Transcript Slide 1

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)
1
N
HHeisenberg   J  Sn .Sn1
n1
Symmetries
there is a conserved wavevector K of all the system
 conserved total spin S, conserved Sz 
N  N 
.
2
Separated problems for each set K,S, Sz
N
HHeisenberg  H0  HXY , H0   J  S n .S
n1
HXY
z
z
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
2
Ground state for J>0: all spins up, no spin can be
raised, and so no shift can occur
 ground state energy E( N 0)   J
N
4
1 reversed spin in the chain
(E  E
is solved by
 N 0
 f (n  1)  f (n  1)

) f n  J 
 f ( n) 
2


f (n) 
ik j n
e
N
,
indeed f (n  1)  f (n  1) 
ik j n
e
N
(e  e
ik j
 ik j
3
).
We must insert the periodic boundary condition for the N-spin chain
 E  E ( N 0)  J 1  cos(k j ) 
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
Two- Magnon solution: scattering, bound states
4
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.
First, we solve for n2  n1  1, when the spins are 'far' and move independently.
( H Heisenberg  E( N0) ) n1 , n2  2 J n1 , n2  J
n1  1, n2  n1  1, n2  n1 , n2  1  n1 , n2  1
2
(Later we introduce the case n2  n1  1by means of a boundary condition.)
5
.
( 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
f (n1 , n2 )   E n1 , n2 ,
2
shift
, n2  n1  1
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
satisfied for n1  n2 by the product
E
i ( k1n1  k2 n2 )
f  n1 , n2   e
,
 (1  cos(k1 ))  (1  cos  k2 ).
J
Thus r.r. is
Recall One Magnon Solution :
E
 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).
6
However,there is invariance under global translations
of both reversed spins:
f  n1  N , n2  N   f  n1 , n2  .
We found the solution
f  n1 , n2   exp[i( k1n1  k 2n2 )];
f  n1  N , n2  N  must be the same.
f  n1  N , n2  N  is obtained from f  n1 , n2  by n1  n1  N and n2  n2  N :
f  n1  N , n2  N   f  n1 , n2  e i ( k1  k2 ) N ; so f  n1  N , n2  N   f  n1 , n2  requires :

  21
k 
 eik1N  ei
 k1N    21
 1
i ( k1  k2 ) N
N
e
1
   :  ik N




2
 e  i
 k2 N    22
 k    22
 e
 2
N
The integers 1  , 2 
are Bethe Quantum Numbers.

  21
 k1 
i ( k1  k2 ) N
N
e
1
   :
, 1  , 2 

 k    22
N
 2
Note: 1and 2  (0, N  1) because at N one is adding 2π to k1 or k 2 ,
and this leads to the very same wave function f  exp[i( k1n1  k2 n2 )].
2 j

For 1 magnon, k j 
,so the interaction produces a phase shift  .
N
N
E
 (1  cos(k1 ))  (1  cos  k2 ) is not the sum of independent magnon energies
J
The Schrödinger equation must fix  . This phase shift must contain the magnon interaction.
f  n1  N , n2  N   f  n1 , n2  is not enough to achieve this,
and there is an extra complication: another degenerate solution.
Consider the solution with eigenvalue E.
E
 (1  cos(k1 ))  (1  cos  k2 )
J
f  n1 , n2   exp[i( k1n1  k 2n2 )], for n1  n2 with
 k1N    21
f  n1  N , n2  N   f  n1 , n2   
.
 k2 N    22
We have seen that this solves the r.r.
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
This says that our solution has k1 behind k 2 .
However, exchanging k1 and k 2 one gets a new solution of the same
recurrence relations with k 2 behind.
k1
k2
f  n1 , n2   exp[i (k2 n1  k1n2 )], still for n1  n2
f  n1 , n2  is equally good, still with E  E ( N 0)  J [(1  cos(k1 ))  (1  cos  k2 )]
9
The degenerate solution with the k1 and k2 exchanged and must be
included for generality and also to satisfy the boundary conditions,
as I show in a moment; so one should write
f(n1, n2) = A e i(k1n1+k2n2) + B e i(k2n1+k1n2) , 0 < n1 < n2 < N.
ik1 N

