Spin rotation operator and unitarity

Keywords: spin 1/2 · Pauli · rotation · unitary operator · operator exponential

Exercise 1 : Spin rotation operator and unitarity

Question 1

Consider the space

H = \mathbb{C}^2

(spin

1/2

) with the basis

\{|\!\uparrow\rangle, |\!\downarrow\rangle\}

. The Pauli matrices are:

\begin{aligned} \sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}. \end{aligned}

We define the operator for a rotation about the $z$ -axis through an angle $\theta$ by:

U(\theta) = e^{-i\theta\sigma_z/2}.

Question 2

Using

\sigma_z^2 = \mathbf{1}

, show that:

\begin{aligned} U(\theta) = \cos\tfrac{\theta}{2}\,\mathbf{1} - i\sin\tfrac{\theta}{2}\,\sigma_z = \begin{pmatrix} e^{-i\theta/2} & 0 \\ 0 & e^{i\theta/2} \end{pmatrix}. \end{aligned}

Solution

We expand the exponential as a power series and separate the even and odd powers, using

\sigma_z^{2k} = \mathbf{1}

and

\sigma_z^{2k+1} = \sigma_z

e^{-i\theta\sigma_z/2} = \sum_{k=0}^\infty \frac{(-i\theta/2)^{2k}}{(2k)!}\,\mathbf{1} + \sum_{k=0}^\infty \frac{(-i\theta/2)^{2k+1}}{(2k+1)!}\,\sigma_z = \cos\tfrac{\theta}{2}\,\mathbf{1} - i\sin\tfrac{\theta}{2}\,\sigma_z.

Substituting $\sigma_z = \mathrm{diag}(1,-1)$ gives the stated diagonal matrix.

Question 3

Verify that

U(\theta)

is unitary:

U^\dagger(\theta)\,U(\theta) = \mathbf{1}

Solution

U^\dagger(\theta) = \mathrm{diag}(e^{i\theta/2}, e^{-i\theta/2})

, and therefore:

\begin{aligned} U^\dagger(\theta)\,U(\theta) = \begin{pmatrix} e^{i\theta/2} & 0 \\ 0 & e^{-i\theta/2} \end{pmatrix} \begin{pmatrix} e^{-i\theta/2} & 0 \\ 0 & e^{i\theta/2} \end{pmatrix} = \mathbf{1}. \end{aligned}

Question 4

Compute the operator

U(\theta)\,\sigma_x\,U^\dagger(\theta)

and interpret the result as a rotation in the

(x,y)

plane.

Solution

We compute:

\begin{aligned} U(\theta)\,\sigma_x\,U^\dagger(\theta) = \begin{pmatrix} e^{-i\theta/2} & 0 \\ 0 & e^{i\theta/2} \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} e^{i\theta/2} & 0 \\ 0 & e^{-i\theta/2} \end{pmatrix} = \begin{pmatrix} 0 & e^{-i\theta} \\ e^{i\theta} & 0 \end{pmatrix}. \end{aligned}

Decomposing this matrix gives:

\begin{aligned} \begin{pmatrix} 0 & e^{-i\theta} \\ e^{i\theta} & 0 \end{pmatrix} = \cos\theta\,\sigma_x + \sin\theta\,\sigma_y. \end{aligned}

This is indeed a rotation of $\sigma_x$ towards $\sigma_y$ through an angle $\theta$ in the $(x,y)$ plane of the space of observables, in accordance with the adjoint formula $U A U^\dagger$ for unitary rotations.