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Can you make a definition of velocity in QM ?
I am trying to find a general definition of velocity in QM. QM is totally different world !
I am trying to find a general definition of velocity in QM. QM is totally different world !
Can you make a definition of velocity in QM ?
I am trying to find a general definition of velocity in QM. QM is totally different world !
Can you make a definition of velocity in QM ?
I am trying to find a general definition of velocity in QM. QM is totally different world !
Sorry to maybe be a little pedantic, but are those definitions satisfying? When we write
[tex]
\frac{d\hat{x}}{dt}=\frac{i}{\hbar}[\hat{x},\hat{H}]
[/tex]
we have some idea in our head that something, 'x', is changing by such and such amount in time, 't'. But what is 'x' talking about? It's easy when we talk about a whole set-up and system that we are describing and we say "that system is translating." Are we asking how fast the approximate location of a particle is changing? I think the question can be much deeper, and I'm not sure if the OP had this in mind, but I would love to see a good definition of velocity for QM. For me it seems that the best at hand is actually
[tex]
v^\mu \equiv \frac{p^\mu}{m}
[/tex]
which might seem bland and not anything new to the discussion, but when it comes down to it, in QFT calculations and the study of say, the Dirac equation, momentum is the main quantity one looks at (especially in scattering), and so it seems like a good place to at least look for a velocity. But, in general I also would like to see a solid definition, or if what has been given above is satisfactory, I feel like I haven't been given the whole deal about what it means to 'move' in QM.
The definition of velocity in QM was given above.
Your is [itex]v^i \equiv {p^i}/{m}[/itex] is not valid for a low velocity particle in presence of a magnetic field for instance. It is also wrong for a Dirac electron. The application of the quantum mechanical definition gives the well-known result
[tex]v^i = c \alpha^i[/tex]
for the velocity operator of an electron in QED.
This is new news to me! I'm not even sure what [itex]\alpha^\mu[/itex] is!? care to expand a little for me? or where can I see a reference for what you just wrote?
Okay, thanks, I'll try and read up a little more on that.
Edit: wait! you mean the gamma matrices [itex]\gamma^\mu[/itex]! Im completely familiar with those! your referencing [itex]\bar{\Psi}\gamma^\mu \Psi[/itex] as the current 4-vector with [itex]\bar{\Psi}\Psi[/itex] as the charge density correct? In which case we agree completely. I meant that in calculating scattering amplitudes (say the H diagram) the final amplitude [itex]\mathcal{M}[/itex] is expressed in terms of initial momenta and energy, and that we normally associate these with how the electron is moving. How do you feel about that statement?
This is new news to me! I'm not even sure what [itex]\alpha^\mu[/itex] is!? care to expand a little for me? or where can I see a reference for what you just wrote?
No I said alpha matrix.
The charge density is [itex]q\Psi^\dagger \Psi[/itex]
and the associated current [itex]q\Psi^\dagger c\alpha^i \Psi[/itex]
this is often rewritten as [itex]cq \bar{\Psi} \gamma^i \Psi[/itex], where [itex]\bar{\Psi} \equiv \Psi^\dagger \beta[/itex] and [itex]\gamma^i \equiv \beta \alpha^i[/itex].
[itex]\beta[/itex] is Dirac beta matrix and satisfies [itex]\beta^2 =1[/itex]
The four current is [itex]cq \bar{\Psi} \gamma^\mu \Psi[/itex] if you introduce a [itex]\gamma^0 \equiv \beta[/itex]
The definition of velocity in QM was given above.
Your is [itex]v^i \equiv {p^i}/{m}[/itex] is not valid for a low velocity particle in presence of a magnetic field for instance. It is also wrong for a Dirac electron. The application of the quantum mechanical definition gives the well-known result
[tex]v^i = c \alpha^i[/tex]
for the velocity operator of an electron in QED.
Thanks for your clarification; I agree that this formulation is correct for the relativistic case. Risking the fallacy of "appeal to authority", I would like to offer this excerpt from Bjorken and Drell's Relativistic Quantum Mechanics, p.37:
Now we come to an important difference in the relativistic theory. In the Schrödinger theory the velocity operator appearing in the current is just p/m and is a constant of motion for free particles. The current is not, however, proportional to the momentum in the Dirac theory, and whereas the Ehrenfest relation (1.27) has shown that dp/dt =0 for free-particle motion, the velocity operator cα is not constant, since [α,H]≠0. Indeed in constructing eigenfunctions of cα we have to include both positive- and negative-energy solutions, since the eigenvalues of cαi are ±c whereas |<cαi>+|<c, according to (3.29).
So, just to reiterate: for the (non-relativistic) Schrödinger equation, it is perfectly fine to use p/m for the velocity operator.
BBB