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Green’s functions and one-body quantum problems 1 2 GF can be used to convert ( r V (r ) ) (r ) 0 2 into an integral equation and to convert integral over some volume to surface integrals over boundary S of . Source introduce G satisfying: 1 ( r2 V (r ) )G(r ', r ) (r ' r ) 2 The delta not yet invented at Green’s time 1 2 multiply ( r V (r ) ) (r ) 0 by G(r',r) 2 1 2 G(r',r)( r V (r ) ) (r ) 0. 2 1 2 multiply ( r V (r ) )G(r ', r ) (r ' r ) by (r ) 2 1 2 (r ) ( r V (r ) )G(r ', r ) (r ) (r ' r ) 2 1 1 2 G(r ',r)( r V ( r ) ) ( r ) 0. 2 1 (r ) ( r2 V (r ) )G( r ', r ) ( r ) ( r ' r ) 2 subtract : V disappears 1 2 1 2 (r ) ( r )G(r ', r ) G(r ',r)( r ) ( r ) ( r ) ( r ' r ) 2 2 integrate on r 1 2 2 dr [ ( r ) G ( r ', r ) G(r',r) (r )] dr (r ) (r ' r ) ( r '). r r 2 Now interchange r and r ' 1 (r ) dr '[G(r,r')r2' (r ') (r ') r2'G(r , r ')] 2 This is the sought integral equation: next, we convert volume integral -> surface integral 2 1 2 2 dr '[G(r,r') ( r ') ( r ') G(r , r ')] r' r' 2 convert volume -> surface integral using the divergence theorem. Starting from (r ) divergence theorem : divA dr A.n dS S Insert A .( ) dr ( ).n dS S ((r )2 (r ) 2 (r )) dr ((r ) (r )).n dS S Now r r ', then (r') G( r,r ' ). In this way the result (r ) 1 dr '[G(r,r') r2' (r ') (r ') r2'G(r , r ')], r ' 2 is conveniently rewritten as a surface integral (r ) 1 [G(r,r') r ' ( r ') (r ') r 'G(r , r ')]n dS , r ' S. 2 S 3 Simple use of Green’s functions for Schroedinger’s equation H E , E Introduce 1 Green's operator G such that ( H )G 1 H H0 H1 with easy H0 . operator identities 1 1 1 ( H 0 ) ( H 0 H1 H1 ) 1 H1 H H H 1 1 1 1 H1 H H0 H0 H 1 1 1 1 H1 H H0 H H0 Example: H1=V(r): then, in the coordinate representation 1 1 1 1 r r' r r' r H1 r' H H0 H0 H 4 Next introduce complete set in 1 1 1 1 r r' r r' r H1 r' H H0 H0 H 1 1 r r' r r' H H0 r '' where 1 1 r H1 r '' r '' r' H0 H H1 r '' V (r '') r '' Lippmann-Schwinger equation G r , r ' G0 r , r ' d 3r '' G0 r , r '' V r '' G r '', r ' The Lippmann-Schwinger equation is most convenient when the perturbation is localized . Typical examples are impurity problems, in metals. The alternative is embedding (Inglesfield 1981) 5 1 We wish to solve the one-body problem Hψ=εψ with H 2 V (r ), 2 but V (r ) is complicated in some region. Suppose we can solve except inside the surface S. We can also solve 1 ( ( 2r V (r )) g (r ', r , ) (r r '), r and r ' outside 2 S and wave functions outside are related to g by Green’s theorem (r ) 1 [G(r,r')r ' ( r ') ( r ') r 'G( r , r ')]n dS , r outside, r ' on S. 2S It is assumed that G outside S is not modified by the presence of the localized perturbation. 1 We can also solve ( ( 2r V (r )) g (r ', r , ) (r r ') 2 with the boundary condition on S g (r , r ') 0. r' Then for r outside S 1 (r ) g (r , r ') ( r ')·ndS , r ' on S. 6 2 S 1 g (r , r ') (r ')·ndS , r outside S S 2 also fixes a relation between and on S through g. Let us see how. The relation (r ) S Suppose (r ) is a known function for r S. We call it (r ). Then, 1 (r ) g (r , r ') (r ')·ndS , for r , r ' S . 