Skip to main content
Physics LibreTexts

10.4: Helium Atom

Consider the helium atom, which is a good example of a two-electron system. The Hamiltonian is written

\[H = \frac

ParseError: invalid DekiScript (click for details)
Callstack:
    at (Core/Quantum_Mechanics/Fitzpatrick's_Quantum_Mechanics/10:_Identical_Particles/10.4:_Helium_Atom), /content/body/p[2]/span[1], line 1, column 1
{2\,m_1} + \frac
ParseError: invalid DekiScript (click for details)
Callstack:
    at (Core/Quantum_Mechanics/Fitzpatrick's_Quantum_Mechanics/10:_Identical_Particles/10.4:_Helium_Atom), /content/body/p[2]/span[2], line 1, column 1
{Z\,m_2} - \frac{Z\,e^2}{4\pi\,\epsilon_0\,r_1} -\frac{Z\,e^2}{4\pi\,\epsilon_0\,r_2} + \frac{e^2}{4\pi\,\epsilon_0\,r_{12}},\tag{1081}\]

$\displaystyle H = \frac{{\bf p}_1^{\,2}}{2\,m_1} + \frac{{\bf p}_2^{\,2}}{Z\,m_...
...} -\frac{Z\,e^2}{4\pi\,\epsilon_0\,r_2} + \frac{e^2}{4\pi\,\epsilon_0\,r_{12}},$

 

where $ Z=2$ , $ r_1=\vert{\bf x}_1\vert$ , $ r_2=\vert{\bf x}_2\vert$ , and $ r_{12} = \vert{\bf x}_1-{\bf x}_2\vert$ . Suppose that the final term on the right-hand side of the above expression were absent. In this case, the overall spatial wavefunction can be formed from products of hydrogen atom wavefunctions calculated with $ Z=2$ , instead of $ Z=1$ . Each of these wavefunctions is characterized by the usual triplet of quantum numbers, \(n\) , \(\l\) , and \(m\). Now, the total spin of the system is a constant of the motion (since $ {\bf S}$ obviously commutes with the Hamiltonian), so the overall spin state is either the singlet or the triplet state. The corresponding spatial wavefunction is symmetric in the former case, and antisymmetric in the latter. Suppose that one electron has the quantum numbers $ n$ , $ l$ , $ m$ whereas the other has the quantum numbers $ n'$ , $ l'$ , $ m'$ . The corresponding spatial wavefunction is

\[\phi({\bf x}_1,{\bf x}_2)=\frac{1}{\sqrt{2}}\left[\psi_{nlm}({\bf x}_1) \psi_{n'l'm'}({\bf x}_2)\pm \psi_{nlm}({\bf x}_2)\,\psi_{n'l'm'}({\bf x}_1)\right],\tag{1082}\]

where the plus and minus signs correspond to the singlet and triplet spin states, respectively. Here, $ \psi_{nlm}({\bf x})$ is a standard hydrogen atom wavefunction (calculated with $ Z=2$ ). For the special case in which the two sets of spatial quantum numbers, $ n$ , $ l$ , $ m$ and $ n'$ ,$ l'$ ,$ m'$ , are the same, the triplet spin state does not exist (because the associated spatial wavefunction is null). Hence, only singlet spin state is allowed, and the spatial wavefunction reduces to

 

$\displaystyle \phi({\bf x}_1,{\bf x}_2)=\psi_{nlm}({\bf x}_1)\,\psi_{nlm}({\bf x}_2).$ (1083)

 

 

In particular, the ground state ($ n=n'=1$ , $ l=l'=0$ , $ m=m'=0$ ) can only exist as a singlet spin state (i.e., a state of overall spin 0), and has the spatial wavefunction

 

$\displaystyle \phi({\bf x}_1,{\bf x}_2) =\psi_{100}({\bf x}_1)\,\psi_{100}({\bf...
...2)= \frac{Z^{\,3}}{\pi\,a_0^{\,3}}\,\exp\left[\frac{-Z\,(r_1+r_2)}{a_0}\right],$ (1084)

 

 

where $ a_0$ is the Bohr radius. This follows because

 

$\displaystyle \psi_{100}({\bf x})= \frac{1}{\sqrt{4\pi}}\left(\frac{Z}{a_0}\right)^{3/2}2\,\exp\left(\frac{-Z\,r}{a_0}\right).$ (1085)

 

 

The energy of this state is

 

$\displaystyle E = 2\,Z^{\,2}\,E_0=-108.8\,{\rm eV},$ (1086)

 

 

where $ E_0=-13.6\,{\rm eV}$ is the ground state energy of a hydrogen atom. In the above expression, the factor of $ 2$ comes from the fact that there are two electrons in a helium atom.

The above estimate for the ground state energy of a helium atom completely ignores the final term on the right-hand side of Equation (1081), which describes the mutual interaction between the two electrons. We can obtain a better estimate for the ground state energy by treating (1084) as the unperturbed wavefunction, and $ e^2/(4\pi\,\epsilon_0\,r_{12})$ as a perturbation. According to standard first-order perturbation theory, the correction to the ground state energy is

 

$\displaystyle {\mit\Delta} E = \left\langle \frac{e^2}{4\pi\,\epsilon_0\,r_{12}...
...p\left[\frac{-2\,Z\,(r_1+r_2)}{a_0}\right]\frac{e^2}{4\pi\,\epsilon_0\,r_{12}}.$ (1087)

 

 

This can be written

 

$\displaystyle \frac{{\mit\Delta E}}{\vert E_0\vert} = \frac{2\,Z^{\,6}}{\pi^2}\...
..._2}{a_0^{\,3}}\,\exp\left(-2\,Z\,\frac{r_1+r_2}{a_0}\right) \frac{a_0}{r_{12}},$ (1088)

 

 

since $ E_0=-e^2/(8\pi\,\epsilon_0\,a_0)$ . Now,

 

$\displaystyle \frac{1}{r_{12}} = \frac{1}{(r_1^{\,2}+r_2^{\,2}-2\,r_1\,r_2\,\co...
...mma)^{1/2} }= \sum_{l=0,\infty} \frac{r_<^{\,l}}{r_>^{\,l+1}}\,P_l(\cos\gamma),$ (1089)

 

 

where $ r_>$ ($ r_<$ ) is the larger (smaller) of $ r_1$ and $ r_2$ , and $ \gamma$ is the angle subtended between $ {\bf x}_1$ and $ {\bf x}_2$ . Moreover, the so-called addition theorem for spherical harmonics states that

 

$\displaystyle P_l(\cos\gamma) = \frac{4\pi}{2\,l+1}\sum_{m=-l,+l} Y_{l\,m}^\ast(\theta_1,\varphi_1)\,Y_{l\,m}(\theta_2,\varphi_2).$ (1090)

 

 

However,

 

$\displaystyle \int d{\mit\Omega} \,Y_{l\,m}(\theta,\phi) = \sqrt{4\pi}\,\delta_{l\,0}\,\delta_{m\,0},$ (1091)

 

 

so we obtain

 

$\displaystyle \frac{{\mit\Delta E}}{\vert E_0\vert}$ $\displaystyle = 32\,Z\int_0^\infty d x_1\,x_1^{\,2}\left(\int_0^{x_1} d x_2\,\f...
...e}^{-2\,(x_1+x_2)} +\int_{x_1}^\infty dx_2\,x_2\,{\rm e}^{-2\,(x_1+x_2)}\right)$    
  $\displaystyle = \frac{5\,Z}{4} = \frac{5}{2}.$ (1092)

 

 

Here, $ x_1=Z\,r_1/a_0$ and $ x_2=Z\,r_2/a_0$ , and $ Z=2$ . Thus, our improved estimate for the ground state energy of the helium atom is

 

$\displaystyle E = \left(8 - \frac{5}{2}\right)E_0 = -74.8\,{\rm eV}.$ (1093)

 

 

This is much closer to the experimental value of $ -78.8\,{\rm eV}$ than our previous estimate.

Consider an excited state of the helium atom in which one electron is in the ground state, while the other is in a state characterized by the quantum numbers $ n$ , $ l$ , $ m$ . We can write the energy of this state as

 

$\displaystyle E = Z^{\,2}\,E_{100} + Z^{\,2}\,E_{nlm} + {\mit\Delta E},$ (1094)

 

 

where $ E_{nlm}$ is the energy of a hydrogen atom electron whose quantum numbers are $ n$ , $ l$ , $ m$ . According to first-order perturbation theory, $ {\mit\Delta E}$ is the expectation value of $ e^2/(4\pi\,\epsilon_0\,r_{12})$ . It follows from (1082) (with $ n,l,m=1,0,0$ and $ n',l',m'=n,l,m$ ) that

 

$\displaystyle {\mit\Delta}E = I\pm J,$ (1095)

 

where

 

$\displaystyle I$ $\displaystyle = \int d^3 x_1\int d^3 x_2\, \vert\psi_{100}({\bf x}_1)\vert^{\,2}\,\vert\psi_{nlm}({\bf x}_2)\vert^{\,2}\,\frac{e^2}{4\pi\,\epsilon_0\,r_{12}},$ (1096)
$\displaystyle J$ $\displaystyle = \int d^3 x_1\int d^3 x_2 \,\psi_{100}({\bf x}_1)\,\psi_{nlm}({\...
...lon_0\,r_{12}}\,\psi_{100}^{\,\ast}({\bf x}_2)\,\psi_{nlm}^{\,\ast}({\bf x}_1).$ (1097)

 

Here, the plus sign in (1095) corresponds to the spin singlet state, whereas the minus sign corresponds to the spin triplet state. The integral $ I$ --which is known as the direct integral--is obviously positive. The integral $ J$ --which is known as the exchange integral--can be shown to also be positive. Hence, we conclude that in excited states of helium the spin singlet state has a higher energy than the spin triplet state. Incidentally, helium in the spin singlet state is known as para-helium, whereas helium in the triplet state is called ortho-helium. As we have seen, for the ground state, only para-helium is possible.

The fact that para-helium energy levels lie slightly above corresponding ortho-helium levels is interesting because our original Hamiltonian does not depend on spin. Nevertheless, there is a spin dependent effect--i.e., a helium atom has a lower energy when its electrons possess parallel spins--as a consequence of Fermi-Dirac statistics. To be more exact, the energy is lower in the spin triplet state because the corresponding spatial wavefunction is antisymmetric, causing the electrons to tend to avoid one another (thereby reducing their electrostatic repulsion).

Contributors