Helium atom

Helium-4
Names
Systematic IUPAC name Helium[1]
Identifiers
CAS Number	7440-59-7 Y
3D model (JSmol)	Interactive image
ChEBI	CHEBI:33681 Y
ChemSpider	22423 Y
EC Number	231-168-5
Gmelin Reference	16294
KEGG	D04420 N
MeSH	Helium
PubChem CID	23987
RTECS number	MH6520000
UNII	206GF3GB41 Y
UN number	1046
InChI InChI=1S/He Y Key: SWQJXJOGLNCZEY-UHFFFAOYSA-N Y
SMILES [He]
Properties
Chemical formula	He
Molar mass	4.002602 g·mol−1
Appearance	Colourless gas
Boiling point	−269 °C (−452.20 °F; 4.15 K)
Thermochemistry
Std molar entropy (S⦵298)	126.151-126.155 J K−1 mol−1
Pharmacology
ATC code	V03AN03 (WHO)
Except where otherwise noted, data are given for materials in their standard state (at 25 °C [77 °F], 100 kPa). N verify (what is YN ?) Infobox references

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 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. Historically, the first such helium spectrum calculation was done by Albrecht Unsöld in 1927.^[2] Its success was considered to be one of the earliest signs of validity of Schrödinger's wave mechanics.^[3]

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^{[citation needed]}

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

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 ${\displaystyle r_{12}=|\mathbf {r} _{1}-\mathbf {r} _{2}|}$. It is important to note, that it operates not in normal space, but in a 6-dimensional configuration space ${\displaystyle (\mathbf {r} _{1},\,\mathbf {r} _{2})}$. 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, so that in operator language, the Hamiltonian simplifies to:

$${\displaystyle H=-{\frac {1}{2m))\mathbf {p} _{1}^{2}-{\frac {1}{2m))\mathbf {p} _{2}^{2}-{\frac {kZe^{2)){\mathbf {r} _{1))}-{\frac {kZe^{2)){\mathbf {r} _{2))}+{\frac {ke^{2)){|\mathbf {r} _{1}-\mathbf {r} _{2}|)).}$$

The wavefunction belongs to tensor product of combined spin states and combined spatial wavefunctions, and since Hamiltonian only acts on spatial wavefunctions, we can neglect spin states until after solving the spatial wavefunction. This is possible since for any general vector: ${\textstyle |\Phi \rangle =\sum _{ij}c_{ij}|\varphi _{i}\rangle |\chi _{j}\rangle }$ where ${\textstyle |\varphi _{i}\rangle }$ is combined spatial wavefunction and ${\textstyle |\chi _{i}\rangle }$ is the combined spin component, the Hamiltonian operator since it only acts on the spatial component gives eigenvector equation:

${\displaystyle H|\Phi \rangle =\sum _{ij}c_{ij}H|\varphi _{i}\rangle |\chi _{j}\rangle =\sum _{ij}c_{ij}(H|\varphi _{i}\rangle )|\chi _{j}\rangle =E\sum _{ij}c_{ij}|\varphi _{i}\rangle |\chi _{j}\rangle =\sum _{ij}c_{ij}E|\varphi _{i}\rangle |\chi _{j}\rangle }$

which implies finding solutions for ${\displaystyle H|\psi _{j}\rangle =E|\psi _{j}\rangle }$ where ${\textstyle |\psi _{j}\rangle =\sum _{i}c_{ij}|\varphi _{i}\rangle }$ is a general combined spatial wavefunction. This energy, however, is not degenerate by the dimension of combined spin state because of symmetrization postulate which requires that physical solutions for identical fermions should be totally antisymmetric, imposing restriction on choice of ${\textstyle |\chi _{i}\rangle }$ based on solutions ${\textstyle |\psi _{i}\rangle }$. Hence the solutions are of form: ${\displaystyle |\psi _{i}\rangle |\chi _{i}\rangle }$ where ${\textstyle |\psi _{i}\rangle }$ is the energy eigenket spatial wavefunction and ${\textstyle |\chi _{i}\rangle }$ is spin wavefunction such that ${\displaystyle |\psi _{i}\rangle |\chi _{i}\rangle }$ is antisymmetric and ${\textstyle |\Phi \rangle }$ is merely some superpostion of these states.

