Lectures #12 and #13 First & Second Quantization

October 17 Lectures #12 and #13 October 21

Second quantization Evaluation of one and two -particle matrix elements Coulomb matrix elements: example of two -particle operator (second quantization form) Helium atom: LS coupled states

Chapter 4, pages 95 -99, 102 -106, Lectures on Atomic Physics Chapter 11, pages 241 -246, Atomic many -body theory I. Lindgren and J. Morrison

First & second quantization

First quantization Second quantization

Classical particles are assigned wave amplitudes Wave fields are “quantized” to describe the problem in terms of “quanta” or particles.

1 Second quantization: atomic electrons (fermions)

One-electron state |k> Ψ Described by the wave function k(ri)

= † k a k 0

Creation operator Vacuum state (no electrons) 0 0= 1

Second quantization: atomic electrons (fermions)

One-electron state |k> Vacuum state Ψ Described by the wave function k(q i) (no electrons)

= † 0 0= 1 k a k 0 Creation operator

= k0 a k Annihilation operator

= “there are no electrons to annihilate in a vacuum” ak 0 0 : † = 0ak 0

2 Anticommutation relations

††= †† + †† = {aaij,} aa ij aa ji 0 = + = {}aaij, aa ij aa ji 0 †= † + † = δ {}aai, j aa ij aa ji ij = ak 0 0 † = Orthonormality of one-electron states: 0ak 0

=† =−δ † = δ ij000 aaij ij aa ji 0 ij

Two -electron states

First quantizaiton : Second quantizaiton Slater determinant

ψ ψ 1 j()q1 k () q 2 Ψ(q , q ) = jk 1 2 ψ ψ = † † 2 j()q2 k () q 2 jk aaj k 0

Orthonormality of two-electron states:

Ψ( ) Ψ( ) =−δδ δδ = † † ijqq12, kl qq 12 , ik jl il jk ijkl0 aaaaj i k l 0

NOTE: reversed order Need to evaluate

3 Orthonormality of two-electron states

=† † = ijkl0 aaaaj i k l 0 ? † † Lets transform a j aa i k a l to “normal form ”:

Move: Creation operators Annihilation operators left right

Use anticommutation relations to move ai to the right

†††=δ − †† =− δδ † δδ − ††† + aaaajikl aa jlik aaaa jkil ikjl aa ljik jkil aa aaaa kjil =−δδ† δδδδ −+ †††† + δ − = ikjlaa l jik jkil jkli aaaa kjil aaaa kjli =−δδ† δδδδ −+ †† + δδ − ††† + ikjlaa l jik jkil jkli aaaa kjil jlki aaaaaa kl ji

Orthonormality of two-electron states =† † = ijkl0 aaaaj i k l 0 ? † † Lets transform a j aa i k a l to “normal order ”: Normal order with Creation operators Annihilaiton operators respect to left right vacuum

Use anticommutation relations to move ai and aj to the right

4 = † † ijkl0 aaaaj i k l 0 Normal order of operators

Why do we want to transform to normal order? To calculate expectation values: expectation value of normal ordered operators is zero. ††=−δδ † δδδδ −+ †† + δδ − ††† + aaaajikl ikjl aa ljik jkil jkli aaaa kjil jlki aaaaaa klji 0 0 †††=−δδ δδδδ −+ † 0aaaajikl 0 ikjl 00 aa lj ik jkil jk 00 aa li 0 0 0 +†δ − δ † + †† 0aakj 0 il jl 0 aa ki 0 0 aaaa klji 0 =δδ − δδ ik jl jk il

Exactly the result we had in first quantization

Many -particle operators in second quantization † ai 0 one-particle state † † States ai a j 0 two-particle state described by Slater determinants † †> † N-particle state in first quantization ai a j a n 0 Operators First quantizaiton Second quantizaiton

N One-particle = = † Fƒ f (ri ) Fƒ faaij i j operator i=1 i, j 1 Two-particle G= g( r ) = 1 † † ƒ ij Gƒ gijkl aaaa i j l k operator 2 ≠ i j 2 i, j = gijkl g jilk

5 Many -particle operators in second quantization

First and second quantization description must produce identical results; however, is it more convenient to use second quantizati on for calculation of properties of many -electro systems as the antisymmetrization properties are carried by the operators rather than the wave functions as in the case of Slater determinants.