e
 e i

k1 and k2 satisfy :  ik N
 i
2
e

e


 k1 N    21

 k2 N    22
The two contributions must enter with the same probability, A and B
differ by a phase a and therefore
 
a 
 
a 
f (n1 , n2 )  exp i  k1n1  k2 n2    exp i  k2 n1  k1n2    ,
2 
2 
 
 
0<n1  n 2 <N
Two phases are too many. Next we show that a = 
10
 
a 

a 
f (n1 , n2 )  exp i  k1n1  k2 n2    exp i  k2 n1  k1n2   , 0<n1  n 2 <N
2 
2 
 

We must be free to decide that the numbering of spins runs in the range n2,…n2+N-1
rather than 1,…N; when we do, n2<n1+N but the wave function does not care:
Therefore we have to impose
f (n1, n2 )  f (n2 , n1  N )
f (n2 , n1  N ) 
 
a 
 
a 
 exp i  k1n2  k2 (n1  N )     exp i  k2 n2  k1 ( n1  N )   
2 
2 
 
 
or, expanding (),
 
a
 
a


 exp i  k1n2  k2 n1   k2 N    exp i  k2 n2  k1n1   k1 N   .
2
2


 
 

a 
 
a 
Equating with f (n1 , n2 )  exp i  k1n1  k2 n2    exp i  k2 n1  k1n2   , we find:
2 
2 

 
 
a 
 
a
 a

 a

exp i  k2 n1  k1n2     exp i  k1n2  k2 n1   k2 N    exp  i   exp i   k2 N  
2 
2

 2

 
 
 2
 
a 
 
a
  a

 a

exp i  k1n1  k2 n2     exp i  k2 n2  k1n1   k1 N    exp i   exp i    k1 N  
2 
2

 2

 
 
  2
From this result
 a
  a
 a

 a

exp  i   exp i   k2 N   , exp i   exp i    k1 N  
 2

 2

 2
  2
we conclude:
ik1N
i

e

e
eik2 N  e  ia and eik1N  eia and recalling  ik N
, we find
 i
2
e  e
e i  e ia and ei  eia . Therefore, a   .
The phase shift a between the two orderings coincides
with the phase collected by k1 in a round trip!
a 
a 
 
 
 We rewrite f (n1 , n2 )  exp i  k1n1  k2n2    exp i  k2n1  k1n2   ,
2 
2 
 
 
in the form:
f (n1 , n2 )  e

i ( k1n1  k2n2  )
2
e
Next, find θ by considering the

i ( k2n1  k1n2  )
2
.

configurations.
The n1, n2  n1 1
spin configurations

N  number of reversed scalar products  2
N
For H0   J  Sz n .Sz n1 , H0  1 , 2 ,  N  E ( 1 , 2 ,  N )  1 , 2 ,  N with
n1
J
E( 1 , 2 ,  N )  2 N
4
The Schrödinger equation, which in general reads:
N
E
f (n1  1, n2 )  f ( n1  1, n2 )  f ( n1 , n2  1)  f ( n1, n2  1)
f  n1 , n2    f  n1 , n2  
J
2
2
for this n2  n1  1case has N   2, not 4; moreover, setting n2  n1  1we find:
E
f (n1  1, n1  1)  f (n1  1, n1  1)  f ( n1 , n1  1  1)  f ( n1 , n1  1  1)
f  n1 , n1  1  f  n1 , n1  1 
J
2
but f ( n, n )  0 (two reversed spins at same position is impossible). So, we are left with:
E
f (n1  1, n1  1)  f (n1 , n1  2)
13
f  n1 , n1  1  f  n1 , n1  1 
.
J
2
Recall : N  number of reversed scalar products, N  4
when the spins are far. However theansatz is still f (n1 , n2 )  e


