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The Dirac equation $(i\gamma^{\mu}\partial_{\mu}-m)\psi = 0$ has plane wave solutions of the form: $$\psi_{+,k}(t,x) = e^{-iE(k)t+ik\cdot x}u(k) \quad \mbox{and} \quad \psi_{-,k}(t,x) = e^{iE(k)t -ik\cdot x}v(-k)$$ where $k, x \in \mathbb{R}^{3}$.

Question: The above expressions are functions of $t,x$ for each fixed $k$. How do I interpret these solutions? Does it mean that the Dirac equation have solutions with constant well-defined momentum? Each such pair of solutions represent a fermion with constant momentum $k$?

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The Dirac equation is linear, meaning that sums of solutions with arbitrary weights are also solutions.

You can take multiple plane waves with different $k$ and add them together.

Or you can take a distribution in $k$-space and integrate your plane wave over $k$-space weighted by that distribution.

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I'd just like to say that, in physics and field theory, whenever you have a solution to a linear theory which is proportional to $e^{-i k_\mu x^\mu}$ (with maybe some constant n dimensional vector out front) where $k^2 = m^2$, you always interpret this quantum mechanically as a particle of four momentum $k_\mu$. This is justified when you go to quantum field theory, as you can build an oscillator in the mode of this solution with integer occupation numbers.

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