Helium atom

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Names Identifiers Helium-4 Systematic IUPAC name Helium [1] 3D model (JSmol) ChEBI ChemSpider EC Number 231-168-5 16294 KEGG MeSH PubChem CID RTECS number MH6520000 UNII UN number 1046 InChI=1S/He Key: SWQJXJOGLNCZEY-UHFFFAOYSA-N [He] He Molar mass 4.002602 g·mol−1 Appearance Colourless gas Boiling point −269 °C (−452.20 °F; 4.15 K) Std molarentropy (So298) 126.151-126.155 J K−1 mol−1 V03AN03 () Except where otherwise noted, data are given for materials in their standard state (at 25 °C [77 °F], 100 kPa).

A helium atom is an atom of the chemical element helium. Helium is composed of two electrons bound by the electromagnetic force to a nucleus containing two protons along with either one or two neutrons, depending on the isotope, held together by the strong force. Unlike for hydrogen, a closed-form solution to the Schrödinger equation for the helium atom has not been found. However, various approximations, such as the Hartree–Fock method, can be used to estimate the ground state energy and wavefunction of the atom.

Introduction

The quantum mechanical description of the helium atom is of special interest, because it is the simplest multi-electron system and can be used to understand the concept of quantum entanglement. The Hamiltonian of helium, considered as a three-body system of two electrons and a nucleus and after separating out the centre-of-mass motion, can be written as

${\displaystyle H(\mathbf {r} _{1},\,\mathbf {r} _{2})=\sum _{i=1,2}\left(-{\frac {\hbar ^{2}}{2\mu }}\nabla _{r_{i}}^{2}-{\frac {Ze^{2}}{4\pi \varepsilon _{0}r_{i}}}\right)-{\frac {\hbar ^{2}}{M}}\nabla _{r_{1}}\cdot \nabla _{r_{2}}+{\frac {e^{2}}{4\pi \varepsilon _{0}r_{12}}}}$

where ${\displaystyle \mu ={\frac {mM}{m+M}}}$ is the reduced mass of an electron with respect to the nucleus, ${\displaystyle \mathbf {r} _{1}}$ and ${\displaystyle \mathbf {r} _{2}}$ are the electron-nucleus distance vectors and ${\displaystyle r_{12}=|\mathbf {r} _{1}-\mathbf {r} _{2}|}$. The nuclear charge, ${\displaystyle Z}$ is 2 for helium. In the approximation of an infinitely heavy nucleus, ${\displaystyle M=\infty }$ we have ${\displaystyle \mu =m}$ and the mass polarization term ${\textstyle {\frac {\hbar ^{2}}{M}}\nabla _{r_{1}}\cdot \nabla _{r_{2}}}$ disappears. In atomic units the Hamiltonian simplifies to

${\displaystyle H(\mathbf {r} _{1},\,\mathbf {r} _{2})=-{\frac {1}{2}}\nabla _{r_{1}}^{2}-{\frac {1}{2}}\nabla _{r_{2}}^{2}-{\frac {Z}{r_{1}}}-{\frac {Z}{r_{2}}}+{\frac {1}{r_{12}}}.}$

It is important to note, that it operates not in normal space, but in a 6-dimensional configuration space ${\displaystyle (\mathbf {r} _{1},\,\mathbf {r} _{2})}$. In this approximation (Pauli approximation) the wave function is a second order spinor with 4 components ${\displaystyle \psi _{ij}(\mathbf {r} _{1},\,\mathbf {r} _{2})}$, where the indices ${\displaystyle i,j=\,\uparrow ,\downarrow }$ describe the spin projection of both electrons (z-direction up or down) in some coordinate system. [2] [ better source needed ] It has to obey the usual normalization condition ${\textstyle \sum _{ij}\int d\mathbf {r} _{1}d\mathbf {r} _{2}|\psi _{ij}|^{2}=1}$. This general spinor can be written as 2×2 matrix ${\textstyle {\boldsymbol {\psi }}={\begin{pmatrix}\psi _{\uparrow \uparrow }&\psi _{\uparrow \downarrow }\\\psi _{\downarrow \uparrow }&\psi _{\downarrow \downarrow }\end{pmatrix}}}$ and consequently also as linear combination of any given basis of four orthogonal (in the vector-space of 2×2 matrices) constant matrices ${\displaystyle {\boldsymbol {\sigma }}_{k}^{i}}$ with scalar function coefficients ${\displaystyle \phi _{i}^{k}(\mathbf {r} _{1},\,\mathbf {r} _{2})}$ as ${\textstyle {\boldsymbol {\psi }}=\sum _{ik}\phi _{i}^{k}(\mathbf {r} _{1},\,\mathbf {r} _{2}){\boldsymbol {\sigma }}_{k}^{i}}$. A convenient basis consists of one anti-symmetric matrix (with total spin ${\displaystyle S=0}$, corresponding to a singlet state )

