Preliminaries: you have,
A
Tr[ 0 A] Tr[ 0 ]
Tr[e
H 0
A]
Z 0
; Tr[
n , n
n n A n n ]
n ,n
n A n nn ;
(1.1)
Green function equivalent to wavefunction. Retarded and advanced Green functions are defined as, [8.28] G R (r t; r t ) i(t t ) [ (rt), † (r t )] B , F
; B, F "bo "bosons, fermions"; [,] B, F [,] ; (1.2) [8.30] G (r t ; r t ) i(t t ) [ (rt), (r t )] B , F †
A
Sidenote: Some textbooks (see the Appendix’s reference to SSP 10, (1.27)) write (1.2) as,
G G(r t; r t) G(1, 1) i Tt B,t , F [ ( rt) † ( rt)] B ,F ; GR ( t t ) G; GA ( t t ) G;
(1.3)
For non-interacting particles: (1.2) is equivalent to the single-particle wavefunction, which is,
[8.22] G G R (1, 1) i( t t )
r|
e iH (t t ) | r i( t t)
n
n | r ei En (t t ) ;
r | n
(1.4)
nd
2 type of single-particle Green-function: Green-function: the “greater” and “lesser” Green’s functions,
G G (r t, r t) i (rt)† ( rt) ; G G ( r t, r t) i ( 1) B † ( r t) ( rt) ; F
(1.5)
Looking at (1.2) and (1.5), we see that {G R , G A } and {G , G } span the same solution space. Therefore: there is a linear transformation between them, and it is, G R (t t ) (t t ) G (1.6) A ( t t ) ( t t ) G G † The Green functions (1.6) can propagate can propagate a single particle, nˆk ak ak , using the many particle H.
1
a n arbitrary one by using, Basis switch: you go from a realspace basis to an
G R (r t ; r t )
( r)G R ( t , t ) * ( r); G R ( t , t ) i(t t ) [a (t ),a† (t )] B , F (1.7)
Example: consider a translation-invariant system. For starters, consider the retarded Green function of (1.2), and to study translational-invariance consider the specific functional form G R (r t ; r t ) G R (r r , t ; t ) . Using (1.7), go to a momentum (k ) space representation, R R R R ik r ik r ik ( r r ) i ( k k ) r e G (k t; k t) e G (k t; t ) k ,k e VG VG (r r , t; t) e kk
e
ik ( r r )
R
G
(k , t, t )
k
kk ik ( r r )
e
i( t t) [ a
k
k
†
( t), ak ( t)] B , F
(1.8)
electrons. Then, it’s natural to use (1.7) to go to k -space. -space. In doing Example: compute the G of (1.5) for free electrons. > R † so, we notice that G uses (rt ) (rt ) rather than a commutator (in contrast to G ). The k-basis: The Hamiltonian is diagonal, H
k
k ck† ck . Thus the Green function G G0 G0 kk , and
so, the Green function appears in the k-basis as,
1
Arbitrary in the sense of the representation
[1.69]
*
(r) a†
r |
*
† a† ( r) | r .
G0 G0 (k t , k t ) G0 (k t , k t ) kk G0 (k ;t t ) i ck (t )ck† (t ) i ck ck† e i (t t ) ; k
(1.9)
2
By the Heisenberg equation of motion, we can find time-evolution of the operators, iHt i Ht , H ] e i Ht [a , a† a ]e i Ht e i Ht [a , a† ]a e i Ht ia (t ) [a (t ), H ] [e a e
(1.10)
D .C . e Ht a (t )e Ht a (t ) a (t ) e t a a†e t a† (t ) i
i
i
i
In this rare case, we can write down all four Green functions. Using ck ck† 1 n F ( k ) in (1.5), yielding, G0 (k ; t t ) i ck e
ik t †
ck e i
G0 (k ; t t ) i (1) c e
† k
k t
i 1 n F ( ) e (t t ) [propagator of electrons] ; i
k
k
i k t
ck e
ik t
i n F ( ) e
i k ( t t )
k
[propagator of holes] ;
(1.11)
At zero temperature, the matrix element of G0 (assuming nothing about homogeneity, G0 G0 kk ) is,
G0 (k, k, t t) i G | ck ( t) ck† ( t) | G i G | ck e i H (t t ) ck† ei E0 (t t ) [overlap integral];
G FS ; (1.12)
Transforming G0 to the frequency domain,
G0 G0 (k ; ) F[G0 (k ; t t )] F[ i 1 n F (k ) e i (t k
t )
]
(1 n F ( )) 2 e (t t ) e t dt 2 (1 nF ( )) ( ) i
k
i
k
i
k
i
k
(1.13)
We could also inverse-transform (1.13) from the k domain to the r domain; for this we require the density of states [2.31] d ( )
dn d
2 1 (2m)3/2 ( ) m3/2 2
G0 (r r, ; )
2 i
2
F
G0 ( k ; ) d 3k ik r [ ] (1 n F (k )) (k ) e 2 i (2 )3
1
(1 n F ( )) ( k
G0 (r r, ; )
2 i
/ 2 , yielding,
???
...
(2m)3/ 2 2 2
k2 2m
k 2 2m
)e
ik r
d 3k
(1.14)
(2 )3
( )(1 n F ( k ))sinc(k
r r )
Thus, there are three effects 1) the density of states 2) the availability of empty states 1 n F ( k ) and 3) quantum interference sinc(k r r ) , which is the amplitude of a spherical wave spreading from point r . The Lehmann representation Problem: Fourier transform G G ( , t , t ) to the energy basis. Solution: use G of (1.5); using the definition of an expected-value in a discrete-basis (see preliminaries, (1.1)) this appears as, 1 11 11 G G ( , t, t ) ( t) † ( t ) Tr e H c ( t) c† ( t ) n | 1e H c ( t)1c† ( t ) 1 | n (1.15) n i i Z i Z
† Then, using the time-evolution of the c c , the Green function (1.15) appears as,
2
This is done in SSP 06 - 084 - creation and annihilation operators in heisenberg picture.
