Lectures #10 and #11 Identical Particles: Example

October 9 October 16 Lectures #10 and #11

Identical particles Bosons and fermions Slater determinants Evaluation of one and two -particle matrix elements Helium atom: Applications of perturbation theory and Variational method

Chapter 10, pages 469 -477, 485 -492 Quantum Mechanics B.H. Bransden & C.J. Joachain

Identical particles: example

Ground state of He atom: 1s 2 Quantum numbers

1s 1s 1s 2 n=1 L=0 n=1 l =0 ML=0 l =0 m =0 l S=0 ml=0 s=1/2 MS=0 s=1/2 m =±1/2 s ms=±1/2 Term: 1S

No triplet 3S term for the 1s 2 configuration

1 Quantum mechanical system of N -particle system

Ψ( , ..., ) ∂ iZΨ(,,, qq > qtHqq ,) = Ψ (,,, > qt ,) ∂t 12N 12 N N ∇2 H= −ƒ Z2 i + V i=1 2mi Ψ> = ψ > −iEt / Z (,,,qq12 qtN ,) (,,, qq 12 qe N )

Total energy of the system Solutions of Hψ=Eψ

N L= L ƒ i i=1 J= L + S N = Sƒ S i i=1

N-particle system: a special case

N = Hƒ h i i=1 ψ= ε ψ hiai() q aai () q Ψ> = Ψ > Hqq(,,,)12 qN Eqq (,,,) 12 q N Ψ >ψ ψ > ψ (,,)=qq1n a()() qq 1 b 2 nN ( q ) ε+ ε +> + ε E = a b n

Uncorrelated system

2 Systems of identical particles

Two particles are considered identical if they can not be distinguished by means of any intrinsic property.

Ψ > > > (,,q1 q 2 , qi , , q j , q N )

All observables must be symmetric o Pij with respect to qi qj interchange as N particles are identical Ψ Pij is also an eigenfunction of H » ÿ = H, P ij ⁄ 0

Ψ>>> = Ψ >>> Pqqij (,,,1 2 qqi ,, j , qqqN ) (,,,1 2 qqj ,, i , q N )

Systems of identical particles

Ψ> > > = Ψ > > > Pqqij (,,,1 2 qqi ,, j , qqqN ) (,,,1 2 qqj ,, i , q N ) 2 = ε P ε= Pij 1 ij ±1

ε = 1 : S ymmet ri c wave functions under Pij interchange Ψ> > > = Ψ > > > Pqqij (,,,1 2 qqij ,, , qqqN ) (,,,1 2 qq ij ,, , q N )

ε = − 1 : Anitsymmetric wave functions under Pij interchange Ψ> > > = −Ψ > > > Pqqij (1 , 2 , , qi , , q j , q N ) (,,q1 q 2 , qi , , q j , q N )

3 Symmetric and antisymmetric wave functions

Ψ> = Ψ > Pqq(,,,12 qN) (, qq P 12 P , q PN )

Totally symm etric state Ψ> = Ψ > PqqS (,,)1 NS (,,) qq 1 N

Totally anitsymmetric s tate ÀΨ (q ,> , q ) for eve n permutaito ns PΨ (,,) q> q = Ã A1 N A1 N −Ψ > Õ A(q1 , , q N ) for od d perm utaitons

Pij

Bosons and fermions

Symmetrisation postulate: Ψ Ψ

Bosons Fermions Ψ Ψ

" ā $ %$ ā

!"# & "'

4 Ψ Ψ How to construct A and S? 1 Ψ(,)qq ={}ψ (,) qq + ψ (,) qq ( S 122 12 21 1 Ψ(,)qq ={}ψ (,) qq − ψ (,) qq A 122 12 21

1 Ψ=(,,)qqq{ψ (,,) qqq + ψ (,,) qqq + ψ (,,) qqq (% S 1236 123 231 312 ψ+ ψ + ψ } + (qqq213 , , ) ( qqq 132 , , ) ( qqq 321, , )

1 Ψ=(,,)qqq{ψ (,,) qqq + ψ (,,) qqq + ψ (,,) qqq A 1236 123 231 312 −ψ −ψ − ψ } ( q213,,)qq (,,) qqq 132 (,,) qqq 321

N = Again, a special case: Hƒ h i i=1 Ψ =1 {}ψ ψ + ψ ψ S(,)qq12 abab ()() qq 12 ()() qq 21 ( 2 1 Ψ(,)qq ={}ψ ()() qq ψ − ψ ()() qq ψ A122 abab 12 21 entangled ψ()q ψ () q? ψ () q a1 b 1 n 1 1 ψ()q ψ () q? ψ () q Ψ(,q> , q ) = a2 b 2 n 2 Slater A1 N N ! @ @ @ determinant ψ ψ? ψ aN()q bN () q nN () q

Ψ > S(,q1 , q N ) ÷ø

5 Slater determinants & Pauli exclusion principle

ψ ψ? ψ a()q1 b () q 1 n () q 1 1 ψ()q ψ () q? ψ () q Ψ(,q> , q ) = a2 b 2 n 2 A1 N N ! @ @ @ ψ ψ? ψ aN()q bN () q nN () q

Ω

) * Only one fermion can occupy a certain quantum state.

