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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?

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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.

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  • $\begingroup$ 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! $\endgroup$ – A Quantum Field Day Apr 16 at 15:46

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