SO(3) Group

The SO(3) group, or special orthogonal group in three dimensions, is an infinite group of all rotational symmetries of a three-dimensional real vector space.

An important characteristic of rotational symmetry is the preservation of the lengths and angles between basis vectors in the three-dimensional real vector space. This characteristic is best described by the dot product of two vectors $\boldsymbol{\mathit{u}},\boldsymbol{\mathit{v}}\in\mathbb{R}^{3}$ :

$\boldsymbol{\mathit{u}}\cdot\boldsymbol{\mathit{v}}=\vert\boldsymbol{\mathit{u}}\vert\vert\boldsymbol{\mathit{v}}\vert cos\theta$

where

$\vert\boldsymbol{\mathit{u}}\vert$ is the length of $\boldsymbol{\mathit{u}}$ ,
$\vert\boldsymbol{\mathit{v}}\vert$ is the length of $\boldsymbol{\mathit{v}}$ ,
$\theta$ is the angle between them.

If both the lengths and the angle between the vectors are preserved (retain the same values) after rotation, then the dot product is preserved. In other words, a rotation changes only the orientation of the vectors, not their geometric relationship. Using matrix notation, with $\hat{R}$ being a rotation operator,

$$\hat{R}\boldsymbol{\mathit{u}}$\cdot$\hat{R}\boldsymbol{\mathit{v}}$=$\hat{R}\boldsymbol{\mathit{u}}$^T$\hat{R}\boldsymbol{\mathit{v}}$=\boldsymbol{\mathit{u}}^T\hat{R}^T\hat{R}\boldsymbol{\mathit{v}}$

where we have used the identity $$ABC\cdots$^T=\cdots C^TB^TA^T$ (see Property 13 of this article for the proof).

Since the dot product is preserved, $$\hat{R}\boldsymbol{\mathit{u}}$^T$\hat{R}\boldsymbol{\mathit{v}}$=\boldsymbol{\mathit{u}}^T\boldsymbol{\mathit{v}}$ . This implies that $\hat{R}^T\hat{R}=I$ , which is the defining condition for an orthogonal matrix, and hence the O in SO(3). Furthermore, the determinant $\vert\hat{R}\vert$ is equal to $\pm 1$ for any orthogonal matrix (see Property 11 of this article for the proof), and $\vert\hat{R}\vert=+1$ corresponds to orientation-preserving transformations. Therefore, the S in SO(3) refers to the special orthogonal case in which $\vert\hat{R}\vert=+1$ , while the 3 denotes the dimension of the underlying vector space on which the group is defined.

Question

Why does the special case of $\vert\hat{R}\vert=+1$ correspond to orientation-preserving transformations?

Answer

Consider the matrix $A=\[\boldsymbol{\mathit{e}}_1\;\boldsymbol{\mathit{e}}_2\;\boldsymbol{\mathit{e}}_3\]$ , where the basis vectors $\boldsymbol{\mathit{e}}_{j=1,2,3}$ form a right-handed basis satisfying $\boldsymbol{\mathit{e}}_1\times\boldsymbol{\mathit{e}}_2=\boldsymbol{\mathit{e}}_3$ . If we carry out a transformation by swapping two of the basis vectors to form $B=\[\boldsymbol{\mathit{e}}_2\;\boldsymbol{\mathit{e}}_1\;\boldsymbol{\mathit{e}}_3\]$ , then the ordering of the basis vectors becomes left-handed because $\boldsymbol{\mathit{e}}_2\times\boldsymbol{\mathit{e}}_1=-\boldsymbol{\mathit{e}}_3$ . According to determinant identities, swapping two columns changes the sign of the determinant: $\vert B\vert=-\vert A\vert$ . Therefore, a determinant of +1 corresponds to transformations that preserve the handedness (orientation) of the basis, while a determinant of -1 corresponds to transformations that reverse the orientation. Since a rotational matrix can be constructed using basis vectors $\hat{R}=\[\hat{R}\boldsymbol{\mathit{e}}_1,\hat{R}\boldsymbol{\mathit{e}}_2,\hat{R}\boldsymbol{\mathit{e}}_3\]$ , and a rotation changes the directions of the basis vectors but preserves their right-handed ordering, the rotated basis remains right-handed. Hence, the condition $\vert\hat{R}\vert=+1$ corresponds to orientation-preserving transformations.

Typically, each element of the group corresponds to a rotation operator, characterised by a unit rotation axis $\boldsymbol{\mathit{k}}$ and an angle $\alpha$ . Since there are infinitely many possible rotation angles and axes, SO(3) contains infinitely many elements. These elements satisfy the properties of a group. For example, the binary operation of two rotation operators, each with a specific rotation angle about an axis, results in another rotation operator, demonstrating the closure property.

