# I Need help with Dirac Equation

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1. Dec 23, 2016

### 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 xp 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!

2. Dec 25, 2016

### secur

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".

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.

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.