${\displaystyle {\boldsymbol {\sigma }}_{0}^{0}={\frac {1}{\sqrt {2}}}{\begin{pmatrix}0&1\\-1&0\end{pmatrix}}={\frac {1}{\sqrt {2}}}(\uparrow \downarrow -\downarrow \uparrow )}$

and three symmetric matrices (with total spin ${\displaystyle S=1}$, corresponding to a triplet state )

{\displaystyle {\begin{aligned}&{\boldsymbol {\sigma }}_{0}^{1}={\frac {1}{\sqrt {2}}}{\begin{pmatrix}0&1\\1&0\end{pmatrix}}={\frac {1}{\sqrt {2}}}(\uparrow \downarrow +\downarrow \uparrow )\;;\\[4pt]&{\boldsymbol {\sigma }}_{1}^{1}={\begin{pmatrix}1&0\\0&0\end{pmatrix}}=\;\uparrow \uparrow \;;\;\;{\boldsymbol {\sigma }}_{-1}^{1}={\begin{pmatrix}0&0\\0&1\end{pmatrix}}=\;\downarrow \downarrow \;.\end{aligned}}}

It is easy to show, that the singlet state is invariant under all rotations (a scalar entity), while the triplet can be mapped to an ordinary space vector ${\displaystyle (\sigma _{x},\sigma _{y},\sigma _{z})}$, with the three components

${\displaystyle \sigma _{x}={\frac {1}{\sqrt {2}}}{\begin{pmatrix}1&0\\0&-1\end{pmatrix}},\quad \sigma _{y}={\frac {i}{\sqrt {2}}}{\begin{pmatrix}1&0\\0&1\end{pmatrix}},\quad \sigma _{z}={\frac {1}{\sqrt {2}}}{\begin{pmatrix}0&1\\1&0\end{pmatrix}}.}$

Since all spin interaction terms between the four components of ${\displaystyle {\boldsymbol {\psi }}}$ in the above (scalar) Hamiltonian are neglected (e.g. an external magnetic field, or relativistic effects, like angular momentum coupling), the four Schrödinger equations can be solved independently. [3] [ better source needed ]

The spin here only comes into play through the Pauli exclusion principle, which for fermions (like electrons) requires antisymmetry under simultaneous exchange of spin and coordinates

${\displaystyle {\boldsymbol {\psi }}_{ij}(\mathbf {r} _{1},\,\mathbf {r} _{2})=-{\boldsymbol {\psi }}_{ji}(\mathbf {r} _{2},\,\mathbf {r} _{1}).}$

Parahelium is then the singlet state ${\displaystyle {\boldsymbol {\psi }}=\phi _{0}(\mathbf {r} _{1},\,\mathbf {r} _{2}){\boldsymbol {\sigma }}_{0}^{0}}$ with a symmetric function ${\displaystyle \phi _{0}(\mathbf {r} _{1},\,\mathbf {r} _{2})=\phi _{0}(\mathbf {r} _{2},\,\mathbf {r} _{1})}$ and orthohelium is the triplet state ${\displaystyle {\boldsymbol {\psi }}_{m}=\phi _{1}(\mathbf {r} _{1},\,\mathbf {r} _{2}){\boldsymbol {\sigma }}_{m}^{1},\;m=-1,0,1}$ with an antisymmetric function ${\displaystyle \phi _{1}(\mathbf {r} _{1},\,\mathbf {r} _{2})=-\phi _{1}(\mathbf {r} _{2},\,\mathbf {r} _{1})}$.

If the electron-electron interaction term is ignored, both spatial functions ${\displaystyle \phi _{x},\;x=0,1}$ can be written as linear combination of two arbitrary (orthogonal and normalized) one-electron eigenfunctions ${\displaystyle \varphi _{a},\varphi _{b}}$:

${\displaystyle \phi _{x}={\frac {1}{\sqrt {2}}}(\varphi _{a}(\mathbf {r} _{1})\varphi _{b}(\mathbf {r} _{2})\pm \varphi _{a}(\mathbf {r} _{2})\varphi _{b}(\mathbf {r} _{1}))}$

or for the special cases of ${\displaystyle \varphi _{a}=\varphi _{b}}$ (both electrons have identical quantum numbers, parahelium only): ${\displaystyle \phi _{0}=\varphi _{a}(\mathbf {r} _{1})\varphi _{a}(\mathbf {r} _{2})}$.

