Is Velocity in Quantum Mechanics Meaningless?

[hokhani]

Why velocity in quantum mechanics is meaningless while we can always put v=p/m where p is momentum?

 

[Simon Bridge]

Well then the momentum is meaningless by the same token that velocity is.

But you need to give us a context. Velocity does have meaning in QM, you just have to be careful about what you are talking about. Where did you hear/read that "velocity is meaningless"?

 

[hokhani]

Thanks, my purpose was also to understand the reality. I haven't so far seen that statement so I am nor confident about it. but I heard it in our solid state class from our professor.

 

[Simon Bridge]

You will have to ask your professor than, I have no idea what he was talking about because I wasn't there.

In QM a particle can be prepared in a state with a well defined velocity.
A wavepacket may have a group velocity.
The classical equations of motion hold for the associated QM expectation values.
In solid state, we can talk about the drift velocity of charges.
... and so on.

OTOH: the behavior of particles in a solid is usually handled statistically so properties for individual particles lose their meaning.

Also: It's usually more useful to use momentum directly: it has a handy relationship with position for instance. It also means we don't have to worry about whether the particle has mass or not.

So take your pick... could be any or none of these.

 

[hokhani]

As far as I remember our professor told: Since the velocity in quantum mechanics is meaningless and also we are using velocity to analyse behavior of electrons in solids, our viewpoint is semi-classical.
Could you please tell me what makes semi-classical approach different from quantum approach?

 

[Astrum]

I think what your professor may be saying, is that in QM particles don't have well defined position before measurement, so they can't have well defined velocity in the classical sense. The only thing we can say is the expectation value of these quantities, such as:

$$\left \langle v \right \rangle = \frac{d \left \langle x \right \rangle}{dt}=\frac{d}{dt}\int ^{\infty} _{- \infty} x \left|\Psi (x,t)\right| ^2 dx$$

I'll stress that this is not the velocity of the particle, but rather the expectation value.

 

[Simon Bridge]

I'm with Astrum on this - it sounds like your prof was still in classical mode, and trying to say that the classical velocity of individual particles is not a useful concept for quantum systems.
So the argument he was posing, in using classical ideas about velocity, mixed with quantum mechanical statistics, is neither classical nor quantum. And then he gave you a name.

Semi-classical is where quantum statistics get used to modify classical mechanics to make an approximation to the pure quantum case with more intuitive maths.

Usually there is a sense that half the system is quantized somehow ... i.e. a semi-classical treatment of the photoelectric effect would keep light as a continuous EM wave while quantizing the electron energies.

The pure quantum approach would be solving the Schrödinger equation at some stage.
You didn't do that right?

For more detail, you can look it up.
http://en.wikipedia.org/wiki/Semiclassical_physics
http://link.springer.com/chapter/10.1007/978-3-540-70626-7_197#page-1