Spin Rotations

Spin-$1/2$ rotations are represented by $SU(2)$ matrices. We use an active, right-handed rotation of the state with the laboratory measurement axes held fixed. 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. It describes how states transform; whether a rotation is also a symmetry of a particular Hamiltonian is a separate question. A field can select an axis even though spin rotations about other axes remain well-defined operations.

For example, rotating $|+x\rangle$ about $z$ gives, up to a global phase,

$$U_z(\theta)|+x\rangle\sim \frac{|+z\rangle+e^{i\theta}|-z\rangle}{\sqrt2}.$$

At $\theta=\pi/2$ this is $|+y\rangle$, providing a quick check of the rotation sign. A passive change of coordinate axes uses the inverse transformation.

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.

Exercises

1. Starting from $|+z\rangle$, apply $U_y(\theta)$ and find the probability of $+x$ in a subsequent spin measurement.

Solution

The rotation matrix is

$$U_y(\theta)=\begin{pmatrix}\cos(\theta/2)&-\sin(\theta/2)\\ \sin(\theta/2)&\cos(\theta/2)\end{pmatrix}.$$

Thus the state has components $(\cos(\theta/2),\sin(\theta/2))$, and

$$P(+x)=\frac{[\cos(\theta/2)+\sin(\theta/2)]^2}{2} =\frac{1+\sin\theta}{2}.$$

A positive quarter-turn maps $+z$ to $+x$, so this probability becomes one.

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 $I$.

References

• 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.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.