The quantum mechanical description of the helium atom is of special interest, because it is the simplest multielectron system and can be used to understand the concept of quantum entanglement. The Hamiltonian of helium, considered as a threebody system of two electrons and a nucleus and after separating out the centreofmass motion, can be written as
H
(
r
→
1
,
r
→
2
)
=
∑
i
=
1
,
2
(
−
ℏ
2
2
μ
∇
r
i
2
−
Z
e
2
4
π
ϵ
0
r
i
)
−
ℏ
2
M
∇
r
1
⋅
∇
r
2
+
e
2
4
π
ϵ
0
r
12
where
μ
=
m
M
m
+
M
is the reduced mass of an electron with respect to the nucleus,
r
→
1
and
r
→
2
are the electronnucleus distance vectors and
r
12
=

r
1
→
−
r
2
→

. The nuclear charge,
Z
is 2 for helium. In the approximation of an infinitely heavy nucleus,
M
=
∞
we have
μ
=
m
and the mass polarization term
ℏ
2
M
∇
r
1
⋅
∇
r
2
disappears. In atomic units the Hamiltonian simplifies to
H
(
r
→
1
,
r
→
2
)
=
−
1
2
∇
r
1
2
−
1
2
∇
r
2
2
−
Z
r
1
−
Z
r
2
+
1
r
12
.
It is important to note, that it operates not in normal space, but in a 6dimensionsional configuration space
(
r
→
1
,
r
→
2
)
. In this approximation (Pauli approximation) the wave function is a second order spinor with 4 components
ψ
i
j
(
r
→
1
,
r
→
2
)
, where the indices
i
,
j
=
↑
,
↓
describe the spin projection of both electrons (zdirection up or down) in some coordinate system. It has to obey the usual normalization condition
∑
i
j
∫
d
r
→
1
d
r
→
2

ψ
i
j

2
=
1
. This general spinor can be written as 2x2 matrix
ψ
=
(
ψ
↑↑
ψ
↑↓
ψ
↓↑
ψ
↓↓
)
and consequently also as linear combination of any given basis of four orthogonal (in the vectorspace of 2x2 matrices) constant matrices
σ
k
i
with scalar function coefficients
ϕ
i
k
(
r
→
1
,
r
→
2
)
as
ψ
=
∑
i
k
ϕ
i
k
(
r
→
1
,
r
→
2
)
σ
k
i
. A convenient basis consists of one antisymmetric matrix (with total spin
S
=
0
, corresponding to a singlet state)
σ
0
0
=
1
2
(
0
1
−
1
0
)
=
1
2
(
↑↓
−
↓↑
)
and three symmetric matrices (with total spin
S
=
1
, corresponding to a triplett state)
σ
0
1
=
1
2
(
0
1
1
0
)
=
1
2
(
↑↓
+
↓↑
)
,
σ
1
1
=
(
1
0
0
0
)
=
↑↑
,
σ
−
1
1
=
(
0
0
0
1
)
=
↓↓
.
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. Since all spin interaction terms between the four components of
ψ
in the above (scalar) Hamiltonian are neglected (e.g. an external magnetic field, or relativistic effects, like spinorbit interactions), the four Schrödinger equations can be solved independently.
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
ψ
i
j
(
r
→
1
,
r
→
2
)
=
−
ψ
j
i
(
r
→
2
,
r
→
1
)
_
.
Parahelium is then the singlet state
ψ
=
ϕ
0
(
r
→
1
,
r
→
2
)
σ
0
0
with a symmetric function
ϕ
0
(
r
→
1
,
r
→
2
)
=
ϕ
0
(
r
→
2
,
r
→
1
)
and orthohelium is the triplet state
ψ
m
=
ϕ
1
(
r
→
1
,
r
→
2
)
σ
m
1
,
m
=
−
1
,
0
,
1
with an antisymmetric function
ϕ
1
(
r
→
1
,
r
→
2
)
=
−
ϕ
1
(
r
→
2
,
r
→
1
)
. If the electronelectron interaction term is ignored, both spatial functions
ϕ
x
,
x
=
0
,
1
can be written as linear combination of two arbitrary (orthogonal and normalized) oneelectron eigenfunctions
φ
a
,
φ
b
:
ϕ
x
=
1
2
(
φ
a
(
r
→
1
)
φ
b
(
r
→
2
)
±
φ
a
(
r
→
2
)
φ
b
(
r
→
1
)
)
or for the special cases of
φ
a
=
φ
b
(both electrons have identical quantum numbers, parahelium only):
ϕ
0
=
φ
a
(
r
→
1
)
φ
a
(
r
→
2
)
. The total energy (as eigenvalue of
H
) is then for all cases
E
=
E
a
+
E
b
(independent of the symmetry).
This explains the absence of the
1
3
S
1
state (with
φ
a
=
φ
b
=
φ
1
s
) for orthohelium, where consequently
2
3
S
1
(with
φ
a
=
φ
1
s
,
φ
b
=
φ
2
s
) is the metastable ground state. (A state with the quantum numbers: principal quantum number
n
, total spin
S
, angular quantum number
L
and total angular momentum
J
=

