User:Pakritar

Double groups are important in exploiting the symmetries of quantum systems that involve the spin of the electron or other half-integral angular momentum states. In quantum physics, when a 360° coordinate rotation is performed on an electron state or spinor state, the state is not invariant, but changes sign. However, a rotation by 720° does leave the state invariant. This is a demonstration of the fact that SU(2) is a double cover of SO(3). The consequence is that any rotation compounded with the rotation by 360° is not identical with just the rotation by itself. Any point group is then extended by introducing the 360° rotation as a new group element. The group obtained in this way is called the double group of the corresponding point group (or single group). A primary application of double groups to quantum or solid state physics is in the determination of the splitting of energy levels by spin-orbit coupling.

More abstractly, a double group (or binary polyhedral group) is the preimage of a particular point group under the universal double cover from SU(2) to SO(3). It is a group extension of the point group by the cyclic group of order 2 and hence has twice as many elements as the single point group.

The homomorphism from SU(2) onto SO(3)
The 2-to-1 mapping from SU(2) to SO(3) is given explicitly by Tij = $$\tfrac{1}{2}$$tr[σiuσju-1]. The proof that this is a surjective homomorphism is given below.

From any vector r = (x, y, z) in R3, one can form a 2x2 traceless Hermitian matrix M(r) as follows



\begin{bmatrix} z & x-iy \\ x + iy & -z \\ \end{bmatrix} = x\sigma_1 + y\sigma_2 + z\sigma_3 $$

where the Pauli matrices have been used for later simplifications. If u &isin; SU(2), then uM(r)u-1 = M(r') is also a traceless Hermitian matrix and defines another vector r' in R3. The Euclidean dot product between two vectors is


 * r1 $$\cdot$$ r2 = $$\tfrac{1}{2}$$tr[M(r1)M(r2)]

Since the trace is invariant under unitary transformation, the scalar product is conserved so that u acts as a rotation on vectors in R3. To every u &isin; SU(2), there corresponds a T &isin; SO(3) such that if uM(r)u-1 = M(r'), then T(u)r = T(-u)r = r'. Note that both u and -u are mapped to the same element in SO(3).

To see that this forms a homomorphism, suppose that u2M(r)u2-1 = M(r1) and u1M(r1)u1-1 = M(r2). Then (u1u2)M(r)(u1u2)-1 = M(r2) so that T(u1u2)r = r2. Since T(u2)r = r1 and T(u1)r1 = T(u1)T(u2)r = r2, this implies that T(u1u2) = T(u1)T(u2).

That the homomorphism is surjective can be shown by finding an explicit rule of correspondence between elements of SU(2) and SO(3). First note that the transformed vector M(r') can also be written as



x'\sigma_x + y'\sigma_y + z'\sigma_z = xu\sigma_x u^{-1} + yu\sigma_y u^{-1} + zu\sigma_z u^{-1} $$

Using the property that tr[σiσj] = 2δij i,j = 1,2,3, the components of T(u) are obtained by multiplying the above equation by σi and then taking the trace. This gives Tij = $$\tfrac{1}{2}$$tr[σiuσju-1]

Every element of SU(2) has the form $$ \begin{bmatrix} a & b \\ -b^* & a^* \\ \end{bmatrix} $$ for which $$ \left|a\right|^2+\left|b\right|^2 = 1 $$, so that all of SU(2) may be parameterized as



\begin{bmatrix} \cos(\tfrac{1}{2} \theta) \exp [i \tfrac{1}{2} (\psi + \phi)] & \sin(\tfrac{1}{2} \theta) \exp [i \tfrac{1}{2} (\psi - \phi)] \\ \\ \ -\sin(\tfrac{1}{2} \theta) \exp [-i \tfrac{1}{2} (\psi - \phi)] & \cos(\tfrac{1}{2} \theta) \exp [-i \tfrac{1}{2} (\psi + \phi)] \\ \end{bmatrix} \quad \quad 0\leq \theta \leq \pi ,\quad 0\leq \psi \leq 4\pi ,\quad 0\leq \phi \leq 2\pi $$

This is also just the general rotation matrix for spinors in terms of the Euler angles. Putting this into the rule of correspondence gives the general rotation matrix in R3, also in terms of the Euler angles. Since the Euler angles parameterize the set of all proper rotations in R3, for every element of SO(3) there are two elements of SU(2) from which it is mapped. Therefore the homomorphism is surjective.

Classes, representations and characters of the double group
To each element Aj of a single point group, there corresponds two elements of its double group, Aj and RAj, where R denotes the 2π rotation. The double group therefore has twice as many elements as the single group, but not necessarily twice as many conjugacy classes. Classes and characters for the double group may be obtained by using the following set of rules and the Schur orthogonality relations.

