# Matrix operation in dirac matrices

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 ? $$\alpha_i p_i= \begin{pmatrix} p_z & p_x-ip_y \\ p_x+ip_y & -p_z \end{pmatrix} .$$

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Dirac matrices in 4 dimensions are 4x4. You've written a 2x2 matrix. Where did you find this equation? – Michael Brown Jan 1 '13 at 4:05
Dirac matrices can be written as $2*2$ block form . check this link .nyu.edu/classes/tuckerman/quant.mech/lectures/lecture_7/… – Unlimited Dreamer Jan 1 '13 at 6:40
I see... you're breaking the Dirac equation down to 2x2 blocks. This is the standard way of solving it. Where in the argument are you having trouble? Unfortunately the equations in your source aren't numbered, but I can see you mean $\vec{\sigma}\cdot\vec{p}$ rather than $\alpha_i p_i$. – Michael Brown Jan 1 '13 at 6:47

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.

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

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

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Your final formula is 4x4 on the left and 2x2 on the right. You should have a 4x4 matrix with what you've written as the upper right and lower left blocks, with zeros everywhere else. – Michael Brown Jan 2 '13 at 23:42
In the right $2\times 2$ makes the matrix $4 \times 4$ – Unlimited Dreamer Jan 3 '13 at 5:19
No, the answer you want is $\alpha_i p_i = \left(\begin{matrix} 0 & p_i \sigma_i \\ p_i \sigma_i & 0 \end{matrix}\right)$ which has four 2x2 blocks. Your right hand side is the 2x2 matrix $p_i \sigma_i$, not $\alpha_i p_i$. You had it right until the very last line! – Michael Brown Jan 3 '13 at 5:27