Rotations Preview

Rotations are the spatial symmetries generated by angular momentum. They act on the position dependence of a wavefunction, on vector operators such as $\mathbf R$ and $\mathbf P$, and, for particles with spin, on internal spin degrees of freedom.

This page is a bridge. The full angular-momentum algebra, spherical harmonics, ladder operators, and central-potential applications live in the rotations chapter. Here the goal is to fix the basic transformation law and prevent the most common sign and interpretation mistakes before the algebra begins.

Active Convention

Let $\mathcal R$ be an ordinary three-dimensional rotation matrix, and let $U(\mathcal R)$ be the corresponding unitary operator on the quantum Hilbert space. With the active convention used in this volume,

$$|\psi\rangle \mapsto U(\mathcal R)|\psi\rangle.$$

For a spinless scalar wavefunction in position representation, the rotated state is

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

The inverse appears because the value of the new wavefunction at the old coordinate point $\mathbf r$ came from the old point that rotates into $\mathbf r$. This is the rotational analogue of the translation formula $(T(\mathbf a)\psi)(\mathbf r)=\psi(\mathbf r-\mathbf a)$.

The position expectation value rotates in the ordinary way:

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

Equivalently,

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

where $\mathcal R\mathbf R$ means

$$(\mathcal R\mathbf R)_i = \sum_j\mathcal R_{ij}R_j.$$

The same relation holds for momentum:

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

Infinitesimal Rotations

A rotation by angle $\theta$ about a unit vector $\hat{\mathbf n}$ has the unitary form

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

where $\mathbf J$ is the total angular momentum generator for whatever Hilbert space is being considered.

For a spinless particle, the generator is orbital angular momentum,

$$\mathbf J=\mathbf L, \qquad \mathbf L=\mathbf R\times\mathbf P.$$

For a particle with spin,

$$\mathbf J=\mathbf L+\mathbf S.$$

This distinction matters: $\mathbf L$ rotates the spatial dependence of the wavefunction, while $\mathbf S$ rotates internal spin components.

Rotation About z

For a spinless scalar wavefunction, an active rotation by $\theta$ about the $z$ axis gives

$$\begin{aligned} (U_z(\theta)\psi)(x,y,z) &= \psi( x\cos\theta+y\sin\theta, -x\sin\theta+y\cos\theta, z ). \end{aligned}$$

To first order in $\theta$,

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

Comparing with

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

gives

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

The same operator becomes $L_z=-i\hbar\,\partial/\partial\phi$ in spherical coordinates. That fact is the doorway to the $e^{im\phi}$ angular dependence of orbital angular-momentum eigenfunctions.

Operators, Vectors, and Scalars

Rotations distinguish vector operators from scalar operators. Position and momentum transform as vectors:

$$U^\dagger R_iU = \sum_j\mathcal R_{ij}R_j, \qquad U^\dagger P_iU = \sum_j\mathcal R_{ij}P_j.$$

Dot products such as $\mathbf R^2$, $\mathbf P^2$, and $\mathbf L^2$ are rotational scalars:

$$U^\dagger \mathbf R^2 U=\mathbf R^2, \qquad U^\dagger \mathbf P^2 U=\mathbf P^2.$$

Infinitesimally, vector behavior is encoded in commutators. For angular momentum generators,

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

For orbital angular momentum,

$$[L_i,R_j] = i\hbar\sum_k\epsilon_{ijk}R_k, \qquad [L_i,P_j] = i\hbar\sum_k\epsilon_{ijk}P_k.$$

These relations are the operator version of the statement that rotations carry vectors into vectors.

The general classification of rotational scalars, vector operators, and higher tensor operators is developed in Scalar, Vector, and Tensor Operators.

Rotationally Invariant Hamiltonians

A Hamiltonian is rotationally invariant when

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

for every rotation $\mathcal R$ in the relevant group. For a continuous rotation group this is equivalent, under the usual domain assumptions, to

$$[H,J_i]=0, \qquad i=x,y,z.$$

The most important spinless example is a central potential:

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

Because $\mathbf P^2$ and $r$ are rotational scalars,

$$[H,L_i]=0.$$

Thus angular momentum labels can be used to organize energy eigenstates. The detailed central-potential construction belongs to Central Potentials and Rotational Symmetry.

Spin and Total Angular Momentum

For spinless wavefunctions, rotations act only by moving the spatial argument. For spinful wavefunctions, rotations also mix spin components. If the spin basis is held fixed, a spin-$s$ wavefunction transforms schematically as

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

where $D^{(s)}(\mathcal R)$ is the spin-$s$ rotation matrix.

