Why does the Heisenberg uncertainty principle refer to momentum rather than velocity?

Question

I've been looking at the Heisenberg uncertainty relations, and something that sticks out to me is the use of momentum rather than velocity. Shouldn't electrons have the same mass? And if they do, why is momentum used?

Answer 1

The general uncertainty relation takes the form $$ \Delta A\Delta B\ge \frac{1}{2}\vert\langle [\hat A,\hat B]\rangle\vert\, . \tag{1} $$ Since it is the commutator $[\hat x,\hat p]=i\hbar$ and constant, this specializes (1) to the usual $\Delta x\Delta p\ge \frac{\hbar}{2}$.

Note that $[\hat x,\hat p]=i\hbar$ is the quantum analogue of the classical Poisson bracket $\{x,p\}$, which involves position and momentum.

Moreover, there are closely related Fourier relations connecting the width of a pulse with the wavenumber: $\Delta k\Delta x\sim 1$. Multiplying by $\hbar$ gives you the uncertainty relation, up to a factor of $1/2$, which appears in the formal, quantum mechanical derivation of (1).

Answer 2

Because the uncertainty principle applies to all particles, regardless of their mass (and even for some massless particles). Thus, you could, if you insisted, phrase the uncertainty principle for electrons as $$ \Delta v\,\Delta x\geq \frac12\frac{\hbar}{m_e} $$ and the uncertainty principle for protons as $$ \Delta v\,\Delta x\geq \frac12\frac{\hbar}{m_p} $$ and the uncertainty principle for neutrons as $$ \Delta v\,\Delta x\geq \frac12\frac{\hbar}{m_m} $$ and the uncertainty principle for helium atoms as as $$ \Delta v\,\Delta x\geq \frac12\frac{\hbar}{m_\mathrm{He}} $$ and so on and on and on, but why would you do that when you can just issue a single principle that applies for all particles, $$ \Delta p\,\Delta x\geq \frac12\hbar, $$ and be done with it?