2 S This can be considered as a matrix relation and inverted: (r ')·n 2 dS g 1 (r , r ') ( r '), S r, r ' S with g 1 ( r , r ') a matrix inverse, and so knowledge of ( r ) determines on the surface S . knowledge of (r ) determines outside S . 7 Next, we write S d 3x 2 as a surface integral using Green's theorem. ouside S Vary ε : Hψ=εψ H ψ=ε ψ+ψ ε Divide by and multiply by * : *H ψ ψ =ε * +|ψ|2 Multiply H * ε * by ε * ψ ψ Substituting, H H * =|ψ|2 * ouside S ψ : ψ ψ H * d 3x 2 d 3 x[ *H ouside S ψ ψ - H * ] The term with V cancels and we can use Green's theorem: ( 2 2 ) dr ( ).n dS S 1 2 with *, , H , outside S 2 ouside S d 3x 2 1 ψ * * [ ].ndS (sign change for ouside normal) 2 S 8 We can vary ε keeping (r ) (r ) on S ψ 0 on S . This is clever because we get : 1 * d x .ndS 2 S ouside S 3 2 (sign change for ouside normal) Recall : we have seen that ( r ')·n 2 dS g 1 (r , r ') (r '), r , r ' S S with g 1 (r , r ') a matrix inverse, and so knowledge of (r ) determines on S , hence also d x . 3 2 ouside S John E. Inglesfield, Emeritus Professor ,Cardiff 9 Inglesfield embedding 1 We wish to solve the one-body problem with H 2 V (r ), but V is complicated in some region. 2 The Lippmann-Schwinger equation is less convenient when the perturbation is extended. Typical examples are surface problems, when one wants to treat the effects of surface creation, reconstruction, or contamination. In such cases one can resort to slab models, but there are serious drawbacks in any attempt to represent a bulk by a few atomic layers, with quantized normal momenta. The only practical alternative is the method of embedding. 1) in an extended system, there is a localized perturbing potential. A surface S divides the perturbed region I from the far region II where the potential is negligible. 2) the problem in I alone can be solved 3) the extended unperturbed system could be treated easily because it is highly symmetric 4) the wave functions match those of the extended system on S. II S I (r ) Problem: to find (r ) (r ) (r ) (r ) | H | such that E is stationary | rI r II r S; (that is, H | E | ) 10 II S I (r ) Problem: to find (r ) ( r ) (r ) (r ) | H | such that E is stationary | rI r II r S; (that is, H | E | ) Strategy: write E only in terms of in I and on S . | d 3 x d 3 x 2 I 2 1 * .ndS , 2 S | d 3 x 2 V I and since II d 3x ouside S 2 1 * .ndS , 2 S where (r ')·n 2 dS g 1 (r , r ') (r '), r , r ' S . S So, | is a functional of . 11 II S I (r ) Problem: to find (r ) ( r ) (r ) (r ) | H | such that E is stationary | rI r II r S; (that is, H | E | ) Strategy: write E only in terms of in I and on S . Next,we want | H | d 3 x * H I where d 3 x * II * H d 3 x V extends to a small volume around S . V We have just seen how to write d 3 x * in terms of : II V 1 * d x .ndS , 2 S ouside S 3 2 where (r ')·n 2 dS g 1 (r , r ') (r '), r , r ' S . S 12 Contribution in V * H d 3 x V V * H d 3 x V 1 * ( 2 V (r )) d 3 x, but the term in V vanishes with V 2 V 1 1 * 2 3 * 2 3 d x d x 2 V 2 V 1 1 In each dS , with z dS , * 2 dSdz * div grad dSdz 2 V 2 V S By the divergence theorem 1 1 1 * div grad dSdz * .nouter .ninner dS * .nouter dS . 