Exponential of Pauli Matrices

Question

Let $\vec{v}$ be any real three-dimensional unit vector and $\theta$ a real number. Prove that $\exp(i\theta \vec{v}\cdot\vec{\sigma}) = \cos(\theta)I + i\sin(\theta)\vec{v}\cdot\vec{\sigma}$, where $\vec{v}\cdot\vec{\sigma} \equiv \sum_{k=1}^{3}v_{k}\sigma_{k}$.

$\vec{v}\cdot\vec{\sigma}$ is a scalar so let this scalar be a non-zero integer $n$.

Effectively, this gives $\exp(i\theta n)$.

$\exp(i\theta n) \Rightarrow (\cos(\theta) + i \sin(\theta))^{n}$

I would appreciates if anyone could provide to me a hint to this. The computation becomes messy very quickly.

Thanks in advance.

Answer 1

Here's an option

Commutation rules

I'm not aware of a fast way of proving this, other than actually calculate all the products directly, but in any case it is fairly straightforward to show that ([ab]using Einstein's notation)

\begin{eqnarray} [\sigma_a, \sigma_b] &=& 2i\epsilon_{abc}\sigma_c \\ \{\sigma_a, \sigma_b \} &=& 2 \delta_{ab}I \tag{1} \end{eqnarray}

From here note that

\begin{eqnarray} [\sigma_a,\sigma_b] +\{\sigma_a, \sigma_b \} &=& (\sigma_a\sigma_b - \sigma_b\sigma_a) + (\sigma_a \sigma_b + \sigma_a \sigma_b) \\ 2i\epsilon_{abc}\sigma_c + 2 \delta_{ab}I &=& 2\sigma_a \sigma_b \\ \sigma_a \sigma_b &=& \delta_{ab}I + i\epsilon_{abc}\sigma_c \tag{2} \end{eqnarray}

Vector product

Now consider two vectors ${\bf u}$ and ${\bf v}$ and multiply both sides of Eqn. (2) by their components, contracting indices leads to

\begin{eqnarray} (u_a v_b) \sigma_a \sigma_b &=& (u_a v_b) \delta_{ab} I + i(u_a v_b) \epsilon_{abc} \sigma_c \\ (u_a \sigma_a) (v_b \sigma_b) &=& u_a v_a I + i (\epsilon_{abc} u_a v_b) \sigma_c \\ ({\bf u}\cdot \mathbf{\sigma}) ({\bf v}\cdot \mathbf{\sigma}) &=& ({\bf u}\cdot {\bf v})I + i ({\bf u}\times {\bf v})_c \sigma_c \\ ({\bf u}\cdot \mathbf{\sigma}) ({\bf v}\cdot \mathbf{\sigma})&=& ({\bf u}\cdot {\bf v})I + i ({\bf u}\times {\bf v}) \cdot \sigma \tag{3} \end{eqnarray}

Set ${\bf u} = {\bf v} = \hat{\bf n}$ in the last equation, you will get

$$ ({\bf u}\cdot \mathbf{\sigma})^2 = I \tag{4} $$

And from this it is trivial to see

\begin{eqnarray} ({\bf u}\cdot \mathbf{\sigma})^{2k} &=& I \\ ({\bf u}\cdot \mathbf{\sigma})^{2k + 1} &=& ({\bf u}\cdot \mathbf{\sigma}) ~~~\mbox{for}~~~ k = 0,1,\cdots \tag{5} \end{eqnarray}

Putting everything together

\begin{eqnarray} e^{i\theta \hat{\bf u}\cdot \sigma} &=& \sum_{k = 0}^{+\infty} \frac{(i\theta)^k}{k!}(\hat{\bf u}\cdot \sigma)^k \\ &=& \sum_{k = 0}^{+\infty} \frac{i^{2k}\theta^{2k}}{(2k)!}(\hat{\bf u}\cdot \sigma)^{2k} + \sum_{k = 0}^{+\infty} \frac{i^{2k+1}\theta^{2k+1}}{(2k+1)!}(\hat{\bf u}\cdot \sigma)^{2k+1} \\ &\stackrel{(5)}{=}& \sum_{k = 0}^{+\infty} \frac{(-1)^k\theta^{2k}}{(2k)!}I + \sum_{k = 0}^{+\infty} \frac{i(-1)^k\theta^{2k + 1}}{(2k + 1)!}(\hat{\bf u}\cdot \sigma) \\ &=& \cos \theta I + i(\hat{\bf u}\cdot \sigma) \sin \theta \end{eqnarray}

Answer 2

The title hints at a crucial bit of missing information: the definition of the Pauli matrices, $\vec\sigma$. The most common representation is

\begin{align} \sigma_1 &= \left(\begin{array}{rr} 0&1\\1&0 \end{array}\right) & \sigma_2 &= \left(\begin{array}{rr} 0&i\\-i&0 \end{array}\right) & \sigma_3 &= \left(\begin{array}{rr} 1&0\\0&-1 \end{array}\right) & \end{align}

but the important parts of the definition are the cyclic product $\sigma_1 \sigma_2 = i\sigma_3$ (and permutations) and $\sigma_i\sigma_i = I$. These are equivalent to the commutation relationships in caverac's answer.

