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I About electron spin

  1. Mar 9, 2016 #1

    BiGyElLoWhAt

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    This has always bugged me, but it appears that the answer is out there via this:
    "But deep theoretical reasons having to do with the rotational symmetry of nature lead to the existence of spins for elementary objects and to their quantization."

    The sentence before says this:
    "A simple answer might be, perhaps they are composite, too."
    Which was always what I had assumed, simply because it seems to make sense (not just for explaining spin).

    Can someone link me to these "deep theoretical reasons"? I may not understand them for a while, but at least I'll know what questions I need to be asking, and that's a pretty good start in my book.

    Thanks.
     
  2. jcsd
  3. Mar 9, 2016 #2

    BiGyElLoWhAt

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    Wow, I suppose I should ask the question...
    What is the origin of spin in fundamental particles? I can understand the spin of composites no problem, but for elementary particles, i.e. the electron, I don't see what would be the reason for it to have a magnetic moment (related to the spin). Links to either or, so long as it's not a circular explanation (i.e. electrons have a magnetic moment because they have spin), would be appreciated.
     
  4. Mar 9, 2016 #3
    You might look into Noether's Theorem.
    You might also give Leon Lederman's book about symmetry a try.
    https://www.amazon.com/Symmetry-Beautiful-Universe-Leon-Lederman/dp/1591025753
     
  5. Mar 9, 2016 #4

    BiGyElLoWhAt

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    Thanks, I'll give it a look.
     
  6. Mar 9, 2016 #5

    BiGyElLoWhAt

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    Is there more to Noether's theorem than conservation derived from a symmetry? It seems the final statement of the theorem arrived at is ##\partial_{\mu}A^{\mu} = 0##
     
  7. Mar 9, 2016 #6

    BiGyElLoWhAt

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  8. Mar 9, 2016 #7

    BiGyElLoWhAt

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    I'm not sure how to interpret this. I would assume bad notation, but I'm not sure.
    "The generators of rotations obey the commutation relations:
    ##[J^i,J^j] = i\epsilon^{ijk}J^k##
    So I'm pretty sure that the i, j, and k in the superscripts are indices, however it seems like perhaps the i before epsilon is the imaginary number? What is epsilon? Also, if this is the case, wouldn't the result of the left hand side be something to the effect of ##iA^{ijkk}##? I must be missing something with this. I could see if it were ##J_k##, how this could work out, so is this the inverse spin in k (drawing analogy from metrics)? If not, what does a double index represent? Or am I just completely off?

    Alright, so bear with me, please.
    https://en.wikipedia.org/wiki/Rotation_operator_(quantum_mechanics)#The_translation_operator
    I don't understand why they're doing what they're doing.
    The first line is ##T(0) = I## That's fine, I can follow all the way up through the taylor series, but then when they define p_x, they define it as a product of ##\frac{dT(0)}{da}## where a is the amount translated. Isn't this quantity zero? The rate of change of a constant ##I## with respect to an argument should be zero, from where I'm at. If that's the case, it should follow that ##T(da)=I## (I'm using I instead of 1). Then they get a differential equation, which, following the same logic, gives ##\frac{dT}{da} = 0 \therefore T(a) = constant## This is far from their answer. So ##\frac{dT(0)}{da}## must not be zero. Hmmm... Ok, maybe it's not. If you plug in p_x = 0, you get ##T(a) = 1##, which makes no physical sense. Where did I mess up? How, rather.
     
  9. Mar 9, 2016 #8

    BiGyElLoWhAt

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    Is it reasonable to integrate covectors without an argument?
     
  10. Mar 9, 2016 #9
    ## \epsilon^{ijk} ## is the Levi-Civita tensor a.k.a. the totally anti-symmetric tensor. To connect this what you have for the translation operator, ## \epsilon^{ijk} ## are the structure coefficients of the Lie Algebra group for the generators of rotation. Sorry, if I'm using to many vocab words, but I'm just trying to give you key words to search for to find out more.

