SU(2) versus SO(3)

$SO(3)$ and $SU(2)$ describe closely related rotation structures, but they are not the same group. $SO(3)$ is the group of ordinary proper rotations of three-dimensional real vectors. $SU(2)$ is the group that acts naturally on spinors and double-covers $SO(3)$.

The distinction is not pedantry. It explains why spin-$1/2$ state vectors change sign under a $2\pi$ rotation, why half-integer spin representations exist, and why quantum rotations are often represented using $SU(2)$ even when the physical space being rotated is ordinary three-dimensional space.

The Short Version

The local Lie-algebra structure is the same:

$$\mathfrak{su}(2) \simeq \mathfrak{so}(3).$$

But the global groups differ:

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

is a two-to-one covering map with kernel

$$\{I,-I\}.$$

Thus each ordinary rotation $R\in SO(3)$ corresponds to two elements of $SU(2)$, namely $U$ and $-U$.

The Covering Map

For a real vector $\mathbf v\in\mathbb R^3$, form the Hermitian matrix

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

For $U\in SU(2)$, conjugation by $U$ sends this matrix to another traceless Hermitian matrix:

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

This defines a real linear map $R(U):\mathbb R^3\to\mathbb R^3$. One can show that $R(U)$ preserves lengths, preserves orientation, and depends smoothly on $U$, so

$$R(U)\in SO(3).$$

The assignment

$$U\mapsto R(U)$$

is a group homomorphism:

$$R(U_1U_2) = R(U_1)R(U_2).$$

Its kernel is exactly $\{I,-I\}$. Therefore $U$ and $-U$ define the same ordinary rotation.

Local Agreement

Near the identity, the groups have the same infinitesimal rotation algebra.

In $SU(2)$, a spinor rotation is written

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

In $SO(3)$, the corresponding vector rotation is

$$R(\hat{\mathbf n},\theta) = \exp \left( \theta[\hat{\mathbf n}]_\times \right),$$

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

Differentiating at $\theta=0$ gives Lie algebra generators with the same structure constants. In physics language, the Hermitian angular-momentum generators satisfy

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

This local agreement is why $SO(3)$ and $SU(2)$ often appear to have the same angular-momentum algebra.

Global Difference

The difference appears when one follows rotations around closed loops.

$SO(3)$ has

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

An ordinary vector returns to itself after a full $2\pi$ rotation.

The corresponding $SU(2)$ path gives

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

and only after $4\pi$ does it return to the identity:

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

So a path in $SO(3)$ that closes after one full turn lifts to a path in $SU(2)$ whose endpoint is $-I$, not $I$. It closes only after going around twice.

Topologically, $SU(2)$ is a three-sphere $S^3$, while $SO(3)$ is obtained by identifying antipodal points:

$$SO(3) \simeq S^3/\{U\sim -U\}.$$

Equivalently,

$$SO(3)\simeq \mathbb{RP}^3.$$

This global identification is what the double cover records.

Why Spin-1/2 Changes Sign

A spin-$1/2$ state vector transforms under the defining representation of $SU(2)$ on $\mathbb C^2$. For a rotation by $\theta$ about $\hat{\mathbf n}$,

$$\lvert\psi\rangle \longmapsto U(\hat{\mathbf n},\theta)\lvert\psi\rangle.$$

At $\theta=2\pi$,

$$\lvert\psi\rangle \longmapsto -\lvert\psi\rangle.$$

This does not contradict the ray nature of pure quantum states. The vectors $\lvert\psi\rangle$ and $-\lvert\psi\rangle$ represent the same ray:

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

The sign can still matter in interference, because relative phases between alternatives are observable. The correct statement is therefore:

A $2\pi$ spinor rotation changes the state vector sign, but not the single isolated physical ray.

Representations and Spin

Representations of $SU(2)$ are labeled by

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

with dimension $2j+1$.

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

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

If $j$ is an integer, then $(-1)^{2j}=1$, so the representation descends to an ordinary representation of $SO(3)$. If $j$ is a half-integer, then $(-1)^{2j}=-1$, so it does not descend to an ordinary representation of $SO(3)$. It is instead a projective representation of $SO(3)$, or equivalently an ordinary representation of the double cover $SU(2)$.

This is the group-theoretic reason orbital angular momentum has integer labels while spin can have half-integer labels.

Physical Implications

The distinction has several practical consequences.

Ordinary spatial vectors transform under $SO(3)$. Position, momentum, electric field vectors, and Bloch vectors rotate by the physical angle and return after $2\pi$.

Spinors transform under $SU(2)$. A spin-$1/2$ state vector returns only after $4\pi$, although its ray is already unchanged after $2\pi$.

Hamiltonians with rotational symmetry may be organized by angular-momentum representations. Integer-spin sectors can be treated as honest $SO(3)$ representations; half-integer sectors require the $SU(2)$ cover on the Hilbert-space level.

The Bloch sphere does not contradict this. The Bloch vector of a spin-$1/2$ state rotates as an ordinary $SO(3)$ vector, while the underlying spinor transforms by $SU(2)$ and carries the half-angle behavior.

Common Mistakes

• Saying $SU(2)$ is equal to $SO(3)$ because their Lie algebras are closely related.
• Thinking a $2\pi$ spinor sign change makes the ray physically different from itself.
• Forgetting that $U$ and $-U$ define the same element of $SO(3)$.
• Using $SO(3)$ when the Hilbert-space representation requires half-integer spin.
• Treating the Bloch vector as the spinor.
• Assuming local generator algebra determines all global representation questions.

Cross-Links

• SO(3)
• SU(2)
• Lie Algebras
• Representations
• Unitary Representations
• Angular Momentum Algebra
• Wigner D-Matrices
• SO(3) and SU(2) Preview
• Projective Representations
• Spin Rotations
• Spinors and 2π Rotations
• Bloch Sphere
• Orbital Angular Momentum

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 $U$ and $-U$ define the same $SO(3)$ 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. Hence

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

2. Compute $U(\hat{\mathbf n},2\pi)$ and $U(\hat{\mathbf n},4\pi)$ for spin-$1/2$.

Solution

Use

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

For $\theta=2\pi$,

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

For $\theta=4\pi$,

$$U(\hat{\mathbf n},4\pi) = \cos 2\pi\,I = I.$$

3. Why does a half-integer spin representation not descend to an ordinary representation of $SO(3)$?

Solution

The two elements $I$ and $-I$ of $SU(2)$ map to the same rotation in $SO(3)$. For a representation to descend to $SO(3)$, these two elements must act in the same way. In the spin-$j$ representation,

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

For half-integer $j$, this is $-I$, not $I$. Therefore the representation distinguishes two elements that $SO(3)$ identifies, so it cannot be an ordinary representation of $SO(3)$.

4. Explain why $U(2\pi)=-I$ does not make a single spin state physically different from itself.

Solution

Pure physical states are rays. Multiplying a state vector by a nonzero complex phase does not change the ray. Since $-1=e^{i\pi}$, the vectors $\lvert\psi\rangle$ and $-\lvert\psi\rangle$ describe the same pure state. The sign can still matter as a relative phase in an interference experiment.

5. Why is the Bloch vector not enough to reconstruct the full spinor phase history?

Solution

The Bloch vector transforms as an ordinary $SO(3)$ vector, so it returns after a $2\pi$ rotation. The spinor transforms through $SU(2)$ and changes sign under the same path. Since the Bloch vector represents the ray but not the overall spinor phase, it does not record this sign change.