- #1

- 244

- 5

Given the de Broglie relation, assuming natural units, where ##\hbar = 1##:

\begin{equation}

\begin{split}

\vec{k} &= M \vec{v}

\end{split}

\end{equation}

My question regards velocity and what is meant by ##\vec{v}##. In classical physics, ##\vec{v}=

\frac{d \vec{x}}{dt}##. However, in quantum physics, position and momentum are Fourier conjugates, whose variance is related by the Uncertainty Principle. Mathematically, I don't see how it is possible that a variable and its derivative can be Fourier conjugates. Therefore, it seems ##\vec{v}## represents a different type of velocity. My understanding is that it represents group velocity (##\vec{v}_G##) related to phase velocity ##\vec{v}_P## as follows:

\begin{equation}

\begin{split}

\vec{v}_P &= \frac{\omega}{\vec{k}}

\end{split}

\end{equation}

And:

\begin{equation}

\begin{split}

\vec{v}_G &= \frac{d \omega}{d \vec{k}}

\end{split}

\end{equation}

Such that, in this case, ``velocity'' is a function of momentum (rather than position), and there is no problem with velocity and position as Fourier conjugates.

Unfortunately, I don't think I have ever seen this distinction in velocity mentioned in discussions of quantum physics, and worse, I have often seen classical and group velocity used interchangeably in the mathematics of quantum physics.

What am I missing?