I am confused as according to Heisenberg's uncertainty principle, we cant define the position and momentum of a particle with absolute accuracy simulaneously. Then how can we define velocity at quantum level. Its confusing because some books even mention the velocity of electrons in different energy levels.

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    $\begingroup$ Possible duplicate of why we dont have "direct" velocity operator just as $p$? ( as use $p$ space not $v$? ) in quantum? $\endgroup$ – BioPhysicist Oct 18 '18 at 9:49
  • $\begingroup$ Also, the books you are referring to might just be using the Bohr model of the atom... $\endgroup$ – BioPhysicist Oct 18 '18 at 9:56
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    $\begingroup$ Related question here. $\endgroup$ – knzhou Oct 18 '18 at 10:01
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    $\begingroup$ @Mechanic7 The Bohr model was developed, and then QM was further developed. But the Bohr model is still what it is when it was made. It uses QM in the sense that it assumed quantized energy levels, but it still had some classical pictures. Then QM developed. Bohr could have learned better ideas of QM, but when we talk about the Bohr model, we talk about electrons having defined positions. $\endgroup$ – BioPhysicist Oct 18 '18 at 10:04
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    $\begingroup$ The Bohr model predates quantum mechanics and it does not form a part of the formal structure of QM. (Indeed, given what we know about QM, the Bohr model is strictly wrong. It is a useful tool to develop intuition in some situations, but it's still wrong.) Bohr went on to do lots of work within QM that's not broadly known as "the Bohr model", including many aspects of the interpretation of measurements within QM that have nothing to do with the previous model. $\endgroup$ – Emilio Pisanty Oct 18 '18 at 10:28

Here is a pedantic and unhelpful answer to the question asked:

Is there a concept of velocity at quantum level?

Answer: Yes, given that you accept the the universe is quantum-y.

The reason being that if you accept that this quantum stuff is true then everything is "at the quantum level," including concepts. Far out dude. That is not a helpful answer, however.

Here is a (only slightly) less annoying answer to a rephrased version of the question:

Question: Is there a concept of velocity in the quantum theory?

Answer: Non-relativistically speaking - no, not really.

To see why, let's answer the question you asked in the body of your post:

...how can we define velocity at quantum level?

The short answer: We can't.

The slightly longer answer: We certainly can't do it in the same way that we carry over our definition of position from the classical world (by making it an operator that will return the value of position if the quantum system is an eigenstate of position). These operators - specifically momentum and position - are deeply linked and investigated well elsewhere. A reason that velocity does not crop up like this could be related to the fact that a system can be completely described by some basic properties, and velocity is not one of those properties (it is derived from more basic properties). In terms of Lagrangian dynamics (which is the language of modern quantum field theory and our best understand of what "quantum level" might even mean) a system is described in terms of coordinates (e.g. position, time, angle, etc.) and velocities can be derived from those. Time holds a special place in our concept and manipulating the Lagrangian with respect to time will give us momentum which (thanks to the Heisenberg Uncertainty Principle) governs the unique quantum-y behaviour of our universe and we don't really need anything else (like velocity) to make predictions about the outcome of measurement - which is really what all this quantum business does.

It was relativity and the so called 'energy-momentum relation' which required us to work with relativistic quantum fields in the first place. Such field theories are commonly, but not necessarily, described within the Lagrangian framework. But is all this necessary? Can we conceive of or define some 'quantum velocity' without it?

Forget relativity, field theories and Lagrangians. We can have a quantum theory without them, so can we can define velocity or not?

Prior to all this there was a non-relativistic quantum theory which was just called, logically, 'the quantum theory'. So lets talk about velocity in terms of that.

The definition of velocity (the classical one that we all know and love) is defined as the (time) rate of change of the position of a particle. We can stop here, as other answers have done, and say that a particle has no well defined position in the quantum theory and therefore we cannot define velocity. This is arguably correct and arguably the end of the story. However we can say more, and we should, if only to deepen our understanding of the tension between the quantum theory and the classical concept of velocity. We carried over the pre-quantum understanding of position, so why can't we do the same with time rate of change of position? Surely it's just another ill-defined, but reasonable, and usable concept. Reasonable - perhaps; usable - probably less so.

What is classical velocity anyway?