L
−
S

…
L
+
S
is denoted by
n
2
S
+
1
L
J
.)
If the electronelectron interaction term
1
r
12
is included, the Schrödinger equation is non separable. However, also if is neglected, all states described above (even with two identical quantum numbers, like
1
1
S
0
with
ψ
=
φ
1
s
(
r
→
1
)
φ
1
s
(
r
→
2
)
σ
0
0
) cannot be written as a product of oneelectron wave functions:
ψ
i
k
(
r
→
1
,
r
→
2
)
≠
χ
i
(
r
→
1
)
ξ
k
(
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 "manybody 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 two electrons can be written as a sum of the Hamiltonians for each electron:
H
=
∑
i
=
1
2
h
(
i
)
=
H
0
+
H
′
where the zeroorder unperturbed Hamiltonian is
H
0
=
−
1
2
∇
r
1
2
−
1
2
∇
r
2
2
−
Z
r
1
−
Z
r
2
while the perturbation term:
H
′
=
1
r
12
is the electronelectron interaction. H_{0} is just the sum of the two hydrogenic Hamiltonians:
H
0
=
h
^
1
+
h
^
2
where
h
^
i
=
−
1
2
∇
r
i
2
−
Z
r
i
,
i
=
1
,
2
E_{ni}, the energy eigenvalues and
ψ
n
i
,
l
i
,
m
i
(
r
→
i
)
, the corresponding eigenfunctions of the hydrogenic Hamiltonian will denote the normalized energy eigenvalues and the normalized eigenfunctions. So:
h
^
i
ψ
n
i
,
l
i
,
m
i
(
r
i
→
)
=
E
n
i
ψ
n
i
,
l
i
,
m
i
(
r
i
→
)
where
E
n
i
=
−
1
2
Z
2
n
i
2
in a.u.
Neglecting the electronelectron repulsion term, the Schrödinger equation for the spatial part of the twoelectron wave function will reduce to the 'zeroorder' equation
H
0
ψ
(
0
)
(
r
→
1
,
r
→
2
)
=
E
(
0
)
ψ
(
0
)
(
r
→
1
,
r
→
2
)
This equation is separable and the eigenfunctions can be written in the form of single products of hydrogenic wave functions:
ψ
(
0
)
(
r
→
1
,
r
→
2
)
=
ψ
n
1
,
l
1
,
m
1
(
r
→
1
)
ψ
n
2
,
l
2
,
m
2
(
r
→
2
)
The corresponding energies are (in atomic units, hereafter a.u.):
E
n
1
,
n
2
(
0
)
=
E
n
1
+
E
n
2
=
−
Z
2
2
[
1
n
1
2
+
1
n
2
2
]
Note that the wave function
ψ
(
0
)
(
r
→
2
,
r
→
1
)
=
ψ
n
2
,
l
2
,
m
2
(
r
→
1
)
ψ
n
1
,
l
1
,
m
1
(
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 twoelectron atoms must either be symmetric or antisymmetric with respect to the interchange of the coordinates
r
→
1
and
r
→
2
of the two electrons. The proper wave function then must be composed of the symmetric (+) and antisymmetric() linear combinations:
ψ
±
(
0
)
(
r
→
1
,
r
→
2
)
=
1
2
[
ψ
n
1
,
l
1
,
m
1
(
r
→
1
)
ψ
n
2
,
l
2
,
m
2
(
r
→
2
)
±
ψ
n
2
,
l
2
,
m
2
(
r
→
1
)
ψ
n
1
,
l
1
,
m
1
(
r
→
2
)
]
This comes from Slater determinants.
The factor
1
2
normalizes
ψ
±
(
0
)
. In order to get this wave function into a single product of oneparticle 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
ψ
−
(
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
ψ
0
(
0
)
(
r
→
1
,
r
→
2
)
=
ψ
1
(
r
1
→
)
ψ
1
(
r
2
→
)
=
Z
3
π
e
−
Z
(
r
1
+
r
2
)
Where
ψ
1
and
ψ
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
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. (
≃
54.4
eV). The experimental values are E
0
=
−
2.90
a.u. (
≃
−
79.0
eV) and V
p
=
.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 electronelectron repulsion term will get smaller.
So far a very crude independentparticle approximation has been used, in which the electronelectron repulsion term is completely omitted. Splitting the Hamiltonian showed below will improve the results:
H
=
H
0
¯
+
H
′
¯
where
H
0
¯
=
−
1
2
∇
r
1
2
+
V
(
r
1
)
−
1
2
∇
r
2
2
+
V
(
r
2
)
and
H
′
¯
=
1
r
12
−
Z
r
1
−
V
(
r
1
)
−
Z
r
2
−
V
(
r
2
)
V(r) is a central potential which is chosen so that the effect of the perturbation
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
)
=
−
Z
−
S
r
=
−
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
ψ
0
(
r
1
r
2
)
=
Z
e
3
π
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 E_{0} = 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
1
3
of the electronic charge.
Not long after Schrödinger developed the wave equation, the Thomas–Fermi model was developed. Density functional theory is used to describe the particle density
ρ
(
r
→
)
,
r
ϵ
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 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:
ξ
=
3
5
γ
∫
R
3
ρ
5
/
3
(
r
→
)
d
3
r
→
+
∫
R
3
V
(
r
→
)
ρ
(
r
→
)
d
3
r
→
+
e
2
2
∫
R
3
ρ
(
r
→
)
ρ
(
r
′
→
)

