quantum mechanics - Electron at rest

David Griffiths suggested a website in his book where I got this paper

http://www.hep.princeton.edu/~mcdonald/examples/electronatrest.pdf

Here the author says classically a particle at rest(in some frame) when its velocity and momentum vanish. In quantum mechanics simply requiring the expectation value of momentum to be zero implies only particle's average velocity is zero. Something closer to the classical meaning of rest is achieved only when one requires that the expectation value of the square of velocity be vanishingly small as well. As indicated in the statement of the question, the uncertainly principle then tell us that this condition can only be achieved if the particle is in an arbitrarily large box.

As I am new in quantum mechanics I don't understand why he said expectation value of the square of velocity? And also Griffiths in the 1st chapter calculated expectation of value of position and here took the time derivative and said this is not velocity of the particle ,this is the velocity of expectation value of the position.

Wait for 3rd chapter to know what exactly velocity of particle mean? Now I haven't read the 3rd chapter yet. I have seen the older questions, I think this is not duplicate of them.

Answer

This has to do with what we really mean when we ask if something is "at rest." Specifically, we usually mean something like this, "Is the particle's momentum in our frame of reference exactly zero?"

As it turns out, we can't actually know a particle's momentum exactly in quantum mechanics because of the Uncertainty Principle. Supposing that we could would require $\sigma_p$ (or, if you like, $\Delta p$) to be zero which would in turn require $\sigma_x$ to be infinity to satisfy the Uncertainty Principle (this is why he says the particle can only achieve "rest" in an "arbitrarily large box").

But $\sigma_p$ is a Standard Deviation defined as $\sqrt{\langle p^2\rangle - \langle p\rangle^2}$. As you can see, the expectation value of $p^2$ needs to be zero as well if the whole expression is to be zero (which it must be if we are to know the particle's momentum exactly).