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Im studying the (stationary) free 1D-Dirac equation

$$H\Psi(x) =(mc^2\sigma_1-i\hbar c\sigma_3\frac{\partial}{\partial x})\Psi(x) = E\Psi(x),$$

where $\sigma_1$ and $\sigma_3$ are the pauli matrices.

I was able to determine the (unnormalized) solutions

$$\Psi_1(x) = \binom{1}{\frac{mc^2}{E-pc}}e^{-ipx/\hbar} \quad \text{and} \quad \Psi_2(x) = \binom{\frac{mc^2}{pc+E}}{1}e^{-ipx/\hbar},$$

where $E=\sqrt{m^2c^4+c^2p^2}$.

I'm wondering, why do I get $H\Psi_1 = +E\Psi_1$ and $H\Psi_2 = +E\Psi_2$? Shouldn't one of the two eigenstates have a negative energy eigenvalue, i.e. $H\Psi = -E\Psi$?

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    $\begingroup$ Your equation only has 2 components. The Dirac equation is a 4-component equation. $\endgroup$ May 20, 2022 at 16:19
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    $\begingroup$ yes this is due to the fact that this is the one dimensional equation. the gamma matrices are in the 1d case the 2x2 pauli metrices. Therfore the Eigenstates are 2 dimensional vectors. $\endgroup$
    – Aralian
    May 20, 2022 at 16:22
  • $\begingroup$ I was also wondering, why there are only 2 components. I think this is because in the 1d case there is no angular momentum, and therefore there is no such thing as spin. So i thought the 2 orthogonal vectors represent each the positive and negative energy case. $\endgroup$
    – Aralian
    May 20, 2022 at 17:31
  • $\begingroup$ Isn't this more like Weyl equation? $\endgroup$
    – Mauricio
    May 20, 2022 at 22:56

2 Answers 2

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Your solutions are actually identical up to normalization. If you multiply your $\Psi_1$ by $\frac{mc^2}{pc+E}$ you get what you are calling $\Psi_2$.

In this solution $E$ and $p$ are only fixed by the condition $E^2= p^2c^2+m^2c^4$, so for any given $p$ there are two solutions for $E$ of differing sign. This is the sense in which there are two solutions.

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I have an Idea on how to get the negative energy states. When I calculate the Energy, I get the condition $E^2 = p^2c^2 + m^2c^4$.

So I can define $E = \pm \sqrt{p^2c^2+m^2c^4} = \pm E_p$ and therefore I have two different differential equations:

$H\Psi_+ = +E_p\Psi_+$ and

$H\Psi_- = -E_p\Psi_-$

the second equation should give me the negative energy states.

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  • $\begingroup$ Did you read my answer? Yes your idea is correct $\endgroup$
    – octonion
    May 21, 2022 at 9:10
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    $\begingroup$ Yes I did I just didn't saw the fact that I actualy get two equations in your comment. As you may noticed, I'm not very experienced so sorry for not seeing this :) $\endgroup$
    – Aralian
    May 21, 2022 at 9:21

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