= Free-particle momentum convention
{title2=$\psi=Ae^{i(\mathbf k\cdot\mathbf r-\omega t)},\quad\mathbf p=\hbar\mathbf k$}
For a <free particle> of <mass> $m>0$ in <nonrelativistic quantum mechanics>, choose the <potential energy> zero and write
$$
\psi(\mathbf r,t)=A\exp\!\left[\frac{i}{\hbar}(\mathbf p\cdot\mathbf r-Et)\right],\qquad E=\frac{|\mathbf p|^2}{2m},\qquad \omega=\frac{E}{\hbar}.
$$
Applying the <momentum operator> $-i\hbar\nabla$ returns $\mathbf p\psi$; this fixes the sign convention. The <angular frequency> is measured relative to this chosen <energy> zero. Adding a constant <energy> changes only a time-dependent <global phase>, leaving the spatial <de Broglie wavelength> unchanged. No rest-energy term is included in this <nonrelativistic quantum mechanics> convention.
The <plane wave> has $|\psi|^2=|A|^2$ and <probability current> $\mathbf j=(\mathbf p/m)|A|^2$. Its real part can be drawn as a <sine> or <cosine>, but that graph is a component of a <probability amplitude>, not a transverse displacement. On infinite space this is a generalized <momentum eigenstate>, not a state satisfying <wavefunction normalization>. A narrow <wave packet> instead moves with <group velocity> $\nabla_{\mathbf k}\omega=\mathbf p/m$; the <phase velocity> $\omega/|\mathbf k|=|\mathbf p|/(2m)$ is not its transport speed.
The equality of <canonical momentum> and <kinetic momentum> here assumes a field-free region with the <magnetic vector potential> chosen zero. For a charged particle in a <magnetic field>, the <kinetic momentum> is $\boldsymbol\Pi=\mathbf p-q\mathbf A$. An arbitrary canonical phase gradient must not then be substituted into $m\mathbf v$; the <free particle> formula needs its stated assumptions.
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