green’s function of a dressed particle (today: holstein polaron) mona berciu, ubc collaborators:...
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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