1. 11 Green’s functions and one-body quantum problems The delta not yet invented at Green’s time Source

2. 22 This is the sought integral equation: next, we convert volume integral -> surface integral

3. 33 convert volume -> surface integral using the divergence theorem.

4. 44 operator identities Example: H 1 =  V(r): then, in the coordinate representation Simple use of Green’s functions for Schroedinger’s equation

5. 555 Lippmann-Schwinger equation 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)

6. 6 S Suppose we can solve except inside the surface S. We can also solve and wave functions outside are related to g by Green’s theorem It is assumed that G outside S is not modified by the presence of the localized perturbation.

7. 7 S

8. 8 S

9. 9 John E. Inglesfield, Emeritus Professor,Cardiff

10. 10 Inglesfield embedding 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. S II I

11. 11 S II I VV

12. 12 S II I VV

13. 13 By the divergence theorem 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. VV

14. 14 Bohr-Van Leeuwen Theorem (1911) Classical electron in magnetic field A change of variable removes the field. Absence of Magnetism in classical physics magnetism cannot exist (catastrophe of classical physics) Partition function for a Classical electron in canonical ensemble Ampere Postulated “molecular-ring” currents to explain the phenomenon of magnetism

15. Langevin paramagnetism Paul Langevin (1872-1946) 15

16. 16

17. Orders of magnitude estimate Can we explain ferromagnetism as due to dipole-dipole interactions?

18. 18 Heisenberg chain: a solvable 1d model of magnetism

19. 19 BornJuly 2July 2, 1906 Strassburg, Germany1906 StrassburgGermany DiedMarch 6March 6, 2005 Ithaca, NY, USA2005 IthacaNYUSA ResidenceUSA Nationality American FieldPhysicist InstitutionUniversity of Tübingen Cornell University Alma materUniversity of Frankfurt University of Munich Academic advisorArnold Sommerfeld Zur Theorie der Metalle. I. Eigenwerte und Eigenfunktionen der linearen Atomkette Zeitschrift für Physik A, Vol. 71, pp. 205-226 (1931) Hans Bethe solved the linear chain Heisenberg model

20. 20 N spins ½ on a ring with a configuration Symmetries

21. 21 Ground state for J>0: all spins up, no spin can be raised, and so no shift can occur

22. 22 1 reversed spin in the chain Consider for the moment no hopping, only H 0 A configuration with r=1 is conveniently denoted by |n>, n=position of down spin. They are all degenerate.

23. 23 Now introduce the hopping term.

24. 24 One- Magnon solution: spin wave We must insert the periodic boundary condition for the N-spin chain

25. One- Magnon solution: the spin wave is a boson (spin 1) This magnon dispersion law looks like a free quasi-particle dispersion for small k 25 Two- Magnon solution: scattering, bound states

26. Several reversed spins

27. 27 provided that adiacent sites are up spins: if the reversed spins are close the energy is different and this is an effective interaction. Two Magnons a configuration is denoted by |n 1,n 2 >, n 1 <n 2 reversed spins

28. 28 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