Helium atom

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Helium is an element and the next simplest atom to solve next to the Hydrogen Atom. The hydrogen atom is used extensively to aid in solving the helilum 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. The solution to the Schrödinger equation for Helium yields Helium's energy levels from which the frequencies of the spectral lines can be calculated. Also, Thomas-Fermi Theory also known as Density Functional Theory is used to solve the helium atom along with the Hartree-Fock method.

Thomas-Fermi Method

Not long after Schrödinger developed the wave equation, density functional theory was invented. 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. Density functional 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:

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

Solution to the Schrödinger equation for Helium

The helium atom is made up of two electrons in orbit around a nucleus containing two protons and some neutrons. The Hamiltonian for this system is:

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

There are experimental results that are of great precision that give the ground state energy E_gs = -78.95 eV (electron Volts). However producing this theoretically is a little bit tricky.

Solving this theoretically will be troublesome because of the elecetron - electron repulsion given by:

$V_{ee}={\frac {e^{2}}{4\pi \epsilon }}{\frac {1}{|{\vec {r_{1}}}-{\vec {r_{2}}}|}}$

Ignoring this electron-electron potential the Hamiltonian for helium will be two independent hydrogen Hamiltonians, and the wave function for helium will be the product of two hydrogen ground state wave functions given by:

$\psi _{0}({\vec {r}}_{1},{\vec {r}}_{2})=\psi _{100}({\vec {r}}_{2})\psi _{100}({\vec {r}}_{2})={\frac {8}{\pi a^{3}}}e^{-2(r_{1}+r_{2})/a}$

The energy is 8E₁ = -109 eV. To obtain a more accurate energy the variational principle can be applied to the electron-electron potential V _ee using the above wave function:

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

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.

References:http://www.sjsu.edu/faculty/watkins/helium.htm, David I. Griffiths Introduction to Quantum Mechanics Second edition year 2005 Pearson Education, Inc