Consider the rotation of the spherical harmonics around the $z$ -axis. Even though spherical harmonics are complex-valued functions, they form a representation of SO(3) because the arguments of these functions, which are points in the real vector space $\mathbb{R}^{3}$ , are transformed by rotations. When the rotation operator $\hat{C}_{\alpha}$ acts on $Y^{m_l}_l(\theta,\phi)=P(\theta)e^{im_l\phi}$ by an angle $\alpha$ around the $z$ -axis, only the azimuthal angle $\phi$ is affected and each basis is transformed into $P(\theta)e^{im_l$\phi-\alpha$}$ . The transformation for all basis functions is summarised as:

$\hat{C}_{\alpha}\begin{pmatrix}Y^l_l\\Y^{l-1}_l\\\vdots\\Y^{-l}_l\end{pmatrix}=\begin{bmatrix}e^{-il\alpha} &0&\cdots &0\\0&e^{-i(l-1)\alpha}&\cdots &0\\\vdots &\vdots &\ddots &\vdots\\0&0&\cdots &e^{il\alpha}\end{bmatrix}\begin{pmatrix}Y^l_l\\Y^{l-1}_l\\\vdots\\Y^{-l}_l\end{pmatrix}\;\;\;\;\;\;\;\;120$

Question

Prove that if $\hat{A}\vert a\rangle=\lambda\vert a\rangle$ , then $e^{\hat{A}}\vert a\rangle=e^{\lambda}\vert a\rangle$ .

Answer

Consider the Taylor series of the exponential function $f$\hat{A}$=e^{\hat{A}}=\sum_{n=0}^{\infty}\frac{\hat{A}^n}{n!}$ . Since

$\hat{A}^n\vert a\rangle=\hat{A}^{n-1}\hat{A}\vert a\rangle=\lambda\hat{A}^{n-1}\vert a\rangle=\cdots=\lambda^n\vert a\rangle$

We have

$e^{\hat{A}}\vert a\rangle=\sum_{n=0}^{\infty}\frac{\hat{A}^n}{n!}\vert a\rangle=\sum_{n=0}^{\infty}\frac{\lambda^n}{n!}\vert a\rangle=e^{\lambda}\vert a\rangle$

In other words, $\hat{C}_{\alpha}Y^{m_l}_l(\theta,\phi)=Y^{m_l}_l(\theta,\phi-\alpha)=e^{-im_l\alpha}Y^{m_l}_l(\theta,\phi)$ , and according to the above Q&A,

$\hat{C}_{\alpha}=e^{-i\alpha\hat{l}_z}\;\;\;\;\;\;\;\;121$

where $\hat{l}_z$ is the $z$ -component of the angular momentum operator, expressed in $\hbar$ units.

Since an irreducible representation of a group is expressed by a set of matrices (or matrix-valued functions) that represent the group elements as linear operators on a vector space, an irreducible representation of the SO(3) group is $(2l+1)$ –dimensional (see eq120), with the set $\{Y^{m_l}_l(\theta,\phi)\}^l_{m_l=-l}$ forming a basis for the irreducible representation for a fixed $l$ .

Question

Why does the matrix representation of $\hat{C}_{\alpha}$ about the $z$ -axis belong to an irreducible representation of SO(3) when it is diagonal?

Answer

An irreducible representation is a group representation whose matrices cannot be simultaneously transformed, via the same invertible similarity transformation, into block diagonal form. For SO(3), an irreducible representation consists of matrices representing rotation operators about different axes. Even though the matrix of $\hat{C}_{\alpha}$ about the $z$ -axis is diagonal, which implies reducibility on its own, other matrices in the same representation, such as $\hat{C}_{\alpha}$ about the $y$ -axis, contain non-zero off-diagonal elements. These non-diagonal matrices cannot all be simultaneously diagonalised by the same similarity transformation. Therefore, the matrix representation of $\hat{C}_{\alpha}$ about the $z$ -axis belongs to an irreducible representation of SO(3).

The character of the rotation matrix is $\chi(C_{\alpha})=e^{-il\alpha}+e^{-i(l-1)\alpha}+\cdots+e^{il\alpha}$ , which is a geometric series $a+ar+ar^2+\cdots +ar^n=\frac{a(r^{n+1}-1)}{r-1}$ , where $a=e^{-il\alpha}$ , $r=e^{i\alpha}$ and $n=2l$ . This implies that

$\chi (C_{\alpha})=\frac{e^{i(l+1/2)\alpha}-e^{-i(l+1/2)\alpha}}{e^{i\alpha/2}-e^{-i\alpha/2}}$

Using Euler’s formula of $e^{ix}=cosx+isinx$ ,

$\chi (C_{\alpha})=\frac{sin(l+\frac{1}{2})\alpha}{sin\frac{\alpha}{2}}\;\;\;\;\;\;\;\;122$

The SO(3) group does not have a standard character table in the same way that finite point groups do. Its irreducible representations are labelled by a non-negative integer $l=0,1,2,\cdots$ , which form a discrete set. However, the group elements and their associated characters are continuous functions that depend on the rotation angle.

