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PART 1 Identical Particles, Fermions and Bosons. Pauli Exclusion

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PART 1 Identical Particles, Fermions and Bosons. Pauli Exclusion O. P. Sushkov School of Physics, The University of New South Wales, Sydney, NSW 2052, Australia (Dated: March 1, 2016) PART 1 Identical particles, fermions and bosons. Pauli exclusion principle. Slater determinant. Variational method. He atom. Multielectron atoms, effective potential. Exchange interaction. 2 Identical particles and quantum statistics. Consider two identical particles. 2 electrons 2 protons 2 12C nuclei 2 protons .... The wave function of the pair reads ψ = ψ(~r1, ~s1,~t1...; ~r2, ~s2,~t2...) where r1,s1, t1, ... are variables of the 1st particle. r2,s2, t2, ... are variables of the 2nd particle. r - spatial coordinate s - spin t - isospin . - other internal quantum numbers, if any. Omit isospin and other internal quantum numbers for now. The particles are identical hence the quantum state is not changed under the permutation : ψ (r2,s2; r1,s1) = Rψ(r1,s1; r2,s2) , where R is a coefficient. Double permutation returns the wave function back, R2 = 1, hence R = 1. ± The spin-statistics theorem claims: * Particles with integer spin have R =1, they are called bosons (Bose - Einstein statistics). * Particles with half-integer spins have R = 1, they are called fermions (Fermi − - Dirac statistics). 3 The spin-statistics theorem can be proven in relativistic quantum mechanics. Technically the theorem is based on the fact that due to the structure of the Lorentz transformation the wave equation for a particle with spin 1/2 is of the first order in time derivative (see discussion of the Dirac equation later in the course). At the same time the wave equation for a particle with integer spin is of the second order in time derivative. An example: The vector potential in electrodynamics is to some extent equivalent to the wave equation of photon. The photon has spin S = 1. Maxwell’s equation for the vector potential A~ reads 2 1 ∂ 2 2 2 ~ A~ =4π~j. c ∂t − ∇ h i here ~j is electric current. The equation contains the second time derivative. Comment: Do not mix permutation with parity, these are different operations. permutation : ψ (r ,s ; r ,s ) ψ(r ,s ; r ,s ) 1 1 2 2 → 2 2 1 1 space reflection: ψ (r ,s ; r ,s ) ψ( r ,s ; r ,s ) 1 1 2 2 → − 1 1 − 2 2 4 Example: Statistics influence rotational spectra of diatomic molecules. 12 Consider rotational spectrum of Carbon2 molecule that consists of two C isotops. A 12C nucleus has spin S = 0, hence it is a boson. 12C 12 C r 1 r2 FIG. 1: Rotation of Carbon2 molecule The wave function of the molecule reads Ψ(1, 2) = U [(~r + ~r )/2] V ~r ~r ϕ ,ϕ 1 2 1 − ] 1 2 Here ϕ1 and ϕ2 are spin wave functions of the first and the second nucleus respectively. U is the wave function of the center of mass motion. V is the wave function of the relative motion. Spin of the nucleus is zero, S = 0. Hence ϕ1 = ϕ2 =1. V (~r ~r )= χ( ~r ~r )Y (~r ~r ) 1 − 2 | 1 − 2 | lm 1 − 2 where Ylm is spherical harmonic. Let us perform permutation of the particles V (2, 1) = χ ( ~r ~r ) Y (~r ~r )=( 1)l χ( ~r ~r )Y (~r ~r )=( 1)l V (1, 2) . | 2 − 1| lm 2 − 1 − | 1 − 2 | lm 1 − 2 − Here I have used the exact mathematical relation (see 3d year quantum mechanics) Y ( ~r)=( 1)l Y (~r) . lm − − lm Requirement of Bose statistics: ψ (2, 1) = ψ (1, 2) Hence only even values of l are allowed in the rotational spectrum of C2 consisting of two 12C isotops. 5 In a molecule consisting of two identical 12C isotops only even values of l are allowed. In a molecule consisting of two different isotops, say 12C and 13C or 12C and 14C all values of l are allowed. l=4 l=4 l=3 l=2 l=2 l=1 l=0 l=0 12CC12 12CC14 FIG. 2: Rotational spectra of C2 molecule. Left: two identical nuclei each with spin S=0. Right: two distinguishable nuclei each with spin S=0. 