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Momentum operator in curvilinear coordinates

  1. Jan 29, 2015 #1

    ShayanJ

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    This paper is about momentum operator in curvilinear coordinates. The author says that using [itex] \vec p=\frac{\hbar}{i} \vec \nabla [/itex] is wrong and this form is only limited to Cartesian coordinates. Then he tries to find expressions for momentum operator in curvilinear coordinates. He's starting point is uncertainty principle in curvilinear coordinates [itex] [q_i,p_j]=i\hbar \delta_{ij} [/itex] and it becomes obvious that by [itex] q_i [/itex], he means the coordinates themselves, e.g. [itex] r, \theta, \varphi [/itex] in spherical coordinates. But both intuition and dimensional analysis tell us that qs should be (for e.g. spherical coordinates)[itex] r, r\theta, r\sin\theta \varphi [/itex]. So I think because of this wrong starting point, the paper is going wrong all the way to the end and its initial claim is wrong. I want to know others' ideas. Any comment is welcome.
     
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  3. Jan 29, 2015 #2
    The conjugate momentum for angular variables will itself have units of angular momentum as opposed to just momentum, so the units of the commutator are fine.
     
  4. Jan 29, 2015 #3

    ShayanJ

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    But even if we take that into account, then we should note that angular variables don't represent a constant spread in length for all distances from origin. For example for a constant [itex] \Delta \varphi [/itex], if [itex] R_1 > R_2 [/itex] then [itex] R_1 \Delta \varphi>R_2 \Delta \varphi [/itex] so [itex] \Delta \varphi [/itex] alone can't represent the uncertainty in position! But by using the angle operator alone in the commutation relation, the corresponding uncertainty relation will be in terms of spread in angle only.
     
    Last edited: Jan 29, 2015
  5. Jan 29, 2015 #4

    dextercioby

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    I don't find anything in that article related to uncertainty relations of any kind.
     
  6. Jan 29, 2015 #5

    strangerep

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    +1 on that.

    @Shyan: the paper is talking about generalized commutation relations. Given these, one derives uncertainty (aka indeterminacy) relations. Cf. Ballentine section 8.4.
     
  7. Jan 29, 2015 #6

    ShayanJ

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    That's exactly what I mean. If we say e.g. [itex] [\varphi,m r^2 \dot \varphi]=i\hbar [/itex], then we'll have an uncertainty relation [itex] \Delta \varphi \Delta (mr^2 \dot\varphi) \geq \frac \hbar 2 [/itex]. What I'm saying is a particular value of [itex] \Delta\varphi [/itex] alone can't give us enough information about uncertainty in position, because such an angle uncertainty gives a larger length uncertainty the further we are from the origin. So if we have points 1 and 2 such that [itex] r_1>r_2 [/itex] and we somehow have a constant amount of [itex] \Delta\varphi [/itex], we know that the particle will have a larger amount of uncertainty in its position if its in position 1 than it had if it was in position 2 because [itex] r_1 \Delta \varphi> r_2 \Delta \varphi [/itex], so I think this should somehow be taken into account in the commutation relations so that the resulting uncertainty relations take this into account too.
    Anyway, despite my argument, the response of you science advosors makes me think I'm missing something here. So I'm ready to hear!
     
  8. Jan 30, 2015 #7

    DrDu

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    The transformation of the momentum operator to other coordinate systems has been studied intensively since the very beginning of QM and every detail of it has been made watertight by mathematical physicists. I don't understand how you can write an article on it citing only some elementary text books on QM nor why one should read this article.
     
  9. Jan 30, 2015 #8

    ShayanJ

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    I came to this paper by accident and because it seemed to me that [itex] \vec p=\frac \hbar i \vec \nabla [/itex] is well established and also I saw the mentioned problem with the paper, I came here to ask others' ideas about this paper. Actually I didn't read the paper, I just glanced at the equations.
    The main point is, the guy is either right or wrong. If he's wrong, then there should be some reason and I want to find it out because it seems to me it doesn't need a lot of time and effort so its OK. But if we can find no reason to say he's wrong, then maybe he has a point!
    Can you mention some papers or books that actually prove [itex] \vec p=\frac \hbar i \vec \nabla [/itex] is OK in general? I always thought its somehow trivial but now I think its good to see at least a hand wavy proof.
     
    Last edited: Jan 30, 2015
  10. Jan 30, 2015 #9

    DrDu

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    The paper is completely scrap. Already the starting point Eq. 4 is wrong as q is not a well defined operator in the case of angular momentum, i.e. an angle operator does not exist.
     
  11. Jan 30, 2015 #10

    Demystifier

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    And yet, you can measure the angle by measuring the position, which, in turn, is an operator. Or is it? How would you interpret the idea that in quantum mechanics you can measure something which is not described by an operator?
     
