* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download E618: Pertubation theory for Helium atom
Relativistic quantum mechanics wikipedia , lookup
Renormalization wikipedia , lookup
Renormalization group wikipedia , lookup
Bremsstrahlung wikipedia , lookup
Elementary particle wikipedia , lookup
Density functional theory wikipedia , lookup
Nitrogen-vacancy center wikipedia , lookup
Quantum electrodynamics wikipedia , lookup
X-ray fluorescence wikipedia , lookup
Ferromagnetism wikipedia , lookup
Chemical bond wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Wave–particle duality wikipedia , lookup
Rutherford backscattering spectrometry wikipedia , lookup
Electron scattering wikipedia , lookup
Auger electron spectroscopy wikipedia , lookup
X-ray photoelectron spectroscopy wikipedia , lookup
Molecular Hamiltonian wikipedia , lookup
Tight binding wikipedia , lookup
Hydrogen atom wikipedia , lookup
Electron-beam lithography wikipedia , lookup
Atomic orbital wikipedia , lookup
1 E618: Pertubation theory for Helium atom Submitted by: Amir Waxman The problem: The Helium Atom includes 2 electrons with half spin. In this problem, the interaction between the electrons should be treated as a pertubation (1) Write the Hamiltonian that describes the electrons in the atom. (2) Find the stable states of the atom, neglecting the interaction between the electrons. (3) What is the Ionization energy of the Helium atom in the zeroth order of the pertubation theory? (4) What is the first order correction for the ionization energy? (5) What is the leading order correction for the bound excited states? The solution: (1) The unpertubed hamiltonian of the helium atom includes the interactins between the two electrons and the nuclei. the interaction between the electrons is refered to as a pertubation. the total hamiltonian looks like that: H= P2 2 α α e2 P1 2 + − − + 2M 2M r1 r2 |r1 − r2 | while α = ze2 and z = 2 for Helium. (2) first let us define the base we are working in: base = |ν1 l1 m1 σ1 i ⊗ |ν2 l2 m2 σ2 i These are the same quntum numbers that describe the electron in the hydrogen atom, only here we have two electrons. the eigen functions are thus, the same like in those of hydrogen: ϕνlm (r, Ω) = hrΩ|lmνi = Rlν (r)Y lm (Ω) The energies of the unpertubed hamiltonian will be: El1m1ν1,l2m2ν2 = −M α2 1 2 (l1 + ν1 ) + ! 1 2 (l2 + ν2 ) The stable states are those with the energies between Eground = −M α2 (1 + 1) andE∞ = −M α2 (1 + 0) as above E∞ , the atom can be ionized (one of the electrons is taken to infinity). all the states below E∞ are stable, and all the states above it belong to the continum. Thus, the stable states are those with at least one electron in the ground state (l + ν = 1) if we have, for example, two electrons in the excited state, this is not a stable state, as the energy is greater than E∞ . (3) The ionzation energy in the zeroth order will thus be: Eionization = E∞ − Eground = M α2 (4) The first order correction of the energy is: ∆E = hΨ|V |Ψi And in the ground state: Ψ = [|l = 0, ν = 1i ⊗ |l = 0, ν = 1i] ⊗ [singlet] When singlet = √12 (| ↑↓i − | ↓↑i) is the unsymetric spin state. The reason we took only the singlet state (and not the triplet) is