- #1

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Squaring, in the ultra-relativistic limit one obtains the dispersion relation

$$(E-A)^{2}-p^{2} \simeq mm^{\dagger}$$ i.e.

$$p \simeq E -(\frac{mm^{\dagger}}{2E}+A).$$

What I have is $$(i\gamma^{\mu}\partial_{\mu}-A\gamma_{0})(-i(\gamma^{\mu}\partial_{\mu})^{\dagger}-A(\gamma_{0})^{\dagger})$$ and I know $$\gamma_{0}^{\dagger} = \gamma_{0}$$ and $$(\gamma^{\mu})^{\dagger} = \gamma^{0}\gamma^{\mu}\gamma^{0}.$$

But I am not seeing how to get $$(E-A)^{2}-p^{2} \simeq mm^{\dagger}$$ from this. Any hints would be greatly appreciated!