G
G
1 i Z
1 i Z
1 i Z
n , n,n , n
n|e
En iEn t
e
n , n
e
n | n n | e H e iHt c e iHt n n e i Ht c† e i Ht | n n | n c e
i En t
n n e
En iEn t i En t i En t i En t
e
e
e
e
n , n
i En t † i E t c e n
|n
(1.16)
n | c n n c† | n
1 i Z
e
En
ei
( En En )( t t )
n , n
n | c | n n | c† | n ;
Fourier-trnasforming (1.16) into the -domain is easy (just use the kernel e i (t t ) and ( x )
G G ( ; ) Ft [G ( ; t t)]
1 i Z
n ,n
e
E n
1 iZ
En
n ,n
e
i ( En En )(t t )
[ e
i ( t t )
e
1 2
e kx dx ), i
dt] n | c | n n | c † | n (1.17)
[2 ( En En )] n | c | n n | c | n ; †
And: notice what happens when we similarly construct G ( ; t , t ) , 1 En † i ( En En )( t t ) G G ( ; t, t ) i1 n | I† (rt ) I ( rt) I | n e n | c | n n | c | n e n,n Z i
Next, transform to G ( ; ) , again using the kernel e
i ( t t )
(1.18)
. Interchanging sum indices as n n and
anticipating the effect of the Dirac delta function b y replacing En E n , we get, G G ( ; ) Ft [G ( ; t t )]
1
e i Z
E n
G
Z 2 i Z
e i Z
En
n , n
n | c† | n n | c | n [ e
i ( En En )( t t )
e i (t t )dt ]
n | c† | n n | c | n [2 (En En )]
n , n
2 i
1
(1.19)
e
E n
e
( E n )
n , n
n | c† | n n | c | n [2 ( En En )]
n , n
n | c † | n n | c | n [2 ( En En )] G ( ; )e G ( ; ) ;
Problem: compute the retarded Green function for fermions in the Lehmann representation.
G R G R ( ; ) Ft [G R ( ; )] Ft [ G R ( ; t t)] Ft 1i (t t ) [ a ( t), a† ( t)] B , F
(t t ) [a (t ), a (t)] B, F e †
1 i
i ( i )t
dt
0
1 1 i Z
H
Tr e
†
[ a ( t), a
( t)] B , F e
i ( i )t
(1.20) dt
† , denoting a c , computing the trace in Specializing to fermions, expanding the commutator [ a (t ), a (t )]
the energy basis Tr
G R Z 1i
nn
e En n , n
0
n | | n n | | n , and distributing the Fourier-transform-integral.
ei ( i )t i (En En )t n | c | n n | c† | n dt
0
ei ( i )t i (En En )t n | c† | n n | c | n dt ; (1.21)
Effecting the integrals, we get the matrix elements, G R
1 Zi
e
E n
nn
1
Z nn
e
n | c | n n | c† | n (0 1) n | c† | n n | c | n (0 1) i ( i ) i ( En En ) i( i ) i( En E n )
En
e
E n
n | c | n n | c | n †
En E n i
G R ( ; )
; B F
(1.22)
Problem: compute the imaginary part of the Lehmann (energy) Green function (1.22), Im G R ( ; ) . Solution: Using ( x i ) 1 P x1 i ( x ) f ( x ) , setting En E n , and using (1.19) to abbreviate the sum as G ,
Im G R Z1
nn
2 Z
e
nn
En
e
En
Im
n| c | n n| c † | n En En i
n | c | n n | c | n e †
E n
1
e
2 Z
n|c
| n n | c † | n e
nn
En
e
E n
( E
n
E n ) (1.23)
( E
n
E n ) i (1 e
)G ( ; ) ;
Fluctuation-dissipation theorem
A( , ) 2Im G R ( , ); iG ( , ) A( , ) 1 n F( ) ;
iG(, ) A(, ) nF ( );
(1.24)
The spectral function R Problem: let A 2 Im G [spectral function] . This new object behaves like a probability-distribution. Show,
then, that A 0 . Solution: Using (1.23), taking the expected value G ( ; ) † using n | c | n n | c | n n | c | n
2
2(1 e
)
nn
n | c | n
in the n, n -basis,
0 , and using e x cosh x 0 (and e x sinh x 0 ) we compute,
A 2i(1 e )G ( ; ) 2i(1 e )( i †
,
2
) 2(1 e )
,
nn
n | c | n n | c | n
e x 4sinh x n | c | n 2 0 nn ; x 12 ; 2 e x 4 cosh x nn n | c | n 0
[8.62] n n c† c iG ( ,0) i
1 2
G ( , ) d
1
2
A( , ) d
(1.25)
(1.26)
Appendix I – Green function according to SSP 10
Green functions: define the following function, ˆ ( X ) ˆ † ( X )] | 0 i T [ ˆ (X ) ˆ † ( X )] G G ( X 1 , t1 , X 2 , t2 ) i 0 | Tt1t2 [ 1 2 t1t 2 1 2 ˆ † ( X ) ˆ ( X ) i (t t ) ˆ ( X ) ˆ † (X ) i(t2 t1 ) 2 1 1 2 1 2
(1.27)
