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My job is to prove: $$\exp(i\theta \vec{v} \cdot \vec{ \sigma })=\cos(\theta)I+i\sin(\theta)\vec{v} \cdot \vec{ \sigma }$$

where $\theta \in \mathbb{R}$ and $\vec{v} \cdot \vec{ \sigma }=\Sigma^3_{i=1}v_i\sigma_i$ such that $\sigma_i$ are the Pauli matrices, and $\vec{v}$ is a three dimensional real vector.

My attempt:

$$\vec{v} \cdot \vec{ \sigma }=v_1 \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} +v_2 \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} +v_3 \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} v_3 & v_1-iv_2 \\ v_1+iv_2 & -v_3 \end{bmatrix} \tag{1} $$

This is an Hermitian matrix as it is the sum of 3 Hermitian matrices because $v_1,v_2,v_3 \in \mathbb{R}$. Here's my plan: as $\vec{v} \cdot \vec{ \sigma }$ is Hermitian, then it is also diagonalizable, which by the spectral decomposition:

$$\vec{v} \cdot \vec{ \sigma }=\Sigma_i\lambda_i|i\rangle\langle i|\\ \exp(i\theta \vec{v} \cdot \vec{ \sigma })=\Sigma_i\exp(i\theta\lambda_i)|i\rangle\langle i|$$

where $\lambda_i$ are the eigenvalues of $\vec{v} \cdot \vec{ \sigma }$ and $|i\rangle\langle i|$ the outer product of its eigenvectors with themselves. Then work out from there. But from $(1)$ I get to:

$$ \exp(i\theta \vec{v} \cdot \vec{ \sigma })=e^{i\theta||\vec{v}||} \begin{bmatrix} \frac{iv_2-v_1}{v_3-||\vec{v}||} \\ 1 \end{bmatrix} \begin{bmatrix} \frac{iv_2-v_1}{v_3-||\vec{v}||} & 1 \end{bmatrix} +e^{-i\theta||\vec{v}||} \begin{bmatrix} \frac{iv_2-v_1}{v_3+||\vec{v}||} \\ 1 \end{bmatrix} \begin{bmatrix} \frac{iv_2-v_1}{v_3+||\vec{v}||} & 1 \end{bmatrix} $$

But this seems a bit excessive, and I don't know where I am mistaken, any help is appreciated.

Notes:

  1. This is the exercise 2.35 from Quantum Computation and Quantum Information by Nielsen and Chuang;
  2. I have seen this solution but I was unable to follow their reasoning. Also, could someone tell me which fundamentals I am missing in order to understand this solution?
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closed as off-topic by ZeroTheHero, G. Smith, Cosmas Zachos, Jon Custer, Kyle Kanos Feb 1 at 11:08

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  • $\begingroup$ duplicate of physics.stackexchange.com/questions/41697/… $\endgroup$ – ZeroTheHero Jan 30 at 20:58
  • $\begingroup$ It is much simpler than that (assuming the space is flat Minkowskian). You basically just need to show that: $$(\overrightarrow{v}\cdot\overrightarrow{\sigma})^{2n}=I$$ Where n is an integer and I the identity. $\endgroup$ – R. Rankin Jan 30 at 21:43
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    $\begingroup$ Taylor series is the fundamental concept that you’re missing. $\endgroup$ – G. Smith Jan 30 at 21:50
  • $\begingroup$ @G.Smith I know the Taylor series but only for numbers, one dimensional objects, I wasn't aware that it expanded for matrix functions. Guess I'll have to check that $\endgroup$ – Bidon Jan 30 at 21:54
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    $\begingroup$ WP. $\endgroup$ – Cosmas Zachos Jan 31 at 0:55
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If I had read the exercise carefully I would have noticed that $||\vec{v}||=1$, thus resulting in the eigenvalues $\lambda_\pm=\pm1$. Now the problem reduces to:

$$\exp(i\theta\vec{v} \cdot\vec{\sigma})=e^{i\theta}|\lambda_+\rangle\langle\lambda_+|+e^{-i\theta}|\lambda_-\rangle\langle\lambda_-|\\ =\cos(\theta)(|\lambda_+\rangle\langle\lambda_+|+|\lambda_-\rangle\langle\lambda_-|)+i\sin(\theta)((|\lambda_+\rangle\langle\lambda_+|-|\lambda_-\rangle\langle\lambda_-|)$$

Now, as $I$ is an hermitian operator then it is also diagonalizable. $I$ has only $1$ eigenvalue($\lambda=1$), and $|\lambda_+\rangle,|\lambda_-\rangle$ are orthogonal, then: $$I=|\lambda_+\rangle\langle\lambda_+|+|\lambda_-\rangle\langle\lambda_-|$$

As we have seen $\vec{v} \cdot\vec{\sigma}$ is Hermitian and has eigenvalues $\lambda_\pm=\pm1$, then:

$$\vec{v} \cdot\vec{\sigma}=|\lambda_+\rangle\langle\lambda_+|-|\lambda_-\rangle\langle\lambda_-|$$

Therefore:

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

Credits to: goropikari

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