Prove $\exp{i\frac{\pi}{2}(-1+\sigma_{i})}=\sigma_{i}$

Question

How do we prove

$e^{{i\frac{\pi}{2}(-1+\sigma_{i})}}=\sigma_{i}$ ?

where $\sigma_{i}:$Pauli matrix and $1=$ Identity matrix

Note: I understand that $i\frac{\pi}{2}(-1+\sigma_{i})$ is anti-hermitian since $(-1+\sigma_{i})$ is hermitian, hence the exponential of it is Unitary.

Answer 1

$e^{\hat{X}+\hat{Y}}=e^{\hat{X}}.e^{\hat{Y}}.e^{-\frac{1}{2}[\hat{X},\hat{Y}]}$ if $[[\hat{X},\hat{Y}],\hat{X}]=0$ and $[[\hat{X},\hat{Y}],\hat{Y}]=0$.

Since $1$ and $\sigma_{i}$ commute, i.e, $i\frac{\pi}{2}[-1,\sigma_{i}]=0$, $$ e^{i\frac{\pi}{2}(-1+\sigma_{i})}=e^{-i\frac{\pi}{2}+i\frac{\pi}{2}\sigma_{i}}=e^{-i\frac{\pi}{2}}.e^{i\frac{\pi}{2}\sigma_{i}} $$

$$ e^{-i\frac{\pi}{2}}=cos({\pi}/{2})-isin(\pi/2)=-i $$

note that $\sigma_{i}^{2}=1$, hence

$$e^{i\frac{\pi}{2}\sigma_{i}}=cos(\pi/2)+i\sigma_{i}.sin(\pi/2)=i.\sigma_{i}$$

Substituting the terms, $$ e^{i\frac{\pi}{2}(-1+\sigma_{i})}=(-i).(i.\sigma_{i})=\sigma_{i} $$

Answer 2

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\exp\pars{\ic\,{\pi \over 2}\,\bracks{-1 + \sigma_{k}}} = \sigma_{k}:\ ?}$

Note that $\ds{\exp\pars{\ic\,{\pi \over 2}\,\bracks{-1 + \sigma_{k}}} = -\ic\exp\pars{\ic\,{\pi \over 2}\,\sigma_{k}}}$. Lets consider $\ds{\mrm{f}\pars{\mu} \equiv \exp\pars{\ic\mu\sigma_{k}}}$. Then, \begin{align} \mrm{f}'\pars{\mu} & = \ic\sigma_{k}\,\mrm{f}\pars{\mu}\,,\qquad \left\{\begin{array}{rcl} \ds{\mrm{f}\pars{0}} & \ds{=} & \ds{\mathbf{1}\ \pars{~identity~}} \\[2mm] \ds{\mrm{f}'\pars{0}} & \ds{=} & \ds{\ic\sigma_{k}} \end{array}\right. \\[5mm] \mrm{f}''\pars{\mu} & = \ic\sigma_{k}\,\mrm{f}'\pars{\mu} = -\,\mrm{f}\pars{\mu} \quad\implies\quad\mrm{f}''\pars{\mu} + \mrm{f}\pars{\mu} = 0 \end{align}

---

Then, \begin{align} \mrm{f}\pars{\mu} & = A\sin\pars{\mu} + B\cos\pars{\mu}\quad\implies\quad \left\{\begin{array}{rcrcl} \ds{0\,A}& \ds{+} &\ds{B} & \ds{=} & \ds{\mathbf{1}} \\[2mm] \ds{A}& \ds{+} &\ds{0\,B} & \ds{=} & \ds{\ic\sigma_{k}} \end{array}\right. \\[5mm] \implies\,\mrm{f}\pars{\mu} & = \ic\sigma_{k}\sin\pars{\mu} + \mathbf{1}\cos\pars{\mu} \implies \color{#f00}{\exp\pars{\ic\,{\pi \over 2}\,\bracks{-1 + \sigma_{k}}}} = -\ic\,\mrm{f}\pars{\pi \over 2} = -\ic\pars{\ic\sigma_{k}} = \color{#f00}{\sigma_{k}} \end{align}