SU(2)

$SU(2)$ is the group of two-by-two unitary complex matrices with determinant one. It is the basic compact Lie group behind spin- $1/2$ , two-level systems, qubit rotations, and the double-cover relation between spinors and ordinary three-dimensional rotations.

The group $SO(3)$ rotates ordinary real vectors. The group $SU(2)$ acts naturally on spinors in $\mathbb C^2$ . The two groups have closely related Lie algebras but different global topology, and that difference is visible in the sign change of spin- $1/2$ state vectors under a $2\pi$ rotation.

Definition

The group is

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

It is a subgroup of $U(2)$ and a compact three-dimensional Lie group. The group operation is matrix multiplication, the identity is $I$ , and the inverse is

U^{-1}=U^\dagger.

The word “special” means determinant one. The determinant condition removes the overall $U(1)$ phase present in $U(2)$ .

Matrix Form

Every element of $SU(2)$ can be written as

U = \begin{pmatrix} \alpha & \beta\\ -\beta^* & \alpha^* \end{pmatrix}, \qquad \lvert\alpha\rvert^2+\lvert\beta\rvert^2=1.

Conversely, every matrix of this form is unitary and has determinant one. The pair $(\alpha,\beta)\in\mathbb C^2$ with unit norm shows that $SU(2)$ is, as a manifold, a three-sphere $S^3$ .

This global shape is one reason $SU(2)$ behaves differently from $SO(3)$ even though their infinitesimal algebras are closely related.

Why this is the general form

The first column of a unitary matrix is a unit vector in $\mathbb C^2$ . Once that column is written as $(\alpha,-\beta^*)^{\mathsf T}$ , its normalized orthogonal complement is fixed up to a phase. Requiring determinant one fixes that phase and gives the second column $(\beta,\alpha^*)^{\mathsf T}$ . Thus unitarity supplies the norm condition and special unitarity removes the remaining phase freedom.

The inverse and trace are especially easy to read:

U^{-1} = \begin{pmatrix} \alpha^* & -\beta\\ \beta^* & \alpha \end{pmatrix}, \qquad \operatorname{Tr}U=2\operatorname{Re}\alpha.

In particular, the trace of an $SU(2)$ matrix is always real. This provides a quick diagnostic when a symbolic or numerical matrix is claimed to lie in $SU(2)$ .

Pauli-Matrix Parameterization

Using the Pauli matrices, every element can also be written as

U = a_0I-i\,\mathbf a\cdot\boldsymbol\sigma,

where $a_0\in\mathbb R$ , $\mathbf a\in\mathbb R^3$ , and

a_0^2+\mathbf a\cdot\mathbf a=1.

In components,

\alpha=a_0-ia_3, \qquad \beta=-a_2-ia_1.

The four real coordinates $(a_0,a_1,a_2,a_3)$ therefore lie on the unit three-sphere. They also identify $SU(2)$ with the multiplicative group of unit quaternions. If

U(a_0,\mathbf a)=a_0I-i\mathbf a\cdot\boldsymbol\sigma, \qquad U(b_0,\mathbf b)=b_0I-i\mathbf b\cdot\boldsymbol\sigma,

then the Pauli identity gives

\begin{aligned} U(a_0,\mathbf a)U(b_0,\mathbf b) ={}&U(c_0,\mathbf c),\\ c_0={}&a_0b_0-\mathbf a\cdot\mathbf b,\\ \mathbf c={}&a_0\mathbf b+b_0\mathbf a +\mathbf a\times\mathbf b. \end{aligned}

The inverse corresponds to $(a_0,\mathbf a)\mapsto(a_0,-\mathbf a)$ . The cross product in the composition law makes the noncommutativity visible without multiplying complex matrices.

For rotations, one often uses the exponential form

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

Since

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

the exponential 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 not a notation accident. It is the local sign of the double-cover relation between $SU(2)$ and $SO(3)$ .

Exponential Coordinates and Periodicity

Every $U\in SU(2)$ has an axis-angle form. Choose $\chi\in[0,\pi]$ so that

a_0=\cos\chi, \qquad \lVert\mathbf a\rVert=\sin\chi.

When $\sin\chi\ne0$ , set $\hat{\mathbf n}=\mathbf a/\sin\chi$ . Then

U = \cos\chi\,I-i\sin\chi\,\hat{\mathbf n}\cdot\boldsymbol\sigma = e^{-i\chi\hat{\mathbf n}\cdot\boldsymbol\sigma}.

For a physical rotation, $\chi=\theta/2$ . Consequently,

U(\hat{\mathbf n},\theta+2\pi) =-U(\hat{\mathbf n},\theta), \qquad U(\hat{\mathbf n},\theta+4\pi) =U(\hat{\mathbf n},\theta).

