Topological solitons in the Heisenberg ferromagnetic

The n-field theory

anti-instanton
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 .
stereograpic 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) = ctg(θ/2) e.
Note that all infinite points are mapped into one point - the "north pole" of the sphere. By means of substitution
    n1 = 2u/(1 + u2 + v2),   n2 = 2v/(1 + u2 + v2),   n3 = 1 - 2/(1 + u2 + v2),
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 (φxθy - 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 solutions are
    w = ∏i=1,n (z - ai )/(z - bi )     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 - e / z = ctg(θ/2) e ,
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 Why VRML?): instanton with α=0, anti-instantons with α=0, α=1.5, "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 = e/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.

Controls: 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).

Below different HUE colors are used for different spin phases φ. "Top" spins (θ = 0) are white and "bottom" ones (θ = π) are black.

Instantons (topological solitons) are looked as color round defects in these pictures.

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 Sn2.
charge Three vectors n(x), n(x+dx1) and n(x+dx2) "occupy" on the sphere Sn2 an area (or space angle) . For small dx1, dx2 the space angle is proportional to the volume between these vectors (n ·[∂1n , ∂2n]) dx1 dx2 .
Therefore the total topological charge is
    Q = 1/ dΩ(x) = 1/ sin θ dθ(x)dφ(x) = 1/ d2x εμν(n [∂μn, ∂μn]) .

I'm grateful to D.E.Burlankov for discussions.


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updated 12 June 2004