Spinors and rotations · Quantum Mechanics

Intuition

Rotations act on spin states through the exponentials of the Pauli matrices, and those exponentials have a surprise: because $(\mathbf{n}\cdot\vec{\sigma})^{2}=1$ , the exponential of a rotation by $\theta$ is $\cos(\theta/2)-i\mathbf{n}\cdot\vec{\sigma}\sin(\theta/2)$ , with half angles. A full turn gives $-1$ . Objects that change sign under a full turn and need two turns to come back are spinors. The sign cannot be seen on one spin alone, since it is an overall phase, but it shows in interference, and neutrons sent round a magnetic field showed it in 1975.

Hold a cup of water flat on your palm and turn it one full turn under your arm: your arm ends twisted. A second full turn in the same direction untwists it. The arm is a spinor: it needs two turns to come back.

Against the angle of rotation, from 0 to $4\pi$ : the component of an ordinary vector returns after one turn, $2\pi$ ; the amplitude $\cos\frac{\theta}{2}$ of a spinor has only reached at and returns after two turns, .

Rotating a spin

A rotation by $\theta$ about the axis $\mathbf{n}$ acts on spin- $\tfrac12$ states as the exponential of $- i θ n \cdot \overset{}{}$ . Since , the series sums in closed form.

\hat{R}_{\mathbf{n}}(\theta)=e^{-i\theta\,\mathbf{n}\cdot\vec{\sigma}/2}=\cos\frac{\theta}{2}-i\,\mathbf{n}\cdot\vec{\sigma}\,\sin\frac{\theta}{2}

Properties

$\hat{R}_{\mathbf{n}}(2\pi)=-1$ and $\hat{R}_{\mathbf{n}}(4\pi)=+1$ , for every axis.

The exponential of a Pauli matrix

e^{-i\frac{\theta}{2}\mathbf{n}\cdot\vec{\sigma}}=\cos\frac{\theta}{2}-i\,\mathbf{n}\cdot\vec{\sigma}\,\sin\frac{\theta}{2}

Expand the exponential. Every even power of $\mathbf{n}\cdot\vec{\sigma}$ is the identity and every odd power is $\mathbf{n}\cdot\vec{\sigma}$ itself, so the even terms sum to a cosine and the odd ones to a sine.

Proof steps

(\mathbf{n}\cdot\vec{\sigma})^{2}=|\mathbf{n}|^{2}=1

The Pauli matrices anticommute and square to one.

e^{-i\frac{\theta}{2}\mathbf{n}\cdot\vec{\sigma}}=\sum_{k}\frac{1}{k!}\Bigl(-i\frac{\theta}{2}\Bigr)^{k}(\mathbf{n}\cdot\vec{\sigma})^{k}

The exponential series.

(\mathbf{n}\cdot\vec{\sigma})^{2k}=1,\qquad (\mathbf{n}\cdot\vec{\sigma})^{2k+1}=\mathbf{n}\cdot\vec{\sigma}

Even and odd powers.

=\cos\frac{\theta}{2}-i\,\mathbf{n}\cdot\vec{\sigma}\,\sin\frac{\theta}{2}

The even terms are the series of the cosine, the odd ones of the sine.

\theta=2\pi\ \Rightarrow\ \cos\pi-i\,\mathbf{n}\cdot\vec{\sigma}\sin\pi=-1

A full turn is minus the identity.

Applications

Practice

Minus One After a Turn

A spin one-half rotated by a full turn of $2\pi$ about any axis comes back multiplied by $-1$ . Only a second full turn restores it.

\hat{R}_{\mathbf{n}}(2\pi)=-1

Try it

What number multiplies a spin- $\tfrac12$ state after a rotation by $2\pi$ about any axis?