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I Spin quantum numbers: correct names and formulae?

  1. Mar 25, 2017 #1

    N88

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    I'm having trouble finding the correct names for the 4 standard labels: n, l, m, s.
    Where might I find, free online: their correct names (ie, no colloquialisms or short cuts) and the accepted formula for each (to be sure I use them correctly)?
    Thanks.
     
  2. jcsd
  3. Mar 25, 2017 #2

    blue_leaf77

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    You can denote any quantum numbers in any system by any character - you can for instance denote the first four quantum numbers out of 4N quantum numbers for a N-electron atom by n, l, m, s. So, without specifying which system is being considered it's actually impossible to give thoughts on it.
    If by those 4 numbers, you mean the usual notation for hydrogen-like system, they are usually called "principal", "angular momentum", "magnetic", and "spin" quantum numbers respectively. If you had taken a chemistry class in highschool, I think it's hard to miss this.
    There is no explicit formula of the form ##n=\ldots##, ##l=\ldots##, ##m=\ldots##, and so on for them since they are actually eigenvalues of some operators. To find them, you need to solve the corresponding eigenvalue equation.
     
  4. Mar 26, 2017 #3

    jtbell

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    The Hyperphysics web site is usually reliable for basic stuff like this:

    http://hyperphysics.phy-astr.gsu.edu/hbase/qunoh.html

    [added] Note that Hyperphysics uses a different set of labels (##n,\ell,m_\ell,m_s##) from yours (##n,\ell,m,s##). I think perhaps physics books (at least quantum-physics books) tend to use the first set, and chemistry books the second. In the first set, one could also include ##s## but it's always the same for electrons (1/2). ##\ell## and ##m_\ell## have the same relationship to each other, with respect to orbital angular momentum, that ##s## and ##m_s## have with respect to spin angular momentum, as described in one of my posts that you've seen elsewhere.
     
    Last edited: Mar 26, 2017
  5. Mar 26, 2017 #4

    N88

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    Again, many thanks; that's exactly what I wanted! As a QM beginner, I need to be very sure about these sort of details, especially re my new favourite formula:
    spin3.gif

    My thanks again; N88
     
  6. Mar 26, 2017 #5

    blue_leaf77

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    I don't think the square root of the operator ##S^2## has a useful meaning in quantum mechanics, physicists rarely address this quantity. For one thing, given the matrix of ##S^2## there are multiple possibility to construct an operator ##A## defined to follow ##A^2 = S^2## due to the multiple choice of the sign of each diagonal element of ##A##. Second, if a ket ##|v\rangle## is an eigenvector of ##S^2##, it doesn't necessarily lead to ##|v\rangle## being an eigenvector of ##A##.
     
  7. Mar 26, 2017 #6

    N88

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    I'd be pleased if you and jtbell would discuss your point in detail.
    Reason: In my private studies (allowing the spin of a polarised spin-half particle to be ##\pm\hbar/2## in relation to a given orientation) I find the expectation of
    spin3.gif
    very useful as the average total angular momentum.

    In QM (without averaging) it is taken to be the total angular momentum http://hyperphysics.phy-astr.gsu.edu/hbase/spin.html#c1

    It looks to me that this is related to the point that you are making? Your critical comments would be very welcome.
     
  8. Mar 26, 2017 #7

    blue_leaf77

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    Looking at the hyperphysics link, it looks to me that the quantity ##L## is just defined to be the positive square root of the eigenvalue of the operator ##L^2##. That is, ##L## is just a number while ##L^2## is an operator. Since ##L## is aroused as a defined quantity, it does not necessarily have any physical meaning.
    I think you have been misled on that part. If you had calculated an expectation value, you must have the operator whose averaged measured value you calculated. I want to know which operator whose expectation value you calculated resulting in ##\sqrt{0.5(0.5+1)}\hbar##?
     
  9. Mar 26, 2017 #8

    N88

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    Thanks for this. I'm an engineer (familiar with vectors and vector-products). So I'm privately studying QM from that perspective (to see how far it takes me). That means that the operators that I work with are the detectors/polarisers (as in EPRB), operating on pristine particles and polarising them.

    So all my variables have physical meaning and I derive the RHS of the relation we are discussing; the LHS follows from s = 1/2.

    A similar idea (I have some different ideas) is to be found in the peer-reviewed http://zfn.mpdl.mpg.de/data/Reihe_A/53/ZNA-1998-53a-0637.pdf -- page 647 and eqns (74)-(75).

    EDIT: In case this helps: I arrive at ##\tfrac{\left\langle \lambda^{2}\right\rangle }{3}=1.## Since ##\lambda## is a non-negative magnitude, I write: ##\left\langle \lambda\right\rangle= \sqrt{3}##.

    Then, since ##\boldsymbol\lambda=
    \tfrac{\hbar}{2}\lambda##: ##\left\langle\boldsymbol\lambda\right\rangle= \sqrt{3}\tfrac{\hbar}{2}## equals the average of the total angular momentum under QM via the relation for total angular momentum ##\sqrt{0.5(0.5+1)}\hbar##: which is what is under discussion.

    Thanks.
     
    Last edited: Mar 27, 2017
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