i ( k1n1  k2n2  )
2
e

i ( k2n1  k1n2  )
2
f (n, n)  0 (is not the amplitude,in this case):this obeys
N  4 . That is, f obeys
the general equation for
E
f (n1  1, n2 )  f (n1  1, n2 )  f (n1 , n2  1)  f (n1 , n2  1)
f  n1 , n2   2 f  n1 , n2  
J
2
where, setting formally n2  n1  1, we get:
E
f (n1  1, n1  1)  f (n1  1, n1  1)  f (n1 , n1  2)  f (n1 , n1 )
f  n1 , n1  1  2 f  n1 , n1  1 
J
2
In summary,
since N   2, we must solve
E
f (n1  1, n1  1)  f (n1 , n1  2)
f  n1 , n1  1  f  n1 , n1  1 
.
J
2
However, f (n1 , n2 )  e

i ( k1n1  k2 n2  )
2
e

i ( k2 n1  k1n2  )
2
solves the equation for N   4.
Both equations must be true!
14
To find  such that: f (n1 , n2 )  e

i ( k1n1  k2n2  )
2
e

i ( k2n1  k1n2  )
2
f (n1  1, n1  1)  f (n1  1, n1  1)  f (n1 , n1  2)  f (n1 , n1 )
E
f  n1 , n1  1  2 f  n1 , n1  1 
J
2
and
f (n1  1, n1  1)  f (n1 , n1  2)
E
f  n1 , n1  1  f  n1 , n1  1 
.
J
2
Equating the rhs,
f (n1  1, n1  1)  f (n1  1, n1  1)  f (n1 , n1  2)  f (n1 , n1 )
2
f (n1  1, n1  1)  f (n1 , n1  2)
 f  n1 , n1  1 
.
2
1
Simplify  f (n1 , n1  1)   f (n1 , n1 )  f (n1  1, n1  1) .
2
  must be such that this is true.
2 f  n1 , n1  1 
15
1
f  n1 , n1  1   f  n1  1, n1  1  f  n1 , n1  
2
must be combined with ansatz:
f (n1 , n2 )  e

i ( k1n1  k2 n2  )
2
e

i ( k2 n1  k1n2  )
2
togive a condition which fixes  . To this end,set n1  0 and n2  1.
f (0,1)  e

i ( k2  )
2
e

i ( k1  )
2


i 
1
1  i ( k1  k2  2 ) i ( k2  k1 2 ) i 2
e
e e 2
 f  1,1  f  0,0    e
2
2

So, the relation is:

i ( k2  )
2

i ( k1  )
2

i ( k1  k2  )
2
2e
 2e
e
e
and we must solve for  .

i ( k2  k1  )
2
i

e 2 e
i

2
16
In order to solve

2e
i ( k2  )
2
 2e

i ( k1  )
2
e

i ( k1  k2  )
2
e
i ( k2  k1 
multiply by e
i

2
)
i

e 2 e
i

2
,

2
2e i e ik2  2e ik1  e i e i ( k1  k2 )  e i ( k2  k1 )  e i  1
Bring  to the lhs:
2e i e ik2  e i e i ( k1  k2 )  e i  e i ( k2  k1 )  1  2e ik1
i ( k2  k1 )
ik1
e

1

2
e
 e i  ik
.
i
(
k

k
)
2e 2  e 1 2  1
This must be solved with Nk1    21
Nk2    22
which express the condition f  n1  N , n2  N   f  n1 , n2  .
17
Before putting this equation in real form we check that this is really a phase factor.
Really a phase?
i ( k2  k1 )
ik1
e

1

2
e
N
i
e  ik2

i ( k1  k2 )
2e  e
1 D
| N |  (e
2
 1 e
 i ( k2  k1 )
 i ( k2  k1 )
 2e
 1  2e
 ik2
e
 ik1
)( e
i ( k2  k1 )
i ( k2  k1 )
 1  2e )
ik1
 1  2e  2e  2e
ik1
ik2
 ik1
4
 6  2 cos( k1  k2 )  6 cos( k1 )  cos( k2 )
is symmetric in
k1 , k2
 | N |2 | D|2
N
| | 1. So,   .
D
18
i ( k2  k1 )
ik1
e

