SO(3) and SU(2) Preview

Ordinary spatial rotations of three-dimensional vectors form $SO(3)$. Spin-$1/2$ state vectors, however, transform naturally under $SU(2)$. The two groups have the same local angular-momentum algebra but different global structure.

This preview explains the physical distinction before the angular-momentum machinery begins. The mathematical double-cover construction lives in SU(2) versus SO(3); the spinor rotation formulas are used concretely in Spin Rotations, and the sign-change interpretation is isolated in Spinors and 2π Rotations.

Ordinary Vector Rotations

$SO(3)$ is the group of proper rotations of ordinary real three-dimensional vectors:

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

It acts on vectors by

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

A full $2\pi$ rotation is the identity on ordinary vectors:

$$\mathcal R(\hat{\mathbf n},2\pi) = I_3.$$

This is the rotation group behind positions, momenta, ordinary polar vectors, and orbital wavefunctions. The geometric action is reviewed in Rotations in Three Dimensions.

Why SU(2) Appears

Quantum states are rays, not individual vectors with fixed overall phase. A physical symmetry of rays may therefore be represented on Hilbert-space vectors only up to phase. Spin is the central elementary example.

$SU(2)$ is the group of two-by-two unitary complex matrices with determinant one:

$$SU(2) = \{U\in M_2(\mathbb C): U^\dagger U=I,\ \det U=1\}.$$

It acts naturally on two-component spinors:

$$\chi\mapsto U\chi, \qquad \chi\in\mathbb C^2.$$

Spinors are not ordinary three-dimensional vectors. They are Hilbert-space objects whose transformation law is tied to the double cover

$$SU(2)\to SO(3).$$

Each ordinary rotation in $SO(3)$ corresponds to two elements of $SU(2)$, conventionally written $U$ and $-U$.

The Double-Cover Map

The useful concrete bridge uses Pauli matrices. For a real vector $\mathbf v$, form

$$\mathbf v\cdot\boldsymbol\sigma = v_x\sigma_x+v_y\sigma_y+v_z\sigma_z.$$

For $U\in SU(2)$, conjugation gives another matrix of the same form:

$$U(\mathbf v\cdot\boldsymbol\sigma)U^\dagger = (R(U)\mathbf v)\cdot\boldsymbol\sigma.$$

This defines an ordinary rotation $R(U)\in SO(3)$. The two $SU(2)$ matrices $U$ and $-U$ give the same $SO(3)$ rotation, because the two minus signs cancel in the conjugation. The kernel is exactly $\{I,-I\}$: a matrix commuting with all three Pauli matrices is scalar, and the determinant-one condition leaves only those two scalars.

That is the practical meaning of “double cover” here:

$$R(U)=R(-U).$$

Spin-Half Rotation Formula

For spin-$1/2$, a rotation by angle $\theta$ about a unit vector $\hat{\mathbf n}$ is represented by

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

Using

$$(\hat{\mathbf n}\cdot\boldsymbol\sigma)^2=I,$$

this becomes

$$U(\hat{\mathbf n},\theta) = \cos\frac{\theta}{2}\,I - i\sin\frac{\theta}{2}\, \hat{\mathbf n}\cdot\boldsymbol\sigma.$$

The half-angle is the visible signature of the double cover. The associated ordinary vector rotation uses angle $\theta$, while the spinor matrix uses $\theta/2$.

The 2π Sign

At a full $2\pi$ rotation,

$$U(\hat{\mathbf n},2\pi) = -I.$$

At $4\pi$,

$$U(\hat{\mathbf n},4\pi) = I.$$

Thus a spin-$1/2$ state vector changes sign under a $2\pi$ rotation:

$$|\psi\rangle\mapsto-|\psi\rangle.$$

This does not make a single isolated ray physically different from itself, because

$$|\psi\rangle \sim e^{i\alpha}|\psi\rangle.$$

The sign can still matter in interference if only one coherent path undergoes the $2\pi$ spin rotation. Then it is a relative sign between paths. Applying the rotation identically to the entire state still produces only an overall sign. The careful language is: a $2\pi$ spinor rotation changes the representative vector, not the isolated ray.