2 V 2 2 To sum up, * H d 3 x V 1 * .nouter dS . 2S But recall: (r ')·n 2 dS g 1 (r , r ') (r '), S r, r 'S so everything is written in terms of and one obtains a localized variational problem in region I equivalent to a Schrödinger equation with a potential at S. 13 Absence of Magnetism in classical physics Ampere Postulated “molecular-ring” currents to explain the phenomenon of magnetism Bohr-Van Leeuwen Theorem (1911) Partition function for a Classical electron in canonical ensemble px 2 p y 2 pz 2 Z d x d p exp 2mK BT 3 3 Classical electron in magnetic field eH 2 2 2 ( p y ) p p y z x c 3 3 Z d x d pexp 2 mK T B A change of variable removes the field. dpx d ( px eH y) c Z V 2 mK BT 3 2 14 magnetism cannot exist (catastrophe of classical physics) Langevin paramagnetism "gas" of ions with magnetic moments g B J in field B. Hamiltonian H g B J .B magnetization M N z Volume Quantum partition function Z J m J (2 J 1) g B B sinh 2 K BT mg B B exp[ ] K BT g B sinh B 2 K BT Mean moment z K BT Paul Langevin (1872-1946) g BJ ln( Z ) g B JBJ ( B ), B K BT where BJ ( x) is known as Brillouin function BJ ( x) 2J 1 (2 J 1) x 1 x coth( ) coth( ) 2J 2J 2J 2J 15 Case J 1 (electron), g 2: 2 the partition function is Z J m J B H N M B tanh V K BT exp[ mg B B B ] 2 cosh( B ) K BT K BT B H High temperature behaviour: K BT 1 N B2 H M Curie law (for paramagnetism) V K BT 16 Can we explain ferromagnetism as due to dipole-dipole interactions? Orders of magnitude estimate e Interaction energy of two dipoles (Bohr Magneton) 2mc at distance alattice several Bohr radii : Edip B2 3 alattice B2 e 2 1 e 2 2 m 3e 6 1 2 4 ( ) mc , 3 3 2 2 6 2 aB 2mc 4m c 4 2 me e2 1 Since , mc 2 0.5MeV , 1 0 K 0.025 eV c 137 if alattice 4aB Edip B2 64aB3 0.10 K . dipole-dipole interaction far too small to explain ferromagnets! Heisenberg chain: a solvable 1d model of magnetism N HHeisenberg J Sn .Sn1 , J 0 n1 ferromagnetic 18 Hans Bethe solved the linear chain Heisenberg model H , , E ( , , 1 2 N 1 2 N ) 1 , 2 , Born July 2, 1906 Strassburg, Germany Died March 6, 2005 Ithaca, NY, USA Residence USA Nationality American Field Physicist Institution University of Tübingen Cornell University Alma mater University of Frankfurt University of Munich Academic advisor N Arnold Sommerfeld Zur Theorie der Metalle. I. Eigenwerte und Eigenfunktionen der linearen Atomkette Zeitschrift für Physik A, Vol. 71, pp. 205-226 (1931) 19 N spins ½ on a ring with a configuration 1 , 2 , , N Symmetries N H Heisenberg J Sn .Sn1 is invariant for translations of all spins n 1 there is a conserved wavevector K of all the system N H Heisenberg J Sn .Sn1 is invariant for spin rotations n 1 (since scalar products are invariant) N N conserved total spin S, conserved Sz . 2 Separated problems for each set K ,S , Sz 20 N HHeisenberg J Sn .Sn1 can be rewritten n1 N HHeisenberg H0 HXY , H0 J S n .S z z n1 HXY n 1 , HXY J N (SnSn1 SnSn1 ) 2 n1 J N (SnSn1 SnSn1 ) moves , to right or left 2 n1 Ground state for J>0: all spins up, no spin can be raised, and so no shift can occur J N HXY (SnSn1 SnSn1 ) HXY ... 