Veliocity in QM

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  • #1
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How do we define veliocity of a particle in QM?
How it's related to the momentum.
Are there any ways to cimpute quickly mean value of veliocity.
 

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  • #2
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one way is still with the same old fashion [tex] \bf{u} [/tex] :)

however, another way is with the group velocity of a wave

[tex]v_{g}=\frac{\partial\omega}{\partial k}[/tex]

and this way can also be thought of as

[tex]v_{g}=\frac{\partial\omega}{\partial k}=\frac{\partial E}{\partial p}[/tex]

with [tex] E [/tex] and [tex] p [/tex] being the particle's energy and momentum respectively.

for more information on this go to:

http://en.wikipedia.org/wiki/Group_velocity" [Broken]
 
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  • #3
Demystifier
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The velocity operator is the momentum operator divided by mass.
 
  • #4
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Hi.

The velocity operator is the momentum operator divided by mass.
Yes, you may define velocity as this. However in case you think of Zitterbewegung motion, velocity has nothing to do with momentum and its magnitude is c, velocity of light.

Regards.
 
  • #5
K^2
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however, another way is with the group velocity of a wave

[tex]v_{g}=\frac{\partial\omega}{\partial k}[/tex]

and this way can also be thought of as

[tex]v_{g}=\frac{\partial\omega}{\partial k}=\frac{\partial E}{\partial p}[/tex]

with [tex] E [/tex] and [tex] p [/tex] being the particle's energy and momentum respectively.
That is the proper way of defining it. Keep in mind that for any momentum-independent potential:

[tex]\frac{\partial E}{\partial p} = \frac{\partial H}{\partial p} = \frac{p}{m}[/tex]

Which is the same thing that Demystifier suggested.

In general, don't forget that Hamiltonian mechanics still works.

[tex]\dot{q_i} = \frac{\partial H}{\partial p_i}[/tex]
 
  • #6
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Thanks for answers. I'm wonder what's the proper definition in case of magnetic field?
 
  • #7
reilly
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Velocity is just the time derivative of the coordinate vector. If X is the x coordinate, then

V(X) = dX/dt = -i [X,H], the Heisenberg Equation of motion for the x position coordinate. With a conserved NR system, V=P/m, where P is the momentum operator, and m the mass.

The Dirac eq. is not so easy for V; H is linear in momentum, so up to constant factors,V = GAMMA x, the Dirac matrix that multiplies Px in the Hamiltonian.

V/dt=dD/dt is non zero, V does not commute with the free Hamiltonian.

This shows one of the substantial difference between relativistic and non-relativistic QM; the interaction between spatial coords and spin. Dirac in his book QM, gives a good explanation of velocity for the Dirac E.

Regards,
Reilly Atkinson
 

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