Topological solitons in the Heisenberg
The n-field theory
In the Heisenberg model spin is a 3D vector n = (n1,
n2, n3) with unit length
n2 = 1 . We consider n(x,y) field
on 2D plane. In continuum limit the system energy
E = 1/2 ∫ dx dy
(nx· nx +
ny· ny ) ,
nx= ∂n/∂x .
It is necessary that n(∞) → no
in order to the total energy remains finite. The trivial field
n(x,y) = no corresponds to the global
energy minimum E = 0 .
n1 = 2u/(1 + u2 + v2),
n2 = 2v/(1 + u2 + v2),
n3 = 1 - 2/(1 + u2 + v2),
By means of the stereogrphic projection one can map a point on the sphere
n = (sin θ cos φ, sin θ sin φ, cos θ)
into a point on the plane w = (u, v)
w = u + i v =
(n1 + i n2)/(1 - n3) =
Note that all infinite points are mapped into one point - the "north
pole" of the sphere. By means of substitution
we get (see also continuum limit)
E = 2 ∫ dx dy
(ux2 + vx2 +
uy2 + vy2) /
(1 + u 2 + v 2)2 =
2 ∫ dx dy
[(ux - vy )2 +
(uy + vx )2 +
2(ux vy - uy vx )] /
(1 + u 2 + v 2)2. (*)
In the spherical coordinate system
u = ctg θ/2 cos φ , v = ctg θ/2 sin φ
the third term in the bracket is reduced to the topological charge
∫ sin θ
φyθx) dx dy =
∫ sin θ dφ dθ =
∫ dΩ = 4π Q .
Thus from (*) it follows that
E ≥ 4π Q
and minimum is reached under the Koshi-Riemannian condishions
ux - vy = 0,
uy + vx = 0 ,
i.e. when w(z) = u(x,y) + iv(x,y) is an analitic function of
z = x + iy.
Instantons and anti-instantons
If w(∞) → 1 , then n-instantons and n-antiinstantons
w = ∏i=1,n (z - ai )/(z -
è w = ∏i=1,n (z* - ai
)/(z* - bi ) ,
where |ai| and αi = Arg(ai )
are radius and phase of ith instanton, and complex number
bi determines its position.
One-instanton solution is
w = u + i v =
1 - eiα / z = ctg(θ/2) eiφ ,
As since E is inariant under scaling transphormations x' = ax,
then instanton energy do not depend on its radius and |a| = 1 ,
b = (0 + 0i) are used. Anti-instanton differs by conjugation z*.
3D VRML field models (see
instanton with α=0, anti-instantons with
"chupa-chups". 3D are usefull for introduction but
2D Java visualization seems me more informaive.
Applet below makes view from above on the vector of anti-instanton with
the phase α = 0 . At the right and bottom borders you see
cross-sections along the vertical and horizontal grey lines. Red arrows are
directed to observer and blue ones - backwards.
w(∞) → 1 corresponds to the direction towards observer.
The backwards direction corresponds to w(zo) = -1 .
Therefore the "pole" (the blue point) is placed at
zo = eiα/2 . At last for vectors placed
in the picture plane Re(w) = 0 , it corresponds to the circle centered
in the pole with radius 1/2 and passing through the coordinate
origine. Inside the circle (blue) arrows are directed towards observer
and outside the circle (red) arrows are directed backwards.
Click mouse with Alt (Ctrl) to zoom in (out) the picture two times.
Draw the gray crossed lines to move them. You can see coordinates of the
crossing in the Status bar (below).
Press Enter to set new phase (a field).
The sphere to sphere maps
By means of the stereographic projection one can "wrap" the plane
(x1, x2) into the sphere Sx2
(all infinite spins are parallel as since all infinite points of the plane
are mapped into the "north pole" of the sphere).
Thus n-field makes a sphere to sphere map
Sx2 → Sn2 .
As like circle to circle maps, non-equivalent maps differ by topological
charge, i.e. how much times the sphere
Sx2 is wraped on the sphere
Therefore the total topological charge is
Three vectors n(x),
n(x+dx2) "occupy" on the sphere
Sn2 an area (or space angle)
dΩ. For small dx1, dx2
the space angle is proportional to the volume between these vectors
(n ·[∂1n ,
∂2n]) dx1 dx2 .
1/4π ∫ dΩ(x) =
1/4π ∫ sin θ
dθ(x)dφ(x) = 1/8π
∫ d2x εμν(n
[∂μn, ∂μn]) .
What does hedgehog hide?
Topologically different maps Sx3 →
Sn2 are determined by the Hopf invariant.
Therefore there are localized topological solitons (with n(∞)
→ no ) in the 3D Heisenberg magnetics.
But if (similar to vorteces) an 3D area with the Heisenberg magnetic
is surrounded by a sphere with nonzero topological charge
(the simplest case is when all vectors are directed outward the shpere,
i.e. it is a "hedgehog") then:
What is inside the circle - monopole?
Do Q = 1 and Q = -1 attract each other and how? ...
I'm grateful to D.E.Burlankov for discussions.
Excitations in 1D spin chain
updated 12 June 2004