Monday 22 October 2018

homework and exercises - Need help with solution of the Dirac equation


$$\left(\vec\sigma \cdot \vec{p} \right)^2=\left(\vec\sigma \cdot \vec{p}\right) \left(\vec\sigma \cdot \vec{p} \right)=\vec{p} \cdot \vec{p}+\mathrm{i}\left(\vec\sigma \cdot \left[ \vec{p} \times \vec{p} \right] \right)=p^2$$ 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.



Answer



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


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...