= † = 1 † † Fƒ faaij i j Gƒ gijkl aaaa i j l k i, j 2 i, j = = 3ψ † ψ fab af b — d r a()r f (r ) b ()r = = 3 3†ψψ † ψψ gabcd abg cd— d r1212 — d r ab ()rr () g (r 12 ) cd( rr 1 ) (2 ) The rules for calculating matrix elements are equivalent, for example:

=1 †† †† = − abG abƒ gijkl0 aaaaaaaa b a i j l k a b 0 g abab g abba 2 ijkl

Examples of one and two -particle operators

First quantizaiton Second quantizaiton

N N = = ε † 1 H0 ƒ h i H0 ƒ i a i a i i=1 i=1 =−1 ∇2 − Z hi i 2 ri eigenvalue of hi

2 Coulomb (1/r 12 ) interaction: two-particle operator 1 1 H'= g aaaa† † H ' = ƒ ijkl i j l k 2 ijkl r12 two-electron matrix element

of the Coulomb potential 1/r 12 = 3 3†ψψ † 1 ψψ gabcd— drdr1212 — a()()rr b c ()() rr 12 d r12

6 He -like systems revisited

N Two-electron state ab = a † a † 0 is an eigenfunction of = ε † a b H0 ƒ i a i a i (0) H0 with an eigenvalueE =ε + ε : i=1 ab a b =(ε + ε ) H0 aba b ab

However, it is not necessarily an angular momentum eigenstate.

L2 S2 Lets construct eigenstates of { H, , Lz, ,S z } from | ab > states.

= + L la l b First, we couple la and lb to construct eigenstates L2 = + of , Lz with eigenvalues L(L+1) and M L S sa s b Second, we couple s and s to construct eigenstates S=0: singlet state a b of S2, S with eigenvalues S(S+1) and M S=1: triplet states z S

sa=sb=1/2 S=0,1

He -like systems: construction of coupled states = † † ab aa a b 0 Therefore, the coupled states are given by the sum over magnetic moments which contains two Clebsch-Gordon coefficients: µ lama ½ a =η † † ab, LMLS , SM ƒ ƒ − − aa a b 0 LM SM ma m bµ ab µ L S l m ½ µ Normalization factor b b b

ab, LM , SM ab , LM , SM =η2 ( 1(1) + − L+ S δ δ ) LS LS nnllab ab

≠ ≠ À 1na nl ba , l b η = Œ Ã 1 = = Œ Identical orbitals na n ba , l l b Õ 2

7 He -like systems: First -order energy

µ lama ½ a =η † † ab, LMLS , SM ƒ ƒ − − aa a b 0 LM SM ma m bµ ab µ L S µ lbmb ½ b

The first-order energy is given by

(1) = E ab,, LMLS SM H ',, ab LM LS SM

= 1 † † H'ƒ gijkl aaaa i j l k 2 ijkl

= 3 3†ψψ † 1 ψψ gabcd— drdr1212 — a()()rr b c ()() rr 12 d r12

Coulomb matrix element

First, lets consider Coulomb matrix element = 3 3†ψψ † 1 ψψ gabcd— drdr1212 — a()()rr b c ()() rr 12 d r12 ψ ψ= θ φ The functions are given by nlm()r R nl () r Y lm (,)

The 1/r 12 can be expanded as 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 This expression may be re-written using C-tensors defined by 4π Ck () rˆ = Y (,)θ φ q(2k + 1) kq 1 1 ∞ r k k = =< ( − 1)q C k ( rCˆ ) k ( r ˆ ) − ƒk +1 ƒ q1− q 2 r12r 1 r 2 k=0 r > q =− k

8 Coulomb matrix element ψ We now substitute the expressions for and 1/r 12 back into our matrix element and separate dr and dΩ integrals 1 ∞ r k k = < − q kˆ k ˆ ψ (r) = R( r )Y (θ, φ ) ƒ k +1 ƒ ( 1) Cq ( r1) C −q (r 2 ) nlm nl lm r12 k =0 r> q=− k

= 3 3ψψ † † 1 ψψ R( abcd ) gabcd— drd1212 — r a()()rr b c ()() rr 12 d k r12 ∞ ∞ ∞ r k radial = drr 2 drrR 2 () rR () r< R () r R ()r integral ƒ —1122 — nlaa 1 nl bb 2k +1 nl cc 1 nld d 2 k=0 0 0 r>

k × (− 1)q dΩ Y (θφ , ) C k ( θφ , ) Y ( θ ,φ ) k ƒ — 1lma a 11 q11 lm c c 11 lmaa C qcc lm q=− k

k k dΩ Y (θ ,φ ) C − ( θφ , ) Y ( θφ , ) — 2lmb b 22 q 22 lm d d 22 lmbb C − qdd lm

Coulomb matrix element

∞ k = − q k k gabcdƒ Rabcd k( ) ƒ ( 1) lmClm a a q c c lmC b b − q lm d d k=0 q =− k

Next, we use Wigner-Eckart theorem for both of the matrix elements:

lama lbmb ∞ k = − q k k gabcdƒ Rabcd k ( ) ƒ ( 1) − − lCllCla cb d k=0 q =− k k q k -q

lcmc ldmd

lbmb lbmb

k− q We use − = (− 1) − Note: k and q are integers k -q k q

ldmd ldmd

9 Coulomb matrix element

lbmb lama ∞ k g= Rabcd( ) ( − 1)k − lCllClk k abcdƒ k ƒ − k q a cb d k=0 q =− k k q

ldmd lcmc lama lbmb ∞ = − k k k ƒ ( 1)Rk ( abcd ) l a C l cb l C l d − + k =0 k

lcmc ldmd

Xk ( abcd )

lama lbmb ∞ g= − + X ( abcd ) abcd ƒ k k k =0

lcmc ldmd

Summary: Coulomb matrix element (non -relativistic case)

lama lbmb ∞ g= − + X ( abcd ) abcd ƒ k k k =0

lcmc ldmd

= − k k k Xk( abcd )(1) R k ( abcd ) l acbd C l l C l

∞ ∞ r k Rabcd( )= drr 2 drrR 2 ()() rR r< R ()() rR r k —1122 — nlaa 1 nl bb 2k +1 nl cc 1 nl dd 2 0 0 r>

