What's the difference between spin-unpolarized, spin-polarized and non-colinear calculation?

Question

The central goal of the first-principles simulation with DFT is to solve the Kohn-Sham equation:

$$[-\dfrac{1}{2}\nabla^2+v_{\textit{eff}}(\vec{r})]\phi_n(\vec{r})=E_n\psi_n(\vec{r}) \tag{1}$$

Here the atomic unit has been adopted. But the EQ.(1) can only be utilized to predict the properties of materials without magnetic properties and strong spin-orbit coupling (SOC). When the spin degree is considered, the simulation can be classified into three types according to the properties of materials:

• Spin-unpolarized [for non-magnetic materials with weak SOC];
• Spin-polarized [for magnetic materials with weak SOC];
• Noncolinear [for non-magnetic/magnetic materials with strong SOC];

But what's the essential difference between them? How can I understand them from the viewpoint of solving the Kohn-Sham equation in the framework of density functional theory?

Answer 1

Noncollinear magnetism means that the orientation of the magnetization varies in space. Examples for such structures are magnetic domain walls, spin spirals, or magnetic skyrmions. To describe these systems one has to consider the Kohn-Sham wave functions as spinors

$$\Psi_\nu(\mathbf{r}) = \begin{pmatrix} \psi_\nu^{\uparrow}(\mathbf{r}) \\ \psi_\nu^{\downarrow}(\mathbf{r})\end{pmatrix},$$

where $\nu$ is the index of the wave function, and $\uparrow$ and $\downarrow$ indicate up and down spins. From these objects one constructs the density matrix

$$\rho_{\sigma,\sigma'}(\mathbf{r}) = \sum_\nu^\text{occ} \left(\psi_\nu^{\sigma}(\mathbf{r})\right)^\ast \psi_\nu^{\sigma'}(\mathbf{r}) = \frac{1}{2} \begin{pmatrix} n(\mathbf{r}) + m_\text{z}(\mathbf{r}) & m_\text{x}(\mathbf{r}) - im_\text{y}(\mathbf{r}) \\ m_\text{x}(\mathbf{r}) + im_\text{y}(\mathbf{r}) & n(\mathbf{r}) - m_\text{z}(\mathbf{r})\end{pmatrix},$$

where $n(\mathbf{r})$ is the charge density and $\mathbf{m}(\mathbf{r}) = \begin{pmatrix} m_\text{x}(\mathbf{r}) \\ m_\text{y}(\mathbf{r}) \\ m_\text{z}(\mathbf{r}) \end{pmatrix}$ is the magnetization density.

Magnetism is due to the exchange interaction, so when constructing a potential from this density the exact expression depends on the choice of the exchange-correlation functional. In general one obtains an exchange $\mathbf{B}$ field and an effective potential matrix of the form

$$ V_{\sigma,\sigma'}(\mathbf{r}) = \begin{pmatrix} V(\mathbf{r}) + \mu_\text{B} B_\text{z}(\mathbf{r}) & \mu_\text{B}(B_\text{x}(\mathbf{r}) - i B_\text{y}(\mathbf{r})) \\ \mu_\text{B}(B_\text{x}(\mathbf{r}) + i B_\text{y}(\mathbf{r})) & V(\mathbf{r}) - \mu_\text{B} B_\text{z}(\mathbf{r}) \end{pmatrix},$$

where $V(\mathbf{r})$ is the effective potential averaged over the spins. For the case of the local density approximation $\mathbf{B}(\mathbf{r})$ points in the same direction as $\mathbf{m}(\mathbf{r})$.

On the basis of this potential matrix one can then solve the Kohn-Sham equations for such noncollinear systems

$$\left\lbrace -\frac{1}{2}\nabla^2 \mathbf{I}_2 + V_{\sigma,\sigma'}(\mathbf{r}) \right\rbrace \Psi_\nu(\mathbf{r}) = \epsilon_\nu \Psi_\nu(\mathbf{r}).$$

In this expression $\mathbf{I}_2$ is the $2 \times 2$ identity matrix.

Note that the kinetic energy operator is spin-diagonal in this equation. In a relativistic description where one uses the Kohn-Sham-Dirac equation this is not the case due to the then included spin-orbit coupling. Spin-orbit coupling will also yield spin-offdiagonal matrix elements. This implies that spin-orbit coupling can cause noncollinear magnetism, but such structures can also have other causes, for example magnetic frustration.

In the spin-polarized case in which a magnetic system only features collinear magnetism, i.e., the magnetization is oriented along the same axis everywhere in space, one can always find a $\text{z}$ axis in the description above that coincides with this spin-quantization axis. If we consider such a setup the offdiagonal terms in the matrices vanish and the Kohn-Sham equations can be solved for the two spins separately.

In the non-spinpolarized case the expressions for the two spins are identical: The Kohn-Sham states feature a spin-degeneracy and therefore it is enough to solve the Kohn-Sham equations for one of the spins and occupy each state with up to two electrons.

Finally, please also note that spin-orbit coupling is not only relevant for systems featuring noncollinear magnetism. For example, in systems with collinear magnetism it causes magnetocrystalline anisotropy. Effects due to spin-orbit coupling in nonmagnetic systems include the Rashba effect and the Dresselhaus effect.

Answer 2

Basically, this is the split between restricted, unrestricted, and generalized Hartree-Fock (or Kohn-Sham) theory.

In the restricted theory, both the spin-up and spin-down electrons occupy the same spatial orbital: $\psi_{2n}({\bf r}) = \phi_n ({\bf r}) |\uparrow \rangle$, $\psi_{2n+1}({\bf r}) = \phi_n ({\bf r}) |\downarrow \rangle$. This is the Ansatz you use to solve the Kohn-Sham equations. (In the restricted open-shell theory, the number of spin-up and spin-down electrons may not match.)

In the unrestricted theory, the spin-up and spin-down electrons can have different spatial parts, for which the Ansatz reads: $\psi_{2n}({\bf r}) = \phi^\uparrow_n ({\bf r}) |\uparrow \rangle$, $\psi_{2n+1}({\bf r}) = \phi^\downarrow _n ({\bf r}) |\downarrow \rangle$. If you have a well-behaved system near the equilibrium, the unrestricted solution typically collapses to the restricted solution.

Most non-relativistic quantum chemistry is based on these two approaches. However, the generalized theory, in which the orbitals aren't eigenfunctions of $\hat{S}_z$ anymore, as you also allow mixing of spin-up and spin-down character, has also been found to be useful in e.g. bond breaking J. Chem. Theory Comput. 7, 2667 (2011), but you really need it if you have an operator in your Hamiltonian that couples the spins.

Formally, the Ansatz of the generalized approach looks something like $\psi_{n}({\bf r}) = c_\uparrow \phi^\uparrow _n ({\bf r}) |\uparrow \rangle + c_\downarrow \phi^\downarrow _n ({\bf r}) |\downarrow \rangle$.

Answer 3

As far I understand, to search for spin-polarized solutions, the DFT code will have to solve two coupled Kohn-Sham equations, one for each of the two spin species. There is both direct and indirect coupling between the two Kohn-Sham equations .

The direct coupling comes about by the dependence of the effective potential in the Kohn-Sham equation corresponding to spin-up/down electrons on the number densities (or charge densities) of both spin-up and spin-down electrons (most evidently, the Hartree part of the effective potential depends on the total number density, which is the sum of the number densities of spin-up and spin-down electrons).

The indirect coupling comes about by the requirement that the total numbers of spin-up and spin-down electrons must add up to a given integer N, the total number of electrons in the system.

Hope this helps :)