1
$\begingroup$

The following unnormalized vectors are solutions to the Dirac equation.

\begin{align*} u_1&=\begin{pmatrix}E+m\\0\\p_z\\p_x+ip_y\end{pmatrix} \exp\left(\frac{i\phi}{\hbar}\right) % & v_1&=\begin{pmatrix}E-m\\0\\p_z\\p_x+ip_y\end{pmatrix} \exp\left(-\frac{i\phi}{\hbar}\right) \\[1ex] u_2&=\begin{pmatrix}0\\E+m\\p_x-ip_y\\-p_z\end{pmatrix} \exp\left(\frac{i\phi}{\hbar}\right) % & v_2&=\begin{pmatrix}0\\E-m\\p_x-ip_y\\-p_z\end{pmatrix} \exp\left(-\frac{i\phi}{\hbar}\right) \\[1ex] u_3&=\begin{pmatrix}p_z\\p_x+ip_y\\E-m\\0\end{pmatrix} \exp\left(\frac{i\phi}{\hbar}\right) % & v_3&=\begin{pmatrix}p_z\\p_x+ip_y\\E+m\\0\end{pmatrix} \exp\left(-\frac{i\phi}{\hbar}\right) \\[1ex] u_4&=\begin{pmatrix}p_x-ip_y\\-p_z\\0\\E-m\end{pmatrix} \exp\left(\frac{i\phi}{\hbar}\right) % & v_4&=\begin{pmatrix}p_x-ip_y\\-p_z\\0\\E+m\end{pmatrix} \exp\left(-\frac{i\phi}{\hbar}\right) \end{align*}

where

\begin{equation*} E=\sqrt{p_x^2+p_y^2+p_z^2+m^2} \end{equation*}

and

\begin{equation*} \phi=p_xx+p_yy+p_zz-Et \end{equation*}

Of course, quantum electrodynamics uses the following solutions (in their normalized form) because the mass term $m$ is positive.

\begin{align*} u_1 &\quad \text{fermion, spin up} \\ u_2 &\quad \text{fermion, spin down} \\ v_3 &\quad \text{anti-fermion, spin up} \\ v_4 &\quad \text{anti-fermion, spin down} \end{align*}

What other combinations of $u$ and $v$ could be used as a basis for quantum electrodynamics? Are there any theories that use a different basis?

There are ${8\choose4}=70$ ways to choose four vectors, but not all of them are linearly independent.

$\endgroup$
1
  • 1
    $\begingroup$ I don't think it's right to make the general statement that QED uses a specific choice of basis. It may be the case that one choice is conventional or particularly convenient for a particular application, of course. And since a choice of basis is ultimately unphysical (in the sense that it has no physical ramifications), picking a different one doesn't give you a new theory in any meaningful sense. $\endgroup$
    – J. Murray
    Commented Sep 23, 2023 at 15:44

1 Answer 1

2
$\begingroup$

By brute force checking of all 70 determinants, there are 16 choices of basis vectors.

\begin{equation*} \begin{matrix} u_1 u_2 v_1 v_2 & u_1 u_4 v_1 v_2 & u_2 u_3 v_1 v_2 & u_3 u_4 v_1 v_2 \\ u_1 u_2 v_1 v_4 & u_1 u_4 v_1 v_4 & u_2 u_3 v_1 v_4 & u_3 u_4 v_1 v_4 \\ u_1 u_2 v_2 v_3 & u_1 u_4 v_2 v_3 & u_2 u_3 v_2 v_3 & u_3 u_4 v_2 v_3 \\ u_1 u_2 v_3 v_4 & u_1 u_4 v_3 v_4 & u_2 u_3 v_3 v_4 & u_3 u_4 v_3 v_4 \end{matrix} \end{equation*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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