Several-Electron Atoms

Identical Particles: Bosons and Fermions

Consider N identical particles (e.g. electrons).Hamiltonian:

H = H(1, 2, 3, ...) 1 = ~x1, �1|{z}spin

; 2 = ~x2,�2

symmetric in the variables 1,2,...

wave function: = (1, 2, ...,N) (not symmetric yet)

Permutation operator Pij (transposition):

Pij (1, ..., i , ..., j , ...) = (..., j , ...i , ...)

P2ij = 1 ) EV of Pij is ± 1

Each permutation P can be represented as a product of transpositions Pij .

P is

(odd if number of Pij is odd

even if number of Pij is even

symmetry of the Hamiltonian: =) PH = HPand thus

(1, ...,N) EF of H with EV E ) P (1, ...,N) EF of H with EV E

H = E ) HP = PH = EP Xconsider arbitrary states and ':

h |'i = hP |P'i = h |P†P'i

) P† = P�1 unitary

physics: permuation of identical particles does not generate considerable conse-quences, thus for operator O:

h |O| i != hP |O|P i = h |P†OP | i ) P†OP = O ) OP = PO

) O must be symmetric und the states

(1, ...,N) and P (1, ...,N)

cannot be distinguished from one another.

Two symmetries are privileged:

- completely symmetric

- completely antisymmetric

defined via:

Pij (...i ...j ...) = ± (...i ...j ...)

example:

s(1, 2) = (1, 2) + (2, 1) ; a(1, 2) = (1, 2)� (2, 1)

a(n) (anti)symmetric state stays (antisymmetric) for all time

results from Schrodinger equation and [H,P] = 0.

(t) =T exp

n

i

Z t

0dt 0H(t 0)/~

o

(0)

P (t) =T exp

n

i

Z t

0dt 0H(t 0)/~

o

P (0)

of course, other symmetries might be possible, but in experiments only two kinds

of particles show up:

Bosons: symmetric, integral spin

Fermions: antisymmetric, half-integral spin

Relation between spin and symmetrie follows from quantum field theory.

Result: di↵erent statistics:

Fermi-Dirac: Fermions

Bose-Einstein: Bosonen

fermions:

leptons: ⌫e , ⌫µ, ⌫⌧e, µ, ⌧

baryons p, n⇤, ⌃, ⌅, ⌦, ...

bosons:

mesons: ⇡, K , ⇢, !, ...photon: �

hadrons = baryons and mesons

combination of quarks

n :ddu

p :uud

⇡+ : du

At this level the fundamental particles are:

leptons Yangs-Mills gauge bosons (⇤⇤)quarks Higgs bosons

(⇤) W+,W�,Z 0 and photon

very heavy, thus (⇤⇤) necessary.

Fermions:

Pij (..., i , ..., j , ...) = (..., j , ..., i , ...)

=� (..., i , ..., j , ...)

(...,~x�, ...,~x�, ...) = 0

Two fermions in the same spin state cannot occupy the same position

(Pauli exclusion principle).

Noninteracting Particles

N identical noninteracting particles.Hamiltonian:

H =NX

i=1

H(i) sum of one-particle Hamiltonians

solution of the H(i):

H(i)'↵i (i) = E↵i'↵i (i) ↵i numbers the one-particle states

) product states: '↵1(1)'↵2(2)'↵3(3)...'↵N (N) (⇤)

are eigen states of H with EV:

E = E↵1 + ... + E↵N

The states are neither symmetric nor antisymmetric.

For fermions

two particles:

a(1, 2) =1p2

('↵1(1)'↵2(2)� '↵2(1)'↵1(2))

N particles:

a(1, ...,N) =

1pN!

X

P

(�1)

P'↵1(1)...'↵N (N)

=

1pN!

�������

'↵1(1) ... '↵1(N)

.

.

.

.

.

.

'↵N (1) ... '↵N (N)

�������| {z }

Slater determinant

(�1)

p= ±1

even

odd

permutations

Thus a(1, ...,N) = 0 for '↵i = '↵j . (Pauli exclusion principle).

For bosons

s(1, . . . ,N) =

rN1!N2!...

N!

X

P0

P 0'↵1(1)...'↵N (N)

where N1 is the multiplicity of ↵1 etc.

P 0: permutations leading to distinct terms.

Free particles in a box

Volume V = L3

one-particle wave function for free particles:

~p

⇠ e i~p·~x/~

E~p

=

~p2

2m; E =

NX

1

~p2i

2m

periodic boundary conditions:

) ~p = ~2⇡L(n1, n2, n3) ; n integer

ground state for bosons:

~pi

= 0 for all N ) E = 0

volume

volume of the cellspin

Fermions: Pauli exclusion principle.

