# Role of momentum in Heisenberg Uncertainty Principle

I am currently studying the Heisenberg Uncertainty Principle, and - while I have understood the general idea - I am struggling with the role of momentum within the principle.

I know momentum as mass times velocity, or - for photons - energy over velocity. I have always related to momentum as an indirect indicator of energy, as both mass and velocity are related to energy (although that interpretation might be wrong, because I was never forced to analyse it until stumbling upon the Heisenberg Uncertainty Principle, and rather simply used it as a comfortable tool to perform certain Classical computations).

Intuitively, though, it feels more logical to use velocity within the context of the Heisenberg Uncertainty Principle. I know intuition is a very bad methodology with Quantum Physics, and often one simply has to accept the way nature works. Nonetheless, I would like to know if there is any logical explanation that could make the role of momentum here a bit easier to understand. What is it an indicator of? Does it indicate energy levels? Or does momentum here stand for something else?

• The "momentum" is just the momentum operator $\hat{p}$. – Avantgarde Sep 30 '18 at 8:46
• You can always use a "velocity operator" $p/m$ instead of the momentum if you like it better. I'm not really sure what sort of answer you're looking for here, can you be a bit more specific what's wrong from your viewpoint with just using momentum "as is" here? – ACuriousMind Sep 30 '18 at 9:17
• @ACuriousMind I am not sure what momentum stands for, in the sense that knowledge about position precludes knowledge about /it/. I always used momentum as a useful tool to make computations, but within this setting it must have a deeper meaning. When we say that precise knowledge of position precludes precise knowledge about momentum, what exactly can we not know? Mass x velocity is a computation, not something that I understand as a natural phenomenon; and for photons the definition is totally different. I do not understand what knowledge is actually precluded. – Pregunto Sep 30 '18 at 9:39

## 1 Answer

Indeed, I don't think it's a good idea. Personally, I just think of momentum as "something that can be measured", and it is extremely helpful, as it appears in the kinetic energy, so you need it almost always.

However, if your desire is finding a significance, my advice is: go back to the beginning": de Broglie's duality, wavefunctions, EM waves.

Momentum is $$p=\hbar k$$, so it is another way to measure the wave number $$k$$.

When you have a wave, it is of the form

$$A\cdot e^{i(\vec{k}\cdot\vec{x}-\omega t)}$$

or a linear combination of many, which we call "Fourier transform".

Measuring $$p$$ means measuring $$k$$.

If $$x$$ is your variable, you'll be able to measure positions easily. If you do a Fourier-transform to work with $$k$$ (or $$p/\hbar$$), momentum will be easy to measure.

Since we dont like switching from one to another all the time, we just use the reverse-fourier transform of the direct momentum operator, which is $$-i\hbar \vec\nabla$$, so that we can measure momentum in the position representation without changing thigs.

So, the conclusion is that measuring momentum is just another way to measure $$k$$.

And the interpretation of the uncertainty principle is easy: the largest uncertainty in position of the wave packet, the less wavenumbers it has. This is a general property of the Fourier-transform: the more frequencies, the less width in position.