r
→
−
r
′
→

d
3
r
→
d
3
r
′
→
Where
γ
=
(
3
π
2
)
2
/
3
ℏ
2
2
m
The electron density needs to be greater than or equal to 0,
∫
R
3
ρ
=
N
, and
ρ
→
ξ
is convex.
In the energy functional, each term holds a certain meaning. The first term describes the minimum quantummechanical kinetic energy required to create the electron density
ρ
(
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
(
r
→
)
. The final term is the electronelectron repulsion potential energy.
So the Hamiltonian for a system of many electrons can be written:
H
=
∑
i
=
1
N
[
−
ℏ
2
2
m
∇
i
2
+
V
(
r
i
→
)
]
+
∫
e
2
ρ
(
r
′
→
)

r
→
−
r
′
→

d
3
r
′
For helium, N = 2, so the Hamiltonian is given by:
H
=
−
ℏ
2
2
m
(
∇
1
2
+
∇
2
2
)
+
V
(
r
1
→
,
r
2
→
)
+
∫
e
2
ρ
(
r
′
→
)

r
→
−
r
′
→

d
3
r
′
Where
∫
e
2
ρ
(
r
′
→
)

r
→
−
r
′
→

d
3
r
′
=
e
2
4
π
ϵ
0
1

r
→
1
−
r
→
2

,
and
V
(
r
1
→
,
r
2
→
)
=
e
2
4
π
ϵ
0
[
2
r
1
+
2
r
2
]
yielding
H
=
−
ℏ
2
2
m
(
∇
1
2
+
∇
2
2
)
+
e
2
4
π
ϵ
0
[
2
r
1
+
2
r
2
−
1

r
→
1
−
r
→
2

]
From the Hartree–Fock method, it is known that ignoring the electronelectron repulsion term, the energy is 8E_{1} = 109 eV.
To obtain a more accurate energy the variational principle can be applied to the electronelectron potential V_{ee} using the wave function
ψ
0
(
r
→
1
,
r
→
2
)
=
8
π
a
3
e
−
2
(
r
1
+
r
2
)
/
a
:
⟨
H
⟩
=
8
E
1
+
⟨
V
e
e
⟩
=
8
E
1
+
(
e
2
4
π
ϵ
0
)
(
8
π
a
3
)
2
∫
e
−
4
(
r
1
+
r
2
)
/
a

r
1
→
−
r
2
→

d
3
r
→
1
d
3
r
→
2
After integrating this, the result is:
⟨
H
⟩
=
8
E
1
+
5
4
a
(
e
2
4
π
ϵ
0
)
=
8
E
1
−
5
2
E
1
=
−
109
+
34
=
−
75
e
V
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:
ψ
(
r
→
1
,
r
→
2
)
=
Z
3
π
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:
⟨
H
⟩
=
2
Z
2
E
1
+
2
(
Z
−
2
)
(
e
2
4
π
ϵ
0
)
⟨
1
r
⟩
+
⟨
V
e
e
⟩
After calculating the expectation value of
1
r
and V_{ee} the expectation value of the Hamiltonian becomes:
⟨
H
⟩
=
[
−
2
Z
2
+
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:
d
d
Z
(
[
−
2
Z
2
+
27
4
Z
]
E
1
)
=
0
Z
=
27
16
∼
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:
1
2
(
3
2
)
6
E
1
=
−
77.5
e
V
Where again, E_{1} represents the ionization energy of hydrogen.
By contrast, we can also use the formula to obtain the best result:
−
49
17
⋅
(
1
+
α
)
=
−
2.903386486
a
.
u
.
=
−
79.005153
e
V
Where
α
is the finestructure constant.
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 for a comprehensive regime of quantum states by G.W.F. Drake and coworkers as well as J.D. Morgan III, Jonathan Baker and Robert Hill using Hylleraas or FrankowskiPekeris basis functions. It should be noted that one needs to include relativistic and quantum electrodynamic corrections to get full agreement with experiment to spectroscopic accuracy.
Helium's first ionization energy is 24.587387936(25)eV. This value was derived by experiment. The theoretic value of Helium atom's second ionization energy is 54.41776311(2)eV. The total ground state energy of the helium atom is 79.005151042（40）eV, or 2.90338583(13）a.u.