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What is the operator for velocity in spherical coordinates?

  1. Jun 17, 2008 #1
    *Before you say anything, this isn't homework. I'm not in school. This is just an independent study.

    Here's the problem statement:
    Calculate the velocity of an electron in the n = 1 state of a hydrogen atom.

    I know the wavefunction and I know HOW to set up and solve the problem, but I don't know what to write for the velocity operator. In other words: ∫ψ(?)ψdτ.

    Thanks
     
  2. jcsd
  3. Jun 17, 2008 #2
    What makes you think that there is such a thing as a velocity operator? There may be such a thing but I've never seen it in any text on quantum mechancis. If there was one then perhaps its equal to the momentum operator divided by the particle's mass. To express this in spherical coordinates one merely expresses the velocity in Cartesian coordinates and does a coordinate transformation ... I think. :rolleyes: Then replace the physical values by their associated observables (which are operators). Let me get back to you on this.

    Pete
     
  4. Jun 17, 2008 #3
    Oh, ok. I guess I'm going about the problem incorrectly. Do you have any ideas?
     
  5. Jun 17, 2008 #4
    Of course there's a velocity operator... Sheesh! =) It is simply the time derivative of the position operator, which you can get from Heisenberg's equation of motion (in the Heisenberg picture)
    [tex]\vec{v} = \frac{1}{i\hbar} \left[ \vec{x}, H \right][/tex].
    For most problems in question, this'll be [itex]\frac{\vec{p}}{m}[/itex].

    I have a sneaking suspicion that since [tex]\ell = 0[/tex], you're going to find zero in any case. Maybe you'd like to do it in the [tex]n=2, \ell = 1, m_\ell = 1[/tex] state instead. Just a word of caution, what you're calculating is the expectation value of velocity, not velocity itself, since these states aren't eigenstates of the velocity (or momentum) operator. Now, if you use the expression for the gradient [tex]\vec\nabla[/tex] in Spherical coordinates, then using [tex]\vec{p} \to -i \hbar \vec\nabla[/tex], you'll find all you need.
     
  6. Jun 17, 2008 #5
    Ahem! "Sheesh"? I hope you don't think I'md umb because I never heard of it before? :tongue:
    Thanks. Very intersting. May I ask where you found/learned/read this definition? I'd like to read a QM text that discusses this operator (Please note that I'm not saying that you're wrong. I'm merely interested in reading more about it). If this operator is different than the one I mentioned then that would seem to be a problem. What do you think in this regard?
    What is [itex]\ell[/itex]? The only time I've seen that used is in regards to angular momentum. However the velocity operator in spherical coordinates should be a different operator than the angular momentum operator. It shouldn't even be proportional.
    Best wishes

    Pete
     
  7. Jun 17, 2008 #6
    No offense intended. The velocity operator is simply [itex]\frac{d\vec{x}}{dt}[/itex]. It is hardly mentioned anywhere. One place that it does surface is when coupling to an electromagnetic field is introduced. There the canonical momentum of the particle is not simply mass times velocity, and most QM books make minor mention of this. I can dig up a ref if you're reeeeallly interested but there isn't much more than what I've said.

    [tex]\ell[/tex] is indeed the angular momentum quantum number. I realize that velocity and angular momentum aren't proportionally related, but the [tex]\ell = 0[/tex] wavefunction is spherically symmetric. Since there is no preferred direction, there is nowhere for [tex]\vec{v}[/tex] for which to point =)
     
  8. Jun 17, 2008 #7
    Yeah. I didn't think so. I just wanted to make sure. Thanks for letting me know. I appreciate it.
    I would greatly appreciate it. I'll owe you one! The reason I want to learn more about it is because I've seen people mention this before and I always was lacking in an understanding of what it is. The reason being that I can't imagine an experiment where the velocity is measured. Especially for photons. Thank you very much for doing this for me. Remember that I owe you one! :smile:

    Best wishes

    Pete
     
    Last edited: Jun 17, 2008
  9. Jun 18, 2008 #8

    Haelfix

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    There is a bit of an ambiguity with a velocity operator, in the sense that you have several candidates. If you take the above definition, it doesn't generalize at all to field theory for instance. Its even worse in curved spacetimes.

    Meanwhile the naive operator (divide by M, where M is suitably normalized) seems to work better, even though its properties are manifestly weird w.r.t to nonrelativistic qm.

    In general textbooks avoid the subject entirely, b/c its controversial and unnecessary. Its trivial once you have an other good observable to translate the system into a language that gives unambigous velocity and position numbers when properly phrased.
     
  10. Jun 18, 2008 #9
    I've been bothered by this since you posted it but have been unable until now to express what it was that was bothering me until this morning. What does it mean to take the time derivative of an operator? I tried to look this up in the QM texts I have and none of them (including Cohen-Tanoudji et al, Sakuari and Liboff) define such an operation. Can it be shown that it commutes with the momentum operator. If not then that would be a problem. It would seem that a neccesary condition to take the time derivative of an operator would be that it must be an explicit function of time. Perhaps you could give me a solid example such as finding the result of the velocity operator acting on a wave packet which has a Gaussian distribution.

    Can you post the definition of the derivative of an operator. Thanks.

    Pete
     
    Last edited: Jun 18, 2008
  11. Jun 18, 2008 #10

    malawi_glenn

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    pmb_phy:

    The time derivative of an operator is basic.. e.g look at an derivation of the Heisenberg equation of motion in Sakurai chapter 2.
     
  12. Jun 18, 2008 #11
    Thank you! That chapter seems only to define the derivative in the Heisenberg picture. Are you saying that it can't be defined in the Schrodinger picture?

    Pete
     
  13. Jun 18, 2008 #12

    malawi_glenn

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    That is the definition of Heisenberg vs. Schrodinger picture, in heisenberg the time evolution is on the operators, while in the schrodinger picture, the time evolutions is with the state.
     
  14. Jun 18, 2008 #13
    You don't need to be in the Heisenberg picture to calculate time derivatives of operators. You can also compute them in Shcrödigner picture, and get zeros, like [tex]\frac{d\hat{x}}{dt} = 0.[/tex] If you want, you can use some other picture, and get more different time derivatives. The time derivatives of operators naturally depend on the used picture.

    The reason why you choose to use the Heisenberg picture is probably, that you get the desired [tex]\hat{p}/m[/tex] that way. But if you already know in advance that that is what you want, why not just postulate [tex]\hat{v} := \hat{p}/m[/tex]? If you did not know in advance that this is what you want, how do you then conclude that the Heisenberg picture is the "right" picture for calculating time derivatives for operators?
     
  15. Jun 18, 2008 #14
    Thanks. So in the schrodinger picture one does not use the time derivative of the position operator but probably uses the momentum operator times 1/m?

    Pete
     
  16. Jun 21, 2008 #15

    Gokul43201

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    To bring this all back on topic, the question in the OP is most likely looking for the square root of the expectation value of v^2.
     
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