The Pauli exclusion principle, quarks & colors

Hadrons: u and d quarks flavor charge spin ÷ø e ! + ÷ ø "# e Three quarks Quark & antiquark (fermions) + proton (uud) π ud neutron(udd) Quarks must have extra ∆++ baryon: spin J=3/2, charge 2 e internal quantum number: expected to be (uuu) color and each of the three But that will violate Pauli exclusion principle! quarks has different color

M. Gell-Mann & G. Zweig

6 System of N particles: many -particle operators

N = One -particle operators N Fƒ f (ri ) = = i 1 Example: Hƒ h i i=1 Ψ Lets designate our Slater determinate functions ab…n How to evaluate the corresponding matrix elements?

If the indices { a’b’…n ’} and { ab…n } Ψ>F Ψ > = 0 a' b ' n ' ab n differ in more than one place Ψ Ψ = a' b '> n ' F ab > n f k' k If only the indices k and k ’ differ N Ψ Ψ = If the sets of indices { a’b’…n ’} and { ab…n } a' b '> n ' F ab > n ƒ f ii are the same. i= a = = 3ψ † ψ fab af b — d r a()r f (r ) b ()r

System of N particles: many -particle operators

= 1 Two -particle operators Gƒ g( r ij ) 1 2 i≠ j Example: r12 Ψ Ψ = a' b '> n ' G ab > n 0 If the indices { a’b’…n ’} and { ab…n } differ in more than two places Ψ>G Ψ= > g − g abn''' abn klkl '' kllk '' If only the pairs of indices kl and k’l’ differ n Ψ Ψ=() − a''' b> nG ab > nƒ g k '' iki g k iik If only the indices k and k ’ differ i= a Ψ Ψ=1 − ’ ’ ’ a' b '> n ' G ab > n ƒ() gijij g ijji If the sets of indices { a b …n } and { ab…n } 2 i, j are the same.

= = 3 3†ψψ † ψψ gabcd abg cd— d r1212 — d r ab ()rr () g (r 12 ) cd( rr 1 ) (2 )

7 Z=2: He Two -electron systems Z=3: Li + Z=4: Be ++ … Origin of the coordinate system: nucleus, r1 and r2 are the position vectors of the two electrons = + H H0 H ' e2 H ' = = + (4πε ) r H0 h 1 h 2 0 12 Z2Ze 2 h =− ∇−2 i i πε 2m (40 ) r i

In atomic units: ≈ ’ = −∇−12 Z + 1 H ƒ ∆i ÷ i=1,2 «2 ri ◊ r 12

Z=2: He Z=3: Li + Two -electron wave functions Z=4: Be ++ … ≈ ’ = −∇−12 Z + 1 H ƒ ∆i ÷ i=1,2 «2 ri ◊ r 12 Ψ = Ψ H(,) qq12 E (,) qq 12

Ψ(,q q ) = φ (,)r r χ (1,2) Eigenvalues of { L2 S2 12 12S , M S H, , Lz, ,S z }

Lets introduce space-symmetric and space-antisymmetric wave functions:

φ+= φ + φ−= − φ − (,)rr12 (,) rr 21 (,)rr12 (,) rr 21

Need to construct symmetric and antisymmetric spin functions

8 Z=2: He Z=3: Li + Two -electron wave fuctions Z=4: Be ++ ÀS=0 M = 0 singlet state (antisymmetric) s=1/ 2 and s = 1/ 2 → Ã S 1 2 = = ± ÕS1 MS 0, 1 triplet state (symmetric) Lets construct the corresponding symmetric and antisymmetric spin functions

= =1111 −=−− 1111 SM S 112222 , 1 1 2222 ,

= =111 −+− 1 1 111 S− 11 2 10222 , 2 2 222 , 1 =111 −+− 1 1 111 10{}22 , 2 2 2 2, 22 2 =111 −+− 1 1 111 00 a222 , 2 b 2 222 , 1 10 00= 0 →ba =− ; 00 00= 1→ 2 aa2 = 1 → = 2 1 = 111− 1 − 1 − 111 00 {}222 , 2 2 222 , 2

Z=2: He Z=3: Li + Two -electron wave fuctions Z=4: Be ++ … Electrons are fermions: the total wave function must be antisymmetric

Space-symmetric Spin singlet state (antisymmetric)

= + 1 ΨS 0 =φ 111 −−− 1 1 111 (,)q q (,)r r {}222 , 2 2 222 , 12 12 2

Antisymmetric À 11 , 11 Œ 22 22 = − Œ 1 ΨS 1 =φ ×111 −+− 1 1 111 (,)q12 q (,)r 12 r Ã {}222 , 2 2 222 , Œ 2 Œ 1− 11 − 1 Õ 2 22 , 2

Space-antisymmetric Spin triplet state (symmetric)

9 Ground state: Z=2: He Z=3: Li + identical electrons Z=4: Be ++ Space-symmetric Spin singlet state (anti-symmetric)

= + 1 ΨS 0 =φ 111 −−− 1 1 111 (,)q q (,)r r {}222 , 2 2 222 , 12 12 2

Spin triplet state (symmetric) Antisymmetric À 11 , 11 Œ 22 22 = − Œ 1 ΨS 1 =φ ×111 −+− 1 1 111 (,)q12 q (,)r 12 r Ã {}222 , 2 2 222 , Œ 2 Œ 1− 11 − 1 Õ 2 22 , 2

Space-antisymmetric There is no triplet state for function is zero if 1s 2 configuration. particles are identical

Ground state: Z=2: He Z=3: Li + identical electrons Z=4: Be ++ …

Ground state of He (or He-like ion): 1s 2 1S

= + 1 ΨS 0 =φ 111 −−− 1 1 111 (,)q q (,)r r {}222 , 2 2 222 , 12 12 2

Space-antisymmetric There is no triplet state for function is zero if 1s 2 configuration. particles are identical

10 Two -electron systems: Z=2: He Lowest order Z=3: Li + Z=4: Be ++ ψ= ε ψ … hiai()r aai () r H= h + h 0 1 2 Solutions: Coulomb (H-like) wave functions 1 Z h =− ∇2 − i i ψ= 1 θ φ 2 ri nlm()r P nl () r Y lm (,) r Z 2 ɣ ε = − (in a.u.) n 2n2 Lowest order:

φ(0) (,)rr= ψψ ()() rr ± ψψ ()() rr “±”: space-(anti)symmetric ±12a 1 b 2 b 1 a 2 wave functions Z 2 ≈1 1 ’ E = −∆ + ÷ (in a.u.) 0 2 n2 n 2 «1 2 ◊ 2== (0) =− 2 1snn : 1 2 1 E Z Ground state: =(0) =− ≈−× =− He (Z 2) : E1s 1 s 4 a . u . 4 27.2 eV 109 eV

Two -electron systems: Z=2: He Perturbation theory Z=3: Li + Z=4: Be ++ Lowest order: … 2 ≈ ’ = −Z 1 + 1 E0 ∆2 2 ÷ (in a.u.) 2 «n1 n 2 ◊ 2== (0) =− 2 1snn : 1 2 1 E Z =(0) =− ≈−× = − Ground state: He (Z 2) : E1s 1 s 4 au . . 4 27.2 e V 109 eV expt ≈ − E1s1s 79 eV

Not a very good result! Lets calculate first order energy:

E(1)= Ψ (0) H ' Ψ (0)

11 Z=2: He First -order energy Z=3: Li + Z=4: Be ++ … 1 E(1)= Ψ (0) H ' Ψ (0) H ' = r12

1 − Single-electron (hydrogenic) wave functions are ψ ( r ) = 2 Z 3/ 2 e Zr . 1s 4π Normalized lowest-order space-symmetric two-electron wave function is:

1 − + φ(0) (,)rr= ψ () r ψ () r = Ze3 Z( r1 r 2 ) + 12 1112s s π

2 1 1 1 E(1)= φ (0) (,) r r ddr r = — +12 12 What to do with − ? r12 r12r 1 r 2

Irreducible tensor operators Coulomb interaction

1/ r12 can be expanded in partial waves ∞ k 1 = r< θ ƒ k +1 Pk (cos ) r− r = r 1 2 k 0 > Legendre polynomial

