In collinear density functional theory, Kohn-Sham equation for spin-dependent wave functions is $$ \left[\left(-\frac{\hbar^{2}}{2 m} \nabla^{2}+\int \frac{n\left(\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|} d \mathbf{r}^{\prime}\right) \underline{\mathrm{I}}+\underline{v}(\mathbf{r})+\underline{V}_{\mathrm{xc}}(\mathbf{r})\right]\left(\begin{array}{c} \phi_{i}^{\uparrow}(\mathbf{r}) \\ \phi_{i}^{\downarrow}(\mathbf{r}) \end{array}\right)=\varepsilon_{i}\left(\begin{array}{c} \phi_{i}^{\uparrow}(\mathbf{r}) \\ \phi_{i}^{\downarrow}(\mathbf{r}) \end{array}\right). $$ Here $n(\mathbf{r'})$ is electron density and can be obtained as $n(\mathbf{r}) = \sum_{i,\sigma = \uparrow,\downarrow} |\phi_{i,\sigma}(\mathbf{r})|^2$. $\underline{\mathrm{I}}$ is a $2\times2 $ identity matrix. If external magnetic field is ignored, $\underline{v}(\mathbf{r}) = v(\mathbf{r})\underline{\mathrm{I}}$ where $v(\mathbf{r})$ is the field from background ions. $\underline{V}_{\mathrm{xc}}$ is the functional derivative of the exchange-correlation energy with respect to the density matrix $n_{\alpha,\beta}(r) = \phi_{i,\alpha}^*(\mathbf{r})\phi_{i,\beta}(\mathbf{r})$, where $\alpha,\beta$ are spin labels.
I read that Hamiltonian above is invariant under spin-rotation in Density Functional Theory for Magnetism and Magnetic Anisotropy in Handbook of Materials Modeling around Eq.10.
Although modern exchange correlation functionals are more sophisticated, many properties of spin-polarized DFT calculations can be studied already in the simple LSDA form. Suppose, there is a collinear magnet with the orientation of the magnetization in z-direction (actually, the Hamiltonian in Eq. (9) is invariant under spin-rotations). Then, the density matrix is diagonal and $\underline{V}_{xc}$ has only two terms $V_{\mathrm{xc}}^{\uparrow \uparrow}=\frac{\delta E_{\mathrm{xc}}}{\delta n_{\uparrow \uparrow}} \propto\left[n_{\uparrow \uparrow}(\mathbf{r})\right]^{1 / 3}$ and $V_{\mathrm{xc}}^{\downarrow \downarrow}=\frac{\delta E_{\mathrm{xc}}}{\delta n_{\downarrow \downarrow}} \propto\left[n_{\downarrow \downarrow}(\mathbf{r})\right]^{1 / 3}$ . This means that Eq. (9) consists of two equations, one for $\varphi_{\uparrow}$ and one for $\varphi_{\downarrow}$ that are identical if $n_{\uparrow \uparrow} = n_{\downarrow \downarrow}$ .
I consider a spin rotation as a $SU(2)$ transformation. However, if a $SU(2)$ transformation is applied to the Hamiltonian, it transforms into other form except that the Hamiltonian is proportional to identity matrix. In spin-polarization case (collinear magnetism), it can not always be true even if $\underline{V}(\mathbf{r})$ is diagonal.
So why is this Hamiltonian invariant under spin-rotation?
