Need help with Dirac Equation

[TimeRip496]

I just started learning this so I am a bit lost. This is where I am lost http://www.nyu.edu/classes/tuckerman/quant.mech/lectures/lecture_7/node1.html .

Why when E>0, we use $$\phi_p=
\begin{pmatrix}
1 \\
0 \\
\end{pmatrix}
$$ or $$
\begin{pmatrix}
0 \\
1 \\
\end{pmatrix}
$$

while when E<0, we use this instead
$$x_p=
\begin{pmatrix}
1 \\
0 \\
\end{pmatrix}
$$ or $$
\begin{pmatrix}
0 \\
1 \\
\end{pmatrix}
$$
where ∅_p is the upper component while x_p is the lower component of the bispinor in Dirac equation.
Can we do it the other way round or
$$\phi_p=
\begin{pmatrix}
1 \\
0 \\
0 \\
0 \\
\end{pmatrix}
...$$ instead?Secondly, how did the author convert $$\phi_p = \frac{c \sigma .p}{E_p -mc^2}x_p=?=\frac{-c \sigma .p}{|E_p| +mc^2}x_p$$? Does the mod sign means anything?

Can someone help me or point me in the right direction cause this is my first time learning this. Thanks a lot!

 

[secur]

Why when E>0, we use

$$\phi_p= \begin{pmatrix}

1 \\

0 \\

\end{pmatrix}
or
\begin{pmatrix}

0 \\

1 \\

\end{pmatrix}
$$
while when E<0, we use this instead
$$\chi_p=\begin{pmatrix}

1 \\

0 \\

\end{pmatrix}
or
\begin{pmatrix}

0 \\

1 \\

\end{pmatrix}
$$

where ∅_p is the upper component while x_p is the lower component of the bispinor in Dirac equation.

Can we do it the other way round ... ?

It's because of the $\beta$ matrix (as it's called in the notation you're using, from Dirac; see lesson 6). The top two rows are +1 (on the diagonal), the bottom two -1. The corresponding eigenvalues are pos and neg, obviously, when setting the momentum to zero, as shown in lesson 7. If you wrote the $\beta$ matrix "the other way round" then the E>0 and E<0 cases would also be switched, that is, $\phi_p$ and $\chi_p$ would play opposite roles. There are many other valid ways to write $\beta$ and $\alpha$ matrix (called "representations") all physically equivalent. BTW that second component is "chi" not "x".

or $$\phi_p=\begin{pmatrix}

1 \\

0 \\

0 \\

0 \\

\end{pmatrix}
...$$ instead?

In my answer above I assumed you meant, can we switch the roles of $\phi_p$ (E>0) and $\chi_p$ (E<0), and ignored this. For one thing $\phi_p$ is a 2-vector not 4 but even if you meant $u_p$ it still doesn't make sense, AFAIK.

Secondly, how did the author convert $$\phi_p = \frac{c \sigma .p}{E_p -mc^2}x_p=?=\frac{-c \sigma .p}{|E_p| +mc^2}x_p$$? Does the mod sign means anything?

The term $|E_p|$ is not a mod but an absolute value, since $E_p$ is just a real number. So you get the RHS simply by multiplying LHS numerator and denominator each by -1, remembering that $E_p$ is negative in this case.