0
$\begingroup$

$$\left(\vec\sigma \cdot \vec{p} \right)^2=\left(\vec\sigma \cdot \vec{p}\right) \left(\vec\sigma \cdot \vec{p} \right)=\vec{p} \cdot \vec{p}+\mathrm{i}\left(\vec\sigma \cdot \left[ \vec{p} \times \vec{p} \right] \right)=p^2$$ Where does this $\left(\vec\sigma \cdot \left[ \vec{p} \times \vec{p} \right] \right)$ come from? Cause isn't $\sigma^2=\mathbb{I}$? It will really be a great help if someone can point me in the right direction.

It does not explain anything from here.

$\endgroup$
3
$\begingroup$

In general : $$ \left(\boldsymbol{\sigma}\boldsymbol{\cdot}\mathbf{a}\right)\left(\boldsymbol{\sigma}\boldsymbol{\cdot}\mathbf{b}\right)= \left(\mathbf{a}\boldsymbol{\cdot}\mathbf{b}\right)\mathrm{I}+i\left[\boldsymbol{\sigma}\boldsymbol{\cdot}\left(\mathbf{a}\boldsymbol{\times}\mathbf{b}\right)\right] \tag{01} $$ since

\begin{align} \left(\boldsymbol{\sigma}\boldsymbol{\cdot}\mathbf{a}\right)\left(\boldsymbol{\sigma}\boldsymbol{\cdot}\mathbf{b}\right)&= \left(a_{1}\sigma_{1}+a_{2}\sigma_{2}+a_{3}\sigma_{3}\right) \left(b_{1}\sigma_{1}+b_{2}\sigma_{2}+b_{3}\sigma_{3}\right)\\ & = a_{1}b_{1}\sigma_{1}^{2}+a_{2}b_{2}\sigma_{2}^{2}+a_{3}b_{3}\sigma_{3}^{2}+\\ & \quad \:\: \left(a_{2}b_{3}-a_{3}b_{2}\right)\sigma_{2}\sigma_{3}+\left(a_{3}b_{1}-a_{1}b_{3}\right)\sigma_{3}\sigma_{1}+ \left(a_{1}b_{2}-a_{2}b_{1}\right)\sigma_{1}\sigma_{2}\\ & =\underbrace{\left(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}\right)\mathrm{I}}_{\sigma_{1}^{2}\boldsymbol{=}\sigma_{2}^{2}\boldsymbol{=}\sigma_{3}^{2}\boldsymbol{=}\mathrm{I}}+\\ &\quad \underbrace{i\Biggl(\begin{vmatrix}a_{2}&a_{3}\\b_{2}&b_{3}\end{vmatrix} \sigma_{1} + \begin{vmatrix}a_{3}&a_{1}\\b_{3}&b_{1}\end{vmatrix}\sigma_{2}+ \begin{vmatrix}a_{1}&a_{2}\\b_{1}&b_{2}\end{vmatrix}\sigma_{3}\Biggr)}_{ \sigma_{2}\sigma_{3}\boldsymbol{=}i\sigma_{1}\boldsymbol{=}\boldsymbol{-}\sigma_{3}\sigma_{2}\:,\: \sigma_{3}\sigma_{1}\boldsymbol{=}i\sigma_{2}\boldsymbol{=}\boldsymbol{-}\sigma_{1}\sigma_{3}\:,\: \sigma_{1}\sigma_{2} \boldsymbol{=}i\sigma_{3}\boldsymbol{=}\boldsymbol{-}\sigma_{2}\sigma_{1}}\\ &=\left(\mathbf{a}\boldsymbol{\cdot}\mathbf{b}\right)\mathrm{I}+i\left[\boldsymbol{\sigma}\boldsymbol{\cdot}\left(\mathbf{a}\boldsymbol{\times}\mathbf{b}\right)\right] \tag{02} \end{align}


