The representation matrix for alpha and beta in Dirac equation

Haorong Wu
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Homework Statement
Prove that in the presentation of ##\beta##,

##\mathbf \alpha=\begin{pmatrix}\mathbf 0 & \mathbf \sigma \\ \mathbf \sigma & \mathbf 0\end{pmatrix} ## and ## \beta=\begin{pmatrix}\mathbf I & \mathbf 0 \\ \mathbf0 & -\mathbf I\end{pmatrix} ,##

where ##\mathbf \alpha## and ##\beta## are in the Dirac equation, in which ##H=c \mathbf \alpha \cdot \mathbf p +\beta m c^2##.
Relevant Equations
1. ##\mathbf \alpha## and ##\beta## are Hermitian.
2. ##\{ \mathbf \alpha_i, \mathbf \alpha_j \}=0##, if ##i\ne j##.
3. ##\{ \mathbf \alpha_i, \beta\}=0##.
4. ##\alpha_i^2=\beta^2=1##.
5. The traces of ##\mathbf \alpha## and ##\beta## are zero.
6. The eigenvalues of them are ##1## or ##-1##.
In the 4-dimensional representation of ##\beta##, ## \beta=\begin{pmatrix}\mathbf I & \mathbf 0 \\ \mathbf0 & -\mathbf I\end{pmatrix} ,## and we can suppose ## \alpha_i=\begin{pmatrix}\mathbf A_i & \mathbf B_i \\ \mathbf C_i & \mathbf D_i\end{pmatrix} ##.

From the anti-commutation relation ##\{ \mathbf \alpha_i, \beta\}=0##, I can derive ##A_i=D_i=0##.

From ##\alpha_i^2=1##, I can have ##C_i=B_i^{-1}##. Furthermore, from the Hermiticity, I can have ##C_i=B_i^{\dagger}##.

But I could not find a way to prove that ##C_i=B_i##. The relation ##\{ \mathbf \alpha_i, \mathbf \alpha_j \}=0## for ##i\ne j## does not help.

Should ##\alpha_i## and ##\beta## always be symmetric? This is not given in the problem. Is there any other properties of ##\alpha_i##?

I have looked in some books. These matrices are just given directly without proof.
 
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Say
<br /> B=<br /> \begin{pmatrix}<br /> a &amp; b \\<br /> c &amp; -a \\<br /> \end{pmatrix}<br />
with trace=0 used.
<br /> C=B^{-1}=\frac{-1}{det \ B}<br /> \begin{pmatrix}<br /> a &amp; b \\<br /> c &amp; -a \\<br /> \end{pmatrix}<br /> =\frac{-1}{det \ B} B
det \ B=-a^2-bc=-1
because B has eigenvalue 1 and -1 so
(a-\lambda)(-a-\lambda)-bc=0
for ##\lambda=\pm1##. As ##\alpha## is Hermitian we observe a is real and ## c=b^*##. so
a^2+|b|^2=1
 
Last edited:
@anuttarasammyak Thanks! I forgot the determinant. I will try to generalize it to 4 dimensional case.
 
Further to post #2 as general expression
<br /> B=C=<br /> \begin{pmatrix}<br /> cos\theta &amp; sin\theta e^{-i\phi} \\<br /> sin\theta e^{i\phi} &amp; -cos\theta \\<br /> \end{pmatrix}<br /> =\cos\theta\ \sigma_z+sin\theta cos\phi\ \sigma_x+sin\theta sin\phi\ \sigma_y
that seems like unit vector in polar coordinates.
 
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