Integer and Half-Integer Spin

Angular-momentum representations are labeled by

$$j=0,\frac12,1,\frac32,\ldots$$

with dimension

$$2j+1.$$

The element $-I\in SU(2)$ acts in the spin-$j$ representation as

$$(-I)\mapsto(-1)^{2j}I.$$

If $j$ is an integer, this action is $+I$, so the representation can be regarded as an ordinary representation of $SO(3)$. If $j$ is a half-integer, this action is $-I$, so the representation does not descend to an ordinary representation of $SO(3)$.

This is why orbital angular momentum of ordinary single-valued scalar wavefunctions has integer $\ell$, while intrinsic spin may have half-integer $j$.

Orbital Versus Spin Rotations

Orbital wavefunctions rotate by changing their spatial argument:

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

For ordinary scalar wavefunctions on $\mathbb R^3$, a $2\pi$ spatial rotation returns the wavefunction itself. The allowed orbital angular-momentum labels are integers.

Spin-$1/2$ states rotate by $SU(2)$ matrices. A two-component spinor can acquire a minus sign under the same physical $2\pi$ rotation. For a spinful particle in space, the total rotation acts on both orbital and spin parts, and the total generator is

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

Bloch Vector Caution

Write a spin-$1/2$ density matrix as $\rho=(I+\mathbf b\cdot\boldsymbol\sigma)/2$. The conjugation map above implies $U\rho U^\dagger=(I+R(U)\mathbf b\cdot\boldsymbol\sigma)/2$, so its Bloch vector behaves like an ordinary three-dimensional vector. It rotates by the physical angle $\theta$ and returns after $2\pi$.

The spinor underneath uses the $SU(2)$ half-angle formula. Confusing the Bloch vector with the spinor is a common source of mistakes: the Bloch vector lives in an $SO(3)$ picture, while the state vector lives in a two-dimensional Hilbert space.

Exercises

1. Use the spin-$1/2$ formula to compute $U(\hat{\mathbf n},2\pi)$ and $U(\hat{\mathbf n},4\pi)$.

Solution

The formula is

$$U(\hat{\mathbf n},\theta) = \cos\frac{\theta}{2}\,I - i\sin\frac{\theta}{2}\, \hat{\mathbf n}\cdot\boldsymbol\sigma.$$

At $\theta=2\pi$,

$$\cos\pi=-1, \qquad \sin\pi=0,$$

so

$$U(\hat{\mathbf n},2\pi)=-I.$$

At $\theta=4\pi$,

$$\cos 2\pi=1, \qquad \sin 2\pi=0,$$

so

$$U(\hat{\mathbf n},4\pi)=I.$$

2. Show that $U$ and $-U$ define the same ordinary vector rotation under the conjugation map.

Solution

The map is defined by

$$U(\mathbf v\cdot\boldsymbol\sigma)U^\dagger = (R(U)\mathbf v)\cdot\boldsymbol\sigma.$$

For $-U$,

$$(-U)(\mathbf v\cdot\boldsymbol\sigma)(-U)^\dagger = U(\mathbf v\cdot\boldsymbol\sigma)U^\dagger,$$

because the two minus signs cancel. Therefore

$$R(-U)=R(U).$$

3. Which of $j=1$, $j=3/2$, and $j=2$ descend to ordinary $SO(3)$ representations?

Solution

The test is how $-I\in SU(2)$ acts:

$$(-I)\mapsto(-1)^{2j}I.$$

For $j=1$, $2j=2$, so the action is $+I$. For $j=3/2$, $2j=3$, so the action is $-I$. For $j=2$, $2j=4$, so the action is $+I$.

Thus $j=1$ and $j=2$ descend to ordinary $SO(3)$ representations, while $j=3/2$ requires the $SU(2)$ cover.

References

• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• B. C. Hall, Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, 2nd ed., Springer, 2015.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, chapter 3.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.