Since the Hamiltonian is independent of spin, it commutes with all spin operators. Since it is also rotationally invariant, the total x, y or z component of angular momentum operator also commutes with the Hamiltonian. From these commutation relations, ${\textstyle S_{\pm ))$ and ${\textstyle L_{\pm ))$ also commutes with the Hamiltonian which implies that energy is independent of ${\textstyle m_{l))$ and ${\textstyle m_{s))$. Although, the purely spatial form of the Hamiltonian implies that the energy is independent of ${\textstyle s}$, this would only be true in the absence of symmetrization postulate. Due to symmetrization postulate, the choice of ${\textstyle s}$ will influence the type of wavefunction required by symmetrization postulate which would in turn influence the energy of the state.^[4]

Other operators that commute with Hamiltonian are spatial exchange operator and parity operator. However a good combination of mutually commuting operators are: ${\textstyle L^{2))$, ${\textstyle L_{z))$, ${\textstyle S^{2))$, and ${\textstyle S_{z))$. Hence the final solutions are given as:

$${\displaystyle H|N,L,m_{l},S,m_{s}\rangle =E_{nls}|N,L,m_{l},S,m_{s}\rangle }$$

Where the energy is ${\textstyle (2l+1)(2s+1)}$ fold degenerate. For electrons, the total spin can have values of 0 or 1. 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 ${\displaystyle n^{2S+1}L_{J))$.

States corresponding to ${\displaystyle S=0}$, are called parahelium (singlet state, so called as there exist ${\textstyle 2s+1=1}$ state) and ${\displaystyle S=1}$ are called orthohelium (triplet state, so called as there exist ${\textstyle 2s+1=3}$ states). Since spin exchange operator can be expressed in terms of dot product of spin vectors, eigenkets of spin exchange operators are also eigenkets of ${\displaystyle S=(S_{1}+S_{2})^{2))$. Hence parahelium can also be said to be the spin anti-symmetric state (singlet state) or orthohelium to be spin symmetric state (triplet state).^[4]

The singlet state is given as:

$${\displaystyle |\chi _{s=0,m=0}\rangle ={\frac {1}{\sqrt {2))}(|\uparrow \rangle |\downarrow \rangle -|\downarrow \rangle |\uparrow \rangle )}$$

$${\displaystyle \psi (\mathbf {r} _{1},\mathbf {r} _{2})=\psi (\mathbf {r} _{2}\,,\mathbf {r} _{1})}$$

and triplet states are given as:

$${\displaystyle {\begin{aligned}&|\chi _{s=1,m=0}\rangle ={\frac {1}{\sqrt {2))}(|\uparrow \rangle |\downarrow \rangle +|\downarrow \rangle |\uparrow \rangle )\;;\\[4pt]&|\chi _{s=1,m=1}\rangle =\;|\uparrow \rangle |\uparrow \rangle \;;\;\;|\chi _{s=1,m=-1}\rangle =\;|\downarrow \rangle |\downarrow \rangle \;.\end{aligned))}$$

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

as per symmetrization and total spin number requirement. It is observed that triplet states are symmetric and singlet states are antisymmetric. Since the total wavefunction is antisymmetric, a symmetric spatial wavefunction can only be paired with antisymmetric wavefunction and vice versa. Hence orthohelium (triplet state) has symmetric spin wavefunction but antisymmetric spatial wavefunction and parahelium (singlet state) has antisymmetric spin wavefunction but symmetric spatial wavefunction. Hence the type of wavefunction of each state is given above. The ${\textstyle (2l+1)}$ degeneracy solely comes from this spatial wavefunction. Note that for ${\textstyle l=0}$, there is no degeneracy in spatial wavefunction.

Alternatively, a more generalized representation of the above can be provided without considering the spatial and spin parts separately. This method is useful in situations where such manipulation is not possible, however, it can be applied wherever needed. Since the spin part is tensor product of spin Hilbert vector spaces, its basis can be represented by tensor product of each of the set, ${\displaystyle \{|\uparrow \rangle _{1}\,,|\downarrow \rangle _{1}\))$ with each of the set, ${\displaystyle \{|\uparrow \rangle _{2}\,,|\downarrow \rangle _{2}\))$. Note that here ${\displaystyle |\uparrow \rangle _{1}|\downarrow \rangle _{2}\neq |\downarrow \rangle _{1}|\uparrow \rangle _{2))$ but are infact orthogonal. In the considered approximation (Pauli approximation), the wave function can be represented as 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 in this coordinate system.^{[5][better source needed]} The usual normalization condition, ${\textstyle \sum _{ij}\int d\mathbf {r} _{1}d\mathbf {r} _{2}|\psi _{ij}|^{2}=1}$, follows from the orthogonality of all ${\displaystyle |i\rangle _{1}|j\rangle _{2))$ elements. This general spinor can be written as 2×2 matrix:

$${\displaystyle {\boldsymbol {\psi ))={\begin{pmatrix}\psi _{\uparrow \uparrow }&\psi _{\uparrow \downarrow }\\\psi _{\downarrow \uparrow }&\psi _{\downarrow \downarrow }\end{pmatrix))}$$

If the Hamiltonian had been spin dependent, we would not have been able to treat each these components independently as shown previously since the Hamiltonian need not act in the same manner for all four components.

