{y}}{4}}+2Z_{\mbox{eff}}^{2}R_{y}-4ZZ_{\mbox{eff}}R_{y}\\&\approx -77.5eV\end{aligned}} Comparison with experimental value and the one obtained with perturbation theory lead us to conclude that this is a better approximation

SUMMARY:

The hamiltonian of this system can be written as:

${\hat {H}}={\frac {\overrightarrow {p_{1}^{2}}}{2m}}+{\frac {\overrightarrow {p_{2}^{2}}}{2m}}-{\frac {Ze^{2}}{4\pi \epsilon _{0}r_{1}}}-{\frac {Ze^{2}}{4\pi \epsilon _{0}r_{2}}}+{\frac {e^{2}}{4\pi \epsilon _{0}\mid {\overrightarrow {r_{1}}}-{\overrightarrow {r_{2}}}\mid }}$ Where the last term can be seen as a perturbation of a two electrons-system hamiltonian:
${\hat {H}}={\hat {H}}_{1}+{\hat {H}}_{2}+{\frac {e^{2}}{4\pi \epsilon _{0}\mid {\overrightarrow {r_{1}}}-{\overrightarrow {r_{2}}}\mid }}={\hat {H}}_{0}+{\hat {H}}_{pert}$ Using the perturbation theory we find:
$E_{0}^{(1)}=\left\langle \psi _{0}\mid {\hat {H}}_{int}\mid \psi _{0}\right\rangle =e^{2}\int d^{3}{\overrightarrow {r}}_{1}d^{3}{\overrightarrow {r}}_{2}{\frac {\left|\psi _{1s}({\overrightarrow {r}}_{1})\right|^{2}\left|\psi _{1s}({\overrightarrow {r}}_{2})\right|^{2}}{4\pi \epsilon _{0}\mid {\overrightarrow {r}}_{1}-{\overrightarrow {r}}_{2}\mid }}$ Solving the integral we get:
$E_{0}^{(1)}={\frac {5}{4}}ZR_{y}$ While the unperturbed energy is:
$E_{0}=\epsilon _{1}^{(1s)}+\epsilon _{2}^{(1s)}=-2Z^{2}R_{y}=-108eV$ Let us state one important principle, called variational principle:
$\left\langle \psi \mid {\hat {H}}\mid \psi \right\rangle \geq E_{GS}$ Where $E_{GS}$ stands for the ground state energy of the system. A more precise result for the energy $E_{0}$ can be found looking for an "effective" charge $Z_{eff}$ and by evaluating $\left\langle {\hat {H}}\right\rangle$ at this $Z_{eff}$ . We get:
$\left\langle {\hat {H}}\right\rangle ={\frac {5}{4}}Z_{eff}R_{y}+2Z_{eff}^{2}R_{y}-4ZZ_{eff}R_{y}$ And the value $Z_{eff}$ can be derived looking for a minimum in the energy. Hence:
$Z_{eff}=Z-{\frac {5}{16}}$ For this particular hamiltonian we have that $\left\langle {\frac {{\hat {\overrightarrow {p}}}^{2}}{2m}}\right\rangle =-{\frac {1}{2}}\left\langle {\mathcal {U}}\right\rangle$ , which is just a consequence of the virial theorem:
$\left\langle n\left|{\frac {{\hat {\overrightarrow {p}}}^{2}}{2m}}\right|n\right\rangle =-\left\langle {\overrightarrow {x}}\nabla {\mathcal {U}}\right\rangle$ Where $\left|n\right\rangle$ is any eigenstate of the hamiltonian. Remember that the ground state wf of the He atom is given by the product of two hydrogen-like, ground state, wave functions. That is:
$\psi _{100}=\psi _{1s}({\overrightarrow {r}}_{1})\psi _{1s}({\overrightarrow {r}}_{2})$ Hence, $\psi _{100}$ is an hydrogen-like wf and its expression is:
$\psi _{100}=\left({\frac {Z}{a_{0}}}\right)^{\frac {3}{2}}{\frac {1}{\sqrt {\pi }}}e^{-{\frac {Zr}{a_{0}}}}$ Actually, this is just the radial part of the wf but for most exercises and applications this is what we need.

1. This expression holds for any He-like hamiltonian.