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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?

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  • $\begingroup$ It comes from using Hamiltonian mechanics instead of Lagrangian mechanics. Lagrangian mechanics is unfavored for the tangent bundle has no Poisson bracket nor a Dirac one. $\endgroup$
    – DanielC
    Nov 15, 2017 at 17:48
  • $\begingroup$ Simplest answer : the principle applies to any particle, including massless ones. They have momentum even if they're massless. Long answer involves some mathematics and you can see it here and is a mathematical consequence of how momentum, position and the wavefunction are defined in quantum theory. $\endgroup$ Nov 15, 2017 at 17:49

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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).

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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?

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  • $\begingroup$ "Because the uncertainty principle applies to all particles" [citation needed]. Massless particles require special-relativity, where $\hat P,\hat X$ may not be well-defined objects. $\endgroup$ Nov 15, 2017 at 18:22
  • $\begingroup$ @AccidentalFourierTransform If there are particles to which the HUP does not apply, I'm happy to learn about it. Otherwise, the choice to reduce special relativity to only a reference is a didactic choice based on the level of OP's question; you're welcome to post a separate answer if you think this oversimplifies crucial aspects that OP would otherwise be able to understand and appreciate. $\endgroup$ Nov 15, 2017 at 18:28
  • $\begingroup$ My point is that you should not claim that the HUP applies to massless particles unless you have a reference that argues that one can do so. I agree with the spirit of the answer though -- that the HUP is oblivious to the actual value of $m$ (as long as $m>0$). Were you to remove the sentence "(even if they don't have mass)", I'd be happy to upvote the answer. $\endgroup$ Nov 15, 2017 at 18:34
  • $\begingroup$ Fine, here's a rewording for the nitpickers ;-). $\endgroup$ Nov 15, 2017 at 18:36

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