The total energy (as eigenvalue of ${\displaystyle H}$) is then for all cases ${\displaystyle E=E_{a}+E_{b}}$ (independent of the symmetry).

This explains the absence of the ${\displaystyle 1^{3}S_{1}}$ state (with ${\displaystyle \varphi _{a}=\varphi _{b}=\varphi _{1s}}$) for orthohelium, where consequently ${\displaystyle 2^{3}S_{1}}$ (with ${\displaystyle \varphi _{a}=\varphi _{1s},\varphi _{b}=\varphi _{2s}}$) is the metastable ground state. (A state with the quantum numbers: principal quantum number ${\displaystyle n}$, total spin ${\displaystyle S}$, angular quantum number ${\displaystyle L}$ and total angular momentum ${\displaystyle J=|L-S|,\dots ,L+S}$ is denoted by ${\displaystyle n^{2S+1}L_{J}}$.)

If the electron-electron interaction term ${\textstyle {\frac {1}{r_{12}}}}$ is included, the Schrödinger equation is non separable. However, even if it is neglected, all states described above (even with two identical quantum numbers, like ${\displaystyle 1^{1}S_{0}}$ with ${\displaystyle {\boldsymbol {\psi }}=\varphi _{1s}(\mathbf {r} _{1})\varphi _{1s}(\mathbf {r} _{2}){\boldsymbol {\sigma }}_{0}^{0}}$) cannot be written as a product of one-electron wave functions: ${\displaystyle \psi _{ik}(\mathbf {r} _{1},\,\mathbf {r} _{2})\neq \chi _{i}(\mathbf {r} _{1})\xi _{k}(\mathbf {r} _{2})}$ — the wave function is entangled. One cannot say, particle 1 is in state 1 and the other in state 2, and measurements cannot be made on one particle without affecting the other.

Nevertheless, quite good theoretical descriptions of helium can be obtained within the Hartree–Fock and Thomas–Fermi approximations (see below).

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.

Perturbation method

The Hamiltonian for helium with two electrons can be written as a sum of the Hamiltonians for each electron:

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

where the zero-order unperturbed Hamiltonian is

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

while the perturbation term:

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

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

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

where

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

Eni, the energy eigenvalues and ${\displaystyle \psi _{n_{i},\ell _{i},m_{i}}(\mathbf {r} _{i})}$, the corresponding eigenfunctions of the hydrogenic Hamiltonian will denote the normalized energy eigenvalues and the normalized eigenfunctions. So:

${\displaystyle {\hat {h}}_{i}\psi _{n_{i},\ell _{i},m_{i}}(\mathbf {r} _{i})=E_{n_{i}}\psi _{n_{i},\ell _{i},m_{i}}(\mathbf {r} _{i})}$

where

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

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

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

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

${\displaystyle \psi ^{(0)}(\mathbf {r} _{1},\mathbf {r} _{2})=\psi _{n_{1},\ell _{1},m_{1}}(\mathbf {r} _{1})\psi _{n_{2},\ell _{2},m_{2}}(\mathbf {r} _{2})}$

The corresponding energies are (in atomic units, hereafter a.u.):

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

Note that the wave function

${\displaystyle \psi ^{(0)}(\mathbf {r} _{2},\mathbf {r} _{1})=\psi _{n_{2},\ell _{2},m_{2}}(\mathbf {r} _{1})\psi _{n_{1},\ell _{1},m_{1}}(\mathbf {r} _{2})}$

An exchange of electron labels corresponds to the same energy ${\displaystyle E_{n_{1},n_{2}}^{(0)}}$. 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 ${\displaystyle \mathbf {r} _{1}}$ and ${\displaystyle \mathbf {r} _{2}}$ of the two electrons. The proper wave function then must be composed of the symmetric (+) and antisymmetric(−) linear combinations:

${\displaystyle \psi _{\pm }^{(0)}(\mathbf {r} _{1},\mathbf {r} _{2})={\frac {1}{\sqrt {2}}}\left[\psi _{n_{1},\ell _{1},m_{1}}(\mathbf {r} _{1})\psi _{n_{2},\ell _{2},m_{2}}(\mathbf {r} _{2})\pm \psi _{n_{2},\ell _{2},m_{2}}(\mathbf {r} _{1})\psi _{n_{1},\ell _{1},m_{1}}(\mathbf {r} _{2})\right]}$

This comes from Slater determinants.