Nevertheless, for illustrative purposes, we can express the relationships between group elements, their irreducible representations and corresponding characters in the following way:

where each $\hat{R}$\alpha,\boldsymbol{\mathit{k}}_n$$ can be further expanded as follows:

Question

Why is $\chi$C_{\alpha=0}$=3$ for $l=1$ ?

Answer

From eq122, $\chi$C_{\alpha}$=\frac{sin(3\alpha/2)}{sin(\alpha/2)}$ for $l=1$ . At $\alpha=0$ , this expression is indeterminate, but we can evaluate the limit. Substituting $x=\alpha/2$ into the trigonometric identity $sin3x=3sinx-4sin^3x$ gives $sin(3\alpha/2)=3sin(\alpha/2)-4sin^3(\alpha/2)$ . So,

$\chi$C_{\alpha}$=\frac{3sin(\alpha/2)-4sin^3(\alpha/2)}{sin(\alpha/2)}=3-4sin^2(\alpha/2)$

Substituting the half-angle formula $sin^2(\alpha/2)=\frac{1-cos\alpha}{2}$ into the above equation yields $\chi(C_{\alpha})=1+2cos\alpha$ . Therefore, $\chi(C_{\alpha=0})=3$ . Since $\hat{R}(\alpha=0)$ is represented by the identity matrix, the trace $\chi(C_{\alpha=0})=3$ equals the dimension of the representation of $2l+1$ . Using the same logic, $\chi(C_{\alpha=0})=5$ for $l=2$ .

In general, a rotation operator of the SO(3) group is given by (see this article for derivation):

$\hat{D}(\phi,\theta,\chi)=e^{-i\phi\hat{J}_z}e^{-i\theta\hat{J}_y}e^{-i\chi\hat{J}_z}\;\;\;\;\;\;\;\;123$

where $0\leq\phi\leq 2\pi$ , $0\leq\theta\leq \pi$ and $0\leq\chi\leq 2\pi$ are the Euler angles.

The question, then, is whether the table relating the general rotation group elements, their irreducible representations and corresponding characters will be the same as that of a single-axis, single-angle rotation operator such as $\hat{C}_{\alpha}$ ?

The character of an irreducible representation of SO(3) depends only on the total rotation angle, not on the specific rotation axis or the individual Euler angles. Since SO(3) is the infinite group of all possible 3D rotations, there exists a one-to-one correspondence in symmetry between three consecutive rotations described by $\hat{D}(\phi,\theta,\chi)$ and a single rotation described by $\hat{C}_{\alpha}$ . In other words, even though $\hat{D}(\phi,\theta,\chi)$ is expressed in terms of Euler angles, it represents the same group element as some $\hat{C}_{\alpha}=e^{-i\alpha\hat{l}_k}$ , because we can always find a $\hat{C}_{\alpha}$ symmetry operation about an appropriate axis $\boldsymbol{\mathit{k}}$ that produces the same $\hat{D}(\phi,\theta,\chi)$ group element and therefore the same character value. Hence,

$\chi\[\hat{D}(\phi,\theta,\chi)\]=\chi(C_{\alpha})$

Finally, the rotation of $D^J_{M_JK}(\phi,\theta,\chi)=d^J_{M_JK}(\theta)e^{-i\phi M_J}e^{-i\chi K}$ , the Wigner D-matrix elements, for fixed $J$ and fixed $K$ is given by:

$\hat{C}_{\alpha}\begin{pmatrix}D^J_{J,K}\\D^J_{J-1,K}\\\vdots\\D^J_{-J,K}\end{pmatrix}=\begin{bmatrix}e^{iJ\alpha} &0&\cdots &0\\0&e^{i(J-1)\alpha}&\cdots &0\\\vdots &\vdots &\ddots &\vdots\\0&0&\cdots &e^{-iJ\alpha}\end{bmatrix}\begin{pmatrix}D^J_{J,K}\\D^J_{J-1,K}\\\vdots\\D^J_{-J,K}\end{pmatrix}$

This shows that, for fixed $K$ , the set $\{D^J_{M_JK}(\phi,\theta,\chi)\}^l_{m_l=-l}$ forms a basis of the same $(2J+1)$ -dimensional irreducible representation of SO(3) as the set $\{Y^{m_l}_l(\theta,\phi)\}^l_{m_l=-l}$ . Since $Y^{m_l}_l(\theta,\phi)$ are eigenfunctions of the angular momentum operators ( $\hat{L}^2$ and $\hat{l}_z$ ), it follows that $D^J_{M_JK}(\phi,\theta,\chi)$ are also eigenfunctions of the corresponding operators $\hat{J}^2$ and $\hat{J}_z$ . In other words, $D^J_{M_JK}(\phi,\theta,\chi)$ are also rotational wavefunctions.

