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Urb
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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 ModelingDensity 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?

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?

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?

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lsdragon
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Invariance of spin-polarized Kohn-Sham Hamiltonian with respect to spin rotations

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?