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In the framework of relativistic quantum mechanics (not QFT) the Dirac equation in presence of external electromagnetic field is obtained by means of the minimal coupling, i.e. the substitution:

$$p_{\mu} \rightarrow p_{\mu}-eA_{\mu}$$

This substitution is often motivated by saying that it "ensures gauge invariance of the theory" (Greiner "Relativistic quantum mechanics", page 121). The resuting "modified" Dirac equation is:

$$i\frac{\partial\psi}{\partial t}=\left( \vec{\alpha} \left( \vec{p}-e\vec{A} \right)+\beta m + e \phi\right)\psi$$ This equation seems to change if one changes the 4-potential by a gauge transformation $A_{\mu}\rightarrow A_{\mu}+\partial_{\mu}\Lambda$. So what does it mean that the minimal coupling ensures gauge invariance? What am I missing?

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You are missing the phase transformation of $\psi$.

The Dirac equation is indeed not invariant to the gauge transformation $$A_{\mu}\rightarrow A_{\mu}+\partial_{\mu}\Lambda$$ alone.

But it is invariant to the combined gauge/phase transformation $$\begin{align} A_{\mu}&\rightarrow A_{\mu}+\partial_{\mu}\Lambda \\ \psi &\rightarrow \exp\left(-\frac{ie\Lambda}{\hbar}\right)\psi \end{align}$$

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What am I missing?

You are missing the fact that a gauge transformation transforms $\psi$ as well. It changes the complex phase of the spinor at each point in spacetime. See if you can figure out how much of a phase change will leave the equation unchanged.

It’s this local phase change of the matter field that explains why the gauge group for electromagnetism is called $U(1)$.

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