Heisenberg uncertainty principle: If a particle is totally localized, does it go faster?

In Feynman's lectures on physics, volume I, section 6-5, Feynman states: $$\Delta x\cdot \Delta v \ge h/m$$ ($\Delta x$ is the width of the probability distribution of the location, $\Delta v$ is the width of the probability distribution of the velocity, $h$ is the Planck constant and $m$ the mass) and goes on to say the following:

Since the right-hand side of this equation is a constant, it says that if we try to "pin down" a particle by forcing it to be in a particular place, it ends up by having a high speed.

My question: Why should the particle have a high speed if it is very localized? To my understanding the Heisenberg uncertainty principle only states that the velocity probability distribution will be very spread.

clarification: I'm not asking how can the particle be moving if it is localized, as asked before here, so this is not a duplicate to my understanding. My question is why should it go faster if it is more localized.

• @ACuriousMind , this is not a duplicate, I'm asking why the particle should be faster if it is more localized. – Adi Ro Apr 25 '17 at 11:06
• Then I don't understand what you are asking. The "it is faster" is a sloppy phrasing since a quantum object with non-zero uncertainty doesn't have a definite speed. The uncertainty principle just tells you the standard deviation of velocity/momentum is higher if that for position is lower, nothing more. It doesn't make sense to speak of the speed of the quantum object as if it had a definite value. – ACuriousMind Apr 25 '17 at 11:17
• I know! and this is why the quote from Feynman's book makes no sense to me! this is my question... the quote, agian, is: "... Since the right-hand side of this equation is a constant, it says that if we try to "pin down" a particle by forcing it to be in a particular place, it ends up by having a high speed" – Adi Ro Apr 25 '17 at 11:23
• Yes, so the quote is using sloppy (and technically wrong) phrasing, it happens. I'm afraid I don't understand what more you want to know about that - even Feynman's utterances are not the infallible word of some deity. – ACuriousMind Apr 25 '17 at 11:25
• I agree it is not a duplicate, but I don't think this question actually has an answer other than "The quote is non-sensical if taken literally". – ACuriousMind Apr 25 '17 at 11:28

The argument says nothing about the average velocity of the particle, which will remain at $\langle v\rangle = 0$ for bound states; moreover, $|v|=0$ will normally still be the most likely speed in that state. Nevertheless, if you want a distribution with a very large width $\Delta v$, then it will need to have support from very large values, so the probability that the particle has some high velocity will increase.
• I guess speed means $\vert v \vert$ here. Its quantum expectation value is indeed large. – Tony Apr 25 '17 at 11:52