SO(3)

$SO(3)$ is the group of proper rotations of three-dimensional Euclidean space. It is the group that rotates ordinary vectors, axes, rigid bodies, coordinate frames, and orbital wavefunctions in three spatial dimensions.

This page owns the rotation-matrix group itself. The closely related spinor group is SU(2), and the quantum commutator machinery is summarized in Angular Momentum Algebra.

Definition

The group $SO(3)$ is

$$SO(3) = \{R\in M_3(\mathbb R):R^TR=I,\ \det R=1\}.$$

The condition $R^TR=I$ says that $R$ is orthogonal. It preserves Euclidean dot products:

$$(R\mathbf v)\cdot(R\mathbf w) = \mathbf v\cdot\mathbf w.$$

The condition $\det R=1$ selects orientation-preserving orthogonal transformations. Orthogonal matrices with determinant $-1$ include reflections and improper rotations; they belong to $O(3)$ but not to $SO(3)$.

The group operation is matrix multiplication. The identity is $I$, and the inverse of $R$ is

$$R^{-1}=R^T.$$

Rotations About Coordinate Axes

A rotation by angle $\theta$ about the $z$ axis is

$$R_z(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0 & 0 & 1 \end{pmatrix}.$$

Similarly,

$$R_x(\theta) = \begin{pmatrix} 1 & 0 & 0\\ 0 & \cos\theta & -\sin\theta\\ 0 & \sin\theta & \cos\theta \end{pmatrix},$$

and

$$R_y(\theta) = \begin{pmatrix} \cos\theta & 0 & \sin\theta\\ 0 & 1 & 0\\ -\sin\theta & 0 & \cos\theta \end{pmatrix}.$$

These matrices use the active convention: the vector is rotated while the coordinate axes are held fixed. Passive coordinate rotations use inverse matrices, so sign conventions must be stated when comparing formulas.

Noncommutativity

Rotations in three dimensions generally do not commute:

$$R_x(\alpha)R_y(\beta) \ne R_y(\beta)R_x(\alpha)$$

for generic angles $\alpha$ and $\beta$.

This noncommutativity is not a quantum effect. It is already present in classical three-dimensional rotations. Quantum angular-momentum commutators are the infinitesimal operator version of this group-level fact.

Axis-Angle Description

Every nonidentity element of $SO(3)$ is a rotation about some axis by some angle. If $\hat{\mathbf n}$ is a unit vector along the axis, define

$$[\hat{\mathbf n}]_\times = \begin{pmatrix} 0 & -n_z & n_y\\ n_z & 0 & -n_x\\ -n_y & n_x & 0 \end{pmatrix}.$$

This matrix satisfies

$$[\hat{\mathbf n}]_\times\mathbf v = \hat{\mathbf n}\times\mathbf v.$$

Rodrigues’ formula gives the rotation matrix:

$$R(\hat{\mathbf n},\theta) = I\cos\theta + (1-\cos\theta)\hat{\mathbf n}\hat{\mathbf n}^T + \sin\theta\,[\hat{\mathbf n}]_\times.$$

The axis-angle description is useful but not globally unique. The angle is periodic, and

$$R(\hat{\mathbf n},\theta) = R(-\hat{\mathbf n},-\theta).$$

At angle $\pi$, the axis sign ambiguity becomes especially visible. These identifications are part of the global topology that distinguishes $SO(3)$ from $SU(2)$.

Lie Group Structure

$SO(3)$ is a compact, connected, three-dimensional Lie group. Its elements vary smoothly with the rotation angle and axis, and matrix multiplication and inversion are smooth operations.

It is compact because it is a closed and bounded subset of the finite-dimensional matrix space $M_3(\mathbb R)$. It is connected because any rotation can be continuously deformed to the identity by shrinking its rotation angle to zero.

It is not simply connected. The group $SU(2)$ is its double cover, and that global difference is what makes spinor representations possible in quantum mechanics.

Lie Algebra of SO(3)

The Lie algebra $\mathfrak{so}(3)$ is the tangent space at the identity. To find it, take a differentiable curve $R(t)\in SO(3)$ with $R(0)=I$ and derivative

$$X = \left.\frac{dR(t)}{dt}\right|_{t=0}.$$

Differentiating $R(t)^TR(t)=I$ at $t=0$ gives

$$X^T+X=0.$$

Thus

$$\mathfrak{so}(3) = \{X\in M_3(\mathbb R):X^T=-X\}.$$

It is the vector space of real antisymmetric $3\times3$ matrices. A standard basis is

$$A_x = \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & -1\\ 0 & 1 & 0 \end{pmatrix}, \quad A_y = \begin{pmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ -1 & 0 & 0 \end{pmatrix}, \quad A_z = \begin{pmatrix} 0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}.$$

These satisfy

$$[A_i,A_j] = \sum_k\epsilon_{ijk}A_k.$$

For example,

$$R_z(\theta) = e^{\theta A_z}.$$

Relation to Angular Momentum

In quantum mechanics, spatial rotations are represented on Hilbert space by unitary operators. A rotation by angle $\theta$ about $\hat{\mathbf n}$ is commonly written

$$U(R(\hat{\mathbf n},\theta)) = \exp \left( -\frac{i}{\hbar}\theta\,\hat{\mathbf n}\cdot\mathbf J \right),$$

where $\mathbf J=(J_x,J_y,J_z)$ is the angular-momentum operator appropriate to the Hilbert space.

