Spin Rotations

Spin-$1/2$ rotations are represented by $SU(2)$ matrices. A rotation by angle $\theta$ about unit vector $\hat{\mathbf n}$ is

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

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

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

Why the Half-Angle Appears

Physical rotations in space form $SO(3)$, while spinors transform under $SU(2)$, the double cover of $SO(3)$. The half-angle in the spinor rotation is the signature of this double-cover relationship.

A $2\pi$ rotation gives

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

while a $4\pi$ rotation gives

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

The sign change under $2\pi$ does not change a single ray, but it can matter in interference where relative phases are compared.

Rotation About z

For $\hat{\mathbf n}=\hat z$,

$$U(\hat z,\theta) = \begin{pmatrix} e^{-i\theta/2}&0\\ 0&e^{i\theta/2} \end{pmatrix}.$$

Thus

$$U(\hat z,\theta)\lvert\uparrow\rangle = e^{-i\theta/2}\lvert\uparrow\rangle,$$

and

$$U(\hat z,\theta)\lvert\downarrow\rangle = e^{i\theta/2}\lvert\downarrow\rangle.$$

Bloch-Sphere Action

Although the spinor uses half-angles, the Bloch vector rotates by the physical angle $\theta$. For a density matrix

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

the transformed state

$$\rho'=U\rho U^\dagger$$

has a Bloch vector $\mathbf r'$ obtained by rotating $\mathbf r$ in ordinary three-dimensional space.

Relation to Angular Momentum Generators

Since

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

the rotation operator can also be written

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

This matches the general angular momentum rotation formula.

Common Mistakes

• Using $\theta$ instead of $\theta/2$ in the spinor exponential.
• Thinking $U(2\pi)=-I$ makes a single state physically different from itself.
• Confusing rotation of the spinor with rotation of the Bloch vector.
• Forgetting that $SU(2)$ has two elements corresponding to each $SO(3)$ rotation.

Cross-Links

• Spin Problems
• Projective Representations
• Spin as Intrinsic Angular Momentum
• Pauli Matrices
• Bloch Sphere
• Single-Qubit Gates
• Spin Measurements
• Spinors and 2π Rotations
• Spin Coherent States
• Higher Spin Systems
• Stern–Gerlach Revisited
• Spin in Magnetic Fields
• Larmor Precession
• Rotations Preview
• SO(3) and SU(2) Preview
• Angular Momentum Algebra
• SO(3)
• Lie Groups
• SU(2)
• SU(2) versus SO(3)
• Wigner D-Matrices
• From Spin to Relativistic Representations
• From SU(2) Spinors to Lorentz Spinors

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, Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, 2nd ed., Springer, 2015.

Exercises

1. Compute $U(\hat z,2\pi)$.

Solution

Using

$$U(\hat z,\theta) = \begin{pmatrix} e^{-i\theta/2}&0\\ 0&e^{i\theta/2} \end{pmatrix},$$

one finds

$$U(\hat z,2\pi) = \begin{pmatrix} -1&0\\ 0&-1 \end{pmatrix} = -I.$$

2. Show that $(\hat{\mathbf n}\cdot\boldsymbol\sigma)^2=I$ for a unit vector $\hat{\mathbf n}$.

Solution

Use

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

Setting $\mathbf a=\mathbf b=\hat{\mathbf n}$ gives $\hat{\mathbf n}\cdot\hat{\mathbf n}=1$ and $\hat{\mathbf n}\times\hat{\mathbf n}=0$, so the square is $I$.