≈l k l ’ k =−l1 + + 1 2 lCl12( 1) (2 l 12 1)(2 l 1) ∆ ÷ «0 0 0 ◊

10 Back to He -like systems

µ lama ½ a

=η † † ab, LMLS , SM ƒ ƒ − − aa a b 0 LM L SM S ma m bµ ab µ µ Assume lbmb ½ b la=la’ l =l (1) = b b’ E ab,, LMLS SM H ',, ab LM LS SM µ µ lama ½ a lama’ ½ a’

=η 2 ƒ ƒ ƒ ƒ − − − − LM L SM S LM L SM S mmababµµ mm ab' ' µµ ab '' µ µ lbmb ½ b lbmb’ ½ b’ ×()gδδ − g δδ abab''µµµµaabb'' abba '' µµµµ abba ''

=1 † † = − H'ƒ gaaaaijklijlk ; abGab g abab g abba 2 ijkl

Sum over µ’s: Term 1

µ µ ½ a ½ a’ Term 1: δ δ ƒ ƒ µµaa' µµ bb ' µµab µ a' µ b ' − − SM S SM S µ µ ½ ½ b ½ b’

SM S SM =(2S + 1) S = 1 + -

½ We used orthogonality relation to evaluate this diagram

j1 j ‘ m ‘ 3 3 + j3m3 = 1 δ δ + j33' j mm 3 ' 3 − 2j3 1

11 Sum over µ’s: Term 2 µ b µ µ ½ a ½ a’ δ δ Term 2 : ƒ ƒ µµ µµ − − ab' ba ' SM SM µµab µ a' µ b ' S S µ µ ½ b ½ b’ ½ µ a SM S SM =+−(2S 1)( 1)S +1 S =−− ( 1) S + -

Extra phase factor

First -order energy: He -like systems

lama lama’ Assume E(1)=η 2 ƒ ƒ ƒ − − la=la’ mma' b ' mm ab k LM L LM L lb=lb’

lbmb lbmb’

lama’ lbmb’ lama’ lbmb’ ×+ − S − + ( − Xk ( abab ) + ( 1) Xk ( abba ) ) k k

lbmb lama lama lbmb where we substituted Coulomb matrix elements g into our formula

lama lbmb ∞ g= − + X ( abcd ) abcd ƒ k k k =0

lcmc ldmd

12 Sum over m ’s: Term 1

lama lama’ lama’ lbmb’ = ƒ ƒ − − − + LM LM ma' m b ' mm ab L L k l m l m lbmb lbmb’ a a b b rotate lama − lama’ to the right lama lama’ + =(2L+1) − 1) Remove triangle (get 6-j symbol) k LM L LM L 2) Use orthogonality relation

lbmb’ (get factor 1/(2L+1) lbmb + lbmb lbmb’

+ + + Àl l L ¤ =( − 1) Lkla l b Ãa b ‹ Õlb l a k ›

Sum over m ’s: Term 2

lama lama’ lama’ lbmb’ = ƒ ƒ − − − + LM LM ma' m b ' mm ab L L k l m l m lbmb lbmb’ b b a a

+ + Àl l L ¤ =( − 1) k la l b Ãa b ‹ Õla l b k ›

The calculation is very similar to Term1, only need to switch a and b lines in the first angular diagram.

13 ab, LM , SM Final expression: L S First order energy for He -like system

ŒÀ Àl l L ¤ (1)=η 2 − Lkl+ +a + l b a b EƒÃ( 1) Ã ‹ Xk ( abab ) k ÕŒ Õlb l a k › Àl l L ¤ Œ¤ − Skl+ +a + l b a b +( 1)Ã ‹Xk ( abba ) ‹ Õla l b k › ›Œ

= − k k k Xk( abcd )(1) R k ( abcd ) l acbd C l l C l

∞ ∞ r k Rabcd( )= drr 2 drrR 2 ()() rR r< R ()() rR r k —1122 — nlaa 1 nl bb 2k +1 nl cc 1 nl dd 2 0 0 r> ≈l k l ’ k =−l1 + + 1 2 lCl12( 1) (2 l 12 1)(2 l 1) ∆ ÷ «0 0 0 ◊ À0 0 0 ¤ Ã ‹ =1 == →k = δ 0 0 0 Special case : if la l b 0 ( s states ) s C s k 0 Õ ›