Helium atom

Image:Helium-term-scheme.svg

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 |last=Drake |first=Gordon W. F. |title=High Precision Calculations for Helium |date=2023 |work=Springer Handbook of Atomic, Molecular, and Optical Physics |series=Springer Handbooks |pages=199–216 |editor-last=Drake |editor-first=Gordon W. F. |url=https://link.springer.com/chapter/10.1007/978-3-030-73893-8_12 |access-date=2025-03-15 |place=Cham |publisher=Springer International Publishing |language=en |doi=10.1007/978-3-030-73893-8_12 |isbn=978-3-030-73893-8}}

$$H(\backslash mathbf\{r\}\_1,\backslash ,\; \backslash mathbf\{r\}\_2)\; =\; \backslash sum\_\{i=1,2\}\backslash left(-\backslash frac\{\backslash hbar^2\}\{2\backslash mu\}\; \backslash nabla^2\_\{r\_i\}\; -\backslash frac\{Ze^2\}\{4\backslash pi\backslash varepsilon\_0\; r\_i\}\backslash right)\; -\; \backslash frac\{\backslash hbar^2\}\{M\}\; \backslash nabla\_\{r\_1\}\; \backslash cdot\; \backslash nabla\_\{r\_2\}\; +\; \backslash frac\{e^2\}\{4\backslash pi\backslash varepsilon\_0\; r\_\{12\}\}$$

where $\backslash mu\; =\; \backslash frac\{mM\}\{m+M\}$ is the reduced mass of an electron with respect to the nucleus, $\backslash mathbf\{r\}\_1$ and $\backslash mathbf\{r\}\_2$ are the electron-nucleus distance vectors and $r\_\{12\}\; =\; |\backslash mathbf\{r\}\_1\; -\; \backslash mathbf\{r\}\_2|$. It is important to note that it operates not in normal space, but in a 6-dimensional configuration space $(\backslash mathbf\{r\}\_1,\backslash ,\; \backslash mathbf\{r\}\_2)$. The nuclear charge, $Z$ is 2 for helium. In the approximation of an infinitely heavy nucleus, $M\; =\; \backslash infty$ we have $\backslash mu\; =\; m$ and the mass polarization term $\backslash frac\{\backslash hbar^2\}\{M\}\; \backslash nabla\_\{r\_1\}\; \backslash cdot\; \backslash nabla\_\{r\_2\}$ disappears, so that in operator language, the Hamiltonian simplifies to:

$$H\; =\; \backslash frac\{1\}\{2m\}\; \backslash mathbf\; p\_1^2\; +\; \backslash frac\{1\}\{2m\}\; \backslash mathbf\; p\_2^2\; -\; \backslash frac\{kZe^2\}\{\; \backslash mathbf\; r\_1\}\; -\; \backslash frac\{kZe^2\}\{\; \backslash mathbf\; r\_2\}+\backslash frac\{k\; e^2\}$$

The wavefunction belongs to the tensor product of combined spin states and combined spatial wavefunctions, and since this 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, one has that $|\backslash Phi\; \backslash rangle\; =\; \backslash sum\_\{ij\}\; c\_\{ij\}\; |\backslash varphi\_i\; \backslash rangle\; |\backslash chi\_j\; \backslash rangle$ where $|\backslash varphi\_i\; \backslash rangle$ is a combined spatial wavefunction and $|\backslash chi\_i\; \backslash rangle$ is the combined spin component. The Hamiltonian operator, since it only acts on the spatial component, gives the eigenvector equation:

$$H\; |\backslash Phi\; \backslash rangle\; =\; \backslash sum\_\{ij\}\; c\_\{ij\}\; H\; |\backslash varphi\_i\; \backslash rangle\; |\backslash chi\_j\; \backslash rangle\; =\; \backslash sum\_\{ij\}\; c\_\{ij\}\; (H|\; \backslash varphi\_i\; \backslash rangle\; )\; |\backslash chi\_j\; \backslash rangle\; =\; E\; \backslash sum\_\{ij\}\; c\_\{ij\}\; |\backslash varphi\_i\; \backslash rangle\; |\backslash chi\_j\; \backslash rangle\; =\; \backslash sum\_\{ij\}\; c\_\{ij\}\; E\; |\backslash varphi\_i\; \backslash rangle\; |\backslash chi\_j\; \backslash rangle,$$

