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If velocity is not quantized, then do moving objects have 'infinitely decimal place' velocities which we just can't measure to infinite decimal places? From my understanding the quantization of velocity in Bohr's model is just a consequence of the quantization of angular momentum.

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Related: physics.stackexchange.com/q/39208/2451 and links therein. –  Qmechanic Feb 16 '13 at 18:16
    
Given that it appears that no distance can be shorter than the Planck constant then it would seem that velocity is quantized. –  ChrisF Feb 16 '13 at 20:14
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@ChrisF actually that's not true. The Planck length isn't a minimum distance. –  David Z Feb 17 '13 at 0:49
    
@DavidZaslavsky - I'm probably confusing it with the Planck constant. –  ChrisF Feb 17 '13 at 9:50
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3 Answers

There's a fundamental problem with speaking about whether or not "velocity is quantized," say for an electron in the hydrogen atom. If we were to model of time-evolution of the state of the particle as a trajectory in three dimensional space, then we could define its velocity and perhaps start asking questions about it, but in quantum mechanics, the state of an electron in the hydrogen atom is modeled by a function in a function space (the wavefunction) that roughly tells you how likely it will be to find the electron in certain locations around the nucleus if you were to measure its position.

So the main point is that even the notion of velocity at such small scales becomes problematic because the model of particles following trajectories breaks down.

When Bohr performed his quantization procedure for the hydrogen atom, he essentially introduced an ad hoc assumption that the angular momentum of the electron were quantized while also keeping the model that it is in some sense orbiting the nucleus along some trajectory. We now know that this is not quite the right view of how to describe the state of the electron for the reasons just outlined.

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Quantum mechanically, the velocity

$$\hat{{\bf v}}~=~\frac{\hat{{\bf p}}}{m}$$

is (like any physical observable) a Hermitian operator. (In this case a vector-valued Hermitian operator.) Possible values/outcome ${\bf v}\in \mathbb{R}^3$ of the velocity are given by the corresponding eigenvalues of the operator $\hat{{\bf v}}$.

Whether the eigenvalues are continuous or discrete depend on the specific system. See also this Phys.SE question. E.g. in certain periodic systems, the momentum (and hence the velocity) is quantized/discrete. On the other hand, the velocity of the electron in the hydrogen atom is not quantized/discrete.

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Are you saying the electron can have any arbitrary velocity in any of the orbitals? It is true that the orbital radius is 'quantised'. Bohr's condition and Scrodigner equation talk about most probable orbits. I think, in that sense, if you write the equation for the total energy of the electron in any orbital, and bear this in mind, you will find that the momentum is quantised, hence the speed is quantised. –  JKL Feb 16 '13 at 23:14
    
@John: Let me put it this way: If we e.g. prepare the electron in the ground state of the hydrogen atom, then the velocity $\hat{\rm v}$ (or for that matter the position $\hat{\rm r}$) can still take a continuum of values. –  Qmechanic Feb 16 '13 at 23:34
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Yes, that is right. One has to be very careful what they mean by quantised quantities in the context of the hydrogen atom or any bound system. The most probable quantum states have quantised values of their observables, but these are not the whole story. An electron in the hydrogen atom can access any energy level, even if it is not one of the quantised levels! These have extremely short life-time $~10^{-15}s$ and are usually called virtual states. That is the beauty of multiphoton ionisation or excitation with intense lasers. See my answer, and I will appreciate any comments. –  JKL Feb 17 '13 at 0:33
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Let us talk about the first question, first: Can "moving objects have infinitely decimal place velocities?"

Most certainly yes!! The simplest example is the motion of a particle on a circular orbit of radius R. Let us assume the radius of the circle is $R=10m$ exactly (as exact as we can make it,) and let the partcle make ten revolutions per second, i.e. $T=0.1 s$. Then we can find the angular speed of the particle as

$\omega =\frac {2\pi}{T}=20\pi$

and the orbital speed will be

$v=\omega R=200\pi$ .

The last equation is very revealing. The number $\pi$ is irrational (it has infinitely many decimal places) and therefore the speed must have infinitely many decimal places. The problem is, however, that we do not have any facility to access such beautiful sabtleties of nature. We have to round off to an experimentally sensible significant fingure. $\hbar$ has infinitely many decimal places, but we only use as many decimal places as the number of decimal places we know Planck's constant, $h$, in. Otherwise any extra decimal places beyond what we know $h$ in would be experimentally and theoretically meaningless. Alternatively, we give our answers in special units in which $c=h=1$ for example.

As for the second question: I think joshphysics has given a good answer. However, we can talk about the mean speed and mean position like with the electron in the hydrogen atom, and with a little bit algebra we can find the quantisation condition for momentum, hence for the mean speed.

$p_n=\frac {m_ee^2}{4\pi \epsilon_0\hbar}\frac 1n.$

What happens at the planck scale?

In this case we know that distances are quantised, Planck length:

$l_p=\sqrt{\frac {\hbar G}{c^3}}$

$t_p=\sqrt{\frac {\hbar G}{c^5}}$

and we see that any speed defined as distance over length at Planck scale gives the speed of light. Or we can say that an 'object'traveled so many Plank lenghts in so many Planck times as

$v=\frac{nl_p}{mt_p}=\frac mn c$ with $n\ge m$ but I don't really know the exact meaning of such hypothetical results(?).

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What's the second question? –  raindrop Feb 17 '13 at 1:01
    
@Raindrop Yes, you are right. The quantisation equation I have written for momentum of an electron in the hydrogen atom in my answer, is a result of Bohr's quantisation condition for the angular momentum of the electron. In a way, Bohr was the first person to grasp the true meaning of Planck's constant in that sense. –  JKL Feb 17 '13 at 20:07
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