So your statement that

$\vec v\cdot \vec\sigma$ is a scalar

isn't correct. In this representation, $\vec v\cdot\vec\sigma$ would be the $2\times2$ matrix,

$$ \vec v\cdot\vec\sigma = \left(\begin{array}{cc} v_3 & v_1 + i v_2 \\ v_1 - i v_2 & -v_3 \end{array}\right) $$

What does it mean to expenentiate a matrix? Well, the exponential for real numbers is equivalent to a power series:

$$ \exp x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \sum_n \frac{x^n}{n!} $$

That power series is a set of instructions that we can totally apply to a matrix. Let's see what happens if we try it with $\sigma_1$:

\begin{align} \exp i\theta\sigma_1 &= I + i\theta\sigma_1 - \frac{\theta^2 I}{2!} - i\frac{\theta^3 \sigma_1}{3!} + \cdots \\ &= I \times \left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} + \cdots \right) + i\sigma_1 \times \left( \theta - \frac{\theta^3}{3!} + \cdots \right) \\&= I\cos\theta + i\sigma_1\sin\theta = \left(\begin{array}{cc} \cos\theta & i\sin\theta \\ i\sin\theta & \cos\theta \end{array}\right) \end{align}

This shows you the shape of the result to expect as you complete the proof for general $\vec v\cdot\vec\sigma$.

Answer 3

Pauli matrices:

\begin{equation*} \sigma_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \quad \sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \quad \sigma_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \end{equation*}

dot product $\overrightarrow{v}\cdot\overrightarrow{\sigma}$:

\begin{equation*} \overrightarrow{v}\cdot\overrightarrow{\sigma} = v_1\sigma_1 + v_2\sigma_2 + v_3\sigma_3 = \begin{pmatrix} v_3 & v_1 - iv_2 \\ v_1 + iv_2 & -v_3 \end{pmatrix} \end{equation*}

$\overrightarrow{v}\cdot\overrightarrow{\sigma}$ eigenvalues:

\begin{equation*} \begin{vmatrix} v_3 - \lambda & v_1 - iv_2 \\ v_1 + iv_2 & -v_3 - \lambda \end{vmatrix} = 0 \quad \Rightarrow \quad v_3^2 - \lambda^2 + v_1^2 + v_2^2 = 0 \quad \Rightarrow \quad \lambda_1 = 1, \lambda_2 = -1 \end{equation*}

Let $\overrightarrow{v}\cdot\overrightarrow{\sigma} = A$. We can show that $AA^\dagger = A^\dagger A$, which implies that $A$ is normal. Therefore, $A$ is diagonalizable. Let $|\varphi\rangle$ and $|\psi\rangle$ be the associated orthonormal eigenvectors. The diagonal representation of $\overrightarrow{v}\cdot\overrightarrow{\sigma}$ is $|\varphi\rangle\langle\varphi| - |\psi\rangle\langle\psi|$.

Therfore, we have:

\begin{align*} e^{i\theta \overrightarrow{v}\cdot\overrightarrow{\sigma}} & = e^{i\theta}|\varphi\rangle\langle\varphi| + e^{-i\theta}|\psi\rangle\langle\psi| \\ & = \cos(\theta)|\varphi\rangle\langle\varphi| + i\sin(\theta)|\varphi\rangle\langle\varphi| + \cos(\theta)|\psi\rangle\langle\psi| - i\sin(\theta)|\psi\rangle\langle\psi| \\ & = \cos(\theta)(|\varphi\rangle\langle\varphi| + |\psi\rangle\langle\psi|) + i\sin(\theta)(|\varphi\rangle\langle\varphi| - |\psi\rangle\langle\psi|) \\ & = \cos(\theta)I + i\sin(\theta)\overrightarrow{v}\cdot\overrightarrow{\sigma} \end{align*}

Reference: Nielsen, Michael A., and Isaac L. Chuang. Quantum computation and quantum information. Cambridge university press, 2010. pp. 61-71.

Answer 4

Here an other an more general answer. As I am just a student I hope that this is correct.

1- First of all let's note that $ \hat{a} \cdot \vec{ \sigma } = \left(\begin{array}{cc} a_3 & a_1 + i a_2 \\ a_1 - i a_2 & -a_3 \end{array}\right) $ is an Hermitian matrix ( $ (\hat{a} \cdot \vec{ \sigma }) ^ H = \hat{a} \cdot \vec{ \sigma } $ )

2- As $\hat{a} \cdot \vec{ \sigma }$ is hermitian is means that $ \hat{a} \cdot \vec{ \sigma } $ can be decomposed as follow $\hat{a} \cdot \vec{ \sigma } = \lambda_+ P_+ + \lambda_- P_-$.
The two eigen values of $\hat{a} \cdot \vec{ \sigma }$ are $+ |\hat{a}|, -|\hat{a}| $ and that $P_{\pm} = \frac{1 \pm \hat{a} \cdot \vec{ \sigma }}{2} $ see here for justifications. hence here the eigen value are $\pm 1$.

3- More over as $\hat{a} \cdot \vec{ \sigma }$ is hermitian $f(\hat{a} \cdot \vec{ \sigma })= \sum f(eigen \; value) \cdot corresponding \; projector $ it causes that $e^{-i \omega t \hat{a} \vec{ \sigma } }= e^{-i \omega t} \frac{1 + \hat{a} \cdot \vec{ \sigma }}{2} + e^{i \omega t} \frac{1 - \hat{a} \cdot \vec{ \sigma }}{2} = \frac{e^{i \omega t} + e^{-i \omega t} }{2}I - \hat{a} \cdot \vec{ \sigma } \frac{e^{i \omega t} - e^{-i \omega t} }{2}=cos(\omega t)I-i \cdot sin(\omega t) \hat{a} \cdot \vec{ \sigma }$.