which implies that one should find solutions for $H\; |\backslash psi\_j\; \backslash rangle\; =\; E\; |\; \backslash psi\_j\; \backslash rangle,$ where $|\; \backslash psi\_j\; \backslash rangle\; =\backslash sum\_i\; c\_\{ij\}\; |\; \backslash varphi\_i\; \backslash rangle$ is a general combined spatial wavefunction. This energy, however, is not degenerate with multiplicity given by the dimension of the space of combined spin states because of a symmetrization postulate, which requires that physical solutions for identical fermions should be totally antisymmetric, imposing a restriction on the choice of $|\backslash chi\_i\; \backslash rangle$ based on solutions $|\; \backslash psi\_i\; \backslash rangle$. Hence the solutions are of the form: $|\backslash psi\_i\; \backslash rangle|\; \backslash chi\_i\; \backslash rangle$ where $|\; \backslash psi\_i\; \backslash rangle$ is the energy eigenket spatial wavefunction and $|\backslash chi\_i\; \backslash rangle$ is a spin wavefunction such that $|\backslash psi\_i\; \backslash rangle|\; \backslash chi\_i\; \backslash rangle$ is antisymmetric and $|\backslash Phi\; \backslash rangle$ 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, $S\_\{\backslash pm\}$ and $L\_\{\backslash pm\}$ also commutes with the Hamiltonian which implies that energy is independent of $m\_l$ and $m\_s$. Although the purely spatial form of the Hamiltonian implies that the energy is independent of $s$, this would only be true in the absence of symmetrization postulate. Due to the symmetrization postulate, the choice of $s$ will influence the type of wavefunction required by symmetrization postulate which would in turn influence the energy of the state.

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

$$H\; |\; N,\; L,\; m\_l,\; S,\; m\_s\; \backslash rangle\; =\; E\_\{nls\}\; |\; N,\; L,\; m\_l,\; S,\; m\_s\; \backslash rangle,$$

where the energy is $(2l+1)\; (\; 2s\; +\; 1\; )$ fold degenerate. For electrons, the total spin can have values of 0 or 1. A state with the quantum numbers: principal quantum number $n$, total spin $S$, angular quantum number $L$ and total angular momentum $J\; =\; |L-S|,\; \backslash dots,\; L+S$ is denoted by $n^\{2S+1\}L\_J$.

States corresponding to $S\; =\; 0$, are called parahelium (singlet state, so called as there exist $2s+1=1$ state) and $S\; =\; 1$ are called orthohelium (triplet state, so called as there exist $2s+1=3$ states). Since the spin exchange operator can be expressed in terms of dot product of spin vectors, eigenkets of spin exchange operators are also eigenkets of $S\; =\; (S\_1\; +\; S\_2)^2$. Hence parahelium can also be said to be the spin anti-symmetric state (singlet state) or orthohelium to be spin symmetric state (triplet state).{{Cite web |last=Littlejohn |first=Robert |title=Helium |url=https://bohr.physics.berkeley.edu/classes/221/1112/notes/helium.pdf}}

The singlet state is given as:$$|\backslash chi\_\{s=0,m=0\}\; \backslash rangle\; =\; \backslash frac\; 1\{\backslash sqrt\; 2\}\; (|\backslash uparrow\backslash rangle\; |\backslash downarrow\; \backslash rangle-\; |\backslash downarrow\backslash rangle|\backslash uparrow\; \backslash rangle)$$$$\backslash psi(\backslash mathbf\; r\_1,\; \backslash mathbf\; r\_2)\; =\backslash psi(\backslash mathbf\; r\_2\backslash ,,\backslash mathbf\; r\_1)$$

and triplet states are given as:$$\backslash begin\{align\}$$

&|\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{align}$$\backslash psi(\backslash mathbf\; r\_1,\; \backslash mathbf\; r\_2)\; =\; -\backslash psi(\backslash mathbf\; r\_2\; \backslash ,,\; \backslash mathbf\; r\_1)$$