6 Two noninteracting fermions in an external potential. Consider any external potential: Coulomb field of a nucleus, a potential well, etc. Let ϕa and ϕb be single particle states in the potential. For an infinite potential well these are simple standing wave as it is illustrated in the Fig. below ϕ ϕ b a FIG. 3: Two lowest single particle orbital quantum states in an infinite square well potential. Let us put two electrons with parallel spins in the potential. Requirement of Fermi statistics: ψ (1, 2) = ψ (2, 1). − Therefore, the many-body (in this case “many” = 2) wave function is 1 Ψ (1, 2) = [ϕa (r1) ϕb (r2) ϕa (r2) ϕb (r1)] 1 2 (1) √2 − | ↑i | ↑i If ϕ = ϕ then Ψ 0. a b ≡ Thus, one cannot put two fermions in the same single-particle quantum state. This is Pauli exclusion principle. If spins of the electrons are opposite then the single particle states are different, ϕ = ϕ , a↑ 6 a↓ even if the coordinate states are identical. So, such two-electron state is possible. 1 Ψ(1, 2) = ϕa (r1) ϕa (r2) [ 1 2 1 2] (2) √2 | ↑i | ↓i −| ↓i | ↑i 7 Slater determinant Let us introduce spin in the definition of the single particle orbitals, and let us enumerate these orbitals by index i : In these notations orbitals of the previous example are ϕ (r, s)= ϕ ϕ (r) 1 a↑ ≡ a | ↑i ϕ (r, s)= ϕ ϕ (r) 2 a↓ ≡ a | ↓i ϕ (r, s)= ϕ ϕ (r) 3 b↑ ≡ b | ↑i ϕ (r, s)= ϕ ϕ (r) 4 b↓ ≡ b | ↓i Hence the wave function (1), page 6 can be written as determinant 1 1 ϕ1(1) ϕ1(2) Ψ (1, 2) = [ϕ1 (1) ϕ3 (2) ϕ1 (2) ϕ3 (1)] √2 − ≡ √2 ϕ3(1) ϕ3(2) Similarly, the wave function (2), page 6 also can be written as determinant. 1 1 ϕ1(1) ϕ1(2) Ψ (1, 2) = [ϕ1 (1) ϕ2 (2) ϕ1 (2) ϕ2 (1)] √2 − ≡ √2 ϕ2(1) ϕ2(2) I repeat the meaning of the notation ϕi(x)). Here the index i enumerates single particle orbitals and x shows coordinates (spatial, spin, etc.) of a particle. Using these notations we can write the many-body wave function for arbitrary number of noninteracting fermions ϕ1(1) ϕ1(2) .... ϕ1(N) ϕ (1) ϕ (2) .... ϕ (N) 2 2 2 1 Ψ(1, 2, ...N)= .... .... .... .... √ N! .... .... .... .... ϕ (1) ϕ (2) .... ϕ (N) N N N This form was suggested by Slater and it is called Slater determinant. 8 Permutation of two particles is equivalent to permutation of two columns. For Example ϕ1(2) ϕ1(1) .... ϕ1(N) ϕ (2) ϕ (1) .... ϕ (N) 2 2 2 1 Ψ(2, 1, ...N)= .... .... .... .... = Ψ(1, 2, ...N) . √ − N! .... .... .... .... ϕ (2) ϕ (1) .... ϕ (N) N N N So, the Fermi statistics requirement is automatically satisfied. If two orbitals in the determinant coincide: i = j; then the determinant vanishes because there are two identical lines. This describes the Pauli exclusion principle. Slater determinant is a very convenient form for the wave function. Unfortunately this form is exact only for noninteracting fermions. 9 Interacting fermions. Variational solution for He atom ground state Hamiltonian of He atom ~p 2 ~p 2 Ze2 Ze2 e2 Hˆ = 1 + 2 + 2m 2m − r − r ~r ~r 1 2 | 1 − 2| Here we neglect all relativistic/magnetic effects. The relative magnitude of these effects is v2/c2 (Zα)2 =(Z/137)2 1. ∼ ∼ ≪ Schrodinger equation Hψ (1, 2) = Eψ (1, 2) can be solved exactly numerically but: 1) The solution is very involved technically. 2) There is no exact solution for three (Li atom) and more electrons. So, we need an approximate but relatively simple method which can be propagated to multi-electron systems. Variational method Energy of the system reads Ψ Hˆ Ψ = Ψ∗Hˆ Ψd3r d3r d3r ... | | 1 2 3 D E Z The wave function is normalized Ψ Ψ = Ψ∗Ψd3r d3r d3r ... =1 h | i 1 2 3 Z Let us find minimum of energy with respect to variation of Ψ∗. To account for the normalization constraint let us use the Lagrange multiplier method with λ being the Lagrange multiplier δ [ Ψ∗H Ψ λ Ψ∗ Ψ ]=0 δΨ∗ (x) h | i − h | i Hψˆ λψ =0 Hψˆ = λψ λ = E. ⇒ − ⇒ ⇒ So, we end up with usual Schrodinger equation. In this case the number of variational parameters = , Ψ∗(x) at each point x. The variational method does not bring anything ∞ new. 10 For practical applications we choose a finite number of parameters and hence the variational method gives an approximate answer. Example He-like ion ground state Hydrogen-like ion: single electron in Coulomb field of a nucleus with charge Z (see 3d-year quantum mechanics.) The Hamiltonian: Z FIG. 4: Hydrogen-like ion p2 Ze2 H = 2m − r The electron ground state energy: Z2me4 ǫ = − 2¯h4 The electron wave function: ϕ(r)= Ae−Zr/aB Z3 A = 3 sπaB aB is Bohr radius, and A is the normalization constant. Atomic units: E Eatomic = me4/h¯4 me4 = 27.2eV h¯4 ratomic = r/aB h¯2 a = 0.53A˚ =0.53 10−8cm B me2 ≈ 11 Below I omit the subscript “atomic”. The electron Hamiltonian, energy, and wave function in atomic units are p2 Z ∆ Z H = = 2 − r − 2 − r Z2 ǫ = − 2 ϕ(r)= Ae−Zr Z3 A = r π Now consider He-like ion.
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