At $U=I$ and $U=-I$ , the axis is not unique. More generally, exponential coordinates are not global one-to-one coordinates; periodicity and axis reversals identify different parameter pairs. This matters when reconstructing rotations from numerical matrices.

The coordinates can be recovered from traces:

a_0=\frac12\operatorname{Tr}U, \qquad a_k=\frac{i}{2}\operatorname{Tr}(\sigma_kU).

For numerical work, $\chi=\operatorname{atan2}(\lVert\mathbf a\rVert,a_0)$ is usually better conditioned near $a_0=\mathord\pm1$ than using $\arccos a_0$ alone.

Lie Algebra

The mathematical Lie algebra $\mathfrak{su}(2)$ consists of traceless anti-Hermitian matrices:

\mathfrak{su}(2) = \{X\in M_2(\mathbb C):X^\dagger=-X,\ \operatorname{Tr}X=0\}.

A convenient anti-Hermitian basis is

T_i = -\frac{i}{2}\sigma_i.

These obey

[T_i,T_j] = \sum_k\epsilon_{ijk}T_k.

Physics usually uses Hermitian generators instead:

J_i = \frac12\sigma_i.

Then

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

With physical spin operators

S_i = \frac{\hbar}{2}\sigma_i,

the algebra is

[S_i,S_j] = i\hbar\sum_k\epsilon_{ijk}S_k.

This is the spin- $1/2$ version of the angular-momentum algebra.

The Pauli product identity also turns commutators into ordinary vector products:

[\mathbf a\cdot\boldsymbol\sigma, \mathbf b\cdot\boldsymbol\sigma] = 2i(\mathbf a\times\mathbf b)\cdot\boldsymbol\sigma.

Thus parallel generators commute, while generators about different axes generally do not. For two infinitesimal rotations, write

\begin{aligned} A&=-\frac{i\epsilon}{2} \mathbf a\cdot\boldsymbol\sigma,\\ B&=-\frac{i\epsilon}{2} \mathbf b\cdot\boldsymbol\sigma. \end{aligned}

Their group commutator is

\begin{aligned} e^Ae^Be^{-A}e^{-B} &=I-\frac{i\epsilon^2}{2} (\mathbf a\times\mathbf b)\cdot\boldsymbol\sigma +O(\epsilon^3). \end{aligned}

This is the local algebraic content of the statement that finite rotations about different axes do not commute.

Defining Spinor Representation

The defining representation of $SU(2)$ is its natural action on $\mathbb C^2$ :

\chi \longmapsto U\chi, \qquad \chi\in\mathbb C^2.

Vectors in this representation are spinors. They are not ordinary three-dimensional vectors. A spinor can change sign under a $2\pi$ rotation even though the corresponding physical ray is unchanged.

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

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

At $\theta=2\pi$ ,

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

and at $\theta=4\pi$ ,

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

The sign under $2\pi$ is unobservable for a single isolated ray, but it matters when relative phases are compared.

Spectrum and Conjugacy Classes

Because $U$ is unitary and has determinant one, its eigenvalues have the form

\lambda_+=e^{i\chi}, \qquad \lambda_-=e^{-i\chi}.

Equivalently, its characteristic polynomial is

\lambda^2-(\operatorname{Tr}U)\lambda+1=0, \qquad \operatorname{Tr}U=2\cos\chi.

Conjugation changes the axis but not $\chi$ :

VUV^\dagger = \cos\chi\,I -i\sin\chi\,(R(V)\hat{\mathbf n})\cdot\boldsymbol\sigma.

Two elements of $SU(2)$ are conjugate exactly when they have the same trace. The central elements $I$ and $-I$ are singleton conjugacy classes. This trace classification is useful in character theory and when comparing two rotation operators without explicitly finding their axes.

Irreducible Representations: A Preview

The finite-dimensional irreducible unitary representations of $SU(2)$ are labeled by

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

and have dimension $2j+1$ . The defining spinor representation is $j=1/2$ ; the three-dimensional vector representation is $j=1$ . In the spin- $j$ representation, the central element acts as

D^{(j)}(-I)=(-1)^{2j}I_{2j+1}.

Integer- $j$ representations therefore assign the same operator to $U$ and $-U$ and descend to representations of $SO(3)$ . Half-integer- $j$ representations do not. The derivation of the labels, dimensions, and matrix elements belongs to Angular Momentum Algebra and Wigner D-Matrices.

Map to SO(3)

There is a natural homomorphism from $SU(2)$ onto $SO(3)$ . For each real vector $\mathbf v$ , form the traceless Hermitian matrix

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

Given $U\in SU(2)$ , define $R(U)$ by

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

The map

U\mapsto R(U)

is a group homomorphism

SU(2)\to SO(3).

Its kernel is

\{I,-I\}.

