Download Green`s functions and one-body quantum problems

Green’s functions and one-body quantum problems
1 2
GF can be used to convert ( 
r  V (r )   ) (r )  0
2
into an integral equation
and to convert integral over some volume  to surface integrals over boundary S of .
Source
introduce G satisfying:
1
(  r2  V (r )   )G(r ', r )   (r ' r )
2
The delta not
yet invented at
Green’s time
1 2
multiply ( 
r  V (r )   ) (r )  0 by G(r',r)
2
1 2
G(r',r)(  r  V (r )   ) (r )  0.
2
1 2
multiply ( 
r  V (r )   )G(r ', r )   (r ' r ) by  (r )
2
1 2
 (r ) (  r  V (r )   )G(r ', r )   (r )  (r ' r )
2
1
1 2
G(r ',r)(   r  V ( r )   ) ( r )  0.
2
1
 (r ) (   r2  V (r )   )G( r ', r )   ( r )  ( r ' r )
2
subtract : V
disappears
1 2
1 2
 (r ) (  r )G(r ', r )  G(r ',r)( r ) ( r )   ( r )  ( r ' r )
2
2
integrate on r
1
2
2
dr
[


(
r
)

G
(
r
',
r
)

G(r',r)

 (r )]   dr (r )  (r ' r )   ( r ').
r
r

2

Now interchange r and r '

1
 (r )   dr '[G(r,r')r2' (r ')  (r ') r2'G(r , r ')]
2
This is the sought integral equation: next, we convert volume integral ->
surface integral
2
1
2
2
dr
'[G(r,r')


(
r
')


(
r
')

G(r , r ')]
r'
r'

2
convert volume -> surface integral using the divergence theorem.
Starting from
 (r ) 
divergence theorem :


divA dr   A.n dS
S
Insert A    





.(  ) dr   (  ).n dS
S
((r )2 (r )   2 (r )) dr   ((r )  (r )).n dS
S
Now r  r ', then (r')  G( r,r ' ). In this way the result
 (r ) 
1
dr '[G(r,r') r2' (r ')  (r ')  r2'G(r , r ')], r '  

2
is conveniently rewritten as a surface integral
 (r ) 
1
[G(r,r') r ' ( r ')  (r ')  r 'G(r , r ')]n dS , r '  S.
2 S
3
Simple use of Green’s functions for Schroedinger’s equation
H  E , E  
Introduce
1
Green's operator G such that ( H   )G  1
H  H0  H1 with easy H0 .
operator identities
1
1
1
(  H 0 )
 (  H 0  H1  H1 )
 1  H1
H
H
H
1
1
1
1


H1
  H   H0   H0
H
1
1
1
1


H1
  H   H0   H
  H0
Example: H1=V(r): then, in the coordinate representation
1
1
1
1
r
r'  r
r'  r
H1
r'
H
  H0
  H0   H
4
Next introduce complete set in
1
1
1
1
r
r'  r
r'  r
H1
r'
H
  H0
  H0   H
1
1
r
r'  r
r' 
H
  H0
r ''
where
1
1
r
H1 r '' r ''
r'
  H0
H
H1 r ''   V (r '') r ''
Lippmann-Schwinger equation
G  r , r '   G0  r , r '    d 3r '' G0  r , r ''   V  r ''  G  r '', r ' 
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)
5
1
We wish to solve the one-body problem Hψ=εψ with H   2  V (r ),
2
but V (r ) is complicated in some region.
Suppose we can solve except inside the surface S. We can also solve
1
(  (  2r  V (r )) g (r ', r ,  )   (r  r '), r and r ' outside
2
S
and wave functions outside are related to g by Green’s theorem
 (r ) 
1
[G(r,r')r 
' ( r ')  ( r ')  r 'G( r , r ')]n dS , r outside, r ' on S.