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 a symmetric spin wavefunction but an antisymmetric spatial wavefunction and parahelium (singlet state) has an antisymmetric spin wavefunction but a symmetric spatial wavefunction. Hence the type of wavefunction of each state is given above. The $(2l+1)$ degeneracy solely comes from this spatial wavefunction. Note that for $l=0$, 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, $\backslash \{\; |\backslash uparrow\backslash rangle\_1\; \backslash ,,\; |\backslash downarrow\backslash rangle\_1\; \backslash \}$ with each of the set, $\backslash \{\; |\backslash uparrow\backslash rangle\_2\; \backslash ,,\; |\backslash downarrow\backslash rangle\_2\; \backslash \}$. Note that here $|\backslash uparrow\backslash rangle\_1|\backslash downarrow\backslash rangle\_2\; \backslash ne\; |\backslash downarrow\backslash rangle\_1|\backslash uparrow\backslash rangle\_2$ but are in fact orthogonal. In the considered approximation (Pauli approximation), the wave function can be represented as a second order spinor with 4 components $\backslash psi\_\{ij\}(\backslash mathbf\{r\}\_1,\backslash ,\; \backslash mathbf\{r\}\_2)$, where the indices $i,j\; =\; \backslash ,\backslash uparrow,\backslash downarrow$ describe the spin projection of both electrons in this coordinate system.{{cite book |first1=P. |last1=Rennert |first2=H. |last2=Schmiedel |first3=C. |last3=Weißmantel |title=Kleine Enzyklopädie Physik |publisher=VEB Bibliographisches Institut Leipzig |year=1988 |pages=192–194 |isbn=3-323-00011-0 |language=de }}{{Better source needed|reason=An English-language source would be more helpful.|date=January 2021}} The usual normalization condition, $\backslash sum\_\{ij\}\backslash int\; d\backslash mathbf\{r\}\_1\; d\backslash mathbf\{r\}\_2|\backslash psi\_\{ij\}|^2\; =\; 1$, follows from the orthogonality of all $|i\backslash rangle\_1|j\backslash rangle\_2$ elements. This general spinor can be written as 2×2 matrix:

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 $\backslash boldsymbol\; \backslash sigma^i\_k$ with scalar function coefficients $\backslash phi^k\_i\; (\backslash mathbf\{r\}\_1,\backslash ,\; \backslash mathbf\{r\}\_2)$ as $\backslash boldsymbol\; \backslash psi\; =\; \backslash sum\_\{ik\}\backslash phi^k\_i\; (\backslash mathbf\{r\}\_1,\backslash ,\; \backslash mathbf\{r\}\_2)\; \backslash boldsymbol\; \backslash sigma^i\_k$.

A convenient basis consists of one anti-symmetric matrix (with total spin $S\; =\; 0$, corresponding to a singlet state)
and three symmetric matrices (with total spin $S\; =\; 1$, corresponding to a triplet state)

$$\backslash begin\{align\}$$

&\boldsymbol \sigma^1_0 = \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{align}

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 $(\backslash sigma\_x,\backslash sigma\_y,\backslash sigma\_z)$, with the three components:

Since all spin interaction terms between the four components of $\backslash boldsymbol\; \backslash psi$ 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.{{cite book |first1=L. D. |last1=Landau |first2=E. M. |last2=Lifschitz |title=Lehrbuch der Theoretischen Physik |volume=Bd. III (Quantenmechanik) |publisher=Akademie-Verlag |location=Berlin |year=1971 |at=Kap. IX, pp. 218 |oclc=25750516 |language=de }} This is identical to the previously discussed method of finding spatial wavefunction eigenstates independently of the spin states, here spatial wavefunctions of different spin states correspond to the different components of the matrix.

