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

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.

(1) (0) 2( )† †

2††

(1) (0)

( , ) ( ) 0 ( ) 0 ( ) 1 0

1 01( , ) 0 0

ˆ ( )

1spectral weight: ( , ) Im ( , ) measured with ARPES

GSi E E tGS GS GSk k k

GSkGS GSk k

GS

G k t i t c t c i t e c

cG k c c

H i E E i

A k G k

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!

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

† † † †

,

ˆq qq qk k kk k q

qk k q

gH c c b b c c b b

N

† †ˆel ph i i i i

i

H g c c b b 1

2 cosd

kt k a

weak coupling Lang-Firsov impurity limit

0

00 2 2

1( , ) ;

( , )

k

k

k

G ki

A k

2

2

2

20

1 1( , )

!

g n

LFn

gG k e

n gn i

E

k

2

GSg

E

How does the spectral weight evolve between these two very different limits?

2

0 ( 0)2

gg

dt

2

( 0)2

gt

dt

Calculate the Green’s function: use Dyson’s identity repeatedly, generate infinite hierarchy

1

†0 1 1

ˆ( , ) 0 ( ) 0 ( , ) 1 ( , , )GS GSk kq

gG k c G c G k F k q

N

0 0 01ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ and ( ) ( ) ( ) ( ) ( )ˆ

H H V G G G G VGH i

1

1

† † †1

ˆIn general, let ( , ,..., , ) 0 ( ) 0nn

ii

n n GS GSq qkk q

F k q q c G c b b

1

1 0 1 1 1 1 1 11 1

( , ,..., , ) ( , ) ( ,..., , ,..., ) ( , ,..., , )n

n n

n n i n i i n ni i q

gF k q q G k q n F k q q F k q q

N

11

† †1 1

ˆ,with ( , , ) 0 ( ) 0GS GSqk k qF k q c G c b

We can solve these exactly if t=0 or g=0. For finite t and g, make Momentum Average approximation:

1 1 1 1

1 1 1 1 1 1,..., ,...

0 01 ,

( , ,..., , ) ( , ,..., ,, , )n

Tn

n

i Tqi

n n n nq q q q

G k q n G kF k q q F k q qq n

0 ( )g n

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)

The MA end result:: exact for both t=0, g=01

( , )( )MA

MAk

G ki

20

0 020 0

20 0

( ) 1( ) where ( ) ( , )

2 ( ) ( 2 )1

3 ( 2 ) ( 3 )1

...

EMA

k BZE E

E E

g gg G k

Ng g g

g g g

Other aproximations: (a) simple to evaluate:2

2

2

2

2

20

1 1gen. LF: ( , ) also exact for t=0, g=0

!

g n

LF gn

k

gG k e

n ge n i

2

2 2 20 0 0

1SCBA: ( , )

( )

( ) ( , )= g g 2 g 3

SCBASCBAk

SCBA SCBAq

G ki

gG k q g g g

N

(b) numerically intensive

Diagrammatic Quantum Monte Carlo (QMC) – in principle exact summation of all diagrams

exact diagonalizations (various cutoffs for Hilbert space),

Comparisons: (I) GS results in 1D

2

2

g

dt

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

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

How about higher-energy results? (much fewer “exact” numerical results).

numerics: G. De Filippis et al., PRB 72, 014307 (2005)

1; 0.4; 0.04t

0(1) 2

2

0( , ) ( , )q

gk G k g gq

N

Diagrammatic meaning of MA : sums ALL self-energy diagrams, but each free propagator is momentum averaged, i.e. any is replaced by

0 ( , )G k

0 01

( ) ( , )k

g G kN

Example: 1nd order diagram:

40 0 0( ) ( 2 ) ( )g g g g

Higher order MA diagrams can be similarly grouped together and summed exactly.

1 2

4(2, )

0 1 0 1 2 0 22,

( , ) , , 2 ,a

q q

gk G k q G k q q G k q

N

1 2

4(2, )

0 1 0 1 2 0 12,

( , ) , , 2 ,b

q q

gk G k q G k q q G k q

N

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

1( ) ( , ) Im ( , )n n

nM k d A k d G k

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)]

0

1 1( ) Im ( , ) Im ( , 0)

n ni t

n t

d dM k i d e G k i G k t

dt dt

ˆ ˆ † †( , ) 0 0 ( , 0) 0 , , ,..., 0n

iHt iHtGS GS GS GSk kk k

dG k t i e c e c i G k t i c H H H c

dt

(ii) Sum rules must be the same irrespective of strength of coupling.• Mn has units of energyn 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!

2 2 3 2 20 1 2 3. . : ( ) 1; ( ) ; ( ) ; ( ) 2 ;k k k kE g M k M k M k g M k g g

1( , )

( )MAMAk

G ki

20

0 020 0

20 0

( ) 1( ) where ( ) ( , )

2 ( ) ( 2 )1

3 ( 2 ) ( 3 )1

...

EMA

k BZE E

E E

g gg G k

Ng g g

g g g

1( ) ( , ) Im ( , )

n

MA n nMA MAM k d A k d G k

How to calculate the sum rules for the MA approximation?

(1 )0 2 2

sgn( )( )

4

Dgi t

Alternative way to compute sum rules: use perturbational expansion (diagrammatics) for small coupling

( , )G k

+ + + + +

2

2 1

22

2 1

2 .

, , .

n

n

p

nn

n

gA diagram of order n decays asymptotically like

i

it only contributes to M with p n

Contribution to M is exactly g for both exact and for MA diagrams

Contribution to M depend on

2 2

, .

, . .n

the diagram but are also identical for both cases

If exact diagram has noncrossed selfenergy diagrams differences start from contrib to M

( , )MAG k

+ + + + +

MA vs SCBA:-0th order diagram correct M0 and M1 are exact for both approximations;-1st order diagram correct M2 and M3 are exact for both approximations;-2nd order diagrams: SCBA misses 1 M4 missing one g4 term (important at large g) MA still exact for M4, M5. Errors appear in M6, but not in the dominant terms.

( , )SCBAG k

+ + + + +

6 2 4 2 2 3 2 2 2 2 26

3 4 4 2 2 2 6

( ) [5 6 2 4 3 6 ( 2 )

2 ] 12 22 25 1518

k k k k k k

k k k

M k g t d d dt

g dt g

2 46, 6( ) ( ) 2MAM k M k dt g

4 66, 6( ) ( ) .... 10SCBAM k M k g g

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.

n

k

2ng 2ng

Keeping all diagrams, even if none is exact, may be more important than summing exactly a subclass of diagrams.

On to more results ….

Momentum-dependent low-energy results:

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

2

2

g

dt