Helium atom

This article is about the physics of atomic helium. For other properties of helium, see helium.

Helium is an element and the next simplest atom to solve after the hydrogen atom. Helium is composed of two electrons in orbit around a nucleus containing two protons along with either one or two neutrons, depending on the isotope. The hydrogen atom is used extensively to aid in solving the helium atom. The Niels Bohr model of the atom gave a very accurate explanation of the hydrogen spectrum, but when it came to helium it collapsed. Werner Heisenberg developed a modification of Bohr's analysis but it involved half-integral values for the quantum numbers. Thomas-Fermi theory, also known as density functional theory, is used to obtain the ground state energy levels of the helium atom along with the Hartree-Fock method.

Introduction

The Hamiltonian of helium is given by

$H\psi ({\vec {r}}_{1},\,{\vec {r}}_{2})={\Bigg [}\sum _{i=1,2}{\Bigg (}-{\frac {\hbar ^{2}}{2\mu }}\nabla _{r_{i}}^{2}-{\frac {Ze^{2}}{4\pi \epsilon _{0}r_{i}}}{\Bigg )}-{\frac {\hbar ^{2}}{M}}\nabla _{r_{1}}\cdot \nabla _{r_{2}}+{\frac {e^{2}}{4\pi \epsilon _{0}r_{12}}}{\Bigg ]}\psi ({\vec {r}}_{1},\,{\vec {r}}_{2})$

where $\mu ={\frac {mM}{m+M}}$ is the reduced mass of an electron with respect to the nucleus and $r_{12}=|{\vec {r_{1}}}-{\vec {r_{2}}}|$. Consider $M=\infty$ so that $\mu =m$ and the mass polarization term ${\frac {\hbar ^{2}}{M}}\nabla _{r_{1}}\cdot \nabla _{r_{2}}$ disappear. The Hamiltonian in atomic units (a.u.) for simplicity is given by

$H\psi ({\vec {r}}_{1},\,{\vec {r}}_{2})={\Bigg [}-{\frac {1}{2}}\nabla _{r_{1}}^{2}-{\frac {1}{2}}\nabla _{r_{2}}^{2}-{\frac {Z}{r_{1}}}-{\frac {Z}{r_{2}}}+{\frac {1}{r_{12}}}{\Bigg ]}\psi ({\vec {r}}_{1},\,{\vec {r}}_{2}).$

The presence of the electron-electron interaction term 1/r₁₂, makes this equation non separable. This means that $\psi _{0}({\vec {r}}_{1},\,{\vec {r}}_{2})$ can't be written as a single product of one-electron wave functions. This means that the wave function is entangled. Measurements cannot be made on on one particle without affecting the other. This is dealt with in the Hartree-Fock and Thomas-Fermi approximations.

Hartree-Fock Method

The Hartree-Fock method is used for a variety of atomic systems. However it is just an approximation, and there are more accurate and efficient methods used today to solve atomic systems. The "many-body problem" for helium and other few electron systems can be solved quite accurately. For example the ground state of helium is known to fifteen digits. In Hartree-Fock theory, the electrons are assumed to move in a potential created by the nucleus and the other electrons. The Hamiltonian for helium with 2 electrons can be written as a sum of the Hamiltonians for each electron:

$H=\sum _{i=1}^{2}h(i)=H_{0}+H^{'}$

where the zero-order unperturbed Hamiltonian is

$H_{0}=-{\frac {1}{2}}\nabla _{r_{1}}^{2}-{\frac {1}{2}}\nabla _{r_{2}}^{2}-{\frac {Z}{r_{1}}}-{\frac {Z}{r_{2}}}$

while the perturbation term:

$H'={\frac {1}{r_{12}}}$

is the electron-electron interaction. H₀ is just the sum of the two hydrogenic Hamiltonians:

$H_{0}={\hat {h}}_{1}+{\hat {h}}_{2}$

where

${\hat {h}}_{i}=-{\frac {1}{2}}\nabla _{r_{i}}^{2}-{\frac {Z}{r_{i}}},i=1,2$

E_n1, the energy eigenvalues and $\psi _{n,l,m}({\vec {r}}_{i})$, the corresponding eigenfunctions of the hydrogenic Hamiltonian will denote the normalized energy eigenvalues and the normalized eigenfunctions. So:

${\hat {h}}_{i}\psi _{n,l,m}({\vec {r_{i}}})=E_{n_{1}}\psi _{n,l,m}({\vec {r_{i}}})$

where

$E_{n_{1}}=-{\frac {1}{2}}{\frac {Z^{2}}{n_{i}^{2}}}{\text{ in a.u.}}$

Neglecting the electron-electron repulsion term, the Schrödinger equation for the spatial part of the two-electron wave function will reduce to the 'zero-order' equation

$H_{0}\psi ^{(0)}({\vec {r}}_{1},{\vec {r}}_{2})=E^{(0)}\psi ^{(0)}({\vec {r}}_{1},{\vec {r}}_{2})$

This equation is separable and the eigenfunctions can be written in the form of single products of hydrogenic wave functions:

$\psi ^{(0)}({\vec {r}}_{1},{\vec {r}}_{2})=\psi _{n_{1},l_{1},m_{1}}({\vec {r}}_{1})\psi _{n_{2},l_{2},m_{2}}({\vec {r}}_{2})$

The corresponding energies are (in a.u.):

$E_{n_{1},n_{2}}^{(0)}=E_{n_{1}}+E_{n_{2}}=-{\frac {Z^{2}}{2}}{\Bigg [}{\frac {1}{n_{1}^{2}}}+{\frac {1}{n_{2}^{2}}}{\Bigg ]}$

Note that the wave function

$\psi ^{(0)}({\vec {r}}_{2},{\vec {r}}_{1})=\psi _{n_{2},l_{2},m_{2}}({\vec {r}}_{1})\psi _{n_{1},l_{1},m_{1}}({\vec {r}}_{2})$

An exchange of electron labels corresponds to the same energy $E_{n_{1},n_{2}}^{(0)}$. This particular case of degeneracy with respect to exchange of electron labels is called exchange degeneracy. The exact spatial wave functions of two-electron atoms must either be symmetric or antisymmetric with respect to the interchange of the coordinates ${\vec {r}}_{1}$ and ${\vec {r}}_{2}$ of the two electrons. The proper wave function then must be composed of the symmetric (+) and antisymmetric(-) linear combinations:

$\psi _{\pm }^{(0)}({\vec {r}}_{1},{\vec {r}}_{2})={\frac {1}{\sqrt {2}}}[\psi _{n_{1},l_{1},m_{1}}({\vec {r}}_{1})\psi _{n_{2},l_{2},m_{2}}({\vec {r}}_{2})\pm \psi _{n_{2},l_{2},m_{2}}({\vec {r}}_{1})\psi _{n_{1},l_{1},m_{1}}({\vec {r}}_{2})]$

This comes from Slater determinants.

The factor ${\frac {1}{\sqrt {2}}}$ normalizes $\psi _{\pm }^{(0)}$. In order to get this wave function into a single product of one-particle wave functions, we use the fact that this is in the ground state. So $n_{1}=n_{2}=1,\,l_{1}=l_{2}=0,\,m_{1}=m_{2}=0$. So the $\psi _{-}^{(0)}$ will vanish, in agreement with the original formulation of the Pauli exclusion principle, in which two electrons cannot be in the same state. Therefor the wave function for helium can be written as

$\psi _{0}^{(0)}({\vec {r}}_{1},{\vec {r}}_{2})=\psi _{1}({\vec {r_{1}}})\psi _{1}({\vec {r_{2}}})={\frac {Z^{3}}{\pi }}e^{-Z(r_{1}+r_{2})}$

Where $\psi _{1}$ and $\psi _{2}$ use the wave functions for the hydrogen Hamiltonian. For helium, Z = 2 from

$E_{0}^{(0)}=E_{n_{1}=1,\,n_{2}=1}^{(0)}=-Z^{2}{\text{ a.u.}}$

where E$_{0}^{(0)}=-4$ a.u. which is approximately -108.8 eV, which corresponds to an ionization potential V$_{P}^{(0)}=2$ a.u. ($\simeq 54.4$ eV). The experimental values are E$_{0}=-2.90$ a.u. ($\simeq -79.0$ eV) and V$_{p}=.90$ a.u. ($\simeq 24.6$ eV).

The energy that we obtained is too low because the repulsion term between the electrons was ignored, whose affect is to raise the energy levels. As Z gets bigger, our approach should yield better results, since the electron-electron repulsion term will get smaller.