Therefore each ordinary rotation in $SO(3)$ corresponds to two elements of $SU(2)$ . This is the precise meaning of the statement that $SU(2)$ is a double cover of $SO(3)$ .

Bloch Vectors and Spin Directions

For a spin- $1/2$ density matrix

\rho = \frac12(I+\mathbf r\cdot\boldsymbol\sigma),

the transformation

\rho \longmapsto U\rho U^\dagger

rotates the Bloch vector by $R(U)$ :

\mathbf r \longmapsto R(U)\mathbf r.

Thus spinors transform through $SU(2)$ , while their associated Bloch vectors rotate through $SO(3)$ . This is why the Bloch sphere shows ordinary spatial directions but hides the spinor sign change under a full $2\pi$ rotation.

The expectation values make this statement operational. Since

r_i=\operatorname{Tr}(\rho\sigma_i),

conjugating $\rho$ by $U$ produces precisely the adjoint rotation of the three measured Pauli components. The global phase of a state spinor cancels from $\rho$ , so it cannot appear in the Bloch vector.

Quantum Use

$SU(2)$ appears throughout quantum mechanics:

spin- $1/2$ rotations;
angular-momentum multiplets and ladder-operator constructions;
two-level Hamiltonians written with Pauli matrices;
qubit gates and single-qubit rotations;
the local group behind $SO(3)$ rotation generators;
projective representations of spatial rotations.

A general traceless two-level Hamiltonian can be written as

H = \mathbf h\cdot\boldsymbol\sigma

up to an irrelevant multiple of the identity. Its time evolution is an $SU(2)$ rotation up to an overall phase. This is the algebraic reason Pauli matrices and Bloch-sphere rotations are so efficient for two-level systems.

Worked example: constant two-level Hamiltonian

Let

H=h_0I+\mathbf h\cdot\boldsymbol\sigma, \qquad \Omega=\frac{\lVert\mathbf h\rVert}{\hbar}.

For $\mathbf h\ne0$ , the propagator factors into an overall phase and an $SU(2)$ matrix:

\begin{aligned} e^{-iHt/\hbar} ={}&e^{-ih_0t/\hbar}\\ &\times \left[ \cos(\Omega t)I -i\sin(\Omega t)\, \hat{\mathbf h}\cdot\boldsymbol\sigma \right]. \end{aligned}

If the system starts in the $+1$ eigenstate of $\sigma_z$ , the probability of finding the $-1$ eigenstate at time $t$ is

P_{+z\to-z}(t) = \frac{h_x^2+h_y^2}{\lVert\mathbf h\rVert^2} \sin^2(\Omega t).

The longitudinal component $h_z$ tilts the rotation axis away from the equatorial plane and reduces the maximum transfer. The scalar term $h_0I$ changes only the global phase and drops out of this probability.

Worked example: order matters

Quarter-turn spinor rotations about the $x$ and $z$ axes are

U_x=\frac{I-i\sigma_x}{\sqrt2}, \qquad U_z=\frac{I-i\sigma_z}{\sqrt2}.

Their products differ:

\begin{aligned} U_xU_z &=\frac12 \left[I-i\sigma_x+i\sigma_y-i\sigma_z\right],\\ U_zU_x &=\frac12 \left[I-i\sigma_x-i\sigma_y-i\sigma_z\right]. \end{aligned}

The sign of the $\sigma_y$ component records the order of the rotations.

Practical Computation

For analytical work, switch among three equivalent descriptions according to the task:

Description	Best use
Complex pair $(\alpha,\beta)$	Direct matrix constraints and spinor action
Real pair $(a_0,\mathbf a)$	Composition, inversion, and numerical storage
Axis-angle pair $(\hat{\mathbf n},\theta)$	Physical interpretation and exponentiation

Useful numerical checks for a proposed matrix $U$ are

\lVert U^\dagger U-I\rVert\ll1, \qquad \lvert\det U-1\rvert\ll1.

Repeated floating-point multiplication can move $(a_0,\mathbf a)$ slightly off the unit three-sphere. Renormalizing the four-vector controls this drift when exact unitarity is required. When a general $V\in U(2)$ is reduced to an $SU(2)$ part, the square root of $\det V$ has two branches; the resulting matrices differ by the central sign and that sign must not be discarded when coherent spinor phases matter.

Boundary of This Page

This page gives the group, its concrete coordinates, generators, and defining spinor representation. The detailed topology and proof of the double cover belong to SU(2) versus SO(3). The construction of all spin- $j$ multiplets belongs to Angular Momentum Algebra. For ordinary rotation matrices, see SO(3); for the ray-level quantum meaning, see Projective Representations.

