1 Second-order surface-anisotropy energy $ \mathcal {E}_{2}$

In Ref. [74] a spherical multi-spin nanoparticle was considered, the anisotropy in the core was ignored and the surface anisotropy was taken as NSA. Using the continuous approximation for $ N\gg1$ the authors evaluated the surface energy density $ E_{S}(\mathbf{m},\mathbf{n})$, where $ \mathbf{m}$ is the magnetization and $ \mathbf{n}$ is the normal to the surface. The equilibrium magnetization satisfied the Brown's condition [98]:

$\displaystyle \mathbf{m}\times \mathbf{H}_{eff}=0,\:\:\;\qquad \mathbf{H}_{eff}=\mathbf{H}_{A}+J\Delta\mathbf{m}$ (25)

here $ \mathbf{H}_{eff}$ is the total effective field, J is the exchange constant, $ \Delta$ is the Laplace operator and $ \mathbf{H}_{A}$ is the anisotropy field, that in this case is due entirely to the surface.

$\displaystyle \mathbf{H}_{A}=-\frac{dE_{S}}{d\mathbf{m}}\delta(r-R),$ (26)

here R represents the radius of the spherical particle.

For the case of $ Ks\ll J$ we can suppose that the deviations of $ \mathbf{m}$ of the homogeneous state $ \mathbf{m}_{0}$ are small and the problem can be linearized.

$\displaystyle \mathbf{m(r)}\cong\mathbf{m}_{0}+\mathbf{\psi(r,m_{0})}, \:\:\; \qquad \psi=\vert\mathbf{\psi}\vert\ll1$ (27)

The correction $ \psi$ is the solution of the internal Neumann boundary problem for a sphere:

$\displaystyle \Delta\mathbf{\psi}=0, \:\:\qquad \frac{\delta\mathbf{\psi}}{\delta r}\Bigr\vert _{r=R}=\mathbf{f(m,n)}$ (28)

$\displaystyle \mathbf{f(m,n)}=-\frac{1}{J}\Bigl[\frac{dE_{S}(\mathbf{m_{0},r})}...
...S}(\mathbf{m_{0},r})}{d\mathbf{m}}\cdot\mathbf{m_{0}}\Bigr)\mathbf{m_{0}}\Bigr]$ (29)

Then, solving the corresponding Neumann problem by the Green's function technique, we obtain:

$\displaystyle \mathbf{\psi(r,m)}=\frac{1}{4\pi}\int_{S}d^{2}\mathbf{r'G(r,r')}\mathbf{f(m,n)}$ (30)

where $ \mathbf{G(r,r')}$is the Green function. Computing the energy of the multi-spin particle, it was found that the corresponding effective energy is of the $ 4^{\mathrm{th}}-$order in the components of the net magnetic moment $ \mathbf{m}$ and the $ 2^{\mathrm{nd}}-$order in the surface anisotropy constant $ k_s$, that is

$\displaystyle \mathcal{E}_{2} = k_{2}\sum_{\alpha=x,y,z}m_{\alpha}^{4},$ (31)

with

$\displaystyle k_{2} = \kappa\frac{ k_{s}^{2}}{z}.$ (32)

$ \kappa$ is a surface integral that depends on the underlying lattice, shape, and the size of the particle and also on the surface-anisotropy model. For instance, $ \kappa\simeq0.53466$ for a spherical particle cut from an sc lattice, with NSA. We would like to note that the contribution (2.14) scales with the system's volume and thus could renormalize the volume anisotropy of the nanoparticle.

The equation (2.15) was obtained analytically for $ K_{s}\ll J$ in the range of the particle size large enough ( $ \mathcal{N}\gg1$) but small enough so that $ \delta\psi$ remains small. Being $ \delta\psi\sim \mathcal{N}^{1/3}K_{s}/J$, the angle of order which describes the noncollinearity of the spins that results from the competition of the exchange interaction and the surface anisotropy [*].

Since $ k_{2}$ is nearly size independent (i.e. the whole energy of the particle scales with the volume), it is difficult to experimentally distinguish between the core cubic anisotropy and the one due to the second order surface contribution (see discussion later on). The physical reason for the independence of $ k_{2}$ on the system size is the deep penetration of the spin non-collinearities into the core of the particle. This means that the angular dependence of the non-collinearities also contributes to the effective anisotropy. Interestingly this implies that the influence of the surface anisotropy on the overall effective anisotropy is not an isolated surface phenomena and is dependent on the magnetic state of the particle. We note that this effect is quenched by the presence of the core anisotropy which could screen the effect at a distance of the order of domain wall width from the surface.

The energy contribution $ \mathcal {E}_{2}$ has also been derived in the presence of core anisotropy [73] and numerically tested in Ref. [87]. Similar conclusions also apply for the case of the transverse surface anisotropy, Ref. [87].

Rocio Yanes