Each state can only be occupied twice (sz = ±~2 )

ground state: systematically filling the lowest states

these states lie within a sphere in momentum space which is called Fermi sphere.

discrete values have separation:

2⇡~L

Number of states within the Fermi sphere:

2 ·4⇡3 p3F

(2⇡~/L)3 =

(pF~)33⇡2

L3 = N

particle number density: n =

N

L3=

(pF~)33⇡2

ground state energy = sum of the one-particle energies within the fermi-sphere

E = 2

X

~p

~p2

2m= 2

⇣ L

2⇡~

⌘3

Z pF

0

d

3pp2

2m= N · 3

5

✏F

substitute L : n =

NL3

integral over sphere

then substitute L with pF

where ✏F =

p2

F2m Fermi-energy

pF Fermi momentum

Composite Particles

Example: H-atom: consists of two fermions: proton p and electron e

Consider wave function of two H-atoms (p1, e1; p2, e2) under permutation

since p and e are fermions:

(p1, e1; p2, e2) = � (p2, e1; p1, e2) = (p2, e2; p1, e1)

) H-atom is a boson

in general:

number of fermions is odd , fermion

number of fermions is even , boson

example:

baryons: 3 quarks: , fermions

mesons: 2 quarks: , bosons

3He =p,p,n and e,e fermion4He =p,p,n,n and e,e boson

Helium

Z = 2. In a first step, neglect the spin-orbit interaction and the movement of thenucleus (m

nucleus

� melectron

)

) Hamiltonian

H =1

2m~p21

+1

2m~p22

� Ze2

r1

� Ze2

r2

+e2

|~x1

� ~x2

|

notation:

H = H(1) + H(2) + V

H(i) =~p2i2m

� Ze2

ri| {z }hydrogen hamiltonian

; i = 1, 2 ; V =e2

|~x1

� ~x2

|| {z }electron repulsion

Neglecting the Electron-Electron Interaction

H is the sum of two one-particle operators

product states:

| i = |n1l1m1i|n2l2m2i

are eigenstates of H if |ni limi i is eigenfunction of H(i)

)✓H(1) + H(2)

◆| i = (En1 + En2)| i

where En = �Z 2Ry 1n2

and E1(Z = 1) = �13.6eV = 1Ry (Rydberg)

table of energy values E 0

n1

,n2

= En1

+ En2

:

n1

n2

E 0

(Ry) E 0

(eV)

1 1 -8 -108.8

1 2 -5 -68.0

1 3 � 40

9

-60.4

.

.

.

.

.

.

.

.

.

.

.

.

1 1 -4 -54.4

2 2 -2 -27.2

ionization of helium atom in the ground state

E 0

Ion

= (E1

+ E1)� 2E1

= 4Ry = �54.4eV

The (n1

= 2, n2

= 2) state is higher than the ionization energy:

not a bound state, in the continuum

energy spectrum:

energy spectrum:

234 13. Several-Electron Atoms

Table 13.2. Energy values of the He atom for various n1 and n2 (13.21)

n1 n2 E0(Ry) E0(eV)

1 1 −8 −108.81 2 −5 −68.01 3 −40/9 −60.4...

......

...1 ∞ −4 −54.52 2 −2 −27.2

and with energy

E0n1,n2

≡ En1 + En2 . (13.21)

In Table 13.2, several values of the total energy are given for variousprincipal quantum numbers n1 and n2 (for Z = 2).

The ionization energy of a helium atom in the ground state is

E0Ion = (E1 + E∞) − 2E1 = 4 Ry .

The (n1, n2) = (2, 2)-state has a higher energy than the singly ionized(1, ∞)-state and is not a bound state, as is true of all the further ones.In the bound, excited states, one of the electrons has the principal quantumnumber 1. One sees the (2,2)-state however as a resonance in the He+– escattering cross section. The energy spectrum of Fig. 13.2 thus results.

