# Worked Problems in Physics

## Nielsen and Chuang Exercise 2.35

Posted in Nielsen/Chuang by rpmuller on March 10, 2010

Exercise 2.35 Exponential of the Pauli matrices: Let ${\vec{v}}$ be any real three-dimensional unit vector, and ${\theta}$ a real number. Prove that

$\displaystyle \exp(i\theta\vec{v}\cdot\vec{\sigma})=\cos(\theta)I+i\sin(\theta)\vec{v}\cdot\vec{\sigma}$

• The wrong way: Here’s my first wild goose chase, which looks good, and gets the flavor of the even/odd powers, which is key to the solution, but which I couldn’t get to complete:

We will break the integral into ${x}$, ${y}$, and ${z}$ components, and note that the eigenvalues of the Pauli matrices are (-1,1). Each component is then given by:

$\displaystyle E_x = \exp(i\theta v_x\sigma_x)=U_x \left[\begin{array}{rr} \exp(i\theta v_x) & 0\\ 0 & \exp(-i\theta v_x)\end{array}\right]U_x^\dagger$

with analogous expressions for ${y}$ and ${z}$. We can use the standard expansion of the exponent

$\displaystyle E_x = U_x \left[\begin{array}{rr} \cos(\theta v_x) & 0\\ 0 & \cos(-\theta v_x)\end{array}\right]U_x^\dagger +iU_x \left[\begin{array}{rr} \sin(\theta v_x) & 0\\ 0 & \sin(-\theta v_x)\end{array}\right]U_x^\dagger$

We note that the cosine is an even function, and the sine is an odd function, so that this term reduces to

$\displaystyle E_x = U_x \left[\begin{array}{rr} \cos(\theta v_x) & 0\\ 0 & \cos(\theta v_x)\end{array}\right]U_x^\dagger +iU_x \left[\begin{array}{rr} \sin(\theta v_x) & 0\\ 0 & -\sin(\theta v_x)\end{array}\right]U_x^\dagger$

$\displaystyle = \cos(\theta v_x) + i\sin(\theta v_x)\sigma_x$

• The right way: In desperation, I found the web page http://en.wikipedia.org/wiki/Pauli_matrices#Commutation_relations. Here’s a sketch of their proof.

First, note that because

$\displaystyle (\vec a\cdot\vec\sigma) = \sum_{ij}a_i\sigma_ib_j\sigma_j=\sum_{ij}a_ib_j\delta_{ij}I+i\vec\sigma\cdot(\vec a\times\vec b)$

we can show that

$\displaystyle (\vec v\cdot\vec\sigma)^{2n} = I$

and

$\displaystyle (\vec v\cdot\vec\sigma)^{2n+1} = \vec v\cdot\vec\sigma.$

Plugging this into the Taylor expansion for ${e^{ix}}$ (which we will break into even and odd powers) gives

$\displaystyle \exp(i\theta\vec v\cdot\vec\sigma)=\sum\frac{(-1)^n(\theta\vec v\cdot\vec\sigma)^{2n}}{(2n)!}+i\sum\frac{(-1)^n(\theta\vec v\cdot\vec\sigma)^{2n+1}}{(2n+1)!}$

$\displaystyle =\sum\frac{(-1)^n\theta^{2n}}{(2n)!}+i(\vec v\cdot\vec\sigma)\sum\frac{(-1)^n\theta^{2n+1}}{(2n+1)!}$

$\displaystyle =\cos(\theta)+i\sin(\theta)(\vec v\cdot\vec\sigma)$

If anyone has any suggestions on how to finish my original proof, let me know.