PART 1 Identical Particles, Fermions and Bosons. Pauli Exclusion

Home , Exchange interaction, Slater determinant

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

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:

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. He-like ion Hamiltonian in atomic units reads ∆ ∆ Z Z 1 Hˆ = 1 2 + − 2 − 2 − r − r r r 1 2 | 1 − 2| Let us consider the two electron variational wave function of the following form

Ψ= ϕ (1) ϕ (2)Φs 1 Φs = [ 1 2 1 2] √2 | ↑i | ↓i −|↓i | ↑i

where Φs is the spin wave function corresponding to total spin zero. For electron orbital ϕ we use the hydrogen-like ansatz,

Z3 ϕ (r)= eff e−Zeff r , s π where Zeff is a variational parameter.

According to the variational method we have to calculate energy of the system and then minimize it with respect to variation of Zeff . 1 Z Z 1 E = Ψ H Ψ = Ψ ∆ + ∆ Ψ Ψ + Ψ + Ψ Ψ h | | i −2 h | 1 2| i − r r ~r ~r 1 2 1 2 | −

Ψ ∆ Ψ = Φ†ϕ∗ (2) ϕ∗ (1)∆ ϕ (1) ϕ (2)Φ d3r d3r h | 1| i s 1 s 1 2 Z = Φ† Φ ϕ∗ (2) ϕ (2) d3r ϕ∗ (1)∆ ϕ (1) d3r s| s 2 1 1 Z Z ∗ 3 = ϕ (1) ∆1ϕ (1) d r1 Z 12

Assignment problem:

∗ 3 2 ϕ (1)∆1ϕ (1) d r1 = Zeff Z Remember that

1 d d ∆= r2 r2 dr dr

Z Z Ψ Ψ = Φ†ϕ∗ (2) ϕ∗ (1) ϕ (1) ϕ∗ (2)Φ d3r d3r r s r s 1 2 1 Z 1 Z = Φ† Φ ϕ∗ (2) ϕ (2) d3r ϕ∗ (1) ϕ (1) d3r s| s 2 r 1 Z Z 1 Z = ϕ∗ (1) ϕ (1) d3r r 1 Z 1 Assignment problem:

Z ϕ∗ (1) ϕ (1) d3r = ZZ r 1 eff Z 1

1 1 Ψ Ψ = Φ†ϕ∗ (2) ϕ∗ (1) ϕ (1) ϕ (2) Φ d3r d3r ~r ~r r r s 1 2 | 1 − 2| Z | 1 − 2| 1 = Φ∗ Φ ϕ∗ (2) ϕ∗ (1) ϕ (1) ϕ (2) d3r d3r h s| si ~r ~r 1 2 Z | 1 − 2| 1 = ϕ∗ (2) ϕ∗ (1) ϕ (1) ϕ (2) d3r d3r ~r ~r 1 2 Z | 1 − 2| Assignment problem:

1 ϕ∗ (2) ϕ∗ (1) ϕ (1) ϕ (2) d3r d3r ~r ~r 1 2 Z | 1 − 2| Z6 1 = eff e−2Zeff r1 e−2Zeff r2 r2dr dΩ r2dr dΩ π2 r r 1 1 1 2 2 2 Z | 1 − 2| 5 = Z 8 eff 13

Altogether the energy is

5 E = Ψ H Ψ = Z2 2ZZ + Z h | | i eff − eff 8 eff

It is minimum at Z = Z 5 . eff − 16 The physical energy which is the minimum energy is

5 2 27 2 E = Ψ H Ψ = Z h | | i − − 16 → − 16 Z=2

In electron volts 27 2 E = 27.2= 77.38eV . − 16 − Experiment: E = 78.9eV , so the simple variational solution works remarkably well. − 14

Multi-electron atom and eﬀective self-consistent potential

A particular electron

Averaged electron cloud

FIG. 5: A cartoon of multielectron atom. A particular electron which we consider is shown by the black line. Other electrons which together with nucleus produce the eﬀective potential for the “black” electron are shown by red.

