1. Green’s function of a dressed particle (today: Holstein polaron) Mona Berciu, UBC Collaborators: Glen Goodvin, George Sawaztky, Alexandru Macridin More details: M. B., PRL 97, 036402 (2006) G. G., M. B. and G. S., to be posted soon on the archives Funding: NSERC, CIAR Nanoelectronics, Sloan Foundation

2. Motivation: Old problem: try to understand properties of a dressed particle, e.g. electron dressed by phonons (polaron), or spin-waves, or orbitronic deg. of freedom, or combinations of these and other bosonic excitations. For a single particle, the quantity of most interest is its Green’s function: poles give us the whole one-particle spectrum, residues have partial information about the eigenstates. Note: there is a substantial amount of work dedicated to finding only low-energy (GS) properties. We want the full Green’s function; we want a simple yet accurate approximation that works decently for all values of the coupling strength, so that we can understand regimes where perturbation does not work!

3. Holstein Hamiltonian – describes one of the simplest (on-site, linear) electron-phonon couplings

4. weak coupling Lang-Firsov impurity limit E k How does the spectral weight evolve between these two very different limits? 

5. Calculate the Green’s function: use Dyson’s identity repeatedly, generate infinite hierarchy We can solve these exactly if t=0 or g=0. For finite t and g, make Momentum Average approximation: Note: this is exact if t=0 (no k dependence)  MA should work well at least for strong coupling g/t>>1, where there was no good approximation for G (perturbation theory gives only the GS, not the whole G)

6. The MA end result:: exact for both t=0, g=0 Other aproximations: (a) simple to evaluate: (b) numerically intensive  Diagrammatic Quantum Monte Carlo (QMC) – in principle exact summation of all diagrams  exact diagonalizations (various cutoffs for Hilbert space),

7. Comparisons: (I) GS results in 1D

8. Agreement becomes better with increasing d, but MA calculations just as easy (fractions of second)

9. 3D: no QMC results, but data is very persuasive

10. How about higher-energy results? (much fewer “exact” numerical results). numerics: G. De Filippis et al., PRB 72, 014307 (2005)

11. Diagrammatic meaning of MA : sums ALL self-energy diagrams, but each free propagator is momentum averaged, i.e. any is replaced by Example: 1 nd order diagram: Higher order MA diagrams can be similarly grouped together and summed exactly.

12. Why is summing ALL diagrams (even if with approx. expressions) BETTER than summing only some (exact) diagrams: understanding the sum rules of the spectral weight Comments: (i)Knowing all the sum rules  exact Green’s function (ii)Sum rules can be calculated exactly, with enough patience. Traditional method of computation – using equations of motion [ P.E. Kornilovitch, EPL 59, 735 (2002)] (ii) Sum rules must be the same irrespective of strength of coupling. M n has units of energy n  combinations of the various energy scales at the right power. (iv) Traditional wisdom: the more sum rules are obeyed, the better the approximation is. WRONG!

13. How to calculate the sum rules for the MA approximation?

14. Alternative way to compute sum rules: use perturbational expansion (diagrammatics) for small coupling + + + + + + + + + + MA vs SCBA: -0 th order diagram correct  M 0 and M 1 are exact for both approximations; -1 st order diagram correct  M 2 and M 3 are exact for both approximations; -2 nd order diagrams: SCBA misses 1  M 4 missing one g 4 term (important at large g) MA still exact for M 4, M 5. Errors appear in M 6, but not in the dominant terms. + + + + +

15. Conclusions: If t>>g  dominant term is. Both approximations get it correctly  ok behavior If g>>t  dominant term is ~ or -- MA gets is exactly (after all, it is exact for t=0)  ok behavior. On the other hand, SCBA does really poorly for large g. Keeping all diagrams, even if none is exact, may be more important than summing exactly a subclass of diagrams. On to more results ….

17. Momentum-dependent low-energy results:

18.   

19. The answer to my initial question, according to the Momentum Average (MA) approximation: