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Angular momentum Operators and Commutation

  1. Apr 21, 2014 #1
    So I understand the commutation laws etc, but one thing I can't get my head around is the fact that L^2 commutes with Lx,y,z but L does not.

    I mean if you found L^2 couldn't you just take the square root of it and hence know the total angular momentum. It seems completely ridiculous that you could know the square of the total angular momentum is 100, but not know that the angular momentum is therefore 10. Either that, or the textbook explains it horrifically and you can know the total angular momentum squared which is a scalar and hence you can know L scalar but L generally has direction, so they use L^2 to be explicit that it's a scalar.

    Please elaborate. Cheers.
     
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  3. Apr 21, 2014 #2
    You've pretty much got it. The scalar [itex]L = \sqrt{\mathbf{L}^2}[/itex] commutes with everything [itex]\mathbf{L}^2[/itex] does. The vector [itex]\mathbf{L}[/itex] does not.
     
  4. Apr 21, 2014 #3
    Oh okay, then why the hell didn't they just put that in the book, he didn't explain it at all, just said L^2 does but L doesn't. Extremely ambiguous. Stupid Griffiths.
     
  5. Apr 21, 2014 #4

    WannabeNewton

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    You do realize that ##\hat{L}## is an operator right...? Do you understand how "square roots" of operators actually work? No one talks about the "square root" of ##\hat{L}^2## they talk about ##\hat{L}^2## and ##\hat{L}## alone, including Griffiths. Here ##\hat{L} = (\hat{L}_x,\hat{L}_y,\hat{L}_z)## so clearly the different components do not commute with one another.

    As an aside, you calling him "stupid" is extremely unwarranted given that you do not yet understand basic operator theory.
     
  6. Apr 21, 2014 #5
    Good point, but yes I do understand that, but he clearly said, "meaning we can know the square of the angular momentum but not the angular momentum itself". So i'm just thinking well if you can know both the angular momentum squared and a component at the same time, you could square root the square and get angular momentum. Then that must mean given they are both able to be observed simultaneously, that the L operator for total angular momentum and Lz a component must commute.
     
  7. Apr 21, 2014 #6
    I don't know what you're talking about wbn, but clearly you can multiply operators together and get ##\hat{L}^n## and that commutes with all ##\hat{L}^k##.
     
    Last edited by a moderator: Apr 21, 2014
  8. Apr 21, 2014 #7
    Btw, Jimmy page Madison Square Garden nice.
     
  9. Apr 21, 2014 #8
    Yeah the Linear algebra is really not covered enough in this textbook. I mean i've done linear algebra and was decent at it, but it would be nice to have the explanations there to refresh on. I feel like I need to re-read my real analysis and linear algebra textbooks just so I can piece together everything.
     
  10. Apr 21, 2014 #9
    Btw WannabeNation, me calling him stupid was a joke, clearly the man is not stupid he is a well regarded physics author and his EM book is hailed as potentially the best ever. I was just making a joke so you would know which textbook I am using, you do not need to rub my ignorance to operator theory in my face, I'm an undergraduate student and I'm still learning.
     
  11. Apr 21, 2014 #10

    micromass

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    You do realize that ##\mathbf{L}^2## is not the square of ##\mathbf{L}##, right?
     
    Last edited: Apr 21, 2014
  12. Apr 21, 2014 #11

    Matterwave

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    One should realize that:

    $$\sqrt{L^2}=\sqrt{L_x^2+L_y^2+L_z^2} \neq \bf{L}=L_x+L_y+L_z$$

    Taken as an operator, it is not even immediately clear what ##\sqrt{L^2}## even means. One can take the Taylor expansion of the square root function, to try to define this operator, but as the series is not convergent everywhere, one runs into problems immediately.
     
  13. Apr 21, 2014 #12

    micromass

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    The spectral theorem for unbounded operators allows you to define this square root.
     
  14. Apr 21, 2014 #13
    Ahaha thanks for all the replies. Slightly two sided answer. Yes, I know you cannot just say L operator is the square of L^2 operator. But what i'm saying is do L operator and Lz operator commute, if that makes it easier. Can I know the magnitude of the total angular momentum and one component at the same time in other words, given I can know the square of the total angular momentum and a component at the same time, given that this is the definition of commuting operators, they share eigenfunctions and you can know both.
     
  15. Apr 21, 2014 #14

    Matterwave

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    Do you know what it's actually defined as then? Would this operator have the same eigenfunctions as L^2, with eigenvalue equal to sqrt(l(l+1))?
     
  16. Apr 21, 2014 #15
    Yes that's exactly what i'm asking, would the operator L commute with L^2, and would it have the eigenvalue sqrt(h^2l(l+1)
     
  17. Apr 21, 2014 #16

    strangerep

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    (Sigh.) Of course Micromass knows. :devil:

    Did you and cooev769 actually make an effort to look up the "spectral theorem", and see how it answers the question in the 2nd part of your post?

    Hint: how does one represent a self-adjoint operator on Hilbert space in terms of its eigenvectors?
     
    Last edited: Apr 21, 2014
  18. Apr 21, 2014 #17
    I don't have the maths knowledge to even understand what you're saying let alone figure it out for myself. Hence why I've asked here. I just want a yes or no answer, not to go and learn some linear algebra which we don't do in our physics degree's here and aren't expected to understand. I don't have the maths ability to tie the two together.
     
  19. Apr 21, 2014 #18

    Matterwave

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    I asked as a point of curiosity since I have not seen, in my QM studies, an operator L=sqrt(L^2) defined. Unfortunately, I don't have time right now to study the spectral theorem and come up with a definition. Well, I can try. On a finite dimensional space, I know that a self-adjoint operator has a set of orthonormal eigenfunctions which span the space, thus given:

    $$\hat{A}\left|\psi_n\right>=A_n\left|\psi_n\right>$$

    For n=1,...,d where d is the dimensionality of the finite dimensional space. We can construct the operator as:

    $$\hat{A}=\sum_{n=1}^d A_n\left|\psi_n\right>\left<\psi_n\right|$$

    For a positive integer p, we have that A commutes with A^p (since A commutes with any function of itself), and therefore:

    $$\hat{A}^p=A_n^p\left|\psi_n\right>$$

    So that:

    $$\hat{A}^p=\sum_n^d A_n^p \left|\psi_n\right>\left<\psi_n\right|$$

    I suppose we can define this formula to work for real p as well? If we make such a definition, then:

    $$\hat{A}^{1/2}=\sum_n^d A_n^{1/2}\left|\psi_n\right>\left<\psi_n\right|$$

    It seems that in order for this operator to be self-adjoint as well, we must demand:

    $$A_n^{1/2}=(A_n^{1/2})^*$$

    Which would be true if all ##A_n\geq 0##. Is this generally how a square root of an operator is defined? Does this extend to the infinite dimensional case without problem? In that case, a direct substitution ##\hat{A}\rightarrow\hat{L^2}## would lead to:

    $$\hat{L}\equiv\sqrt{\hat{L}^2}=\sum_{l,m}^\infty \sqrt{\hbar^2 l(l+1)}\left|l,m\right>\left<l,m\right| $$

    Which would have indeed the same eigenfunctions, with eigenvalue sqrt(l(l+1)) by construction. Now, my question would be, are there no issues with convergences and the like when moving from the finite dimensional case to the infinite dimensional case?
     
    Last edited: Apr 21, 2014
  20. Apr 22, 2014 #19

    strangerep

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    It depends on the details of the operator. This opens a large can of worms.

    Yes, convergence issues are critical -- one must specify the topology in which one is working. But I won't give you a course in General Topology here.

    For compact s.a. operators on (inf-dim) Hilbert space (in which one usually assumes the standard norm topology when discussing convergence issues), one finds a set of eigenvectors that constitute an orthormal basis. Further, the sequence of eigenvalues tends to 0.

    For noncompact bounded s.a. operators on (inf-dim) Hilbert space, things are trickier. Basically, one constructs an integral measure over the space of eigenvalues, which generalizes the formulas you wrote.

    For unbounded operators, things are trickier still, since one must specify domains carefully. (Such operators are not well-defined everywhere on the Hilbert space.) That's why some authors generalize to use a so-called "Rigged Hilbert Space" (kinda like a generalization of the Schwarz theory of distributions). Within that framework, there's another theorem, the so-called "nuclear spectral theorem" that establishes the existence of an integral measure that allows one to do QM.

    Ballentine gives a brief, but useful, introduction to the notions (and motivation for) Rigged Hilbert Space. (Every serious student of QM should have a copy of Ballentine...)
     
  21. Apr 22, 2014 #20

    strangerep

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    Tbh, I'm not really interested in teaching a parrot.

    But... more constructively...

    You aren't expected to understand any linear algebra for your physics degree??? Sorry, but I think that's rubbish. Of course a physics student needs to understand some linear algebra, just as they need to understand some calculus. There is no hope of understanding QM properly if you don't remedy your lack of knowledge of linear algebra, and soon.

    Axler's "Linear Algebra Done Right" might be a good starting, or maybe "Linear Algebra" in the Schaum Outline series. I see Schaum now even has a "Linear Algebra -- Crash Course" volume.


    Anyway, Matterwave has given a sketch of how one represents operators in terms of their eigenvectors.
     
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