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)

$$\backslash boldsymbol\; \backslash psi\_\{ij\}\; (\backslash mathbf\{r\}\_1,\backslash ,\; \backslash mathbf\{r\}\_2)\; =\; -\; \backslash boldsymbol\; \backslash psi\_\{ji\}\; (\backslash mathbf\{r\}\_2,\backslash ,\; \backslash mathbf\{r\}\_1).$$

Parahelium is then the singlet state $\backslash boldsymbol\; \backslash phi\; =\; \backslash psi\_+\; (\backslash mathbf\{r\}\_1,\backslash ,\; \backslash mathbf\{r\}\_2)\; \backslash boldsymbol\; \backslash sigma^0\_0$ with a symmetric spatial function $\backslash psi\_+\; (\backslash mathbf\{r\}\_1,\backslash ,\; \backslash mathbf\{r\}\_2)\; =\; \backslash psi\_+\; (\backslash mathbf\{r\}\_2,\backslash ,\; \backslash mathbf\{r\}\_1)$ and orthohelium is the triplet state $\backslash boldsymbol\; \backslash phi\_m\; =\; \backslash psi\_-\; (\backslash mathbf\{r\}\_1,\backslash ,\; \backslash mathbf\{r\}\_2)\backslash boldsymbol\; \backslash sigma^1\_m,\backslash ;\; m\; =-1,0,1$ with an antisymmetric spatial function $\backslash psi\_-(\backslash mathbf\{r\}\_1,\backslash ,\; \backslash mathbf\{r\}\_2)\; =\; -\backslash psi\_-\; (\backslash mathbf\{r\}\_2,\backslash ,\; \backslash mathbf\{r\}\_1)$.

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

$$H\; =\; \backslash frac\{1\}\{2m\}\; \backslash mathbf\; p\_1^2\; +\; \backslash frac\{1\}\{2m\}\; \backslash mathbf\; p\_2^2\; -\; \backslash frac\{kZe^2\}\{\; \backslash mathbf\; r\_1\}\; -\; \backslash frac\{kZe^2\}\{\; \backslash mathbf\; r\_2\}+\backslash frac\{k\; e^2\}$$

This Hamiltonian for helium with two electrons can be written as a sum of two terms:$$H\; =\; H\_0\; +\; H\text{'}$$

where the zero-order unperturbed Hamiltonian is

$$H\_0\; =\; \backslash frac\{1\}\{2m\}\; \backslash mathbf\; p\_1^2\; +\backslash frac\{1\}\{2m\}\; \backslash mathbf\; p\_2^2\; -\; \backslash frac\{kZe^2\}\{r\_1\}\; -\; \backslash frac\{kZe^2\}\{r\_2\}$$

while the perturbation term:$$H\text{'}\; =\; \backslash frac\{k\; e^2\}$$

is the electron-electron interaction. {{math|H₀}} is just the sum of the two hydrogenic Hamiltonians:

$$H\_0\; =\; \backslash frac\{1\}\{2m\}\; \backslash mathbf\; p\_1^2\; +\backslash frac\{1\}\{2m\}\; \backslash mathbf\; p\_2^2\; -\; \backslash frac\{kZe^2\}\{r\_1\}\; -\; \backslash frac\{kZe^2\}\{r\_2\}\; =\; H\_1\; +\; H\_2$$

where$$H\_i\; =\; \backslash frac\{1\}\{2m\}\; \backslash mathbf\; p\_i^2\; -\; \backslash frac\{kZe^2\}\{r\_i\}$$are independent Coulomb field Hamiltonian of each electron. Since the unperturbed Hamiltonian is a sum of two independent Hamiltonians (i.e. are separable), the wavefunction must be of form $|\; \backslash psi\; \backslash rangle\; =\; |\backslash psi\_1\; \backslash rangle\; |\backslash psi\_2\; \backslash rangle$ where $|\; \backslash psi\_1\; \backslash rangle$ and $|\backslash psi\_2\; \backslash rangle$ are eigenkets of $H\_1$ and $H\_2$ respectively.{{Cite book |last=Shankar |first=R. |date=1994 |title=Principles of Quantum Mechanics |url=http://dx.doi.org/10.1007/978-1-4757-0576-8 |page=274 |doi=10.1007/978-1-4757-0576-8|isbn=978-1-4757-0578-2 }} However, the spatial wavefunction of the form $\backslash psi(\backslash mathbf\; r\_1,\; \backslash mathbf\; r\_2)\; =\; \backslash psi\_a(\backslash mathbf\; r\_1)\; \backslash psi\_b(\backslash mathbf\; r\_2)$ need not correspond to physical states of identical electrons as per the symmetrization postulate. Thus, to obtain physical solutions symmetrization of the wavefunctions $|\; \backslash psi\_1\; \backslash rangle$ and $|\backslash psi\_2\; \backslash rangle$ is carried out.