1. If a set elements {Ci} form a class in the single group, then the set of elements {Ci} and {RCi} form two separate classes in the double group with only one exception.

2. The exception to the first rule is if {Ci} forms a class of rotations by π in the single group and there is another class of rotations by π about an axis perpendicular to the axis of the elements {Ci}, then the elements Ci and RCi are in the same class of the double group.

3. If Γ is a representation of the single group, then it also defines a representation of the double group with the matrices representing elements that differ by the 2π rotation being identical, that is, Γ(RAj) = Γ(Aj).

Since the number of irreducible representations of a group is the same as the number of conjugacy classes that the group has, there are additional irreducible representations of the double group. For these extra representations, matrices representing elements that differ by the 2π rotation have opposite sign, that is, Γ(RAj) = -Γ(Aj).

Spin-orbit coupling
If the Hamiltonian of some quantum system is invariant under a group G of coordinate transformations T while neglecting the spin-orbit interaction, then each d-fold degenerate set of eigenfunctions forms a basis for a d-dimensional representation Γp (p is an index) of the symmetry group of the Hamiltonian. Once the spin-orbit term is included, the Hamiltonian is then invariant under the double group of coordinate transformations, and the eigenfunctions form a basis for representation of the double group of G. The Hamiltonian with the spin-orbit interaction is given by



\mathbf{H}(\mathbf{r})=\left[{{-\hbar}^2\over 2 m_e}\nabla^2 + V(\mathbf{r})\right]{\mathbf{1}}_2 + \left[ {{\hbar}^2 \over 4 i m_e c^2} \nabla V(\mathbf{r}) \times \nabla \right] \cdot \vec{\sigma} $$

where $$ {\mathbf{1}}_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \quad $$ and $$ \quad \vec{\sigma} = \sigma_1 \hat{\mathbf{x}} + \sigma_2 \hat{\mathbf{y}} + \sigma_3 \hat{\mathbf{z}} $$

The first term is the spin-independent part of the Hamiltonian. The second term, which will be denoted by HSO, is the spin-orbit interaction. Since it is taken that the first term is invariant under the group G coordinate transformations T, it remains to be shown that the term involving the triple product is also invariant under the group elements T, that is u-1(T)HSO(r)u(T) = HSO(Tr) = HSO(r).

Regard the the scalar triple product as a determinant



\mathbf{H}_{SO}(\mathbf{r})={{\hbar}^2 \over 4 i m_e c^2} \det \begin{bmatrix} \partial_x V(\mathbf{r}) & \partial_y V(\mathbf{r}) & \partial_z V(\mathbf{r}) \\ \partial_x & \partial_y  & \partial_z \\ \sigma_1 & \sigma_2  & \sigma_3 \\ \end{bmatrix} $$

The partial derivatives transforms accordingly



\partial_{x'} = \mathbf{T}_{11}\partial_{x} + \mathbf{T}_{12}\partial_{y} + \mathbf{T}_{13}\partial_{z} $$

and since V(Tr) = V(r') = V(r)



\partial_{x'} V(\mathbf{r'}) = \mathbf{T}_{11}\partial_{x}V(\mathbf{r}) + \mathbf{T}_{12}\partial_{y}V(\mathbf{r}) + \mathbf{T}_{13}\partial_{z}V(\mathbf{r}) $$

The Pauli matrices transform under the elements of SU(2)



u^{-1}(\mathbf{T})\sigma_1 u(\mathbf{T}) = \mathbf{T}_{11}\sigma_1 + \mathbf{T}_{12}\sigma_2 + \mathbf{T}_{13}\sigma_3 $$

and similarly for the other components so that the transformed determinant can be written as the determinant of a product



\mathbf{H}_{SO}(\mathbf{r'})={{\hbar}^2 \over 4 i m_e c^2} \det \begin{bmatrix} \partial_x V(\mathbf{r}) & \partial_y V(\mathbf{r}) & \partial_z V(\mathbf{r}) \\ \partial_x & \partial_y  & \partial_z \\ \sigma_1 & \sigma_2  & \sigma_3 \\ \end{bmatrix} \det \mathbf{T}^{-1} $$

Since T is a proper rotation, the determinant of T-1 is just 1, so that the spin-orbit interaction term is invariant under the double group of coordinate transformations.

If a set of eigenfunctions transform according to an irreducible representation Γ(T) of the symmetry group of the Hamiltonian neglecting spin, then with the spin-orbit interaction term included, the eigenfunctions will transform under the representation u(T)⊗Γ(T) of the double group. However, this representation is not necessarily an irreducible representation. The number and dimensions of the irreducible representations contained in this reducible representation in effect determines splitting of the energy levels due to the spin-orbit interaction.