The Hermitian generators obey

$$[J_i,J_j] = i\hbar \sum_k\epsilon_{ijk}J_k.$$

This is the same Lie algebra structure as $\mathfrak{so}(3)$, translated into the physics convention where self-adjoint operators generate unitary transformations. The corresponding anti-Hermitian infinitesimal operators are $-iJ_i/\hbar$.

For orbital wavefunctions, $\mathbf J$ may be the orbital angular momentum $\mathbf L$. For spin systems, $\mathbf J$ may include spin. The group $SO(3)$ describes ordinary spatial rotations; spin-$1/2$ state vectors require the double cover $SU(2)$ on the Hilbert-space side.

Action on Scalar Wavefunctions

For a scalar wavefunction on $\mathbb R^3$, a rotation $R\in SO(3)$ acts by

$$(U(R)\psi)(\mathbf r) = \psi(R^{-1}\mathbf r).$$

The inverse is needed so that the assignment is a representation:

$$U(R_1)U(R_2) = U(R_1R_2).$$

The action is unitary on $L^2(\mathbb R^3)$ because rotations preserve volume:

$$d^3(R\mathbf r) = d^3\mathbf r.$$

This is the representation-theoretic origin of orbital angular momentum as the generator of rotations of position-dependent wavefunctions.

SO(3) and SU(2)

$SO(3)$ rotates ordinary three-dimensional vectors. $SU(2)$ acts naturally on two-component spinors and maps two-to-one onto $SO(3)$. As Lie algebras, $\mathfrak{su}(2)$ and $\mathfrak{so}(3)$ are closely related; globally, the groups differ.

This distinction matters physically:

• integer angular-momentum representations descend to ordinary representations of $SO(3)$;
• half-integer spin representations are ordinary representations of $SU(2)$ but projective representations of $SO(3)$;
• a $2\pi$ rotation is the identity in $SO(3)$ but can act as $-I$ on spinors.

The double-cover story is developed in SU(2) versus SO(3) and connected to ray phases in Projective Representations.

Common Mistakes

• Treating $SO(3)$ as abelian because rotations have angles.
• Confusing active rotations of vectors with passive rotations of coordinates.
• Including reflections in $SO(3)$ instead of $O(3)$.
• Assuming the axis-angle parameterization is globally unique.
• Identifying $SO(3)$ with $SU(2)$ rather than recognizing the double cover.
• Forgetting the factor of $i$ and $\hbar$ when translating $\mathfrak{so}(3)$ into Hermitian quantum generators.
• Treating spin-$1/2$ transformations as ordinary vector rotations.

Cross-Links

• Groups
• Group Actions
• Lie Groups
• Lie Algebras
• Representations
• Unitary Representations
• SU(2)
• SU(2) versus SO(3)
• Angular Momentum Algebra
• Wigner D-Matrices
• Rotations in Three Dimensions
• SO(3) and SU(2) Preview
• Orbital Angular Momentum
• Spin Rotations
• Projective Representations

References

• B. C. Hall, Lie Groups, Lie Algebras, and Representations, 2nd ed., Springer, 2015.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• J. F. Cornwell, Group Theory in Physics, Vol. 1, Academic Press, 1984.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• D. A. Varshalovich, A. N. Moskalev, and V. K. Khersonskii, Quantum Theory of Angular Momentum, World Scientific, 1988.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.

Exercises

1. Show that if $R^TR=I$, then $R^{-1}=R^T$.

Solution

The equation $R^TR=I$ says that $R^T$ is a left inverse of $R$. Since $R$ is a square matrix, a left inverse is also the inverse. Therefore

$$R^{-1}=R^T.$$

2. Verify that $R_z(\theta)$ is in $SO(3)$.

Solution

The upper-left $2\times2$ block of $R_z(\theta)$ is the ordinary plane rotation matrix. Its columns are orthonormal, and the third column is $(0,0,1)^T$, orthogonal to the first two. Hence

$$R_z(\theta)^TR_z(\theta)=I.$$

The determinant is the determinant of the plane rotation block:

$$\det R_z(\theta) = \cos^2\theta+\sin^2\theta = 1.$$

Thus $R_z(\theta)\in SO(3)$.

3. Differentiate the orthogonality condition to derive $\mathfrak{so}(3)$.

Solution

Let $R(t)\in SO(3)$, $R(0)=I$, and $X=R'(0)$. Since

$$R(t)^TR(t)=I,$$

differentiating gives

$$R'(t)^TR(t)+R(t)^TR'(t)=0.$$

At $t=0$, this becomes

$$X^T+X=0.$$

So $X$ is antisymmetric, and $\mathfrak{so}(3)$ consists of real antisymmetric matrices.

4. Using the basis matrices above, compute $[A_x,A_y]$.

Solution

Direct multiplication gives

$$A_xA_y-A_yA_x = \begin{pmatrix} 0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix} = A_z.$$

This is the $i=x$, $j=y$ case of

$$[A_i,A_j] = \sum_k\epsilon_{ijk}A_k.$$

5. Show that the scalar wavefunction rule is a representation.

Solution

Define

$$(U(R)\psi)(\mathbf r) = \psi(R^{-1}\mathbf r).$$

Then

$$(U(R_1)U(R_2)\psi)(\mathbf r) = (U(R_2)\psi)(R_1^{-1}\mathbf r) = \psi(R_2^{-1}R_1^{-1}\mathbf r).$$

Since

$$R_2^{-1}R_1^{-1} = (R_1R_2)^{-1},$$

we get

$$U(R_1)U(R_2) = U(R_1R_2).$$