The proper wave function then must be composed of the symmetric (+) and antisymmetric(−) linear combinations:$$\backslash psi\_\backslash pm^\{(0)\}(\backslash mathbf\; r\_1\; \backslash ,,\; \backslash mathbf\; r\_2)\; =\; \backslash frac\; 1\; \{\backslash sqrt\; 2\}(\backslash psi\_a(\backslash mathbf\{r\}\_1)\; \backslash psi\_b(\backslash mathbf\{r\}\_2)\; \backslash pm\; \backslash psi\_a(\backslash mathbf\{r\}\_2)\; \backslash psi\_b(\backslash mathbf\{r\}\_1))$$or for the special cases of $\backslash psi\_a\; =\; \backslash psi\_b$ (both electrons have identical quantum numbers, parahelium only): $\backslash psi\_+^\{(0)\}\; =\; \backslash psi\_a(\backslash mathbf\{r\}\_1)\; \backslash psi\_a(\backslash mathbf\{r\}\_2)$.

This explains the absence of the $1^3\; S\_1$ state (with $\backslash psi\_a\; =\; \backslash psi\_b\; =\; \backslash psi\_\{1s\}$) for orthohelium, where consequently $2^3\; S\_1$ (with $\backslash psi\_a\; =\; \backslash psi\_\{1s\},\; \backslash psi\_b\; =\; \backslash psi\_\{2s\}$) 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 $(n\_1\backslash ,,\; l\_1\; \backslash ,,\; m\_1\; )$ and $(n\_2\backslash ,,\; l\_2\; \backslash ,,\; m\_2\; )$ where wavefunction is $\backslash psi^\{(0)\}(\backslash mathbf\{r\}\_1,\; \backslash mathbf\{r\}\_2)\; =\; \backslash psi\_\{n\_1,\backslash ell\_1,m\_1\}(\backslash mathbf\{r\}\_1)\; \backslash psi\_\{n\_2,\backslash ell\_2,m\_2\}(\backslash mathbf\{r\}\_2)$ because then, the spin of the electrons are in a superposition of different spin states: $|\backslash uparrow\backslash rangle|\backslash downarrow\backslash rangle$ and $|\backslash downarrow\backslash rangle|\backslash uparrow\backslash rangle$ from $\backslash sigma^0\_0$) i.e. the wavefunctions are always in superposition of some kind. In other words, one cannot completely determine states $(n\backslash ,,\; l\; \backslash ,,\; m\_l\; \backslash ,,\; m\_s)$ of particle 1 and 2, or measurements of all details, $(n\backslash ,,\; l\; \backslash ,,\; m\_l\; \backslash ,,\; m\_s)$ 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 $(n\backslash ,,\; l\; \backslash ,,\; m\_l\; \backslash ,,\; m\_s)$. This is in agreement with Pauli exclusion principle.

We can infer from these wavefunctions that $E=\; E\_a+E\_b$.

The corresponding energies are:$$E^\{(0)\}\_\{n\_1,n\_2\}\; =\; E\_\{n\_1\}\; +\; E\_\{n\_2\}\; =\; -\; \backslash frac\{kZ^2e^2\}\{2a\_0\}\; \backslash left[\backslash frac\{1\}\{n\_1^2\}\; +\; \backslash frac\{1\}\{n\_2^2\}\; \backslash right]$$