2
e
1
i
We must solve e  ik
for k1 , k2 ,
i
(
k

k
)
2e 2  e 1 2  1
with Nk1    21 Nk2    22 .
Real form of Bethe Ansatz equation for 
i ( k2  k1 )
ik1
e

1

2
e
To cast e i   i( k  k )
in real form,
ik
e 1 2  1  2e 2
 
e i  1
we exploit the identity cot    i i
which is readily shown:
2
e

1
 
i
i

i

i

i

e 1
e 2 e 2
2i
e 2 e 2
i i
i 
 
.


i
i
i
i
2
e 1
e 2 e 2 e 2 e 2
e i  1
e i ( k2  k1 )  2e ik1  1
i
Mathematica simplifies i i
where e  ik
e 1
2e 2  e i ( k1  k2 )  1
k 
k 
1
to give [cot  1   cot  2 ].
2
2
 2
19
Real form of Bethe Ansatz equation for 
 k1 
 k2 
 
2 cot    cot    cot  
2
2
 2
Nk1    21 Nk2    22
Bethe Ansatz equations for 2 magnons
 k1 
 k2 
 
2 cot    cot    cot  
2
2
2
Nk1    21 Nk2    22
The Bethe Ansatz equations can be solved for k1, k2 and θ
by combined analytic and numerical techniques for N = some tens.
Most solutions are scattering states with real momenta and a finite
θ; however when one of the Bethe quantum numbers λ vanishes, the
corresponding k and θ also vanish, and there is no interaction.
Other solutions with |λ1 - λ2|=1 have complex momenta with k1 = k2∗
and represent magnon bound states in which the two flipped spins
propagate at a close distance.
21
Rossi: uno dei  nullo : non c’e’ interazione
Bianchi 2 – 1 >= 2 scattering states
Blu: alcuni sono stati legati
Da Karbach et al. Cond-mat9809162
22
Three Magnons: here the real fun begins!
Basis: a configuration is denoted by |n1, n2, n3>,
n1<n2<n3
N
H0   J  Sz n .Sz n1
J N
HXY    (SnSn1  SnSn1 )
2 n1
moves  to right or to the left
n1
provided that adiacent sites are up spins: this is an effective interaction
Strategy: first solve for unhindered jump region
n3  n2  1, n2  n1  1
then impose boundary conditions.
23
amplitude :
f (n1 , n2 , n3 )   E n1 , n2 , n3 ; Schrödinger Equation 
 N 0
(E  E
)
1
f (n1 , n2 , n3 )  3 f (n1 , n2 , n3 )  [ f (n1  1, n2 , n3 )  f (n1  1, n2 , n3 )
J
2
 f (n1 , n2  1, n3 )  f (n1 , n2  1, n3 )  f (n1 , n2 , n3  1)  f (n1 , n2 , n3  1)]
trivially satisfied by
f  n1 , n2 , n3   ei ( k1n1  k2n2  k3n3 ) ,
E  E ( N 0)  J [(1  cos(k1 ))  (1  cos  k2 )  (1  cos  k3 )]
Exchange 1 and 2:The
by
Schrödinger
Equation
is also satisfied
f  n2 , n1 , n3   ei ( k2n1 k1n2 k3n3 ) ,
still with
E  E ( N 0)  J [(1  cos( k1 ))  (1  cos  k2 )  (1  cos  k3 )]
But we must account for 3! permutations P of 3 objects: (123),(231),(312),(213),(132),(321)
24
and all must enter with same probability  mutual phase shifts:
i
Starting with exp[i (k1n1  k2n2  k3n3 )  (12  13   23 )] with ij   ji mutual phase shifts
2
and including all permutations ,
f  n1 , n2 , n3  
i
i
exp[i (k1n1  k2 n2  k3n3 )  (12  13   23 )]  exp[i (k2 n1  k3n2  k1n3 )  ( 23   21  31 )]
2
2
i
i
 exp[i (k3n1  k1n2  k2 n3 )  (31  32  12 )]  exp[i (k2 n1  k1n2  k3n3 )  ( 21   23  13 )]
2
2
i
i
exp[i (k1n1  k3n2  k2 n3 )  (13  12  32 )]  exp[i (k3n1  k2 n2  k1n3 )  (32  31   21 )]
2
2
25
f  n1 , n2 , n3  
i
i
exp[i (k1n1  k2 n2  k3n3 )  (12  13   23 )]  exp[i (k2 n1  k3n2  k1n3 )  ( 23   21  31 )]
2
2
i
i
 exp[i (k3n1  k1n2  k2 n3 )  (31  32  12 )]  exp[i (k2 n1  k1n2  k3n3 )  ( 21   23  13 )]
2
2
i
i
exp[i (k1n1  k3n2  k2 n3 )  (13  12  32 )]  exp[i (k3n1  k2 n2  k1n3 )  (32  31   21 )]
2
2
Since we can start the numbering from n 2 , this must be equal to f  n2 , n3 , n1  N  ;
With n1  n2 , n2  n3 , n3  n1  N
the first term becomes:
i
exp[i(k1n2  k2 n3  k3n1  k3 N )  (12  13   23 )]
2
that must coincide with the third above
i
exp[i (k3n1  k1n2  k2 n3 )  (31  32  12 )]:
2
26
i
first term : exp[i(k1n2  k2n3  k3n1  k3 N )  (12  13  23 )] must coincide with
2
i
 exp[i(k3n1  k1n2  k2n3 )  (31  32  12 )] 
2
i
i
exp[ik3 N  (12  13  23 )]  exp[ (31  32  12 )].
2
2
12 cancels.
i
i
exp[ik3 N ]  exp[ (13   23 )]exp[ (31  32 )]
2
2
 since 31  13 etc.

