# What is spin stiffness?

I read the defination of spin stiffness here

But I can't understand how to twist an angle. Any help will be appreciated!

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I believe it means rotating each spin by $\theta$. – leongz Dec 5 '12 at 4:51

Let's consider, to fix the ideas, that we have the Hamiltonian of the XY model $${\mathcal{H}}=-J\sum_{\langle i,j \rangle}\mathbf{S}_i \cdot \mathbf{S}_j$$ where $\langle i,j \rangle$ denotes nearest neighbour site summation. This model can be seen as planar rotors of unit length in a two-dimensional lattice and can be therefore written as $${\mathcal{H}}=-J\sum_{\langle i,j \rangle} \cos(\theta_i -\theta_j)$$ where $\theta_i$ denotes the angle of the rotor which lies at site $i$ with respect to sone orthogonal direction to the plane containing the rotors. If the direction of the rotors changes smoothly from site to site we can pass to the continuum limit and the Hamiltonian can be well approximated by $${\mathcal{H}}=E_0+\dfrac{J}{2}\int d\mathbf{r} [\nabla_\mathbf{r} \theta(\mathbf{r})]^2$$ where $E_0$ is the ground state energy [all the spins aligned].

The spin stiffness or helicity modulus, noted $\rho_s$, denotes the energy cost of applying a gradient or twist to the rotors $$\theta(\mathbf{r})\rightarrow \theta'(\mathbf{r})=\theta(\mathbf{r})+\delta\theta(\mathbf{r})$$ Indeed, the twist/rotation of the spins gives an increase of the free energy [per unit surface or in the discrete lattice, per site] of $$F(\delta \theta)-F_0=\dfrac{1}{2}\rho_s \delta \theta^2$$ for $\delta \theta\equiv \delta \theta(\mathbf{r})$ and therefore $$\rho_s =\dfrac{\partial^2 F(\delta \theta)}{\partial(\delta \theta)^2}\Bigg|_{\delta \theta=0}$$ For the XY model (continuum) it is easy to see that $\rho_s =J$.

The spin stiffness is analogous to the shear constant of a material which determines how much the free energy of the system changes when it suffers a [shear] deformation.

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You meant all spin twist the same angle? $\delta\theta(\mathbf{r})$ does nt depend on $\mathbf{r}$? – hlew May 13 '13 at 13:04
No, the angle can generally depend on the position of the rotor. – DaniH Aug 25 '13 at 18:43