The matrix can also be represented as a 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)

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

${\displaystyle S=1}$

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

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

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

${\displaystyle {\boldsymbol {\psi ))}$

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 (totally antisymmetric wavefunction condition)

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

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

Approximation methods

Following from the above approximation, effectively reducing three body problem to two body problem, we have:

$${\displaystyle H=-{\frac {1}{2m))\mathbf {p} _{1}^{2}-{\frac {1}{2m))\mathbf {p} _{2}^{2}-{\frac {kZe^{2)){\mathbf {r} _{1))}-{\frac {kZe^{2)){\mathbf {r} _{2))}+{\frac {ke^{2)){|\mathbf {r} _{1}-\mathbf {r} _{2}|)).}$$

This Hamiltonian for helium with two electrons can be written as a sum of two terms:

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

where the zero-order unperturbed Hamiltonian is

$${\displaystyle H_{0}=-{\frac {1}{2m))\mathbf {p} _{1}^{2}-{\frac {1}{2m))\mathbf {p} _{2}^{2}-{\frac {kZe^{2)){r_{1))}-{\frac {kZe^{2)){r_{2))))$$

while the perturbation term:

$${\displaystyle H'={\frac {ke^{2)){|\mathbf {r} _{1}-\mathbf {r} _{2}|))}$$

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

$${\displaystyle H_{0}=-{\frac {1}{2m))\mathbf {p} _{1}^{2}-{\frac {1}{2m))\mathbf {p} _{2}^{2}-{\frac {kZe^{2)){r_{1))}-{\frac {kZe^{2)){r_{2))}=H_{1}+H_{2))$$

$${\displaystyle H_{i}=-{\frac {1}{2m))\mathbf {p} _{i}^{2}-{\frac {kZe^{2)){r_{i))))$$

${\textstyle |\psi \rangle =|\psi _{1}\rangle |\psi _{2}\rangle }$

${\textstyle |\psi _{1}\rangle }$

${\textstyle |\psi _{2}\rangle }$

${\textstyle H_{1))$

${\textstyle H_{2))$

${\textstyle \psi (\mathbf {r} _{1},\mathbf {r} _{2})=\psi _{a}(\mathbf {r} _{1})\psi _{b}(\mathbf {r} _{2})}$

${\textstyle |\psi _{1}\rangle }$

${\textstyle |\psi _{2}\rangle }$

The proper wave function then must be composed of the symmetric (+) and antisymmetric(−) linear combinations:

$${\displaystyle \psi _{\pm }^{(0)}(\mathbf {r} _{1}\,,\mathbf {r} _{2})={\frac {1}{\sqrt {2))}(\psi _{a}(\mathbf {r} _{1})\psi _{b}(\mathbf {r} _{2})\pm \psi _{a}(\mathbf {r} _{2})\psi _{b}(\mathbf {r} _{1}))}$$

${\displaystyle \psi _{a}=\psi _{b))$

${\displaystyle \psi _{+}^{(0)}=\psi _{a}(\mathbf {r} _{1})\psi _{a}(\mathbf {r} _{2})}$

This explains the absence of the ${\displaystyle 1^{3}S_{1))$ state (with ${\displaystyle \psi _{a}=\psi _{b}=\psi _{1s))$) for orthohelium, where consequently ${\displaystyle 2^{3}S_{1))$ (with ${\displaystyle \psi _{a}=\psi _{1s},\psi _{b}=\psi _{2s))$) is the metastable ground state.

Note that all wavefunction obtained thus far cannot be separated into wavefunctions of each particle (even for electrons with identical ${\textstyle (n_{1}\,,l_{1}\,,m_{1})}$ and ${\textstyle (n_{2}\,,l_{2}\,,m_{2})}$ where wavefunction is ${\textstyle \psi ^{(0)}(\mathbf {r} _{1},\mathbf {r} _{2})=\psi _{n_{1},\ell _{1},m_{1))(\mathbf {r} _{1})\psi _{n_{2},\ell _{2},m_{2))(\mathbf {r} _{2})}$ because then, the spin of the electrons are in a superposition of different spin states: ${\displaystyle |\uparrow \rangle |\downarrow \rangle }$ and ${\displaystyle |\downarrow \rangle |\uparrow \rangle }$ from ${\textstyle \sigma _{0}^{0))$) ie. the wavefunctions are always in superposition of some kind. In other words, one cannot completely determine states ${\textstyle (n\,,l\,,m_{l}\,,m_{s})}$ of particle 1 and 2, or measurements of all details, ${\textstyle (n\,,l\,,m_{l}\,,m_{s})}$ of each electrons cannot be made on one particle without affecting the other. This follows since the wavefunction is always a superposition of different states where each electron has unique ${\textstyle (n\,,l\,,m_{l}\,,m_{s})}$. This is in agreement with Pauli exclusion principle.