Common Mistakes

Identifying $SU(2)$ and $SO(3)$ as the same group.
Forgetting that $SU(2)$ acts on spinors, not ordinary real three-vectors.
Missing the factor of one half in spin- $1/2$ rotations.
Confusing anti-Hermitian mathematical generators with Hermitian physics generators.
Treating $U(2\pi)=-I$ as a contradiction with rotational invariance of rays.
Forgetting that the map $SU(2)\to SO(3)$ has kernel $\{I,-I\}$ .
Assuming the Bloch vector contains the full spinor phase information.
Reading $\theta$ directly from $\operatorname{Tr}U$ without accounting for the half-angle and branch choice.
Multiplying axis-angle parameters componentwise instead of using matrix or quaternion composition.
Removing the central sign $-I$ in a calculation where relative spinor phases can interfere.

Cross-Links

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.
H. Georgi, Lie Algebras in Particle Physics, 2nd ed., Westview Press, 1999.
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.

Exercises

Verify that the standard matrix form lies in $SU(2)$ .

Solution

Let

U = \begin{pmatrix} \alpha & \beta\\ -\beta^* & \alpha^* \end{pmatrix}, \qquad \lvert\alpha\rvert^2+\lvert\beta\rvert^2=1.

The columns are orthonormal:

\lvert\alpha\rvert^2+\lvert\beta\rvert^2=1,

and their inner product is

\alpha^*\beta+(-\beta)\alpha^*=0.

Thus $U^\dagger U=I$ . The determinant is

\det U = \lvert\alpha\rvert^2+\lvert\beta\rvert^2 = 1.

So $U\in SU(2)$ .

Derive the spinor rotation exponential formula.

Solution

For a unit vector $\hat{\mathbf n}$ ,

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

Splitting the exponential power series into even and odd powers gives

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

If $J_i=\sigma_i/2$ , compute $[J_x,J_y]$ .

Solution

Since $[\sigma_x,\sigma_y]=2i\sigma_z$ ,

[J_x,J_y] = \frac14[\sigma_x,\sigma_y] = \frac{i}{2}\sigma_z = iJ_z.

Show that a $2\pi$ spin- $1/2$ rotation gives $-I$ .

Solution

Using the exponential formula,

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

Explain why $I$ and $-I$ define the same rotation in $SO(3)$ under the map above.

Solution

For any $\mathbf v$ ,

I(\mathbf v\cdot\boldsymbol\sigma)I^\dagger = \mathbf v\cdot\boldsymbol\sigma,

and

(-I)(\mathbf v\cdot\boldsymbol\sigma)(-I)^\dagger = (\mathbf v\cdot\boldsymbol\sigma).

Both elements therefore leave every vector $\mathbf v$ unchanged under the defining relation

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

Thus $I$ and $-I$ map to the identity rotation in $SO(3)$ .

Derive the real-coordinate multiplication law for $SU(2)$ .

Solution

Use

(\mathbf a\cdot\boldsymbol\sigma) (\mathbf b\cdot\boldsymbol\sigma) = (\mathbf a\cdot\mathbf b)I +i(\mathbf a\times\mathbf b)\cdot\boldsymbol\sigma.

Multiplying the two matrices gives

\begin{aligned} &(a_0I-i\mathbf a\cdot\boldsymbol\sigma)\\ &\qquad\times (b_0I-i\mathbf b\cdot\boldsymbol\sigma) \\ &=(a_0b_0-\mathbf a\cdot\mathbf b)I -i(a_0\mathbf b+b_0\mathbf a +\mathbf a\times\mathbf b)\cdot\boldsymbol\sigma. \end{aligned}

The scalar and vector components are therefore

c_0=a_0b_0-\mathbf a\cdot\mathbf b, \qquad \mathbf c=a_0\mathbf b+b_0\mathbf a+\mathbf a\times\mathbf b.

For $H=h_0I+\mathbf h\cdot\boldsymbol\sigma$ , derive the transition probability from $\lvert+z\rangle$ to $\lvert-z\rangle$ .

Solution

The identity term contributes only $e^{-ih_0t/\hbar}$ . With $\Omega=\lVert\mathbf h\rVert/\hbar$ ,

U(t) = e^{-ih_0t/\hbar} \left[ \cos(\Omega t)I -i\sin(\Omega t)\hat{\mathbf h}\cdot\boldsymbol\sigma \right].

Because the identity and $\sigma_z$ have no matrix element between the two $\sigma_z$ eigenstates,

\langle-z\rvert U(t)\lvert+z\rangle = -ie^{-ih_0t/\hbar} \sin(\Omega t) \frac{h_x+ih_y}{\lVert\mathbf h\rVert}.

Taking the modulus squared yields

P_{+z\to-z}(t) = \frac{h_x^2+h_y^2}{\lVert\mathbf h\rVert^2} \sin^2(\Omega t).