Let's be a little bit more explicit about what we mean by the term "velocity". Let's call it the time rate of change of the position as a function of time of the particle (AKA 'the derivative with respect to time of the position as a function of time'). A particle has, in classical mechanics, a well defined position as a function of time which we call the particle's trajectory. A trajectory is a well defined function and the differentiation of it is also well defined (thanks to the wonders of calculus). The problem is that in the quantum theory particles have no trajectory! This is because they have no well defined position for every moment of time. To know a trajectory you would have to measure position with infinite precision (something we can't do experimentally) constantly (something that the quantum theory isn't even set up to describe). So we are instantly having to think harder about this whole velocity thing.

Besides these points - both complex issues - a more easy to discuss problem lies ahead. An infinitely precise measurement of position would infinitely perturb the system in terms of its momentum and since the momentum would be completely unknown to us we wouldn't know where to look for the particle at the next instance of time. What does this mean in terms of differentiation and velocity? Well... we cannot easily differentiate a trajectory through space (ie a function) unless we know that function perfectly... and we cannot know that function perfectly without measuring it perfectly at every given point in time... and we can't do that. So the differentiable function doesn't exist according to the quantum theory. And hence defining velocity becomes a bit of a challenge.

Indeed we can consider every trajectory possible, and somehow take the average, this is the realm of Feynman's path integrals. But our classical notion of velocity must be abandoned. Classical velocity has a nice definition (it's a variable describing a trajectory at some given point in time) but take away the classical notion of trajectory and all of a sudden velocity isn't the well behaved thing you thought it once was.

But none of that means we can't define a quantum velocity, right?

No, it doesn't! We can define a quantum velocity and use it - why not? For example the rate of change of the expectation value of position might be a good definition of 'quantum velocity', you could use that... but we don't ...why? The fact remains that we can describe our quantum system fully in a way that does not need any concept of 'quantum velocity'; (whatever that is). Our system is fully described by its position probability density wave function with some additional information about some initial momentum (or, equivalently, its momentum probability density wave function with some information about initial position). If you want to find out it's speed then you could get technical and "Fourier transform" the "position wave function" into a "momentum wave function" then you know the likelihood of finding the particle with some specific momentum and you can divide by the mass and find some 'velocity probability wave function' but this is slight of hand quantum black magic. Be warned, it won't work for massless particles, it will give you no velocity operator with which to act on your so-concieved velocity probability wave function and there is no velocity commutation relation that tells you how even a theoretical measurement of velocity will effect your system. All you've done is dress up your momentum in a velocity disguise by assuming you know the mass - which is far from a safe assumption!

But velocity seems so familiar, should we just forget about it?

Answer: no

It seems that our quantum world should be governed by some universal limit on velocity and so the concept of velocity must have a place in our quantum description of reality.

And what of trajectories, do we abandon them too?

Answer: No

Even testing quantum field theory we use classical trajectories all the time. How do you measure the momentum of an electrically charged quantum system? Usually, pass it through a magnetic field and measure its position a bunch of times with sensors (as at the LHC), reconstruct its curved trajectory (!!!) and calculate it's momentum from the radius of curvature.

The world is not a simple one to understand when gazed at through the eyes of modern theory. You question is a good one because, if you think about hard enough, it raises more questions... questions worth considering...

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In quantum mechanics, we strictly talk about the "velocity" of the expectation value of x, which is not the same thing as the velocity of the particle. Nothing we have seen so far in quantum mechanics which would allow us to calculate the velocity of a particle, in fact it's not even clear what velocity means in quantum mechanics. If the particle doesn't have a determinate position (prior to measurement), neither does it have a well-defined velocity. All we could reasonably ask for is the probability of getting a particular value.

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Yes there is. It is $\vec p/m$, which is equal to $d\vec x/dt = - i[H, \vec x] $ for a sound theory.

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There are momenta in quantum theory, but they appeare as operators...so momentum operator acts on some quantim state and produces momentum eigenvalue, that is, a number. Knowing that classical momentum goes something like p=mv, we could say that after promoting small p (dynamic var) into capital P (operator) we still can say, ok, whatever eigenvalue P produces, lets just devide it with the mass and we have velocity operator because, classicaly, v=p/m, so V (operator) = P (operator) / m . There you have it. Different way and different velocity is connected to the quantum probability current density.

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