We can infer from these wavefunctions that ${\textstyle E=E_{a}+E_{b))$.

The corresponding energies are:

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

A good theoretical descriptions of helium including the perturbation term 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.

Ground state of Helium: Perturbation method

Since ground state corresponds to (1,0,0) state, there can only be one representation of such wavefunction whose spatial wavefunction is:

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

We note that the ground state energy of unperturbed helium atom as:

$${\displaystyle E_{0}=-4{\frac {ke^{2)){a_{0))}=-108.8{\text{ eV))}$$

We can find the first order correction in energy due to electron repulsion in Hamiltonian ${\textstyle {\frac {ke^{2)){|\mathbf {r} _{1}-\mathbf {r} _{2}|))}$:

$${\displaystyle \Delta ^{(1)}=\int \int {\frac {Z^{6)){\pi ^{2}a_{0}^{6))}{\frac {ke^{2)){|\mathbf {r} _{1}-\mathbf {r} _{2}|))e^{-{\frac {2Z(r_{1}+r_{2})}{a_{0))))d^{3}\mathbf {r} _{1}d^{3}\mathbf {r} _{2}={\frac {5}{4)){\frac {ke^{2)){a_{0))))$$

The energy for ground state of helium in first order becomes ${\textstyle E\approx -74.8{\text{ eV))}$ compared to its experimental value of −79.005154539(25) eV.^[8] A better approximation for ground state energy is obtained by choosing better trial wavefunction in variational method.

Screening effect

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.

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

${\displaystyle {\bar {H'))}$

$${\displaystyle V(r)=-{\frac {k(Z-S)e^{2)){r))=-{\frac {kZ_{e}e^{2)){r))}$$

$${\displaystyle E_{0}=-{\frac {k(Z-S)^{2}e^{2)){2a_{0}n^{2))}=-{\frac {kZ_{e}^{2}e^{2)){2a_{0}n^{2))))$$

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

${\textstyle {\frac {1}{3))}$

Ground state of Helium: 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

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

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

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 V_ee 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. This result matches with experimental results closely as well as the calculations of effective Z in screening effect. Hence, 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.^[10]

By using more complicated/accurate wave functions, the ground state energy of helium has been calculated closer and closer to the experimental value −79.005154539(25) eV.^[8] 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^[11][12][13] as well as J.D. Morgan III, Jonathan Baker and Robert Hill^[14][15][16] 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.^[17][18]

Perturbation theory for Helium

Consider the same setting where unperturbed Hamiltonian is:

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

${\textstyle {\frac {ke^{2)){|\mathbf {r} _{1}-\mathbf {r} _{2}|))}$

In general, for (1s)(nl) state, in first order perturbation theory:

$${\displaystyle E=E_{100}+E_{nlm}+\Delta ^{(1)))$$

$${\displaystyle \Delta ^{(1)}=I\pm J}$$

${\textstyle \Delta ^{(1)))$

${\textstyle \Delta ^{(1)))$

They are calculated as:^[3]

$${\displaystyle I=\int \int |\psi _{100}(\mathbf {r} _{1})|^{2}|\psi _{nlm}(\mathbf {r} _{2})|^{2}{\frac {ke^{2)){|\mathbf {r} _{1}-\mathbf {r} _{2}|))d^{3}\mathbf {r} _{1}d^{3}\mathbf {r} _{2))$$

$${\displaystyle J=\int \int \psi _{100}(\mathbf {r} _{1})\psi _{nlm}(\mathbf {r} _{2}){\frac {ke^{2)){|\mathbf {r} _{1}-\mathbf {r} _{2}|))\psi _{nlm}^{\text{*))(\mathbf {r} _{1})\psi _{100}^{\text{*))(\mathbf {r} _{2})d^{3}\mathbf {r} _{1}d^{3}\mathbf {r} _{2))$$

First excited state of Helium

Experimental value of ionization energy

Helium's first ionization energy is −24.587387936(25) eV.^[20] This value was measured experimentally.^[21] The theoretic value of Helium atom's second ionization energy is −54.41776311(2) eV.^[20] The total ground state energy of the helium atom is −79.005154539(25) eV,^[8] or −2.90338583(13) Atomic units a.u., which equals −5.80677166(26) Ry.