Where me , is the mass of electron, r₁ and r₂ is the distances of the electron 1 and 2 from the nucleus, r₁₂ is the distance from electron 1 to 2. The first two terms are operators for the electron's kinetic energy.
The 3rd and 4th terms are the potential energies of attraction between the nucleus and the electron. The final term is potential energy of interelectronic repulsion. The potential of the system of interacting particles can't be write as sum of potential energies of the individual particles. The properties of the potential energy of the system as a whole.
The schrodinger (eqn) involves 6 independent variables, 3 coordinates for each electron,. In spherical coordinates,
φ= φ(r₁,θ₁,ɸ₁,r₂,θ₂ɸ₂).
The operator ∇ ²₁ is given with r₁,θ₁,ɸ₁ replacinɡ r,θ,ɸ. The variable r₁₂ is
r₁₂= [ (x₁-x₂)² +(y₁-y₂)² +(z₁-z₂)²]¹/₂ and usinɡ the relation between spherical and Cartesian coordinates , we can express r₁₂ in terms of r₁,θ₁,ɸ₁,r₂,θ₂,ɸ₂.
Because of the e²/4π∈r₁₂ the term, Schrodinger (eqn) for He can't be separated by coordinate system. We can use approximation methods. The perturbation method separate by Hamiltonian into two parts. Ĥ⁰ and Ĥ ̍, where Ĥ⁰ is the Hamiltonian of exactly solvable problem. If we can consider,
equation.......... (3)
Then, Ĥ⁰ is the sum of two hydrogenlike Hamiltonians, one for each electron,
Ĥ⁰ = Ĥ⁰₁+H₂⁰........equation.. (4)
The unperturbed system of the helium element atom, in which two electrons exert there is no forces on each other. However such system doesn't exist, and this doesn't prevent us from applying perturbation theory to this system.
Since unperturbed Hamiltonian sum of hamiltonian for two independent particles. We can the variables of separation to conclude the unperturbed wave functions has form,
φ(0)(r₁,θ₁,ɸ₁,r₂,θ₂,ɸ₂) = F₁(r₁,θ₁,ɸ₁)F₂(r₂,θ₂,ɸ₂)......equation...(6)
and the unperturbed energies are
E(⁰)= E₁+E₂.........equation.....(7)
Ĥ₁⁰=E¹F₁, Ĥ⁰₂F₂=E₂F₂.....equation...(8)
Since Ĥ₁⁰ and Ĥ⁰₂ are hydrogenlike Hamiltonians, the solution are the hydrogen - like eigenfunction and eigenvalues. (6) we have 
equation.... (9)
The lowest level function have n1= 1,n2=1, and it's zeroth-order wave function is,
equation....... (11)
Where 1s(1) 1(s)2 denotes the product of the hydrogen-like 1s function for electrons 1 and 2, where the subscript indicates both electrons in hydrogen-like 1s orbitals (Note : procedure of assigning electrons to orbitals and writing the atomic wave function as a product of one-electron orbitals function is approximation)
The unperturbed energy of ground state is, 
equation.......... (12)
The quantity - e²/8π∈₀a₀ is the hydrogen atom is the ground state energy level. ( taking the nucleus to be infinitely) and equals - 13.606eV, if the electron mass (me) in a₀ is replaced mass for ⁴He,−e²/8π∈₀a₀ is changed to - 13.604eV. and ew can use this number to partly correct for the nuclear motion in Helium atom. For He, Z=2 and gives eqn(12) gives - 8(13.604eV)= - 108.83eV.
E⁽⁰⁾₁s²= -108.83 eV.....equation.. (13)
This experimental is first ionization energy of Helium is 24.587eV. The second ionization energy of Helium is calculated theoretically, since it's ionization energy of the hydrogen-like ion He+ and is equal to 2²(13.604eV)= 54.416 eV. If we can choose zero of energy as the completely ionised atom then ground state energy of the He atom is - (24.587+54.416)eV =- 79.00 eV. The zeroth order energy eqn (13) in error by 38%. Then we can expected such a large error. Since the perturbation term e²/4π∈₀r₁₂ is a not small.
Next step is to evaluate the first order perturbation correction to the energy. The second unperturbed ground state is always nondegenerate. And it's gives
E⁽¹⁾= ⟨ ψ⟩⁽⁰⁾[Ĥ']ψ⁽⁰⁾⟩
Equation....... (14)
The total volume element for this two-electron problem contains the coordinates of both electrons. dT1, dT2. The integral eqn(14) can be evaluated by using of 1/r₁₂ in term of spherical hormonics, as outlined. 
Equation...... (15)
Recall that 1/2e²/4πε0α0 is equal to 13.604 eV when the ⁴He reduced mass and putting Z=2, we can find the first order perturbation energy correction for the He ground state.
E⁽¹⁾= 10/4(1.3604eV)=34.01eV
Our approximation to the energy is,
E⁽⁰⁾+E⁽¹⁾= - 108.83 eV+ 34.01eV= - 74.82eV..........Equation (16)
Compare with the experimental value of - 79.00eV,is in error by 5.3%.
The evaluate the first order correction to the wavefunction and higher order correcton to the energy requires evaluating the mixture of elements of 1/r₁₂ between the ground unperturbed state and total excited states including the continuum and perform the appropriate summation and integration,. Still no one has yet figured out how to evaluate directly all the contribution to E⁽¹⁾.
That effect of ψ⁽¹⁾ is to mixing into the wavefunction from other contributions besides. 1s² we can call this configuration interaction. The largest contribution to the truly ground state wave function of He comes from the 1s² configuration which is zero- order wave function.
E⁽²⁾ for the He ground state have been calculated using this variation - perturbation method.
Scientists knight and Scherr used 100 term trial functions to get highly accurate approximations to the wave function correction through six order and thus to energy correction through 13th order.
For calculations of the energy correction through order 401, In 1990's J. D. BAKER - The second order corrction E⁽²⁾ turns out to be - 4.29eV and E⁽³⁾ is + 0.12 eV. Through the 3rd order, we have for the ground state energy.
E≈−108.83eV+ 34.01eV-4.29eV+0.12eV=-78.99eV
Which is the close to experiment value - 79.00 eV. Including corrections through 13th order. Knight and scherr obtained a ground state Helium energy of - 2.90372433(e²/4πε0 )whic is close to the value - 2.90372438(e²/4πε0α0) obtained from the pure variational calculations described.
Hook's - law atom or harmonium with Hamiltonian operator 
Here, r1 and r2 are the distances of the electrons from the origin For certain values of the forces-constant k, accurate and exact value ground state wave function and energies has been found. 
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Thanks for reading