eik3N  ei (31 32 )
Let us go on with the same change of the numbering
n1  n2 , n2  n3 , n3  n1  N
f  n1 , n2 , n3  
i
i
exp[i (k1n1  k2 n2  k3n3 )  (12  13   23 )]  exp[i (k2 n1  k3n2  k1n3 )  ( 23   21  31 )]
2
2
i
i
 exp[i (k3n1  k1n2  k2 n3 )  (31  32  12 )]  exp[i (k 2 n1  k1n2  k3n3 )  ( 21   23  13 )]
2
2
i
i
exp[i (k1n1  k3n2  k2 n3 )  (13  12  32 )]  exp[i (k3n1  k2 n2  k1n3 )  (32  31   21 )]
2
2
f  n1 , n2 , n3   f  n2 , n3 , n1  N  : the second
term becomes :
i
exp[i (k2n2  k3n3  k1[n1  N ])  ( 23   21  31 )]
2
i
 exp[i (k2n2  k3n3  k1n1  k1 N )  ( 23   21  31 )]
2
that must coincide with
Finally, of course,
the first above :
eik1N  ei (12 13 ) .
eik2 N  ei (21 23 )
28
 e ik1N  e i (12 13 )
 ik2N
i ( 21  23 )
e

e

 eik3N  ei (31 32 )

integers 1 , 2 , 3
 k1N  12  13  21

  k2 N   21   23  22
 k N      2
31
32
3
 3
with  ji   ij
 Bethe Quantum Numbers  (0, N  1)
3 independent magnons? NO!
Interaction produces phase shift and E is not the sum of independent
magnon energies
29
It is time worry about configurations like 
e.g.
reversed arrows at n1 , n 2 =n1 +1 , n 3 ;
Allowed jumps: while n3 is free, n3
n1 can only jump to left n1
total N  4
n3  1, n3  1,
n1  1, and n2 can only jump to right n2
n2  1,
Schrödinger Equation 
H n1 , n1  1, n3  ( E
N  0
 2 J ) n1 , n1  1, n3
J
 ( n1  1, n1  1, n3  n1 , n1  2, n3  n1 , n1  1, n3  1  n1 , n1  1, n3  1 )
2
30
The special recurrence formula for the amplitude f
accounting for the hindered jumps becomes:

 N 0 
EE
J
However
 f n , n  1, n   2 f n , n  1, n   1  f n  1, n  1, n   f n , n  2, n 


2   f  n1 , n1  1, n3  1  f  n1 , n1  1, n3  1


1
1
1
3
1
1
3
1
3
1
1
3
the general r.r. also holds, like above,since the Ansatz is still
f  n1 , n2 , n3  
i
i
exp[i ( k1n1  k2n2  k3n3 )  (12  13   23 )]  exp[i (k2n1  k3n2  k1n3 )  ( 23   21  31 )]
2
2
i
i
 exp[i ( k3n1  k1n2  k2n3 )  (31  32  12 )]  exp[i ( k2n1  k1n2  k3n3 )  ( 21   23  13 )]
2
2
i
i
exp[i ( k1n1  k3n2  k2n3 )  (13  12  32 )]  exp[i (k3n1  k2n2  k1n3 )  (32  31   21 )]
2
2
Recall the general r.r.:
(E  E  )
1
f (n1 , n2 , n3 )  3 f (n1 , n2 , n3 )  [ f (n1  1, n2 , n3 )  f (n1  1, n2 , n3 )
J
2
 f (n1 , n2  1, n3 )  f (n1 , n2  1, n3 )  f (n1 , n2 , n3  1)  f (n1 , n2 , n3  1)]
N 0
31
comparison 
f  n1 , n1  1, n3  
1
 f  n1  1, n1  1, n3   f  n1 , n1 , n3  
2
this condition (for consecutive n1 and n2 ) must be combined with
the Ansatz: computing f(n1 ,n1  1,n3 ) etc. by the Ansatz,
f  n1 , n2 , n3  
i
i
exp[i (k1n1  k2 n2  k3n3 )  (12  13   23 )]  exp[i (k2 n1  k3n2  k1n3 )  ( 23   21  31 )]
2
2
i
i
 exp[i (k3n1  k1n2  k2 n3 )  (31  32  12 )]  exp[i (k2 n1  k1n2  k3n3 )  ( 21   23  13 )]
2
2
i
i
exp[i (k1n1  k3n2  k2 n3 )  (13  12  32 )]  exp[i (k3n1  k2 n2  k1n3 )  (32  31   21 )]
2
2
i ( k1  k2 )
ik1
e

1

2
e
 ei12  ik
.
i
(
k

k
)
2e 2  e 1 2  1
This is identicalto the condition found
before with just 2 reversed spins!!!
32
   ei  1
Then, to put in real form, use Identity cot    i i
 2  e 1
 12 
 k1 
 k2 
 2 cot 
  cot    cot  
 2 
2
 2
Nk1  12  13  21 Nk2   21   23  22
Nk3   31   32  23
for fixed integers i  Bethe Quantum Numbers  (0, N  1),
coupled transcendental problem with
unknowns ij , ki .
Equations remain the same as in the case of
2 magnons.
With 3 magnons, many more terms, but direct extension of rules.
33
Many Magnons (many reversed spins)
N
H Heisenberg   J  Sn .Sn1 , J  0 ferromagnetic chain
n1
One tries with a product of magnons and everything generalizes!
Basis : |n1,n2,..nr> in growing order :sites with down spins
Each k i
e
ik2 N
 e i
picks up a
e
 ei
ik1N
phase from each k j
generalizes to e
iki N
e
i
 ij
j
, ij   ji
 Nki   ij  2i ,
j
i
integer, 0  i  N  1 Bethe quantum numbers
Bethe Ansatz : 2  spins
f (n1 , n2 )  e

i ( k1n1  k2n2  )
2
e

i ( k2n1  k1n2  )
2
3  spins
f (n1 , n2 , n3 )  e
e
e
i ( k3 n1  k2 n2  k1n3 
i ( k2 n1  k3n2  k1n3 
f (n1 , n2 ,
i ( k1n1  k2 n2  k3n3 
32 31  21
2
23 21 31
2
, nr )  
P
)
)
e
e
12 13  23
2
)
i ( k1n1  k3n2  k2 n3 
i ( k3n1  k1n2  k2 n3 
e
i ( k2 n1  k1n2  k3n3 
13 12  32
2
31 32 12
2
)
21 13  23
2

