0
$\begingroup$

For particles defined with positive energy, we use $$\phi= \begin{pmatrix} 1 \\ 0 \\ \end{pmatrix} $$ or $$ \begin{pmatrix} 0 \\ 1 \\ \end{pmatrix} $$

while for particles defined with negative energy, we use this instead $$\chi= \begin{pmatrix} 1 \\ 0 \\ \end{pmatrix} $$ or $$ \begin{pmatrix} 0 \\ 1 \\ \end{pmatrix} $$ where ∅ is the upper component while $\chi$ 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? I don't understand is why is the 4-spinor split into two components, one for the positive energy and the other for the negative energy. It did not say anything here.

$\endgroup$
1
$\begingroup$

In the Dirac equation:

$$(\beta mc^2+\sum_i \alpha_i p_i)\psi = i\hbar \frac{\partial\psi}{\partial t}$$

there are four 4x4 matrix operators, which can be defined as follows in terms of 2x2 blocks:

$$\alpha_i=\begin{bmatrix} 0 & \sigma_i \\ \sigma_i & 0\end{bmatrix}$$ $$\beta=\begin{bmatrix} I & 0 \\ 0 & -I\end{bmatrix}$$

where the $\sigma_i$ are the standard Pauli matrices, $I$ is the 2x2 identity matrix, and 0 is the 2x2 zero matrix. The fact that the operators in the Dirac equation naturally split into 2x2 blocks should suggest that splitting the 4x1 bispinor $\psi$ into two 2x1 spinors would also behave nicely. Indeed, as developed in your source, it does.

$\endgroup$
1
$\begingroup$

I don't understand is why is the 4-spinor split into two components, one for the positive energy and the other for the negative energy. It did not say anything here.

You are misinterpreting the large and small components of the Dirac 4-spinor for the positive and negative energy solutions (a common mistake). Each Dirac spinor (whether a positive or negative energy solution) has four components. For a positive energy solution the upper 2-spinor is called the large component because the lower 2-spinor is much smaller (of order $\frac {v}{c}$ in comparison, $v$ is the average velocity of the particle described). The same applies to the negative energy solutions except that the relative size of the upper and lower components is reversed.

Your link describes the situation correctly as does the answer by @probably_someone.

$\endgroup$
  • $\begingroup$ I know that each dirac spinor has four components. I just don't get why we use $\phi=(1,0) or (0,1)$ for the positive energy. I think my confusion is due to my lack of understanding about the upper and lower components of the dirac spinor cause I don't really have an intuitive understanding of them. $\endgroup$ – newbie125 Dec 27 '16 at 13:20
  • $\begingroup$ @newbie125 I suggest that you spend some time studing bound state solutions of the Dirac equation (hydrogen atom for example). There the radial and spin/angular functions separate and you will see that the upper component radial wave function is very much like the Schrodinger radial solutions. The small component only contributes 1-2% to the probability and can be thought of as a small relativistic correction. If you let the nuclear charge increase, the relative importance of the lower component increases as relativity corrections grow. $\endgroup$ – Lewis Miller Dec 27 '16 at 15:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.