4π k P(cosθ )= Y* ( θφθφ , ) Y ( , ) r and r are lesser k+ ƒ kq11 kq 22 < > 2k 1 q=− k and greater of r1, r 2 θ is the angle between the vectors

r1 and r 2 1∞ 4 π r k k = < Y* (θφ , ) Y ( θφ , ) ƒ+ k +1 ƒ kq11 kq 22 r12 k=0 2 k 1 r > q =− k

12 r< and r> are lesser and greater of r1, r 2 First -order energy

1∞ 4 π rl k 2 1 = < Y* (θφ , ) Y ( θφ , ) E(1)= φ (0) (,) r r ddr r ƒ+ l+1 ƒ kq11 kq 22 — +12 12 r12 k=0 2 k 1 r > q =− k r12 ∞ k 2 4π r k = φ(0) (,)rr< YY* (,)(,) θφθφ ddr r — + 1 2 ƒ+ k+1 ƒ kq11 kq 22 12 k=0 2k 1 r > q =− k

∞ k ∞ ∞ l 4π r 2 = drrdrr2 2< φ (0)(,) rr dYΩ * (,) θφ dY Ω (,) θφ + ƒ ƒ ——11 22l+1 +12 — 1kq 11 — 2 kq 22 2k 1 k=0 q =− k 0 0 r >

φ (0) Note: + ( r 1 , r 2 ) is spherically symmetric

How to integrate over d Ω1 and d Ω2?

π = 1 Lets multiply our expression by 4 Y 00 Y 00 , Y . 00 4π

r< and r> are lesser and greater of r1, r 2 First -order energy

Ω* θφ Ω θφπ =Ω* θφ Ω θφ ——dY1kq(,) 11 dY 2 kq (,)4 22 — dY 1 kq (,) 1100 YdY — 2 kq (,) 2200 Y

The spherical harmonics are orthonormal on the unit sphere:

Ω* θφ θφ = δδ — d Yk'' q (,) Y kq (,) kk '' qq

Ω* θφ = δδ Therefore: — d 1 Y kq ( 1100 , ) Y k 00 q and only one term k=0 q=0 contributes from the sums over k and q.

∞ ∞ ∞ ∞ 2 12 1 − + (1)=()π 2 2(0) φ = 6 2 2 2Z ( r 1 r 2 ) = E4—— drrdrr11 22 +12 ( rr , ) 16 Zdrrdrr —— 11 22 e (*) 0 0 r> 0 0 r >

13 r< and r> are lesser and greater of r1, r 2 First -order energy

∞ Àr 1 ∞ ¤ Œ1−+ 1 −+ Œ (*)= 16 Z622 dr rÃ r dr e2()Zrr 1 2 + r 2 dr e 2() Zrr 1 2 ‹ —1122 —r — 22 r 0ÕŒ 0 1r 1 2 ›Œ

À∞ r 1 ∞ ∞ ¤ Œ − − − − Œ 5 =16 Z6Ã drre2Zr 1 rdre 2 2 Zr 2 + drre 2 2 Zr 1 rdre 2 Zr 2 ‹ = Z —11 — 22 — 11 — 22 8 ÕŒ 0 0 0r 1 ›Œ

We used Maple to evaluate this integral

> 5 Z~ 8

He ground state energy

5 He : (Z= 2) E(1) = a . u . ≈ 34 eV 4 EEE=(0) + (1) = − 109 + 34= −75 eV v s . E expt = −79 eV

Much better agreement! 5% instead of 38%

Can we improve it without calculating the second order? We can try to use the variational method.

14 He ground state energy: Rayleigh -Ritz variational method

φH φ φ is a trial function depending on the variational E[φ ]= λ ,λ ,.. φ φ parameters 1 2 . The optimum ground-state energy is obtained by minimizing E(λ1, λ 2,… ) with respect to the parameters λ.

1 −λ + Trial function : φ ( λ , r , r )= λ 3 e (r 1 r 2 ) φ φ = 1 1 2 π ≈ ’ = −∇−12 Z + 1 H ƒ ∆i ÷ i=1,2 «2 ri ◊ r 12

5 dE 5 E[φλλλ ]=−+2 2 Z ; = 0 →=− λ Z 8 dλ 16

Same is previous calculation of first-order energy with Z= λ

He isoelectronic sequence

Difference with accurate values E(0) = −Z 2 (0) (0)+ (1) 5 ZE E E E var E(0) +E (0) = −Z 2 + Z 8 He 2 38% 5% 2% + ≈5 ’ 2 Li 3 24% 2% 0.7% E= −∆ Z − ÷ var «16 ◊ Be 2+ 4 17% 1.2% 0.5% B3+ 5 13% 0.7% 0. 3% C4+ 6 11% 0.5% 0.2%