)
r  spins
 r

i r
exp i  k Pj n j   Pi Pj  , P  permutations
2 i j
 j 1

r
E  J  (1  cos k j ) (energy origin at E( N0) )
j 1
35
)
with the result:
 ij 
 kj 
 ki 
2 cot    cot    cot   , i , j  1,....r
 2
2
2


 
 
to be solved with
Nk1    21
Nk2    22
 Nki   ij  2i ,
j
i
int eger , 0  i  N  1 Bethe quantum
numbers
36
The Bethe equations are quite hard to solve for many flipped spins in large systems. In
recent years solutions have been obtained which are far from trivial; among others,
bound states of several magnons have been reported.
The Bethe Ansatz is a rare example of exact solution of an interacting
many-body problem; it keeps the same form independent of the size of the
system.
The root of the (relative) simplicity that allows this solution is onedimensionality: the evolution does not allow overtakings, and spins always
keep a fixed order.
However, without the ingenuity of Hans Bethe, the solution
could have remained undetected; who knows how many important model
problems are solvable, but still unsolved.
37
How magnons arise
Heisenberg model: H   J  Si .S j   B Si
ij 
i
On each site, S=(S x ,S y ,S z ) with S  S  i S
S   S x +iS y
S   S x -iS y
[ S  , S  ]  2 S z .
Intuitively, S  , S  must have to do with magnon a,a †
however magnon must interact and commutation relations
[a,a † ]  1 is not compatible.
Holstein-Primakov transformation to boson operators
a †a
S = 2S 1a S- = 2S a †
2S
a a †  1  a †a =1+n.
+
a †a
S S =2S 1a a†
2S
+ -
S S  2 Sa
- +
†
a †a
1.
2S
a †a
n
1 2 S (1  )(1  n)  2 S  2 Sn  n  n 2 .
2S
2S
a †a
a †a
a †na
11a  2S (n 
)
2S
2S
2S
a †a
S = 2S 1a S- = 2S a †
2S
a a †  1  a †a =1+n.
+
a †a
S S =2S 1a a†
2S
+ -
a †a
1.
2S
a †a
n
1 2 S (1  )(1  n)  2 S  2 Sn  n  n 2 .
2S
2S
a †a
a †a
a †na
S S  2 Sa
11a  2S (n 
)
2S
2S
2S
a †na=a †a †aa=a † ( aa †  1)a  n 2  n
- +
†
S-S+  2 Sn  n 2  n.
[S+ ,S- ]  2( S  a †a). This is OK provided that S z  S  a †a.
S   S x +iS y
Indeed, using
S   S x -iS y
1  
S  S ( S  1)  S  S  S ( S  1)  ( S S  S  S  )
2
 ( S  n ) 2 , which agrees with
2
z
2
x
[S  , S  ]  2 S z .
2
y
This holds on every site of the solid:
Heisenberg model: H   J  Si .S j   B Si
ij 

j
S = 2S 1a j a †j
aj 
a †j a j
aj
2S
 1  a †j a j =1+n.
1
N

k
e
 ikx j
bk

j
S = 2S a
i
†
j
1-
a †j a j
2S
.
bk magnon vaiables [bk , bk†' ]= kk ' .
42
43
AF phase only
exchange field of 0.25 eV,
Predicted
spin flip
excitation
in NiO
44
Some Developments: theoretical physics in 1d
Hultén (1938) modeled 1d antiferromagnet
Lieb-Liniger (1963) solved exactly 1d spin 0 Bose gas
Lieb-Wu (1968) 1d Hubbard model (exact solution, no magnetism)
Wiegmann-Tsvelik solution of Anderson model (with assumptions that V
does not depend on k, U is small, level close to Fermi level): as long as the
Anderson model is taken to be spherically symmetric, it is 1d.
Integrable systems-statistical mechanics, string theory…
45