Now, equation (01) has an interpretation in case that $\:\mathbf{a},\mathbf{b}\:$ are unit vectors. So let the identity (01) with unit vectors \begin{equation} \left(\boldsymbol{\sigma}\boldsymbol{\cdot}\mathbf{n}_{2}\right)\left(\boldsymbol{\sigma}\boldsymbol{\cdot}\mathbf{n}_{1}\right)= \left(\mathbf{n}_{1}\boldsymbol{\cdot}\mathbf{n}_{2}\right)\mathrm{I}-i\left[\boldsymbol{\sigma}\boldsymbol{\cdot}\left(\mathbf{n}_{1}\boldsymbol{\times}\mathbf{n}_{2}\right)\right], \quad \text{where} \:\: \Vert\mathbf{n}_{1}\Vert=1=\Vert \mathbf{n}_{2}\Vert \tag{03} \end{equation} If the angle between $\:\mathbf{n}_{1},\mathbf{n}_{2}\:$ is $\:\phi\:$ and $\:\mathbf{n}\:$ the unit vector normal to the plane of $\:\mathbf{n}_{1},\mathbf{n}_{2}\:$ \begin{align} \cos\phi & = \mathbf{n}_{1}\boldsymbol{\cdot}\mathbf{n}_{2} \tag{04a}\\ \mathbf{n} & = \dfrac{\mathbf{n}_{1}\boldsymbol{\times}\mathbf{n}_{2}}{\Vert\mathbf{n}_{1}\boldsymbol{\times}\mathbf{n}_{2}\Vert}= \dfrac{\mathbf{n}_{1}\boldsymbol{\times}\mathbf{n}_{2}}{\sin\phi} \tag{04b} \end{align} then the rhs of equation (03) is expressed as \begin{equation} \mathrm{Q}= \left(\mathbf{n}_{1}\boldsymbol{\cdot}\mathbf{n}_{2}\right)\mathrm{I}-i\left[\boldsymbol{\sigma}\boldsymbol{\cdot}\left(\mathbf{n}_{1}\boldsymbol{\times}\mathbf{n}_{2}\right)\right]=\cos\left(\dfrac{2\phi}{2}\right)-i\sin\left(\dfrac{2\phi}{2}\right)\left(\boldsymbol{\sigma}\boldsymbol{\cdot}\mathbf{n}\right) \tag{05} \end{equation} that is a special unitary matrix $\:\mathrm{Q} \in \mathrm{SU(2)}\:$ or a unit quaternion, representation of a rotation around the axis $\:\mathbf{n}\:$ through an angle $\:\theta=2\phi$. Note that the matrix $\:-\mathrm{Q} \in \mathrm{SU(2)}\:$ as expressed by \begin{equation} -\mathrm{Q}= \cos\left(\dfrac{2\pi+2\phi}{2}\right)-i\sin\left(\dfrac{2\pi+2\phi}{2}\right)\left(\boldsymbol{\sigma}\boldsymbol{\cdot}\mathbf{n}\right) \tag{06} \end{equation} represents a rotation through $\:\theta'=2\pi+2\phi$, that is the same rotation as $\:+\mathrm{Q}\:$ does.

Now, the special unitary matrix \begin{equation} \mathrm{R_\jmath}=-i\left(\boldsymbol{\sigma}\boldsymbol{\cdot}\mathbf{n_\jmath}\right) =\cos\left(\dfrac{\pi}{2}\right)-i\sin\left(\dfrac{\pi}{2}\right)\left(\boldsymbol{\sigma}\boldsymbol{\cdot}\mathbf{n_\jmath}\right) \tag{07} \end{equation} represents a rotation around the axis $\:\mathbf{n_\jmath}\:$ through an angle $\:\pi$, that is a reflection through the axis $\:\mathbf{n_\jmath}$.

So equation (03) is written as

\begin{equation} \bigl[-i\left(\boldsymbol{\sigma}\boldsymbol{\cdot}\mathbf{n}_{2}\right)\bigr]\bigl[-i\left(\boldsymbol{\sigma}\boldsymbol{\cdot}\mathbf{n}_{1}\right)\bigr]= -\biggl[\left(\mathbf{n}_{1}\boldsymbol{\cdot}\mathbf{n}_{2}\right)\mathrm{I}-i\left[\boldsymbol{\sigma}\boldsymbol{\cdot}\left(\mathbf{n}_{1}\boldsymbol{\times}\mathbf{n}_{2}\right)\right]\biggr] \tag{08} \end{equation} or \begin{equation} \mathrm{R_2}\mathrm{R_1}=-\mathrm{Q} \tag{09} \end{equation} meaning that a reflection through an axis $\:\mathbf{n}_1\:$ followed by a reflection through a second axis $\:\mathbf{n}_2\:$ is a rotation around $\:\mathbf{n}_{1}\boldsymbol{\times}\mathbf{n}_{2}\:$ by an angle $\:2\phi\:$ where $\:\phi\:$ the angle between $\:\mathbf{n}_{1},\mathbf{n}_{2}$ as shown in the Figure below.

enter image description here

$\endgroup$
1
$\begingroup$

It is because $\sigma_i \sigma_j=\delta_{ij}+\mathrm{i} \epsilon_{ijk} \sigma_{k}$. Look at this article on Pauli matrices.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.