De Broglie Wavelengths I have a working knowledge of wave-particle duality, I think. I know the de Broglie wavelength is a sort of probability of finding a particle in a specific position, and is calculated by $\lambda=\frac{h}{\vec{p}}$. I have a couple questions I'm hoping to have cleared up, though.
First, since $\vec{p}$ is momentum, a vector, what happens to the direction in the above formula? Is it meant to be the magnitude of momentum instead? $\lambda$ having a direction doesn't seem to make sense.
Second, clearly if $\vec{v}=\vec{0}$, then $\vec{p}=m \vec{v}=\vec{0}$, and $\lambda$ is undefined. Does this mean particles that are stationary have no wave character? This also doesn't make sense, so I think this is not the case.
 A: I have never seen de Broglie's relation written with vector quantities. A quick search online reveals a lack of vectors as well. In the relation
$$\lambda = \frac{h}{p}$$
it is implied that the quantity $p$ is the magnitude of the momentum $\left | \vec{p} \right |=p.$ Yes, the word momentum in a strict sense refers to a vector quantity, but often physicists will use this term when referring to the well-defined scalar $p$.
On a related note, the term wavelength typically (always, probably) refers to a scalar, not a vector. So trying to ascribe a direction for a wavelength is something one might expect isn't done. Though as David H described in comments, there is a  commonly-used vector quantity that are related to wavelength: wave vector:
$$\vec{k}=\frac{2\pi}{\lambda}\hat{v},$$
where the unit vector $\hat{v}$ typically points in the direction of propagation. If you fiddle around with these definitions you can write something akin to the de Broglie relation with vectors: $\vec{p}=\hbar \vec{k},$ with $\hbar=h/(2\pi)$.