The factor ${\textstyle {\frac {1}{\sqrt {2}}}}$ normalizes ${\displaystyle \psi _{\pm }^{(0)}}$. 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 ${\displaystyle n_{1}=n_{2}=1,\,\ell _{1}=\ell _{2}=0,\,m_{1}=m_{2}=0}$. So the ${\displaystyle \psi _{-}^{(0)}}$ will vanish, in agreement with the original formulation of the Pauli exclusion principle, in which two electrons cannot be in the same state. Therefore, the wave function for helium can be written as

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

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

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

where E(0)
0
= −4 a.u. which is approximately −108.8 eV, which corresponds to an ionization potential V(0)
P
= 2 a.u. (54.4 eV). The experimental values are E0 = −2.90 a.u. ( −79.0 eV) and Vp = 0.90 a.u. ( 24.6 eV).

The energy that we obtained is too low because the repulsion term between the electrons was ignored, whose effect 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:

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

where

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

and

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

V(r) is a central potential which is chosen so that the effect of the perturbation ${\displaystyle {\bar {H'}}}$ 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

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

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

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

and the corresponding wave function is given by

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

If Ze 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 = 0.30. 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 ${\textstyle {\frac {1}{3}}}$ of the electric charge. [5]

The variational method

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

${\displaystyle \psi _{0}(\mathbf {r} _{1},\,\mathbf {r} _{2})={\frac {8}{\pi a^{3}}}e^{-2(r_{1}+r_{2})/a}}$
${\displaystyle \langle H\rangle =8E_{1}+\langle V_{ee}\rangle =8E_{1}+\left({\frac {e^{2}}{4\pi \varepsilon _{0}}}\right)\left({\frac {8}{\pi a^{3}}}\right)^{2}\int {\frac {e^{-4(r_{1}+r_{2})/a}}{|\mathbf {r} _{1}-\mathbf {r} _{2}|}}\,d^{3}\mathbf {r} _{1}\,d^{3}\mathbf {r} _{2}}$

After integrating this, the result is:

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

This is closer to the experimental 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:

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

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

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

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

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

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:

${\displaystyle {\frac {d}{dZ}}\left(\left[-2Z^{2}+{\frac {27}{4}}Z\right]E_{1}\right)=0}$
${\displaystyle Z={\frac {27}{16}}\sim 1.69}$

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:

${\displaystyle {\frac {1}{2}}\left({\frac {3}{2}}\right)^{6}E_{1}=-77.5{\text{ eV}}}$

Where again, E1 represents the ionization energy of hydrogen.

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. [6] The variational approach has been refined to very high accuracy for a comprehensive regime of quantum states by G.W.F. Drake and co-workers [7] [8] [9] as well as J.D. Morgan III, Jonathan Baker and Robert Hill [10] [11] [12] using Hylleraas or Frankowski-Pekeris basis functions. One needs to include relativistic and quantum electrodynamic corrections to get full agreement with experiment to spectroscopic accuracy. [13] [14]

Experimental value of ionization energy

Helium's first ionization energy is −24.587387936(25) eV. [15] This value was derived by experiment. [16] The theoretic value of Helium atom's second ionization energy is −54.41776311(2) eV. [15] The total ground state energy of the helium atom is −79.005151042(40) eV, [15] or −2.90338583(13) Atomic units a.u., which equals −5.80677166 (26) Ry.

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References

1. For n = 1, ℓ = 0 and m = 0, the wavefunction in a spherically symmetric potential for a hydrogen electron is ${\displaystyle \psi (r)={\frac {1}{\sqrt {\pi }}}\left({\frac {Z}{a_{0}}}\right)^{\frac {3}{2}}\ e^{-{\tfrac {Zr}{a_{0}}}}\;}$. [4] In atomic units, the Bohr radius ${\displaystyle {a_{0}}}$ equals 1, and the wavefunction becomes ${\displaystyle \psi (r)={\frac {1}{\sqrt {\pi }}}Z^{\frac {3}{2}}\ e^{-Zr}}$.
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