2S
It is assumed that G outside S is not modified by the presence
of the localized perturbation.
1
We can also solve (  (  2r  V (r )) g (r ', r ,  )   (r  r ')
2
with the boundary condition on S
 g (r , r ')  0.
r'
Then for r outside S
1
 (r )    g (r , r ') ( r ')·ndS , r ' on S.
6
2 S
1
g (r , r ') (r ')·ndS , r outside S

S
2
also fixes a relation between  and  on S through g. Let us see how.
The relation  (r )  
S
Suppose  (r ) is a known function for r  S.
We call it  (r ). Then,
1
 (r )    g (r , r ') (r ')·ndS , for r , r '  S .
2 S
This can be considered as a matrix relation and inverted:
 (r ')·n  2  dS g 1 (r , r ') ( r '),
S
r, r ' S
with g 1 ( r , r ') a matrix inverse, and so knowledge of  ( r )
determines  on the surface S .
 knowledge of  (r ) determines  outside S .
7
Next, we write
S

d 3x 
2
as a surface integral using Green's theorem.
ouside S
Vary ε :
Hψ=εψ  H ψ=ε ψ+ψ ε
Divide by  and multiply by  * :
 *H
ψ
ψ
=ε *
+|ψ|2


Multiply H *  ε * by
 ε *
ψ ψ
Substituting,  H
H * =|ψ|2
 
*

ouside S
ψ
:

ψ ψ

H *
 

d 3x  
2
d 3 x[ *H
ouside S
ψ ψ
- H * ]
 
The term with V cancels and we can use Green's theorem:


( 2    2 ) dr   (   ).n dS
S

1 2
with    *,  
, H    ,   outside S

2

ouside S
d 3x   
2
1
ψ
* 
*
[





].ndS (sign change for ouside normal)

2 S


8
We can vary ε keeping  (r )   (r ) on S
ψ

 0 on S . This is clever because we get :

1
* 
d x   
 .ndS

2 S 
ouside S
3
2
(sign change for ouside normal)
Recall : we have seen that  ( r ')·n  2  dS g 1 (r , r ') (r '), r , r '  S
S
with g 1 (r , r ') a matrix inverse, and so knowledge of  (r )
determines  on S , hence also

d x .
3
2
ouside S
John E. Inglesfield,
Emeritus Professor
,Cardiff
9
Inglesfield embedding
1
We wish to solve the one-body problem with H    2  V (r ), but V is complicated in some region.
2
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.
II
S
I
  (r )

Problem: to find (r )    (r )
 (r )   (r )

 | H | 
such that E 
is stationary
 | 
rI
r  II
r  S;
(that is, H |   E |  )
10
II
S
I
  (r )

Problem: to find (r )    ( r )
 (r )   (r )

 | H | 
such that E 
is stationary
 | 
rI
r  II
r  S;
(that is, H |   E |  )
Strategy: write E only in terms of  in I and on S .
 |    d 3 x    d 3 x 
2
I

2
1
* 

 .ndS ,

2 S 
 |    d 3 x  
2
V
I
and since
II
d 3x   
ouside S
2
1
* 

 .ndS ,

2 S 
where  (r ')·n  2  dS g 1 (r , r ') (r '), r , r '  S .
S
So,  |  is a functional of  .
11
II
S
I
  (r )

Problem: to find (r )    ( r )
 (r )   (r )

 | H | 
such that E 
is stationary
 | 
rI
r  II
r  S;
(that is, H |   E |  )
Strategy: write E only in terms of  in I and on S .
Next,we want  | H |    d 3 x * H   
I
where


d 3 x * 
II

* H d 3 x
V
extends to a small volume around S .
V
We have just seen how to write

d 3 x  * in terms of  :
II
V
1
* 
d x   
 .ndS ,

2 S 
ouside S
3
2
where  (r ')·n  2 dS g 1 (r , r ') (r '), r , r '  S .
S
12
Contribution in  V

* H d 3 x 
V
V

 * H d 3 x  
V

1
* (  2  V (r )) d 3 x, but the term in V vanishes with  V
2
V
1
1
* 2
3
* 2
3



d
x





d
x


2 V
2 V
1
1
In each dS , with z  dS ,    *   2 dSdz    *  div grad dSdz
2 V
2 V
S
By the divergence theorem
1
1
1
  *  div grad dSdz    *  .nouter  .ninner  dS    *     .nouter dS .
2 V
2
2
To sum up,

* H d 3 x  
V
1
*
    .nouter dS .