that the ”‘general”’ wave function has to be unsymetric to the exchange of the electrons (fermions), 2 and as the orbital function is symetric, the spin function has to be unsymetric.The pertubation is:V = correction to the energy will thus be calculated by the integral on the Radial functions: Z e2 1 dr1 dr2 [R01 (r1 )R01 (r2 )]∗ R01 (r1 )R01 (r2 ) ∆E = 2 π |r1 − r2 | 1 z ∆E = ( )2 ( )6 e2 π a0 Z zr1 dr1 r12 e− a0 Z zr2 dr2 r22 e− a0 Z dΩ1 dΩ2 = 2 2r1 r2 The 1 |r1 − r2 | when we use spherical coordinatesand,and the eigen function are those of the hydrogen atom. a0 = Bhor. first, we have to solve the integral over Ω Z Z 1 1 1 dΩ1 dΩ2 = 4π · 2π dcos(θ) p 2 2 − 2r r cos(θ) |r1 − r2 | + r r −1 1 2 2 1 = 8π 2 e2 |r1 −r2 | 1 me2 is radius q r12 + r22 − 2r1 r2 cos(θ) 8π 2 (r1 + r2 − |r1 − r2 |) r1 r2 = 16π 2 · 1 r> ∆E = 16 z 6 e2 a60 Z ∞ dr1 r12 e− 2zr1 a0 0 Z 0 ∞ dr2 r22 e− 2zr2 a0 1 r> we divide the integral over r1 into two integrals. one for r1 < r2 , and one for r1 > r2 . we then get Z r2 Z Z ∞ 2zr2 r2 2zr1 r2 2zr1 z 6 e2 ∞ dr2 r22 e− a0 · dr1 1 e− a0 + dr1 1 e− a0 ∆E = 16 6 a0 0 r2 r1 0 r2 8z 5 e2 =− 5 a0 Z 4z 4 e2 a40 Z =− =− ∞ dr2 r22 e− 2zr2 a0 dr2 r22 e− 4zr2 a0 0 ∞ 0 2zr 2zr a20 a0 a20 a0 − a 2 − a 2 0 0 + · e (r2 + )− −e (r2 + ) z 2zr2 2zr2 2z + 4zr2 a0 a0 2zr2 r2 e − a0 − e − a0 z z 4z 4 e2 −a0 3 a0 −a0 2 a0 −a0 2 · 2( ) + ( ) − ( ) a40 4z z 4z z 2z ∆E = 5 ze2 5 = M α2 . 8 a0 8 (5) In the first excited state of helium, one of the electrons has n = l + ν = 2 the other one got n = 1. The n = 2 electron can have both l = 1 and l = 0 states. first we deal the case in which both of the electrons got l = 0. The symetrical and unsymetrycal states will be: 1 Ψ+ = √ (|l = 0, ν = 2i ⊗ |l = 0, ν = 1i + |l = 0, ν = 1i ⊗ |l = 0, ν = 2i) ⊗ [singlet](degeneration = 1) 2 3 1 Ψ− = √ (|l = 0, ν = 2i ⊗ |l = 0, ν = 1i − |l = 0, ν = 1i ⊗ |l = 0, ν = 2i) ⊗ [triplet](degeneration = 3) 2 and one can notice we attached a singlet (unsymetric) to the symetric orbital function and a triplet to the unsymetric orbital function in order to keep the general wave function unsymetric. in the case that 1 electron has l = 1 and the other has l = 0 we get: 1 Ψ+ = √ (|l = 1, m, ν = 1i ⊗ |l = 0, ν = 1i + |l = 0, ν = 1i ⊗ |l = 1, m, ν = 1i) ⊗ [singlet](degeneration = 3) 2 1 Ψ− = √ (|l = 1, m, ν = 1i ⊗ |l = 0, ν = 1i − |l = 0, ν = 1i ⊗ |l = 1, m, ν = 1i) ⊗ [triplet](degeneration = 9) 2 we will now want to calculate the change in the energy due to pertubation: ∆E = hΨ|V |Ψi and thus: Z ∆E = ∗ dr1 dr2 ϕ100 (r1 , Ω1 )ϕνlm (r2 , Ω2 ) ± ϕ100 (r2 , Ω2 )ϕνlm (r1 , Ω1 ) 1 ϕ100 (r1 , Ω1 )ϕνlm (r2 , Ω2 )± |r1 −2 | ±ϕ100 (r2 , Ω2 )ϕνlm (r1 , Ω1 ) = J ± K when ϕνlm (r, Ω) was defined in section 2 and ϕ100 (r, Ω) goes for the electron in the ground state. J and K are the self and exchange integral that were discussed in class. note that this is just a general form and we get different J and K values for different quantum numbers. All together we get a 16 × 16 diagonal matrix. the reason that we got a diagonal matrix relates to our choice of base ( two particles base ) where between two states, atleast one of the quantum numbers is different. in a one particle base we would have got off diagonal elements.