Matrix operation in dirac matrices

Question

If we define $\alpha_i$ and $\beta$ as Dirac matrices which satisfy all of the conditions of spin 1/2 particles , p defines the momentum of the particle, then how can we get the matrix form ? \begin{equation} \alpha_i p_i= \begin{pmatrix} p_z & p_x-ip_y \\ p_x+ip_y & -p_z \end{pmatrix} . \end{equation}

Answer 1

The equation you wrote only makes one choice that should answer all questions about this context: it chooses a representation of the $\alpha_i$ matrices with $$ \alpha_i = \sigma_i $$ where $\sigma_i$ are the three Pauli matrices. You may check that if you substitute the Pauli matrices (particular $2\times 2$ matrices listed in the Wikipedia article linked in the previous sentence) for $\alpha_i$ on the left hand side of your equation, you obtain the right hand side.

If your formula had the Greek letter $\sigma$ instead of $\alpha$ on the left hand side, it would be uncontroversial. However, with $\alpha$, it is problematic. The $\alpha_i$ matrices are really $4\times 4$, not $2\times 2$, so all the equations above must be interpreted so that each matrix entry of the Pauli matrices is actually a block $$ z \to \pmatrix {z&0 \\ 0&-z }. $$ We say that the Pauli matrices were tensor-multiplied by a $2\times 2$ unit matrix (in certain order). This extra tensor factor actually can't be the unit matrix because one couldn't find any matrix $\beta$ that anticommutes with all the $\alpha_i$ matrices. But it may be another $\sigma_z$, for example, in which case $\beta$ may be chosen to be ${\rm diag}(\sigma_x,\sigma_x)$, for example. Alternatively, you should ignore the source and learn some/all of the standard representations of the Dirac matrices.

At any rate, something is sloppy about the notation in which $\alpha_i$ were written as $2\times 2$ matrices and the simplest recipe to get $4\times 4$ matrices (tensor product with the $2\times 2$ unit matrix) doesn't work. So one should first see what $4\times 4$ matrices your source (if it is correct at all) actually means.

Answer 2

It's just a matrix manipulation. Let $\sigma_i$ pauli matrices. \begin{equation} \alpha_i p_i= \begin{pmatrix} 0& \sigma_i \\ \sigma_i & 0 \end{pmatrix} p_i . \end{equation} $ \alpha_i p_i= \begin{pmatrix} 0& p_1 \sigma_1 \\ p_1\sigma_1 & 0 \end{pmatrix} + \begin{pmatrix} 0& p_2 \sigma_2 \\ p_i\sigma_2 & 0 \end{pmatrix} + \begin{pmatrix} 0& p_3 \sigma_3 \\ p_3\sigma_3 & 0 \end{pmatrix} $

But $ \sigma_1 p_1 = \begin{pmatrix} 0& 1 \\\ 1 & 0 \end{pmatrix}p_1=\begin{pmatrix} 0& p_1 \\\ p_1 & 0 \end{pmatrix}$ ,

$ \sigma_2 p_2= \begin{pmatrix} 0& -i \\\ -i & 0 \end{pmatrix}p_2=\begin{pmatrix} 0& -ip_2 \\\ ip_2 & 0 \end{pmatrix}$

$ \sigma_3 p_3= \begin{pmatrix} 1& 0 \\\ 0 & -1 \end{pmatrix}p_3= \begin{pmatrix} p_3& 0 \\\ 0 & -p_3 \end{pmatrix}$

Now adding these we get ($1\rightarrow x $, $2\rightarrow y $ ,$3\rightarrow z $) , \begin{equation} \alpha_i p_i= \begin{pmatrix} p_z & p_x-ip_y \\ p_x+ip_y & -p_z \end{pmatrix} . \end{equation}