2S
But recall:  (r ')·n  2 dS g 1 (r , r ') (r '),
S
r, r 'S
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.
13
Absence of Magnetism in classical physics
Ampere Postulated
“molecular-ring”
currents to explain the
phenomenon of
magnetism
Bohr-Van Leeuwen Theorem (1911)
Partition function for a Classical electron in canonical ensemble
 px 2  p y 2  pz 2 
Z   d x  d p exp 

2mK BT


3
3
Classical electron in magnetic field
eH 2

2
2
(
p

y
)

p

p
y
z 
 x c
3
3
Z   d x  d pexp 

2
mK
T
B




A change of variable removes the field.
 dpx
  d ( px 
eH
y)
c
Z  V  2 mK BT 
3
2
14
magnetism cannot exist (catastrophe of classical physics)
Langevin paramagnetism
"gas" of ions with magnetic moments   g B J in field B.
Hamiltonian
H  g  B J .B
magnetization M 
N z
Volume
Quantum partition function
Z
J

m  J
 (2 J  1) g  B B 
sinh 

2 K BT
mg  B B


exp[
]
K BT
 g B 
sinh  B 
 2 K BT 
Mean moment
 z  K BT
Paul Langevin
(1872-1946)
g  BJ
 ln( Z )
 g  B JBJ ( B ),
B
K BT
where BJ ( x) is known as Brillouin function
BJ ( x) 
2J 1
(2 J  1) x
1
x
coth(
)
coth( )
2J
2J
2J
2J
15
Case J 
1
(electron), g  2:
2
the partition function is Z 
J

m  J
 B H
N
 M   B tanh 
V
 K BT
exp[ 
mg  B B
 B
]  2 cosh( B )
K BT
K BT

B H

High
temperature
behaviour:

K BT

1
N B2 H
M 
 Curie law (for paramagnetism)
V K BT
16
Can we explain ferromagnetism as due to dipole-dipole
interactions?
Orders of magnitude estimate
e
Interaction energy of two dipoles  
(Bohr Magneton)
2mc
at distance alattice  several Bohr radii : Edip 
 B2
3
alattice
B2
e 2
1
e 2 2 m 3e 6 1 2 4
(
)

 mc  ,
3
3
2 2
6
2
aB
2mc 
4m c
4

 2
 me 
e2
1
Since  

, mc 2  0.5MeV , 1 0 K  0.025 eV
c 137
if
alattice  4aB  Edip 
 B2
64aB3
 0.10 K .
 dipole-dipole interaction far too small to explain ferromagnets!
Heisenberg chain: a solvable 1d model of magnetism
N
HHeisenberg   J  Sn .Sn1 , J  0
n1
ferromagnetic
18
Hans Bethe solved the linear chain
Heisenberg model H  , ,   E ( , , 
1
2
N
1
2
N
)  1 , 2 ,
Born
July 2, 1906
Strassburg, Germany
Died
March 6, 2005
Ithaca, NY, USA
Residence
USA
Nationality
American
Field
Physicist
Institution
University of Tübingen
Cornell University
Alma mater
University of Frankfurt
University of Munich
Academic advisor
N
Arnold Sommerfeld
Zur Theorie der Metalle. I. Eigenwerte und Eigenfunktionen der linearen Atomkette
Zeitschrift für Physik A, Vol. 71, pp. 205-226 (1931)
19
N spins ½ on a ring with a configuration
 1 , 2 ,
, N
Symmetries
N
H Heisenberg   J  Sn .Sn1
is invariant for translations of all spins
n 1
 there is a conserved wavevector K of all the system
N
H Heisenberg   J  Sn .Sn1
is invariant for spin rotations
n 1
(since scalar products are invariant)
N  N 
 conserved total spin S, conserved Sz 
.
2
Separated problems for each set K ,S , Sz
20
N
HHeisenberg   J  Sn .Sn1 can be rewritten
n1
N
HHeisenberg  H0  HXY , H0   J  S n .S
z
z
n1
HXY
n 1
, HXY
J N
   (SnSn1  SnSn1 )
2 n1
J N
   (SnSn1  SnSn1 ) moves ,  to right or left
2 n1
Ground state for J>0: all spins up, no spin can be
raised, and so no shift can occur
J N
HXY    (SnSn1  SnSn1 )  HXY  ...   0
2 n1
N
H Heisenberg  ...    J  S n .S
n 1
z
N
z
n 1
 ground state energy E( N 0)   J
 ...    J 
n 1
1
 ... 
4
N
4
21
1 reversed spin in the chain
Consider for the moment no hopping, only H0
N
H0   J  Sz n .Sz n1
n1
H0  ...   E( ... )
E( ... )  E( ... )  2 * 2 *
 ... 
( J )
 E( ... )  J ,
4
N
,
4
a factor 2 enters because positive  negative,
where E( ... )  E( N 0)   J
extra 2 because 2 of the
scalar products are reversed.
A configuration with r=1 is conveniently denoted by |n>, n=position of down spin.
22
They are all degenerate.
Now introduce the hopping term.
J N
HXY    (SnSn1  SnSn1 ) moves  to right or left, so
2 n1
J
 N 0
( H Heisenberg  E
) n  J n  ( n  1  n  1 ).
2
Introducing the eigenvector  E ,
H Heisenberg  E  E  E
Introduce  E n  f  n  amplitude of
 E ( H Heisenberg  E 
N  0 
 on site n in eigenvector of H.
) n  J E n 
J
(  E n 1   E n 1 )
2
that is,
(E  E
 N 0
 f (n  1)  f (n  1)

