This is very trivial, but it's really bugging me. The ansatz for the Dirac equation in terms of $$\boldsymbol\alpha$$ and $$\beta$$ matrices is $$[i(\partial_t+\boldsymbol\alpha\cdot\boldsymbol\nabla)-\beta m]\psi=0,$$ and to get the standard form in terms of gamma matrices, one defines $$\gamma^\mu=(\beta, \beta\boldsymbol \alpha)$$ so that multiplying the equation by $$\beta$$ gives $$[i(\beta\partial_t+\beta\boldsymbol\alpha\cdot\boldsymbol\nabla)-\beta^2m]=[i\gamma^\mu\partial_\mu-m]=0$$ as long as the index on $$\gamma^\mu$$ is actually a vector index, which we can make sure of by requiring $$\psi$$ to transform in the spinor representation. But considering that the correct expansion of the Minkowski product is $$\gamma^\mu\partial_\mu=\gamma^0\partial_0-\boldsymbol\gamma\cdot\boldsymbol\nabla=\beta\partial_t-\beta\boldsymbol\alpha\cdot\boldsymbol\nabla,$$ shouldn't the expression in the first term be $$\beta\partial_t-\beta\boldsymbol\alpha\cdot\boldsymbol\nabla$$, with a minus sign rather than a plus sign?

This is a classic misuse of the Einstein summation convention. The correct expression in your final equation is $$\gamma^{\mu} \partial_{\mu} = \sum_{\mu = 0}^3 \gamma^{\mu} \partial_{\mu} = \gamma^0 \partial_0 + \vec{\gamma} \cdot \vec{\nabla}$$ The reason for your confusion is because you're used to taking the Minkowski inner product between two contravariant or two covariant vectors. If $$v^{\mu} = (v^0, \vec{v})$$ and $$w^{\mu} = (w^0, \vec{w})$$, then of course $$v^{\mu} w_{\mu} = g_{\mu \nu} v^{\mu} w^{\nu} = v^0 w^0 - \vec{v} \cdot \vec{w}$$. But in your expression, $$\partial_{\mu}$$ is naturally covariant, not contravariant.

• Oh, right. I should have written $\gamma^j\partial_j$ explicitly to see that there was no need to raise the index of $\partial_\mu$. Thanks! Commented Apr 16, 2021 at 15:46

The ansatz originally use was to decompose the mass shell relation $$\frac{E^2}{c^2} = |𝐩|^2 + m^2 c^2$$ into linear form as $$\frac{E}{c} = 𝝰·𝐩 + β m c.$$ Under the operator correspondence $$E ⇔ iħ\frac{∂}{∂t},\quad 𝐩 ⇔ -iħ∇,$$ and upon application to $$ψ$$, this becomes: $$\frac{iħ}{c}\frac{∂ψ}{∂t} = (-i ħ 𝝰·∇ + β m c) ψ,$$ or $$\left(\frac{1}{c}\frac{∂ψ}{∂t} + 𝝰·∇\right) ψ = β m c ψ.$$ Thus, upon multiplication by $$β$$, noting that $$β^2 = 1$$: $$\left(β \frac{1}{c}\frac{∂ψ}{∂t} + β 𝝰·∇\right) ψ = m c ψ,$$ or just: $$\left(γ^0 \frac{1}{c}\frac{∂ψ}{∂t} + 𝝲·∇\right) ψ = m c ψ,$$ where $$γ^0 = β$$ and $$𝝲 = \left(γ^1, γ^2, γ^3\right) = β 𝝰$$.

The signs associated with the operator correspondences reflect the fact that $$E dt - 𝐩·d𝐫,$$ is a Lorentz invariant, where $$𝐫 = (x, y, z)$$, so that $$(E, 𝐩)$$ transforms as $$(∂/∂t, -∇)$$, with the extra sign.