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Pauli spin matrices and hbar/2

  1. Jun 5, 2013 #1
    Why are the Pauli matrices multiplied by 1/2 ?? Why are they represented as σ1/2 σ2/2 and σ3/2 and not just σ1 σ2 σ3 ?
     
  2. jcsd
  3. Jun 5, 2013 #2
    We want the angular momentum operators to satisfy the commutation relation

    ##[\hat{J}_i, \hat{J}_j] = i \epsilon_{i j k} \hat{J}_k##

    For example, you can verify that the orbital angular momentum operators ##\hat{\vec{x}} \times \hat{\vec{p}}## satisfy this commutation relation. We want the spin angular momentum operators to satisfy the same commutation relation. This forces a certain normalization on the 2x2 matrices we use to represent the angular momentum operators for spin-1/2: the Pauli matrices only satisfy this commutation relation after we multiply them by 1/2. The Pauli matrices themselves satisfy a commutation relation with an extra factor of 2:

    ##[\sigma_i, \sigma_j] = 2 i \epsilon_{i j k} \sigma_k##.

    We remove the extra factor of two by defining the spin angular momentum operators as ##\hat{S}_i = \sigma_i / 2##. Then the ##\hat{S}_i## satisfy the correct commutation relations for angular momentum operators.
     
    Last edited: Jun 5, 2013
  4. Jun 5, 2013 #3
    Thanks Duck... so then the 1/2 is not related to fermions having spin of 1/2 ?
     
  5. Jun 5, 2013 #4
    It's because the Pauli matrices are used for spin 1/2 particles, which have intrinsic spins of either [itex]\frac{\hbar}{2}[/itex] or [itex]\frac{-\hbar}{2}[/itex].

    The reason we factor out the [itex]\frac{\hbar}{2}[/itex] is convience. It makes the eigenvalue equation of the [itex]\hat{S} = \frac{\hbar}{2}σ[/itex] operators more clear

    [itex]\hat{S}_{n}\left|n\right\rangle = \frac{\hbar}{2}\left|n\right\rangle[/itex]

    and it also makes it easier to evaluate rotation operators:

    [itex]\hat{R}(\phi\hat{n}) = e^{i\hat{S}_{n}\phi/\hbar}[/itex]

    which need to be taylor expanded to give meaning, and will involve powers of the [itex]\hat{S}[/itex] operator.
     
  6. Jun 5, 2013 #5
    Wotanub, thank you... if you don't mind, maybe you could explain what the bracket notation means? I would GREATLY appreciate it! All I can see from it is that S is an operator, and maybe n is the principal quantum number? What exactly does | signify... "over"?
     
  7. Jun 5, 2013 #6

    Nugatory

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    Staff: Mentor

    Google for "bra-ket notation".
     
  8. Jun 5, 2013 #7
    Sure. I highly recommend Townsend's "A Modern Approach to Quantum Mechanics" If you want to really get a hold of spin and bra-ket notation. It's the first chapter and he presents it in a very intuitive way.

    [itex]\left|ψ\right\rangle[/itex] (called a "ket") is an expression for a state of a system. When talking only about the intrinsic spin degree of freedom, we express the state as as [itex]\left|±n\right\rangle[/itex] (maybe I should have used a ± in my last post) where [itex]n[/itex] is the axis we are measuring the spin along (ie, x, y, z or anything in between).

    So for example a spin 1/2 particle that is "spin up" along the z-axis could be denoted [itex]\left|+z\right\rangle[/itex] and the eigenvalue equation of it with the [itex]S_{z}[/itex] operator would be (I dropped the hats because I don't like the way the look):

    [itex]S_{z}\left|+z\right\rangle = \frac{\hbar}{2}\left|+z\right\rangle[/itex]

    and similarly for a "spin down",

    [itex]S_{z}\left|-z\right\rangle = \frac{-\hbar}{2}\left|-z\right\rangle[/itex]

    since [itex]\left|+z\right\rangle[/itex] and [itex]\left|-z\right\rangle[/itex] are eigenstates of the [itex]S_{z}[/itex] operator.
     
  9. Jun 5, 2013 #8
    :biggrin: Aaaah... a light at the end of the tunnel!!! Thank you very much!!!!
     
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