    As for your question regarding the Translation operator, it is just a straightforward Taylor expansion. So, it is not the derivative of T(0) (that doesn't make any sense) but the derivative of T evaluated at 0.
     
  11. Mar 9, 2016 #10

    BiGyElLoWhAt

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    Ahh, I see. I have actually run into the Levi-Cevita connection many times in the past, but have never actually worked with it. I'm assuming they are related.

    No problem, I've seen all of these before, but haven't really delved into them very deeply.
     
  12. Mar 9, 2016 #11

    BiGyElLoWhAt

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    That still doesn't answer the question about the resultant of the RHS, however. It seems to solidify my result, which I'm not sure how to interpret.
     
  13. Mar 9, 2016 #12
    I do not see how it solidfies your argument. ## p_x ## is a Hermitian operator and can't be just set to zero.
     
  14. Mar 9, 2016 #13

    BiGyElLoWhAt

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    Sorry, not that. The product ##i\epsilon^{ijk}J^k = i\epsilon^{ijk}\frac{1}{2}\sigma^k = [J^i,J^j] = \frac{1}{2}[\sigma^i,\sigma^j]## This was the original dilemma. All the rest just came from me trying to find the answer =/
     
  15. Mar 9, 2016 #14

    BiGyElLoWhAt

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    Also, I'm not sure how this product would work, even. With the levi civita tensor being 3x3x3 and sigma k being 2x2. Is the 'i' out front shorthand for some sort of index collapse?
     
  16. Mar 9, 2016 #15

    BiGyElLoWhAt

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    I'm just trying to wrap my head around this. Hopefully it's coming across that way. I'm clearly missing something.
     
  17. Mar 9, 2016 #16
    Ok, I'll just a couple of quick examples for the commutator. First, ## \epsilon^{123}=1 ## and for all cyclic permutations of {123}, and is equal to zero if any indices are repeated. It's totally antisymmetric, meaning that swapping any two indices gives a minus sign.

    So, for instance ## [J^1,J^2]=i\epsilon^{12k}J^k## . Repeated index implies summation. But, since i,j,k cannot be equal, then it must be that ##k=0##.
    ##\implies\ [J^1,J^2]=i\epsilon^{123}J^3=iJ^3##. Just another quick example, ##[J^1,J^1]=i\epsilon^{11k}J^k=0## since ##\epsilon^{11k}=0## for all k.

    Hopefully that clears up a bit of confusion.

    EDIT: I think this was the problem. The Einstein Summation convention is typically used - meaning that wherever you see a repeated index on the same side of an equation, those are dummy indices that are actually summed over.
    The explicit equation is the ## [J^i,J^j]=i\sum_{k=1}^3\epsilon^{ijk}J^k ##
     
  18. Mar 9, 2016 #17

    BiGyElLoWhAt

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    I thought only contravariant indices that were repeate as covariant indices or vice versa implied summation. That makes sense, though.
     
  19. Mar 11, 2016 #18

    BiGyElLoWhAt

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    Hmmm... So it doesn't explicitly say that that is the case, however, it sure does imply it.
    https://en.wikipedia.org/wiki/Einstein_notation#Common_operations_in_this_notation

    What I'm curious about is if it is indeed acceptable to sum over to contravariant indices, why they would use the Levi-Civita tensor in this manner.
    Why would they use this particular arrangement of indices? I suppose I don't ever really remember reading that one has to be co and the other contravariant, but it seems that this is the first time I've seen it not that way.
     
  20. Mar 11, 2016 #19

    Mentz114

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    If the expressions are set in curved spacetime then the index position is important.
    Except for the Levi-Civita pseudo tensor, where only the index order is relevant.

    So indexes of ##\epsilon## can be adjusted to suit any arrangement of the other indexes.
     
    Last edited: Mar 11, 2016
  21. Mar 11, 2016 #20

    BiGyElLoWhAt

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    Ok. I'll have to look into that. Most of my experience with this notation is GR, so that would explain why I got that impression. Thanks.
     
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