So far a very crude independent-particle approximation has been used, in which the electron-electron repulsion term is completely omitted. Splitting the Hamiltonian showed below will improve the results:

$H={\bar {H_{0}}}+{\bar {H'}}$

where

${\bar {H_{0}}}=-{\frac {1}{2}}\nabla _{r_{1}}^{2}+V(r_{1})-{\frac {1}{2}}\nabla _{r_{2}}^{2}+V(r_{2})$

and

${\bar {H'}}={\frac {1}{r_{12}}}-{\frac {Z}{r_{1}}}-V(r_{1})-{\frac {Z}{r_{2}}}-V(r_{2})$

V(r) is a central potential which is chosen so that the effect of the perturbation ${\bar {H'}}$ is small. The net effect of each electron on the motion of the other one is to screen somewhat the charge of the nucleus, so a simple guess for V(r) is

$V(r)=-{\frac {Z-S}{R}}=-{\frac {Z_{e}}{r}}$

where S is a screening constant and the quantity Z_{_e} is the effective charge. The potential is a Coulomb interaction, so the corresponding individual electron energies are given (in a.u.) by

$E_{0}=-(Z-S)^{2}=-Z_{e}^{2}$

and the corresponding wave function is given by

$\psi _{0}(r_{1}\,r_{2})={\frac {Z_{e}^{3}}{\pi }}e^{-Z_{e}(r_{1}+r_{2})}$

If Z_e was 1.70, that would make the expression above for the ground state energy agree with the experimental value E₀ = -2.903 a.u. of the ground state energy of helium. Since Z = 2 in this case, the screening constant is S = .30. For the ground state of helium, for the average shielding approximation, the screening effect of each electron on the other one is equivalent to about ${\frac {1}{3}}$ of the electronic charge.

Thomas–Fermi method

Not long after Schrödinger developed the wave equation, the Thomas–Fermi model was developed. Density functional theory is used to describe the particle density $\rho ({\vec {r}}),\,r\,\epsilon \,\mathbb {R} ^{3}$, and the ground state energy E(N), where N is the number of electrons in the atom. If there are a large number of electrons, the Schrödinger equation runs into problems, because it gets very very difficult to solve, even in the atoms ground states. This is where density functional theory comes in. Thomas-Fermi theory gives very good intuition of what is happening in the ground states of atoms and molecules with N electrons.

The energy functional for an atom with N electrons is given by:

$\xi ={\frac {3}{5}}{\frac {\hbar ^{2}}{2m}}(3\pi ^{2})^{2/3}\gamma \int _{\mathbb {R} ^{3}}\rho ({\vec {r}})d^{3}{\vec {r}}\,-\int _{\mathbb {R} ^{3}}V({\vec {r}})\rho ({\vec {r}})d^{3}{\vec {r}}\,+{\frac {1}{2}}\int _{\mathbb {R} ^{3}}{\frac {e^{2}\,\rho ({\vec {r'}})}{|{\vec {r}}-{\vec {r'}}|}}\,d^{3}{\vec {r'}}$

Where

$\gamma =(3\pi ^{2})^{2/3}{\frac {\hbar ^{2}}{2m}}$

The electron density needs to be greater than or equal to 0, $\int _{\mathbb {R} ^{3}}\rho =N$, and $\rho \rightarrow \xi$ is convex.

In the energy functional, each term holds a certain meaning. The first term describes the minimum quantum-mechanical kinetic energy required to create the electron density $\rho ({\vec {x}})$ for an N number of electrons. The next term is the attractive interaction of the electrons with the nuclei through the Coulomb potential $V({\vec {r}})$. The final term is the electron-electron repulsion potential energy.

So the Hamiltonian for a system of many electrons can be written:

$H=\sum _{i=1}^{N}{\Bigg [}-{\frac {\hbar ^{2}}{2m}}\nabla _{i}^{2}+V({\vec {r_{i}}}){\Bigg ]}+\int {\frac {e^{2}\rho ({\vec {r'}})}{|{\vec {r}}-{\vec {r'}}|}}d^{3}r'$

For helium, N = 2, so the Hamiltonian is given by:

$H=-{\frac {\hbar ^{2}}{2m}}(\nabla _{1}^{2}+\nabla _{2}^{2})+V({\vec {r_{1}}},\,{\vec {r_{2}}})+\int {\frac {e^{2}\rho ({\vec {r'}})}{|{\vec {r}}-{\vec {r'}}|}}d^{3}r'$