The eﬀective potential method reduces (approximately) the many-body problem to a single particle problem. H many-body H , “sp” stands for single-particle → sp

p H = 2 + V (r) (3) sp 2m eff For a neutral atom

2 aB Ze r << Z : Veff = r − 2 (4)  r >>a : V = e  B eff − r  Veff

−e2 r

−Ze2 r

FIG. 6: A sketch of the eﬀective potential of a neutral atom. 15

The effective potential is different from the simple Coulomb potential of a point-like nucleus. Therefore the hydrogen degeneracy of states with the same principal quantum numbers is lifted. Energy levels in a Coulomb potential Energy levels in the effective atomic potential

3d 3p 3s 3p 3d 3s 2p 2s 2p 2s

1s 1s

FIG. 7:

In the periodic table of elements the levels are ﬁlled from down to up. 16

Exchange interaction

Consider two lowest excitations of He atom. The experimental spectrum is as follows

Spectroscopic Energy notation 1 1s2s S0 ——————— 20.62eV 3 1s2s S1 ——————— 19.82eV

ground state 2 1 1s S0 ——————— 0.eV

2S+1 ˆ ˆ ˆ The spectroscopic notation is LJ , J~ = L~ + S~

1 For example S0 means that

Letter S L =0 ⇒ Left superscript =1 2S +1=1 S =0 spin ⇒ ⇒ Right subscript =0 J = L + S =0+0=0 ⇒

The states 1s2s, 1S > and 1s2s, 3S > diﬀer by total spin only, | 0 | 4 0 S =  1  Question:  The interaction is spin independent (Coulomb). Why the states have diﬀerent energies? 17

Single particle orbitals

Z3 ϕ (r) ϕ (r)= eff e−Zeff r 1 ≡ 1s π ϕ (r)= ϕ (r) someq function with one node  2 2s − 

We do not need an explicit form of ϕ1(r) and ϕ1(r) for our discussion. The two electron wave functions (Slater determinants) are

1 1 1 1s2s, S0 = [ϕ1(r1)ϕ2(r2)+ ϕ1(r2)ϕ2(r1)] [ 1 2 1 2] | i √2 × √2 | ↑i | ↓i −| ↓i | ↑i 3 1 1s2s, S1 = [ϕ1(r1)ϕ2(r2) ϕ1(r2)ϕ2(r1)] 1 2 | i √2 − ×| ↑i | ↑i Reminder: Spin wave functions of two electrons are

1 S =0,Sz =0 = [ 1 2 1 2] | i √2 | ↑i | ↓i −|↓i | ↑i S =1,S =1 = | z i | ↑i1| ↑i2 1 S =1,Sz =0 = [ 1 2 + 1 2] | i √2 | ↑i | ↓i | ↓i | ↑i S =1,S = 1 = | z − i | ↓i1| ↓i2

Let us calculate contributions to energies that come from the Coulomb interaction between electrons

1 1 1 E1 = S S S0 0 ~r ~r 0 | 1 − 2 1 1 = [ϕ ∗(1)ϕ∗(2) + ϕ∗(2)ϕ∗(1)] [ϕ (1)ϕ (2) + ϕ (2)ϕ (1)] 2 1 2 1 2 ~r ~r 1 2 1 2 Z | 1 − 2| ϕ (r ) 2 ϕ (r ) 2 ϕ∗(1)ϕ (2)ϕ∗(2)ϕ (1) = | 1 1 | | 2 2 | d3r d3r + 1 1 2 2 d3r d3r ~r ~r 1 2 ~r ~r 1 2 Z | 1 − 2| Z | 1 − 2|

ϕ ϕ ϕ 1 1 ϕ1 2 r1 r1

ϕ r2 ϕ r2 ϕ 2 2 ϕ2 1

FIG. 8: Feynman-like diagrams for the direct and the exchange interaction 18

3 1 3 E3 = S S S1 1 ~r ~r 1 1 − 2 1 1 = [ϕ ∗(1)ϕ∗(2) ϕ∗(2)ϕ∗(1)] [ϕ (1)ϕ (2) ϕ (2)ϕ (1)] 2 1 2 − 1 2 ~r ~r 1 2 − 1 2 Z | 1 − 2| ϕ (1) 2 ϕ (2) 2 ϕ∗(1)ϕ (2)ϕ∗(2)ϕ (1) = | 1 | | 2 | d3r d3r 1 1 2 2 d3r d3r ~r ~r 1 2 − ~r ~r 1 2 Z | 1 − 2| Z | 1 − 2|

ϕ ϕ ϕ 1 1 ϕ1 2 r1 r1

ϕ r2 ϕ r2 ϕ 2 2 ϕ2 1

FIG. 9: Feynman-like diagrams for the direct and the exchange interaction

Thus, the energy is ϕ ϕ ϕ 1 1 ϕ1 2 r1 r1 E

ϕ r2 ϕ r2 ϕ 2 2 ϕ2 1 direct contribution exchange contribution

Due to statistics the exchange term depends on total spin in spite of the fact that the interaction is spin independent. Comparing with experimental data on He we ﬁnd: ϕ ϕ1 2 r1 2 0.76 eV

r2 ϕ ϕ2 1

Magnetism in solids is due to exchange interaction.