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 with high numerical accuracy.{{Cite journal |last1=Mitroy |first1=Jim |last2=Bubin |first2=Sergiy |last3=Horiuchi |first3=Wataru |last4=Suzuki |first4=Yasuyuki |last5=Adamowicz |first5=Ludwik |last6=Cencek |first6=Wojciech |last7=Szalewicz |first7=Krzysztof |last8=Komasa |first8=Jacek |last9=Blume |first9=D. |last10=Varga |first10=Kálmán |date=2013-05-06 |title=Theory and application of explicitly correlated Gaussians |url=https://journals.aps.org/rmp/abstract/10.1103/RevModPhys.85.693 |journal=Reviews of Modern Physics |volume=85 |issue=2 |pages=693–749 |doi=10.1103/RevModPhys.85.693|bibcode=2013RvMP...85..693M }}{{rp|709}} For example, the ground state energy of helium has been computed to 40 digits, {{val|-2.903724377034119598311159245194404446696925309||u=hartree}}, but the difference between the value and experiment is not understood.{{Cite journal |last=Schwartz |first=Charles |date=June 2006 |title=EXPERIMENT AND THEORY IN COMPUTATIONS OF THE He ATOM GROUND STATE |url=https://www.worldscientific.com/doi/abs/10.1142/S0218301306004648 |journal=International Journal of Modern Physics E |volume=15 |issue=4 |pages=877–888 |doi=10.1142/S0218301306004648 |arxiv=physics/0208004 |bibcode=2006IJMPE..15..877S |issn=0218-3013}}{{Citation |last=Schwartz |first=Charles |title=Further Computations of the He Atom Ground State |date=2006-05-04 |arxiv=math-ph/0605018 |bibcode=2006math.ph...5018S |id=arXiv:math-ph/0605018}}

Since ground state corresponds to (1,0,0) state, there can only be one representation of such wavefunction whose spatial wavefunction is:$$\backslash psi\_0(r\_1\backslash ,,\; r\_2)\; =\backslash psi\_\{1,0,0\}(r\_1)\; \backslash psi\_\{1,0,0\}(r\_2)\; =\; \backslash frac\{Z^3\}\{\backslash pi\; a\_0^3\}\; \backslash mathrm\{e\}^\{-\backslash frac\{Z(r\_1\; +\; r\_2)\}\{a\_0\}\}$$

We note that the ground state energy of unperturbed helium atom as:$$E\_0\; =\; -\; 4\; \backslash frac\{k\; e^2\}\{a\_0\}\; =\; -108.8\; \backslash text\{\; eV\}$$Which is 30% larger than experimental data.

We can find the first order correction in energy due to electron repulsion in Hamiltonian $\backslash frac\{k\; e^2\}$

The energy for ground state of helium in first order becomes $E\; \backslash approx\; -74.8\; \backslash text\{\; eV\}$ compared to its experimental value of {{val|-79.005154539|(25)|u=eV}}.{{Cite journal |date=2009-07-21 |title=Atomic Spectra Database |url=https://www.nist.gov/pml/atomic-spectra-database |journal=NIST |language=en |doi=10.18434/t4w30f |last1=Kramida |first1=Alexander |last2=Ralchenko |first2=Yuri }} A better approximation for ground state energy is obtained by choosing better trial wavefunction in variational method.

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 {{mvar|Z}} gets bigger, our approach should yield better results, since the electron-electron repulsion term will get smaller.$$\backslash bar\{H\text{'}\}\; =\; \backslash frac\{ke^2\}\{r\_\{12\}\}\; -\; \backslash frac\{kZe^2\}\{r\_1\}\; -V(r\_1)\; -\; \backslash frac\{kZe^2\}\{r\_2\}\; -\; V(r\_2)$$

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

$$V(r)\; =\; -\backslash frac\{k(Z-S)e^2\}\{r\}\; =\; -\; \backslash frac\{kZ\_\backslash mathrm\{e\}\; e^2\}\{r\}$$

where {{mvar|S}} is a screening constant and the quantity {{math|Z_e}} is the effective charge. The potential is a Coulomb interaction, so the corresponding individual electron energies are given by

$$E\_0\; =\; -\backslash frac\{k(Z-S)^2e^2\}\{2a\_0n^2\}\; =\; -\; \backslash frac\{kZ\_\backslash mathrm\{e\}^2e^2\}\{2a\_0n^2\}$$

and the corresponding spatial wave function is given by

$$\backslash psi\_0(r\_1\backslash ,,\; r\_2)\; =\backslash sqrt\{\backslash frac\{Z\_\backslash mathrm\{e\}^3\}\{\backslash pi\; a\_0^3\}\}\; e^\{-Z\_\backslash mathrm\{e\}(r\_1\; +\; r\_2)\}$$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 {{math|1=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 $\backslash frac\{1\}\{3\}$ of the electric charge.{{cite book |last1=Bransden |first1=B. H. |title=Physics of Atoms and Molecules |last2=Joachain |first2=C. J. |publisher=Pearson Education |edition=2nd}}