Where

$\int {\frac {e^{2}\rho ({\vec {r'}})}{|{\vec {r}}-{\vec {r'}}|}}d^{3}r'={\frac {e^{2}}{4\pi \epsilon _{0}}}{\frac {1}{|{\vec {r}}_{1}-{\vec {r}}_{2}|}},\,{\text{ and }}\,V({\vec {r_{1}}},\,{\vec {r_{2}}})={\frac {e^{2}}{4\pi \epsilon _{0}}}{\Bigg [}{\frac {2}{r_{1}}}+{\frac {2}{r_{2}}}{\Bigg ]}$

yielding

$H=-{\frac {\hbar ^{2}}{2m}}(\nabla _{1}^{2}+\nabla _{2}^{2})+{\frac {e^{2}}{4\pi \epsilon _{0}}}{\Bigg [}{\frac {2}{r_{1}}}+{\frac {2}{r_{2}}}-{\frac {1}{|{\vec {r}}_{1}-{\vec {r}}_{2}|}}{\Bigg ]}$

From the Hartree-Fock method, it is known that ignoring the electron-electron repulsion term, the energy is 8E₁ = -109 eV.

The variational method

To obtain a more accurate energy the variational principle can be applied to the electron-electron potential V_ee using the wave function

$\psi _{0}({\vec {r}}_{1},\,{\vec {r}}_{2})={\frac {8}{\pi a^{3}}}e^{-2(r_{1}+r_{2})/a}$:

$\langle H\rangle =8E_{1}+\langle V_{ee}\rangle =8E_{1}+{\Bigg (}{\frac {e^{2}}{4\pi \epsilon _{0}}}{\Bigg )}{\Bigg (}{\frac {8}{\pi a^{3}}}{\Bigg )}^{2}\int {\frac {e^{-4(r_{1}+r_{2})/a}}{|{\vec {r_{1}}}-{\vec {r_{2}}}|}}\,d^{3}{\vec {r}}_{1}\,d^{3}{\vec {r}}_{2}$

After integrating this, the result is:

$\langle H\rangle =8E_{1}+{\frac {5}{4a}}{\Bigg (}{\frac {e^{2}}{4\pi \epsilon _{0}}}{\Bigg )}=8E_{1}-{\frac {5}{2}}E_{1}=-109+34=-75eV$

This is closer to the theoretical value, but if a better trial wave function is used, an even more accurate answer could be obtained. An ideal wave function would be one that doesn't ignore the influence of the other electron. In other words, each electron represents a cloud of negative charge which somewhat shields the nucleus so that the other electron actually sees an effective nuclear charge Z that is less than 2. A wave function of this type is given by:

$\psi ({\vec {r}}_{1},{\vec {r}}_{2})={\frac {Z^{3}}{\pi a^{3}}}e^{-Z(r_{1}+r_{2})/a}$

Treating Z as a variational parameter to minimize H. The Hamiltonian using the wave function above is given by:

$\langle H\rangle =2Z^{2}E_{1}+2(Z-2){\Bigg (}{\frac {e^{2}}{4\pi \epsilon _{0}}}{\Bigg )}\langle {\frac {1}{r}}\rangle +\langle V_{ee}\rangle$

After calculating the expectation value of ${\frac {1}{r}}$ and V_ee the expectation value of the Hamiltonian becomes:

$\langle H\rangle =[-2Z^{2}+{\frac {27}{4}}Z]E_{1}$

The minimum value of Z needs to be calculated, so taking a derivative with respect to Z and setting the equation to 0 will give the minimum value of Z:

${\frac {d}{dZ}}{\Bigg (}[-2Z^{2}+{\frac {27}{4}}Z]E_{1}{\Bigg )}=0$

$Z=1.69$

This shows that the other electron somewhat shields the nucleus reducing the effective charge from 2 to 1.69. So we obtain the most accurate result yet:

${\frac {1}{2}}{\Bigg (}{\frac {3}{2}}{\Bigg )}^{6}=-77.5eV$

By using more complicated/accurate wave functions, the ground state energy of helium has been calculated closer and closer to the experimental value -78.95 eV. The variational approach has been refined to very high accuracy by G.W.F. Drake and co-workers as well as J.D. Morgan III, Jonathan Baker and Robert Hill using Hylleraas or Frankowski basis functions.

See also

• Helium
• Quantum Mechanics
• Theoretical and experimental justification for the Schrödinger equation
• Quantum field theory
• Quantum states
• [4] - see "Helium atom" on Wikiversity