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I know there are many questions about this topic and also various answers, but it's never stated explicitly, why there is a certain sign before the mass term in the Dirac Lagrangian. I'm also confused, since the text I'm following, it is stated without proof that

The relative sign between the two Lorentz-Scalars can be deduced, by implying that the equation of motion must satisfy the Klein-Gordon-Equation for every components.

But when I tried to get to this result by keeping the $\pm$ explicitly through all calculations, I saw that it drops out anyways

$$\mathcal{L} = i\bar{\Psi}(\partial_\mu \gamma^\mu \pm m)\Psi$$

Euler-Lagrange equation yields

$$\frac{\partial\mathcal{L}}{\partial\bar{\Psi}} - \partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial\bar{\Psi})} = (i\partial_\mu\gamma^\mu \pm m)\Psi = 0. $$

Multiplying the hermitian conjugate of the operator from the left

$$ \begin{align} 0 &= (-i\partial_\nu\gamma^\nu \pm m)(i\partial_\mu\gamma^\mu \pm m)\Psi \\ &=(\partial_\mu\gamma^\mu\partial_\nu\gamma^\nu \pm im\partial_\mu\gamma^\mu\mp im\partial_\nu\gamma^\nu+m^2)\Psi \\ &= (\frac{1}{2} \left[\partial_\mu\gamma^\mu\partial_\nu\gamma^\nu + \partial_\mu\gamma^\mu\partial_\nu\gamma^\nu\right] + m^2)\Psi \\ &= (\frac{1}{2}\{\gamma^\mu,\gamma^\nu\}\partial_\mu\partial_\nu + m^2)\Psi \\ &= (\partial_\mu\partial^\mu + m^2)\Psi = 0, \end{align}$$

where the identity matrix has been suppressed. So this seems indeed to be the Klein-Gordon-Equation for each component of the spinor, but the $\pm$ drops out in the beginning.

Hence, my question does it matter which sign we choose (at the level of the Lagrangian), and is there a deeper reason of why to choose one or the other?

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  • $\begingroup$ Your Lagrangian density seems to have the i at the wrong place. $\endgroup$ – lalala Sep 20 '18 at 19:32
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Let's say, $\Psi$ is a solution of Dirac equation, that is, $$(i\gamma^{\mu}\partial_{\mu}-m)\Psi=0.$$

Multiplying by $\gamma^5$ and using $\gamma^5\gamma^{\mu}=-\gamma^{\mu}\gamma^5$, $$(i\gamma^{\mu}\partial_{\mu}+m)\gamma^5\Psi=0.$$ Thus, $\gamma^5\Psi$ is also a solution with mass $-m$. The two solutions correspond to the two factors of $E^2=p^2+m^2$. Since Dirac equation is linear, the linear combinations of its solutions: $\psi_L=\frac{1}{2}(1-\gamma^5)\Psi$ and $\psi_R=\frac{1}{2}(1+\gamma^5)\Psi$, will also be solutions.

All this basically means that Dirac equation describes two solutions and depending upon the choice of basis (for $\gamma$-matrices) these solutions could be interpreted as particle and anti-particle (in Dirac basis) or two-component left- and right- handed Weyl spinors (in chiral basis). These components separately are solutions to the Klein-Gordon equation.

EDIT: Notice that in the last expression in the question, there is no $\gamma$-matrix, so each component satisfies the K-G equation individually. As per the disappearance of the $\pm$ sign, I think it has to do with the fact that EOM for both $\Psi$ and $\bar{\Psi}$ (Dirac equation and its adjoint, respectively) can be obtained from the same Lagrangian by varying it w.r.t. to $\bar{\Psi}$ and $\Psi$, respectively. So, in that sense, it doesn't matter which one you begin with (the Dirac Lagrangian with the minus sign or its adjoint with the plus sign). They carry the same information; they're just adjoint of each other!

For detailed explanation, go here.

A similar thread is this one.

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  • $\begingroup$ Oke so changing the sign in the Lagrangian corresponds to a redefinition of what is a right or left handed Weyl-spinor? Hence, it would not be possible to determine the sign of the Lagrangian mass term, just by stating that the EOM have to fulfill the Klein-Gordon equation? $\endgroup$ – Patrick Amrein Jun 12 '17 at 16:42
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    $\begingroup$ @PatrickAmrein see the EDITed answer above. I hope this completes the discussion. $\endgroup$ – phydev Jun 12 '17 at 17:16
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Change of the sign at the mass in the Dirac equation is equivalent to replacing $\gamma^\mu$ with $-\gamma^\mu$, but matrices $-\gamma^\mu$ have the same anti-commutation relations as $\gamma^\mu$, so you get an equivalent equation, if I am not mistaken. Specific solutions may have a different form, but the physics seems to be the same.

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