0 2 n1 N H Heisenberg ... J S n .S n 1 z N z n 1 ground state energy E( N 0) J ... J n 1 1 ... 4 N 4 21 1 reversed spin in the chain Consider for the moment no hopping, only H0 N H0 J Sz n .Sz n1 n1 H0 ... E( ... ) E( ... ) E( ... ) 2 * 2 * ... ( J ) E( ... ) J , 4 N , 4 a factor 2 enters because positive negative, where E( ... ) E( N 0) J extra 2 because 2 of the scalar products are reversed. A configuration with r=1 is conveniently denoted by |n>, n=position of down spin. 22 They are all degenerate. Now introduce the hopping term. J N HXY (SnSn1 SnSn1 ) moves to right or left, so 2 n1 J N 0 ( H Heisenberg E ) n J n ( n 1 n 1 ). 2 Introducing the eigenvector E , H Heisenberg E E E Introduce E n f n amplitude of E ( H Heisenberg E N 0 on site n in eigenvector of H. ) n J E n J ( E n 1 E n 1 ) 2 that is, (E E N 0 f (n 1) f (n 1) ) f n J f (n) 2 23 One- Magnon solution: spin wave (E E N 0 is solved by N f (n 1) f (n 1) ) f n J f (n) , E ( N 0) J , 2 4 f ( n) e ik j n N , indeed f (n 1) f (n 1) e ik j n N (e j e ik We must insert the periodic boundary condition for the N-spin chain f n N 2 j 1 f ( n) k j E N N N 0 e energy : ( E E ) ik j n N J e N e ik j n n n 1 eik j e ik j 1 2 N ik j n E E ( N 0) J 1 cos(k j ) 24 ik j ). N HHeisenberg J Sn .Sn1 , J 0 ferromagnetic n1 One- Magnon solution: the spin wave is a boson (spin 1) E E 0 1 E N N n 1 e ik j n n , 2 j kj N J 2.0 1.5 1.0 E E ( N 0) J 1 cos(k j ) 0.5 k 3 2 1 1 2 3 This magnon dispersion law looks like a free quasi-particle dispersion for small k Two- Magnon solution: scattering, bound states 25 Several reversed spins N Let us first consider H0 J S z n .S z n1 without hopping, n 1 with r N reversed spins. H0 1 , 2 , N E ( 1 , 2 , N ) 1 , 2 , N . Any pattern of spins will do, but E ( 1 , 2 , N ) differs if far or consecutive spins are reversed: ... and ... have different E ( 1 , 2 , N ). J E( 1 , 2 , N ) E( N 0) 2 N , 4 N number of reversed scalar products. Now include H XY . Two Magnons a configuration is denoted by |n1,n2>, HXY n1<n2 reversed spins J N (SnSn1 SnSn1 ) moves to right or left 2 n1 provided that adiacent sites are up spins: if the reversed spins are close the energy is different and this is an effective interaction. We introduce the case n2 n1 1by means of a boundary condition. First, we solve for n2 n1 1, when the spins are 'far' and move independently. ( N 0) ( H Heisenberg E ) n1 , n2 2 J n1 , n2 J n1 1, n2 n1 1, n2 n1 , n2 1 n1 , n2 1 2 27 . ( H Heisenberg E( N 0) ) n1 , n2 2 J n1 , n2 J Let n1 1, n2 n1 1, n2 n1 , n2 1 n1 , n2 1 2 f (n1 , n2 ) E n1 , n2 , , n2 n1 1 shift the energy origin : E E N 0 E f (n1 1, n2 ) f (n1 1, n2 ) f (n1 , n2 1) f (n1 , n2 1) E f n1 , n2 2 f n1 , n2 . J 2 Thus r.r. is f n1 , n2 e satisfied for n1 n2 by the product i ( k1n1 k2 n2 ) , E (1 cos(k1 )) (1 cos k2 ). J E Recall One Magnon Solution : 1 cos(k j ). J Two independent magnons? NO! New k1 and k2 must be found since PBC do not separately apply to n1 and n2 (you cannot translate one magnon across the other). However there is Invariance under global translations of both reversed spins: f n1 N , n2 N f n1 , n2 . 28