) f n  J 
 f (n) 
2


23
One- Magnon solution: spin wave
(E  E
 N 0
is solved by
N
 f (n  1)  f (n  1)

) f n  J 
 f (n)  , E ( N 0)   J ,
2
4


f ( n) 
e
ik j n
N
,
indeed f (n  1)  f (n  1) 
e
ik j n
N
(e j  e
ik
We must insert the periodic boundary condition for the N-spin chain
f n  N  
2 j
1
f ( n)  k j 
 E 
N
N
 N 0 e
energy : ( E  E
)
ik j n
N
 J
e
N

e
ik j n
n
n 1
 eik j  e ik j

 1

2
N 

ik j n
 E  E ( N 0)  J 1  cos(k j ) 
24
 ik j
).
N
HHeisenberg   J  Sn .Sn1 , J  0 ferromagnetic
n1
One- Magnon solution: the spin wave is a boson (spin 1)
E E 0
1
E 
N
N

n 1
e
ik j n
n ,
2 j
kj 
N
J
2.0
1.5
1.0
E  E ( N 0)  J 1  cos(k j ) 
0.5
k
3
2
1
1
2
3
This magnon dispersion law looks like a free quasi-particle
dispersion for small k
Two- Magnon solution: scattering, bound states
25
Several reversed spins
N
Let us first consider H0   J  S z n .S z n1 without hopping,
n 1
with r  N  reversed spins.
H0  1 , 2 ,
 N  E ( 1 , 2 ,  N )  1 , 2 ,  N .
Any pattern of spins will do, but
E ( 1 , 2 ,  N ) differs if far or consecutive spins are reversed:
 ... 
and
 ... 
have different E ( 1 ,  2 ,  N ).
J
E( 1 , 2 ,  N )  E( N 0)  2 N ,
4
N  number of reversed scalar products. Now include H XY .
Two Magnons
a configuration is denoted by |n1,n2>,
HXY
n1<n2 reversed spins
J N
   (SnSn1  SnSn1 ) moves  to right or left
2 n1
provided that adiacent sites are up spins: if the reversed spins are close
the energy is different and this is an effective interaction.
We introduce the case n2  n1  1by means of a boundary condition.
First, we solve for n2  n1  1, when the spins are 'far' and move independently.
( N  0)
( H Heisenberg  E
) n1 , n2  2 J n1 , n2  J
n1  1, n2  n1  1, n2  n1 , n2  1  n1 , n2  1
2
27
.
( H Heisenberg  E( N 0) ) n1 , n2  2 J n1 , n2
J
Let
n1  1, n2  n1  1, n2  n1 , n2  1  n1 , n2  1
2
f (n1 , n2 )   E n1 , n2 ,
, n2  n1  1
shift the energy origin : E  E
 N 0
E
f (n1  1, n2 )  f (n1  1, n2 )  f (n1 , n2  1)  f (n1 , n2  1)
E
f  n1 , n2   2 f  n1 , n2  
.
J
2
Thus r.r. is
f  n1 , n2   e
satisfied for n1  n2 by the product
i ( k1n1  k2 n2 )
,
E
 (1  cos(k1 ))  (1  cos  k2 ).
J
E
Recall One Magnon Solution :
 1  cos(k j ).
J
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
Invariance under global translations of both reversed spins:
f  n1  N , n2  N   f  n1 , n2  .
28