Excitations in 1D spin chain

The XY model

We begin with a very simple example of 1D chain of XY spins si rotating in a plane. The chain energy is
    E = -J ∑ i (si si-1 ) = -J ∑ i cos(φi - φi-1 )
where si = (cos φi , sin φi ) and φi is i-th spin phase. For small δφ = φi - φi-1 one can expand cos up to the -δφ2/2 term. Replacing discrete variabe i by continuous one x we get (see below continuum limit)
    E = L(φ') dx = 1/2 (φ')2 dx ,     φ' = dφ/dx .
If spins rotate perpendicular to x one can imagine them as an elastic band. It follows from δE = 0 that
    ∂L /∂φ - (∂L /∂φ')' = 0   or   φ" = 0 ,   φ = a x .
We fix φo = φN = 0 (one can wrap the chain (band) in a ring too). The trivial solution φ(x) = 0 (i.e. flat band) corresponds to the global minimum of energy E = 0 .
band But it follows from the picture that there are also "twisted" n-times solutions φ(x) = 2πn x/L which correspond to local minima of energy En ~ n2.
It is impossible to deform smoothly one of these solutions to another, therefore they are topologically non-equivalent.

The circle to circle maps

Spin rotating in a plane is equivalent topologically to a circle Sn1 (it is marked by the subscript n). Closed spin chain is equivalent to a circle Sx1 too.
map Topologically non-equivalent maps Sx1 → Sn1 differ by the topological charge Q which determines how many times the circle Sx1 is "wraped" on the circle Sn1.
map To calculate Q one can integrate throughout the chain phase difference dφ(x) = φ'(x) dx between close vectors s(x) and s(x+dx)
    Q = 1/ φ'(x) dx = Δφ/2π .
Note that one can calculate by the area of sector confined by these vectors dφ = (s1 s2' - s1' s2 ) dx .

The Sx2 → Sn1 maps

If at infinity n(x,y) → no , then 2D plane (x, y) is equivalent topologically to a sphere. But one can shrink into a point any map of a sphere to a circle. Therefore any smooth field is equivalent to the trivial field n = no . It is true for any D > 1 too.


Topoligical "deffects" on a plane appear if one take a contour with Q ≠ 0 . Then there are vorteces within it. In three dimensional space one shall draw a surface on this contour and then vortex threads cross it.

The circle to sphere map

band band
For the continuum XY-chain with Q ≠ 0 above it is impossible to move upwords smoothly all "blue" arrows rotating spins in the picture plane. But one makes it easy by rotation around the x axis in 3D space (i.e. for the Heisenberg spins). Therefore Heisenberg chains have no topological excitations. It is equivalent to the statement that one can shrink into a point any map of a circle to a sphere (i.e. any contour on a sphere).
But metastable states will appear in the Heisenberg model on 2D plane.

Appendix: Continuum limit

As we have seen in continuum limit interaction energy of neighboring spins is J/2 δφ2. If ρ is the number of spins per unit length, then the distance between two spins is 1/ρ, and δφ = 1/ρ dφ/dx. Therefore energy of segment Δx is J/2 (1/ρ φ')2ρΔx and if one put J = ρ = 1 , then total energy of chain is
    E = 1/2 (φ')2 dx .
You can see that in the polar coorninates s = (s1 , s2 ) = (cos φ , sin φ ) and dl2 is
    dl2 = ds12 + ds22 = dφ2 .
I.e. one can evaluate δφ2 as δs12 + δs22 and the total energy as
    E = 1/2 (ss') dx .
This formula is correct for the Heisenberg spins when s = (s1 , s2 , s3 ). In the XY model on (x, y) plane spin energy is
    E = 1/2 dx dy (φx2 + φy2 ) ,     φx= ∂φ /∂x .
And Heisenberg ferromagnetic on 2D plane has energy
    E = 1/2 dx dy (sx· sx + sy· sy ) ,     sx= ∂s /∂x .
At last in the stereographic coordinates (u, v)
    dl2 = 4 (du2 + dv2) / (1 + u2 + v2)2,
    E = 2 dx dy (ux2 + vx2 + uy2 + vy2) / (1 + u 2 + v 2)2.
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updated 9 June 2004