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

$$\backslash psi\_0(\backslash mathbf\{r\}\_1,\backslash ,\; \backslash mathbf\{r\}\_2)\; =\; \backslash frac\{8\}\{\backslash pi\; a^3\}\; \backslash text\{e\}^\{-2(r\_1+r\_2)/a\}$$

$$\backslash langle\; H\; \backslash rangle\; =\; 8E\_1\; +\; \backslash langle\; V\_\backslash mathrm\{ee\}\; \backslash rangle\; =\; 8E\_1\; +\; \backslash left(\backslash frac\{e^2\}\{4\backslash pi\backslash varepsilon\_0\}\backslash right)\; \backslash left(\backslash frac\{8\}\{\backslash pi\; a^3\}\backslash right)^2\; \backslash int\; \backslash frac\{\backslash text\{e\}^\{-4(r\_1\; +\; r\_2)/a\}\}$$

After integrating this, the result is:

$$\backslash langle\; H\; \backslash rangle\; =\; 8E\_1\; +\; \backslash frac\{5\}\{4a\}\; \backslash left(\backslash frac\{e^2\}\{4\backslash pi\backslash epsilon\_0\}\backslash right)\; =\; 8E\_1\; -\; \backslash frac\{5\}\{2\}E\_1\; =\; -109\; +\; 34\; =\; -75\; \backslash text\{\; eV\}$$

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:

$$\backslash psi(\backslash mathbf\{r\}\_1,\; \backslash mathbf\{r\}\_2)\; =\; \backslash frac\{Z^3\}\{\backslash pi\; a^3\}\; \backslash text\{e\}^\{-Z(r\_1+r\_2)/a\}$$

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

$$\backslash langle\; H\; \backslash rangle\; =\; 2\; Z^2\; E\_1\; +\; 2(Z-2)\; \backslash left(\backslash frac\{e^2\}\{4\backslash pi\backslash varepsilon\_0\}\backslash right)\; \backslash left\backslash langle\; \backslash frac\{1\}\{r\}\; \backslash right\backslash rangle\; +\; \backslash left\backslash langle\; V\_\backslash mathrm\{ee\}\; \backslash right\backslash rangle$$

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

$$\backslash langle\; H\; \backslash rangle\; =\; \backslash left[-2Z^2\; +\; \backslash frac\{27\}\{4\}Z\backslash right]E\_1$$

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:

$$\backslash frac\{d\}\{dZ\}\; \backslash left(\backslash left[-2Z^2\; +\; \backslash frac\{27\}\{4\}Z\backslash right]\; E\_1\backslash right)\; =\; 0$$

$$Z\; =\; \backslash frac\{27\}\{16\}\; \backslash sim\; 1.69$$

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:

$$\backslash frac\{1\}\{2\}\; \backslash left(\backslash frac\{3\}\{2\}\backslash right)^6\; E\_1\; =\; -77.5\; \backslash text\{\; eV\}$$

Where again, {{math|E₁}} represents the ionization energy of hydrogen.{{cite book |first=David I. |last=Griffiths |title=Introduction to Quantum Mechanics |edition=Second |year=2005 |publisher=Pearson Education }}

Consider the same setting where unperturbed Hamiltonian is:$$H\_0\; =\; -\backslash frac\{1\}\{2\}\; \backslash nabla\_\{r\_1\}^2\; -\; \backslash frac\{1\}\{2\}\; \backslash nabla\_\{r\_2\}^2\; -\; \backslash frac\{kZe^2\}\{r\_1\}\; -\; \backslash frac\{kZe^2\}\{r\_2\}$$and perturbation is electron repulsion: $\backslash frac\{k\; e^2\}$