For spin-$1/2$, the spin part is represented by $SU(2)$:

$$U_S(\hat{\mathbf n},\theta) = \exp\left( -\frac{i}{2}\theta\,\hat{\mathbf n}\cdot\boldsymbol\sigma \right).$$

This is why spin-$1/2$ states use half-angles and can acquire a minus sign under a $2\pi$ rotation. The physical spatial rotation group is $SO(3)$, while spinors are naturally acted on by its double cover $SU(2)$.

Noncommutativity and Multiplets

Translations in ordinary flat space commute:

$$T(\mathbf a)T(\mathbf b) = T(\mathbf b)T(\mathbf a).$$

Three-dimensional rotations do not generally commute. A small rotation about $x$ followed by one about $y$ differs from doing them in the opposite order. Quantum mechanically this noncommutativity becomes

$$[J_x,J_y]=i\hbar J_z$$

and cyclic permutations.

This is why angular momentum has multiplets rather than simultaneous sharp values of $J_x$, $J_y$, and $J_z$. The standard commuting labels are $J^2$ and one chosen component, usually $J_z$.

Rotations Are Not Every Spatial Symmetry

Rotations preserve distances and orientation. They are continuous transformations connected to the identity. Parity, by contrast, sends

$$\mathbf r\mapsto-\mathbf r$$

and reverses orientation. It is a discrete inversion, not a rotation in three dimensions. Reflections, lattice rotations, and screw symmetries are also distinct spatial operations, even when they share some intuition with ordinary rotations.

This distinction matters for selection rules: rotational symmetry controls angular momentum labels, while parity symmetry controls even and odd behavior under inversion.

Common Mistakes

• Confusing active rotation of a state with passive rotation of coordinate axes.
• Forgetting the inverse argument in $(U(\mathcal R)\psi)(\mathbf r)=\psi(\mathcal R^{-1}\mathbf r)$.
• Using orbital angular momentum $\mathbf L$ when the conserved generator is total angular momentum $\mathbf J=\mathbf L+\mathbf S$.
• Assuming full rotational symmetry when a Hamiltonian has only axial symmetry.
• Treating spin as literal spatial circulation because it obeys the same angular momentum algebra.
• Calling parity a rotation.

Cross-Links

• Active and Passive Transformations
• Generators
• Translations and Momentum
• Rotations in Three Dimensions
• Angular Momentum Algebra
• Orbital Angular Momentum
• Central Potentials and Rotational Symmetry
• What Spin Is
• Spin Rotations
• SO(3)
• SU(2)
• SU(2) versus SO(3)
• Parity as Spatial Inversion
• Parity

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.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.

Exercises

1. Use $(U(\mathcal R)\psi)(\mathbf r)=\psi(\mathcal R^{-1}\mathbf r)$ to show that an active rotation sends $\langle\mathbf R\rangle$ to $\mathcal R\langle\mathbf R\rangle$ for a normalized scalar wavefunction.

Solution

The transformed expectation value is

$$\langle\mathbf R\rangle_{U\psi} = \int d^3r\,\mathbf r\, |\psi(\mathcal R^{-1}\mathbf r)|^2.$$

Change variables to $\mathbf r'=\mathcal R^{-1}\mathbf r$, so $\mathbf r=\mathcal R\mathbf r'$ and $d^3r=d^3r'$ because rotations have determinant $1$. Then

$$\langle\mathbf R\rangle_{U\psi} = \mathcal R \int d^3r'\,\mathbf r'\,|\psi(\mathbf r')|^2 = \mathcal R\langle\mathbf R\rangle_\psi.$$

2. Expand the $z$-axis rotation formula to first order in $\theta$ and identify $L_z$.

Solution

For small $\theta$,

$$x\cos\theta+y\sin\theta=x+\theta y+O(\theta^2),$$

and

$$-x\sin\theta+y\cos\theta=y-\theta x+O(\theta^2).$$

Therefore

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

Writing this as $U_z(\theta)=I-i\theta L_z/\hbar+O(\theta^2)$ gives

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

3. A Hamiltonian is invariant under rotations about the $z$ axis but not under rotations about $x$ or $y$. Which angular momentum label is protected, and which full-rotation label should not be assumed?

Solution

Axial symmetry gives $[H,J_z]=0$, so a $J_z$ eigenvalue can be used as a good quantum number. Full rotational invariance would require $[H,J_i]=0$ for all three components and would allow states to be organized by $J^2$ and $J_z$. With only axial symmetry, one should not assume a conserved $J^2$ label or full $(2j+1)$ rotational multiplets.