# Does Heisenberg's uncertainty principle also apply to measuring velocity? Can we measure velocity at an exact time and point?

There will always be an error in measuring velocity because it is average velocity in some range of time not the exact time. So the physical quantity velocity doesn't exist at a moment. So how come momentum exist at a moment?

• The Heisenberg uncertainty principle is not due to imprecise measurements. See Is the uncertainty principle a property of elementary particles or a result of our measurement tools? for more on this. Jul 11, 2022 at 8:58
• @JohnRennie 's comment should be an answer, because it's a crucial part for understanding Heisenberg's uncertainty principle, which is neither about uncertainty (but indeterminacy, it's not a precision problem) nor a principle (it's a proven theorem). Jul 11, 2022 at 9:36

You can talk about instantaneous velocity measurement, you just need to define the velocity observable as an operator. The natural way to do it is using to take the time derivative of the position operator.

It is easiest way to define it in the Heisenberg picture (in the Schrodinger picture observables don’t evolve). In this case, you have: $$v= \dot x = -i[x,H]$$, and you can therefore calculate $$[x,v]$$ and see if there is an uncertainty. Note that it will heavily depend of $$H$$.

In most examples, you have the form $$H = \frac{p^2}{2m}+V(x)$$ so $$v=\frac{p}{m}$$ as in classical mechanics. You therefore have the same uncertainty principle: $$[x,v] = i\frac{\hbar}{m}$$ up to a mass scaling.

Even in th case of a charged particle of charge $$q$$ in a magnetic field of vector potential $$\vec A$$ : $$H = \frac{(\vec p-q\vec A(x))^2}{2m}+V(x)$$. In this case, you have just as in classical mechanics: $$\vec v = \frac{\vec p-q\vec A(x)}{m}$$, you still have $$[x_i,v_j] = i\frac{\hbar\delta_{ij}}{m}$$, so the same reasoning applies.

Hope this helps.

There is instantaneous speed, which is speed at a moment.

Heisenberg's Uncertainty Principle states that it is not possible to accurately measure both the momentum of particle and the position of the particle at the same time.

Therefore, it is possible for us to measure the speed of a particle accurately (p = mv) but that means we can't measure the exact position that particle is at.

• That means we all show wavd nature even a moving car Jul 12, 2022 at 5:07