In general, for (1s)(nl) state, in first order perturbation theory:$$E\; =\; E\_\{100\}\; +\; E\_\{nlm\}\; +\; \backslash Delta^\{(1)\}$$with:$$\backslash Delta^\{(1)\}=\; I\; \backslash pm\; J$$where I is known as direct integral and J is known as exchange integral or exchange energy. If the combined spatial wavefunction is symmetric, its energy level has the + symbol in $\backslash Delta^\{(1)\}$, whereas for the antisymmetric combined spatial wavefunction, $\backslash Delta^\{(1)\}$ has a minus symbol. Since due to the symmetrization postulate, the combined spatial wavefunctions differ on symmetric or antisymmetric nature, the J term is responsible for the splitting of energy levels between ortho and para helium states.

They are calculated as:

$$I\; =\; \backslash int\; \backslash int\; |\backslash psi\_\{100\}(\backslash mathbf\; r\_1)|^2\; |\backslash psi\_\{nlm\}(\backslash mathbf\; r\_2)|^2\; \backslash frac\{ke^2\}$$

The Schrodinger equation for helium, like that of hydrogen, can be solved to accuracies equivalent to the most precise experimental values. Among the additional effects that must be included for these high accuracies include:{{Cite book |last=Drake |first=Gordon W. F. |chapter-url=https://link.springer.com/chapter/10.1007/3-540-45395-4_4 |title=The Hydrogen Atom. Lecture Notes in Physics, vol 570. |date=2001 |publisher=Springer |isbn=978-3-540-45395-6 |editor-last=Karshenboim |editor-first=Savely G. |location=Berlin, Heidelberg |pages=57–78 |language=en |chapter=The Hydrogen Atom: Precision Physics of Simple Atomic Systems |doi=10.1007/3-540-45395-4_4 |editor-last2=Bassani |editor-first2=F. |editor-last3=Pavone |editor-first3=F.S. |editor-last4=Inguscio |editor-first4=M. |editor-last5=Hänsch |editor-first5=T.W.}}G.W.F. Drake (2006). "Springer Handbook of Atomic, molecular, and Optical Physics", Edited by G.W.F. Drake (Springer, New York), 199-219. [https://www.amazon.com/Springer-Handbook-Molecular-Optical-Physics/dp/038720802X]

• mass polarization: the dynamics of nucleus around the atomic center of mass.
• relativity: Breit-Pauli corrections
• quantum electrodynamics effects: the Lamb shift due to electron interaction with vacuum fluctuations.

Helium's first ionization energy is {{val|−24.587387936|(25)|u=eV}}.{{cite web|publisher=NIST| url=http://physics.nist.gov/PhysRefData/ASD/ionEnergy.html|title=NIST Atomic Spectra Database Ionization Energies Data| author1=Kramida, A. | author2 = Ralchenko, Yu. | author3 = Reader, J., and NIST ASD Team | location=Gaithersburg, MD}} This value was measured experimentally.{{ cite journal|date=2010|title=Extreme Ultraviolet Frequency Comb Metrology|journal=Phys. Rev. Lett. | volume=105 | issue=6 | page=063001 | bibcode=2010PhRvL.105f3001K | doi=10.1103/PhysRevLett.105.063001 | pmid=20867977 |author=D. Z. Kandula |author2=C. Gohle |author3=T. J. Pinkert |author4=W. Ubachs |author5=K. S. E. Eikema | arxiv=1004.5110 | s2cid=2499460}} The theoretic value of Helium atom's second ionization energy is {{val|−54.41776311|(2)|u=eV}}. The total ground state energy of the helium atom is {{val|-79.005154539|(25)|u=eV}}, or {{val|−2.90338583|(13)|u=Atomic units a.u.}}, which equals {{val|−5.80677166|(26)|u=Ry}}.

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• Araki–Sucher correction
• Fermi heap and Fermi hole
• Isotopes of helium
• Hydrogen molecular ion
• Lithium atom
• List of quantum-mechanical systems with analytical solutions
• Quantum field theory
• Quantum mechanics
• Quantum states
• "Helium atom" on Wikiversity

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