Question

Elaborate on why $D^J_{M_JK}(\phi,\theta,\chi)$ are also eigenfunctions of $\hat{J}^2$ and $\hat{J}_z$ if $Y^{m_l}_l(\theta,\phi)$ are eigenfunctions of $\hat{L}^2$ and $\hat{l}_z$ .

Answer

As mentioned earlier, $\hat{l}_z$ (or equivalently $\hat{J}_z$ ) is a generator of an irreducible representation of SO(3). Its matrix representation is

$\hat{l}_z =\hbar\begin{pmatrix}l &0&\cdots &0\\0&l-1&\cdots &0\\\vdots &\vdots &\ddots &\vdots\\0&0&\cdots &-l\end{pmatrix}$

However, the matrix itself, is not an element of an irreducible representation of SO(3). $\hat{l}_z$ , or any component $\hat{l}_k$ of the angular momentum operator, is related to the rotation operator $\hat{D}(R)$ , by $\hat{D}(R)=e^{f(\hat{l}_k)}$ (see eq121 for an example). Using the Taylor series definition of the exponential function yields:

$\hat{L}^2,\hat{D}(R)\]=\[\hat{L}^2,e^{f(\hat{l}_k)}\]=\biggr\[\hat{L}^2,\sum_{n=0}^{\infty}\frac{\hat{l}_k^n}{n!}\biggr$

Applying the commutation relation identities $\[\hat{A},\hat{B}+\hat{C}+\cdots\]=\[\hat{A},\hat{B}\]+\[\hat{A},\hat{C}\]+\cdots$ and $\hat{A},\hat{B}\hat{C}\]=\[\hat{A},\hat{B}\]\hat{C}+\hat{B}\[\hat{A},\hat{C}$ gives:

$\begin{align}\[\hat{L}^2,\hat{D}(R)\]&=\[\hat{L}^2,\frac{\hat{l}_k^0}{0!}\]+\[\hat{L}^2,\frac{\hat{l}_k^1}{1!}\]+\[\hat{L}^2,\frac{\hat{l}_k^2}{2!}\]+\cdots\\&=\[\hat{L}^2,1\]+\[\hat{L}^2,\hat{l}_k\]+\[\hat{L}^2,\frac{\hat{l}_k}{2}\]\hat{l}_k+\frac{\hat{l}_k}{2}\[\hat{L}^2,\hat{l}_k\]+\cdots\\&=\[\hat{L}^2,1\]+\[\hat{L}^2,\hat{l}_k\]+\[\hat{L}^2,\hat{l}_k\]\frac{\hat{l}_k}{2}+\[\hat{L}^2,\frac{1}{2}\]\hat{l}_k^2+\frac{\hat{l}_k}{2}\[\hat{L}^2,\hat{l}_k\]+\cdots\end{align}$

Since $\[\hat{L}^2,c\]=0$ , where $c$ is a constant, and $\hat{L}^2$ commutes with $\hat{l}_k$ , i.e. $\[\hat{L}^2,\hat{l}_k\]=0$ , we have:

$\[\hat{L}^2,\hat{D}(R)\]=0$

According to Schur’s first lemma, any non-zero matrix that commutes with all matrices of an irreducible representation of a group is a multiple of the identity matrix. Therefore, $\hat{L}^2$ (or $\hat{J}^2$ ) must be a multiple of the identity operator within the $(2J+1)$ -dimensional subspace:

$\hat{L}^2=\lambda\hat{I}$

When $\hat{L}^2$ (or $\hat{J}^2$ ) acts on any basis function $\vert\psi\rangle$ within this irreducible subspace, we obtain:

$\hat{L}^2\vert\psi\rangle=\lambda\hat{I}\vert\psi\rangle=\lambda\vert\psi\rangle$

This shows that the basis functions $Y^{m_l}_l(\theta,\phi)$ are eigenfunctions of $\hat{L}^2$ and $\hat{l}_z$ . Similarly, $D^J_{M_JK}(\phi,\theta,\chi)$ , are eigenfunctions of $\hat{J}^2$ and $\hat{J}_z$ .

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