Fig. 13.2. The energy spectrum of He,neglecting the interaction of the electrons(schematic); zero at the ionization energy

We must now take into account the Pauli principle, that is, the anti-symmetry of the total wave function, which is the product of the coordinatefunction and the spin state |S, ms〉. The spin states of the two electrons arethe three symmetric triplet states and the antisymmetric singlet state. The

Continuum

Pauli-exclusion principle: antisymmetry of the total wave function

= product of the spatial function and the spin state |S ,msi

remember addition of two spins: (

~S2, SZ , ~S2

1 ,~S22 )

singlet state:

|0, 0i = 1p2

(|"#i � |#"i) antisymmetric

triplet state:

|1, 1i =|""i

|1, 0i = 1p2

(|"#i+ |#"i) symmetric

|1,�1i =|##i

) two classes (historical)

1. Parahelium: spatial wave function symmetric, spin singlet

|0i = |100i|100i|0, 0i ground state

wave function wave function spin

particle 1 particle 2

1p2(|100i|2lmi+ |2lmi|100i)|0, 0i

...

2. Orthohelium: spatial wave function antisymmetric, spin tripletantisymmetrization of |100i|100i = 0)lowest spin triplet state

1p2(|100i|2lmi � |2lmi|100i)|l ,msi ms = �1, 0, 1

Energy Shift due to the Repulsive Electron-Electron Interaction

ground state: 1st order perturbation theory

�E =h0|V |0i = h100|h100|V |100i|100i

=e

2

Zd

3x1d

3x2

| 100(~x1)|2| 100(~x2)|2

|~x1 � ~x2|

wave function 100 already known:

100 =1p⇡

⇣Z

a

⌘ 32e

�Zr/a

) �E =

5

4

Ze2

2a=

5

4

Zmc2↵2

2

(do the calculation as an exercise)

�E =2.5Ry = 34eV for Z = 2

) total ground state energy:

E 0

11

= �8Ry = �108.8eV

E11

= E 0

11

+�E = �5.5Ry = �74.8eV

experiment:

(E11

)

exp

= �5.807Ry = �78.575eV

excited states: 1st order perturbation theory

�E

s,tnl =

1

2

Zd

3x1d

3x2| 100(~x1) nl0(~x2)± nl0(~x1) 100(~x2)|2

e

2

|~x1 � ~x2|

=e

2

"Zd

3x1d

3x2

| 100(~x1)|2| nl0(~x2)|2

|~x1 � ~x2|

±Z

d

3x1d

3x2 ⇤100(~x1)

⇤nl0(~x2) 100(~x2) nl0(~x1)

|~x1 � ~x2|

#

=Jnl ± Knl

Result is independent of m, thus it is enough to calculate �E for m = 0.

Why does the result not depend on m?

we have [

~L,

1

|~x1 � ~x2|

] = 0

~L =

~L1 +

~L2 ) [L±,

1

|~x1 � ~x2|

] = 0

we remember:

L± nlm =~pl(l + 1)�m(m ± 1) nl ,m±1

L�L+ nlm =(

~L

2 � L

2z ⌥ ~Lz) nlm

=~2(l(l + 1)�m

2 ⌥m) nlm

singlet, triplet

) proposition:

h nlm|1

|~x1 � ~x2|

| nlmi =C h nlm�1|L�1

|~x1 � ~x2|

L+| nlm�1i

=C h nlm�1|1

|~x1 � ~x2|

|L�L+| nlm�1i

=h nlm�1|1

|~x1 � ~x2|

| nlm�1i

Jnl : positiveKnl : also positive (exercise)

since 2 ~S1 · ~S2/~2 = S(S + 1)� 32 =

⇢� 3

2 singlet12 triplet

we can write the energy shift

�E

s,tnl = Jnl �

1

2(1 + ~�1 · ~�2)Knl

spin-orbit-interaction: ~S · ~L = (~J2 � ~L

2 � ~S

2)/2

Parahelium: ~S = 0 ) no spin-orbit interaction

) no interaction

Orthohelium: ~S = 1, m = �1, 0, 1 ) triple degeneration

as in first oderperturbation theory

Variational method

variational ansatz: | i = |100i|100i|0, 0i

where 100(~x) =1p⇡

⇣Za

⌘ 32e�Z r/a

screening by the other electron: we expect Z < Z

variational parameter Z : h |H| i minimized

h |H| i = 2E0(Z )� 2h |e2(Z � Z )

|~x1|| i+ h | e2

|~x1 � ~x2|| i

= �2Ry(�Z 2 + 2ZZ � 5

8Z )

minimum: Z = Z � 516 use and insert

E0 = �2(Z � 5

16)2Ry =

h�2Z 2 +

5

4Z

| {z }�2(

5

16)2iRy

Z = 2 : (He) E0 = �5.7Ry =� 77.48eV

=� 78.975eV experiment

=� 74.8eV 1st order perturbation theory