  12. Jan 30, 2015 #11

    ShayanJ

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    Now that I think, its pretty straight forward. We should just consider a wave-function in position representation written in curvilinear coordinates. Then we should translate it by an infinitesimal amount and subtract it from the original wave-function. Then an operator simply pops out which we define to be the momentum operator and this is something you can't argue with. So I wonder why the authors doubt it!
    The important point here is that, in spherical coordinates, such a translation only involves the r coordinate and in cylindrical coordinates it only involves [itex] \rho [/itex] and z coordinates. So this way it seems that we get no canonical momentums conjugate to angle coordinates and so they aren't on an equal footing with the ones conjugate to distance coordinates.
     
  13. Jan 30, 2015 #12

    DrDu

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    Ok, so let us introduce an angle operator as ##\phi=\mathrm{atan2}(x,y) ## this operator will jump from pi to -pi at when x becomes negative for negative y. We already know the correct angular momentum operator to be ##-i (x\partial/\partial y -y \partial/\partial x)##, so this will introduce a delta function into the commutation relations, which are therefore no longer of the canonical form we were assuming.
     
    Last edited: Jan 30, 2015
  14. Jan 30, 2015 #13

    ShayanJ

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    In "An introduction to quantum theory" by F.S. Levin, section 10.4, its explained that the operator [itex] \hat \Phi |\phi\rangle=\phi |\phi\rangle [/itex] is ambiguous and both [itex] \hat \Phi |\phi\rangle=\phi |\phi\rangle [/itex] and [itex] [\hat L_z,\hat \Phi]=-i\hbar [/itex] lead to inconsistencies. But he mentions that it doesn't mean an angle operator does not exist and in fact there are several possibilities among which [itex] e^{i\hat \Phi}, \cos\hat\Phi \ and \ \sin\hat\Phi [/itex] are examples. He then examines [itex] \hat\Upsilon=e^{i\hat \Phi} [/itex]. He says:
    He then calculates [itex] [\hat L_z,\hat \Upsilon]=\hbar \hat \Upsilon [/itex] and [itex] \hat \Upsilon[/itex]'s matrix elements which makes clear that it is actually a raising operator, i.e. [itex]\hat \Upsilon=C_+^{-1} \hat L_+ [/itex] with [itex] C_+=\sqrt{(l-m_l)(l+m_l+1)} [/itex].
     
  15. Jan 30, 2015 #14

    DrDu

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    Interesting, we had a similar discussion only a week ago:
    https://www.physicsforums.com/threads/follow-on-to-is-the-time-derivative-hermitian.792270/
    If we take the angle operator to be a generator of shifts of the angular momentum, we have to take into account that angular momentum is quantized. However, this good, as ##\exp (im\phi)## will only transform a continuous function into a continuous function if m is integer.
    So I think we can write some Weyl commutation relations ##UV=VU\exp(i\alpha m)## where ##U=\exp(i\alpha L_z)##, ##V=\exp(i m \phi)## with alpha real and m integer. For non-integer values of m, the Weyl commutation relations won't hold.
     
  16. Jan 30, 2015 #15

    DrDu

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    So one could define ##\Upsilon## in terms of ##L^2##, ##L_z##, and ##L_+##, although the representation is somewhat nasty.
     
  17. Jan 30, 2015 #16

    Demystifier

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    So there is angle operator, but it does not satisfy the canonical commutation relation. Is that right?
     
  18. Jan 30, 2015 #17

    ShayanJ

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    As I said in my last post, [itex] \Upsilon [/itex] is clearly unitary. So the fact that its proportional to [itex] L_+ [/itex], means that [itex] L_+ L_+^\dagger \propto I [/itex]. But when I calculate [itex] L_+ L_+^\dagger [/itex] using the definitions given here, I find out that [itex] L_+ L_+^\dagger=\hbar^2 \left\{ \frac{\partial^2}{\partial \theta^2}+i\frac{\partial}{\partial \varphi}+\cot\theta(\frac{\partial}{\partial \theta }+\cot\theta \frac{\partial^2}{\partial \varphi^2})\right\} [/itex] which doesn't seem to be proportional to identity! What's wrong here?
     
  19. Jan 30, 2015 #18

    DrDu

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    Why should it be proportional to the identity? However, it should commute with L^2 and L_z.
     
  20. Jan 30, 2015 #19

    ShayanJ

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    Because [itex] \Upsilon [/itex] is unitary, we have [itex] \Upsilon \Upsilon^\dagger=I \Rightarrow C_+^{-2} L_+ L_+^\dagger= I \Rightarrow L_+ L_+^\dagger\propto I [/itex].
     
  21. Jan 30, 2015 #20

    DrDu

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    Ok, but this ##C_+## is an operator, too, which can be expressed in terms of L^2 and L_z.
     
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