- #1

- 255

- 0

How it's related to the momentum.

Are there any ways to cimpute quickly mean value of veliocity.

- Thread starter paweld
- Start date

- #1

- 255

- 0

How it's related to the momentum.

Are there any ways to cimpute quickly mean value of veliocity.

- #2

- 649

- 3

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]

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]

Last edited by a moderator:

- #3

- 10,947

- 3,658

The velocity operator is the momentum operator divided by mass.

- #4

- 1,225

- 75

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.The velocity operator is the momentum operator divided by mass.

Regards.

- #5

K^2

Science Advisor

- 2,469

- 29

That is the proper way of defining it. Keep in mind that for any momentum-independent potential: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.

[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

- 255

- 0

Thanks for answers. I'm wonder what's the proper definition in case of magnetic field?

- #7

reilly

Science Advisor

- 1,075

- 1

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