# Helium atom

## Perturbative method

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

${\displaystyle {\hat {H}}=\underbrace {{\frac {{\hat {\vec {p_{1}}}}^{2}}{2m}}-{\frac {Ze_{0}^{2}}{4\pi \epsilon _{0}{\vec {r}}_{1}}}} _{{\hat {H}}_{1}}+\underbrace {{\frac {{\hat {\vec {p_{2}}}}^{2}}{2m}}-{\frac {Ze_{0}^{2}}{4\pi \epsilon _{0}{\vec {r}}_{2}}}} _{{\hat {H}}_{2}}+\underbrace {\frac {e_{0}^{2}}{4\pi \epsilon _{0}\left|{\vec {r}}_{1}-{\vec {r}}_{2}\right|}} _{{\hat {V}}_{ee}}}$
Let us consider as unperturbed hamiltonian
${\displaystyle {\hat {H}}^{\left(0\right)}={\hat {H}}_{1}+{\hat {H}}_{2}}$
The solutions of each ${\displaystyle H_{i},i=1,2}$ are
{\displaystyle {\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 ${\displaystyle {\hat {H}}^{\left(0\right)}}$ is the sum of ${\displaystyle {\hat {H}}_{1}}$ and ${\displaystyle {\hat {H}}_{2}}$ 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 ${\displaystyle 1s}$ level, with opposite spins because of Pauli's exclusion principle. We can obtain the appropriate antisymmetric wavefunction using Slater determinant

{\displaystyle {\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 ${\displaystyle 0^{th}}$ order energy value is
${\displaystyle E^{\left(0\right)}=\epsilon _{1}+\epsilon _{2}\approx 108{\mbox{eV}}}$
Now we consider the electron-electron repulsion ${\displaystyle {\hat {V}}_{ee}}$ as a perturbation and evalue the first order correction to the ground-state energy:
{\displaystyle {\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:

• ${\displaystyle -e_{0}\left|\psi _{1s}\left({\vec {r}}_{1}\right)\right|^{2}}$ 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 ${\displaystyle -e_{0}\left|\psi _{1s}\left({\vec {r}}_{2}\right)\right|^{2}}$ 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 ${\displaystyle {\frac {5}{4}}ZR_{y}}$, which is positive and thus increases the ground state energy. In the case of helium, with ${\displaystyle Z=2}$, we obtain:

{\displaystyle {\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 ${\displaystyle -79eV}$.

## Variational method

Theorem

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

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

Proof

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

{\displaystyle {\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 ${\displaystyle \left\langle \varphi \left|{\hat {H}}\right|\varphi \right\rangle }$ we have:
{\displaystyle {\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:

${\displaystyle \psi _{1s}\propto e^{\frac {-Z_{eff}r}{a_{0}}}}$
The wavefunction of helium ground state is thus:
${\displaystyle \varphi \left(Z_{eff},{\vec {r}}_{1},{\vec {r}}_{2}\right)\propto e^{\frac {-Z_{eff}r_{1}+r_{2}}{a_{0}}}}$
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:
{\displaystyle {\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 ${\displaystyle E\left(Z_{\mbox{eff}}\right)}$ in order to get the best approximation, that is computing ${\displaystyle {\frac {\partial E\left(Z_{\mbox{eff}}\right)}{\partial Z_{\mbox{eff}}}}}$
{\displaystyle {\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
${\displaystyle 2\left\langle \psi _{1s}\left(Z_{\mbox{eff}}\right)\left|{\hat {H}}_{1}\right|\psi _{1s}\left(Z_{\mbox{eff}}\right)\right\rangle =2\left\langle \psi _{1s}\left(Z_{\mbox{eff}}\right)\left|{\frac {{\hat {\vec {p_{1}}}}^{2}}{2m}}-{\frac {Ze_{0}^{2}}{4\pi \epsilon _{0}{\vec {r}}_{1}}}\right|\psi _{1s}\left(Z_{\mbox{eff}}\right)\right\rangle }$
We need to evaluate two terms:
{\displaystyle {\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
${\displaystyle \left\langle \psi _{1s}\left(Z_{\mbox{eff}}\right)\left|-{\frac {1}{{\vec {r}}_{1}}}\right|\psi _{1s}\left(Z_{\mbox{eff}}\right)\right\rangle ={\frac {Z_{\mbox{eff}}}{a_{0}}}}$
so that this term gives
{\displaystyle {\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
${\displaystyle {\frac {{\hat {\vec {p}}}_{1}^{2}}{2m}}={\hat {H}}_{1}\left(Z_{\mbox{eff}}\right)+{\frac {Z_{\mbox{eff}}e_{0}^{2}}{4\pi \epsilon _{0}r_{1}}}}$
so that
{\displaystyle {\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
{\displaystyle {\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 ${\displaystyle Z_{\mbox{eff}}}$ gives
{\displaystyle {\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:
{\displaystyle {\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 ${\displaystyle Z_{\mbox{eff}}={\frac {27}{16}}}$, which is greater than ${\displaystyle Z=2}$: with this method electron screening has been taken into account. If we put this value of ${\displaystyle Z_{\mbox{eff}}}$ in the function ${\displaystyle \varphi \left(Z_{eff},{\vec {r}}_{1},{\vec {r}}_{2}\right)\propto e^{\frac {-Z_{eff}r_{1}+r_{2}}{a_{0}}}}$ we get the wavefunction of the ground state, and if we compute the energy we get
{\displaystyle {\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:

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