Rotations in Three Dimensions

Three-dimensional rotations are the geometric origin of angular momentum. They rotate vectors, coordinate axes, wavefunctions, and operators, but those four statements do not all use the same formula. Most sign errors in angular momentum begin by silently switching between them.

This page fixes the geometry and conventions. The full angular-momentum commutators, ladder operators, eigenvalues, and spherical harmonics are developed in the later pages of this chapter.

Rotation Matrices

An ordinary proper rotation of three-dimensional Euclidean space is represented by a real matrix $\mathcal R$ satisfying

$$\mathcal R^T\mathcal R=I, \qquad \det\mathcal R=1.$$

The first condition says that dot products and lengths are preserved:

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

The determinant condition says that orientation is preserved. Reflections and full spatial inversion are orthogonal transformations too, but they have determinant $-1$ and are not proper rotations in three dimensions.

The group of all such matrices is $SO(3)$. The group-theoretic home for the matrix group itself is SO(3).

Rotating Vectors

With the active convention, a rotation changes the vector while the coordinate axes are held fixed:

$$\mathbf v \mapsto \mathbf v' = \mathcal R\mathbf v.$$

For a rotation by $\theta$ about the $z$ axis,

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

Thus

$$\begin{pmatrix} v_x'\\ v_y'\\ v_z' \end{pmatrix} = \begin{pmatrix} v_x\cos\theta-v_y\sin\theta\\ v_x\sin\theta+v_y\cos\theta\\ v_z \end{pmatrix}.$$

A vector initially along the positive $x$ axis is carried toward the positive $y$ axis for small positive $\theta$.

Because $\det\mathcal R=1$, cross products transform as axial geometry expects:

$$\mathcal R(\mathbf v\times\mathbf w) = (\mathcal R\mathbf v)\times(\mathcal R\mathbf w).$$

This orientation-preserving property is one reason rotations differ from parity.

Rotating Coordinates

A passive coordinate rotation describes the same geometric vector using rotated axes. If the new axes are related to the old axes by the same active matrix $\mathcal R$, the coordinate column of the same vector changes by the inverse matrix:

$$\mathbf v_{\rm new} = \mathcal R^{-1}\mathbf v_{\rm old} = \mathcal R^T\mathbf v_{\rm old}.$$

For the $z$-axis example,

$$\begin{pmatrix} v_{x'}\\ v_{y'}\\ v_{z'} \end{pmatrix} = \begin{pmatrix} \cos\theta & \sin\theta & 0\\ -\sin\theta & \cos\theta & 0\\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} v_x\\ v_y\\ v_z \end{pmatrix}.$$

The inverse is not a contradiction. Active and passive rotations answer different questions:

• active: where did the physical vector move in fixed axes?
• passive: what are the components of the same vector in rotated axes?

The general convention warning is developed in Active and Passive Transformations.

Rotating Scalar Wavefunctions

For a spinless scalar wavefunction, the active rotation operator $U(\mathcal R)$ acts by

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

The inverse appears for the same reason it appears in translations. The value of the new wavefunction at the old coordinate point $\mathbf r$ comes from the old point that rotates into $\mathbf r$.

This formula preserves normalization because rotations preserve volume:

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

It also rotates the expectation value of position:

$$\langle\mathbf R\rangle_{U\psi} = \mathcal R\langle\mathbf R\rangle_\psi.$$

Equivalently, with the position operator kept as the measured observable,

$$U(\mathcal R)^\dagger\mathbf R\,U(\mathcal R) = \mathcal R\mathbf R.$$

The same equation holds for momentum:

$$U(\mathcal R)^\dagger\mathbf P\,U(\mathcal R) = \mathcal R\mathbf P.$$

Spinful Wavefunctions

If the state has spin, rotating the spatial argument is not enough. A spinful wavefunction has components, and a rotation can mix them:

$$(U(\mathcal R)\psi)_a(\mathbf r) = \sum_b D^{(s)}_{ab}(\mathcal R)\, \psi_b(\mathcal R^{-1}\mathbf r).$$

The matrix $D^{(s)}(\mathcal R)$ is the spin-$s$ representation of the rotation. For spin-$1/2$, it is naturally an $SU(2)$ matrix rather than an ordinary $3\times3$ vector rotation matrix. The detailed spinor story belongs to Spin Rotations and SU(2) versus SO(3).

Infinitesimal Rotation About z

For small $\delta\theta$,

$$\mathcal R_z(\delta\theta) = I+\delta\theta A_z+O(\delta\theta^2),$$

where

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

The inverse rotation is

$$\mathcal R_z(-\delta\theta) = I-\delta\theta A_z+O(\delta\theta^2).$$

For a scalar wavefunction,

$$\begin{aligned} (U_z(\delta\theta)\psi)(x,y,z) &= \psi( x+\delta\theta\,y, y-\delta\theta\,x, z ) +O(\delta\theta^2) \\ &= \psi - \delta\theta \left( x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x} \right)\psi +O(\delta\theta^2). \end{aligned}$$

The corresponding quantum generator satisfies

$$U_z(\delta\theta) = I-\frac{i}{\hbar}\delta\theta\,L_z +O(\delta\theta^2),$$

so

$$L_z = -i\hbar \left( x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x} \right).$$

This is the position-space orbital angular momentum generator about the $z$ axis. The full operator story is continued in Orbital Angular Momentum.

General Axis and Quantum Generator

A finite rotation by angle $\theta$ about a unit vector $\hat{\mathbf n}$ is represented on Hilbert space by

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

The generator $\mathbf J$ depends on the Hilbert space:

• for a scalar spinless particle, $\mathbf J=\mathbf L$;
• for a spin degree of freedom alone, $\mathbf J=\mathbf S$;
• for a spinful particle in space, $\mathbf J=\mathbf L+\mathbf S$.

The shared commutator algebra is the subject of Angular Momentum Algebra.

Rotational Invariance

A Hamiltonian is rotationally invariant when

$$U(\mathcal R)H U(\mathcal R)^\dagger = H$$

for every rotation under consideration. Infinitesimally this gives

$$[H,J_i]=0$$

for the corresponding generators, under the usual domain assumptions.

For a spinless particle in a central potential,

$$H=\frac{\mathbf P^2}{2m}+V(r),$$

rotational invariance is generated by $\mathbf L$. The resulting labels and degeneracies are treated in Central Potentials and Rotational Symmetry.

Common Mistakes

• Using the active vector formula when doing a passive coordinate change.
• Forgetting the inverse argument in $(U(\mathcal R)\psi)(\mathbf r)=\psi(\mathcal R^{-1}\mathbf r)$.
• Treating a spinor rotation as a $3\times3$ rotation of components.
• Calling every orthogonal transformation a rotation; reflections and parity have determinant $-1$.
• Assuming rotations commute because translations commute.
• Using $\mathbf L$ as the generator when spin contributes to the total rotation.

Cross-Links

• Rotations and Orbital Angular Momentum
• Rotations Preview
• Active and Passive Transformations
• SO(3)
• SO(3) and SU(2) Preview
• Angular Momentum Operators
• SU(2) versus SO(3)
• Angular Momentum Algebra
• Orbital Angular Momentum
• Spin Rotations
• Central Potentials and Rotational Symmetry
• Parity as Spatial Inversion

References

• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.

Exercises

1. Verify that $\mathcal R_z(\theta)$ preserves the length of a vector.

Solution

For

$$v_x'=v_x\cos\theta-v_y\sin\theta, \qquad v_y'=v_x\sin\theta+v_y\cos\theta, \qquad v_z'=v_z,$$

one finds

$$\begin{aligned} (v_x')^2+(v_y')^2 &= v_x^2(\cos^2\theta+\sin^2\theta) + v_y^2(\sin^2\theta+\cos^2\theta) \\ &\quad + 2v_xv_y(-\cos\theta\sin\theta+\sin\theta\cos\theta) \\ &= v_x^2+v_y^2. \end{aligned}$$

Adding $v_z^2$ gives $\lvert\mathbf v'\rvert^2=\lvert\mathbf v\rvert^2$.

2. A vector has old components $(1,0,0)$. If the coordinate axes are passively rotated by $+\pi/2$ about $z$, what are the new components?

Solution

For a passive coordinate rotation,

$$\mathbf v_{\rm new} = \mathcal R_z(-\pi/2)\mathbf v_{\rm old}.$$

Since

$$\mathcal R_z(-\pi/2) = \begin{pmatrix} 0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 1 \end{pmatrix},$$

the new components are

$$\begin{pmatrix} 0\\ -1\\ 0 \end{pmatrix}.$$

The vector did not move; the axes did.

3. Expand the active scalar wavefunction rotation about $z$ to first order and identify the generator.

Solution

The active rotation is

$$(U_z(\delta\theta)\psi)(x,y,z) = \psi( x+\delta\theta\,y, y-\delta\theta\,x, z ) +O(\delta\theta^2).$$

Taylor expansion gives

$$(U_z(\delta\theta)\psi) = \psi - \delta\theta \left( x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x} \right)\psi +O(\delta\theta^2).$$

Comparing with

$$U_z(\delta\theta) = I-\frac{i}{\hbar}\delta\theta L_z +O(\delta\theta^2)$$

gives

$$L_z = -i\hbar \left( x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x} \right).$$