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Can you make a definition of velocity in QM ?

  1. Nov 4, 2011 #1
    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 !
     
  2. jcsd
  3. Nov 4, 2011 #2

    martinbn

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    Why not the "same" as in classical mechanics. It is the time derivative of position. So the x-component of the velocity is dx/dt, which by Heisenberg's equations is some constant times the comutator [x,H].
     
  4. Nov 5, 2011 #3
    [tex]v_{QM} \equiv \frac{dx_{QM}}{dt}[/tex]
     
  5. Nov 5, 2011 #4
    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.
     
  6. Nov 6, 2011 #5
    I basically agree with jfy4: if you're studying a particle of mass m, the velocity is p/m, where p is the momentum. I think this relation is true for both the operator [itex]\hat{p}[/itex] and the eigenvalue p. Note that the commutator of position and velocity operators is not zero, so there is an uncertainty relation for x and v, but since the commutator of momentum and velocity is zero, there is no uncertainty relation of momentum and velocity. In other words, velocity acts like momentum in quantum mechanics.

    BBB
     
  7. Nov 6, 2011 #6
    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.
     
    Last edited: Nov 6, 2011
  8. Nov 6, 2011 #7
    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?
     
  9. Nov 6, 2011 #8
    It is Dirac alpha matrix. Any reference in QED uses that velocity operator for the interaction of an electron with the vector potential.
     
  10. Nov 6, 2011 #9
    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?
     
    Last edited: Nov 6, 2011
  11. Nov 7, 2011 #10
    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]
     
    Last edited: Nov 7, 2011
  12. Nov 7, 2011 #11
    Ok, then never heard of them, but they look like they fit in pretty similarly. Thanks again.
     
  13. Nov 7, 2011 #12

    dextercioby

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    The alpha and beta matrices are the original matrices used by Dirac in 1928 in his famous articles. They are part of the so-called non-covariant version of Dirac's equation. Most books dealing in one way or another with the Dirac equation from a historical perspective use them.
     
  14. Nov 7, 2011 #13
    For example, Dirac's book: Principles of quantum mechanics (in the chapter where Zitterbewegung is discussed). Or http://en.wikipedia.org/wiki/Zitterbewegung
     
  15. Nov 7, 2011 #14

    DrDu

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    m is a positive parameter in the Dirac equation. However the actual mass, i.e. the energy of a particle at rest, is -m for a negative energy solution. Hence this paradox can be avoided if we replace m by [itex]m\beta[/itex]. Then instead of the alpha matrices, the gamma matrices appear, as they should.

    Now to the problem of how to define velocity in QM: Beginners often think that it is necessary to perform two position measurements at two (infinitesimally) different times to measure velocity and then wonder how that should be possible as the velocity operator doesn't commute with the position operator. However, only the difference has to be measured which can be done without absolute measurement of position. In fact, this is what happens e.g. in the Doppler effect, which is probably the most routinely used method to measure the velocity of qm particles.
     
  16. Nov 8, 2011 #15
    Given juanrga's definition of QM velocity is correct, that is v^i = c(alpha)^i, is it true that

    <v^i> = <dx/dt> ?

    (I'm talking about averages here...).

    I'm just interested because the other day I went through the Ehrenfest Theorem which was used to show that d/dt<p_x> = -<dV/dx> (partial for V, but I suck at Latex).
     
  17. Nov 8, 2011 #16

    dextercioby

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    Averages follow from the operator equation, obviously.

    Pages 19-->26 of B. Thaller's <The Dirac equation> describe the issues with the velocity and position operators for the Dirac equation.
     
  18. Nov 8, 2011 #17
    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
     
  19. Nov 10, 2011 #18
    It depends what you mean by the non-relativistic Schrödinger equation. If you mean strictly that the equation does not contain c, then you are right. However, usually people calls non-relativistic also to the Schrödinger equation for a charged particle in an electromagnetic field {*}

    [tex]H = \frac{(p - eA/c)^2}{2m} + e\phi[/tex]

    In that case the velocity operator is [itex]\frac{p - eA/c}{m}[/itex].

    {*} They mean that the electron is being treated non-relativistically with a quadratic momenta kinetic Hamiltonian

    http://quantummechanics.ucsd.edu/ph130a/130_notes/node484.html

    Of course rigorously it cannot be non-relativistic because contains c.
     
    Last edited: Nov 10, 2011
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