Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quantum Mechanics: Three Spin Observables

  1. Feb 16, 2015 #1
    When trying to solve ##\mathbb{S}^2 =\hbar^2s(s+1)\mathbb{I},##
    I got that ##\mathbb{S}^2 = \mathbb{S}^2 _x+\mathbb{S}^2_y+\mathbb{S}^2_z = \frac{3\hbar^2}{4}
    \left[\begin{array}{ c c }1 & 0\\0 & 1\end{array} \right] = \frac{3\hbar^2}{4}\mathbb{I},## but how does ##\frac{3\hbar^2}{4} = \hbar^2s(s+1)?##
     
  2. jcsd
  3. Feb 16, 2015 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    I don't understand, It's a matrix equality => 4 equalities.
     
  4. Feb 16, 2015 #3
    I don't understand your comment?
     
  5. Feb 16, 2015 #4

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Well, S^2 is equal to a 2x2 matrix in 2 different ways. One is through the eigenvalue equation, and the other is through the sum of the squares of the 3 Pauli matrices. So you can equal these 2 matrices and obtain 4 equalities *one for each matrix element*.
     
  6. Feb 16, 2015 #5
    Hm, so how does ##\frac{3\hbar^2}{4} = \hbar^2s(s+1)?##
     
  7. Feb 16, 2015 #6

    Strilanc

    User Avatar
    Science Advisor

    Can you explain what ##\mathbb{S}## and ##s## are supposed to be? Is ##\mathbb{S}## required to be unitary? Hermitian? A linear combination of Pauli matrices? Is ##s## a scaling factor? Real? Complex? Known ahead of time?

    If ##\mathbb{S}## is a linear combination of Pauli matrices like ##\hat{v} \cdot \vec{\sigma}##, where ##\hat{v}## is a unit vector, then it is unitary and Hermitian and thus its own inverse. So ##\mathbb{S}^2## will be just ##\mathbb{I}##, meaning ##\hbar^2 s(s+1)## must equal 1, and all that's left is a quadratic equation.
     
  8. Feb 16, 2015 #7
    ##s## is the eigenvalues and ##\mathbb{S}## is Hermitian.
     
  9. Feb 17, 2015 #8

    Strilanc

    User Avatar
    Science Advisor

    In that case I think it reduces to ##\left( s \hat{v} \cdot \vec{\sigma} \right)^2 = \hbar^2 s(s+1) \mathbb{I}##, where ##\hat{v}## is an arbitrary unit vector and ##s## is a root of ##s^2 (1 - \hbar^2) - s \hbar^2 = 0##.
     
  10. Feb 17, 2015 #9

    samalkhaiat

    User Avatar
    Science Advisor

    Because [itex]s = \frac{1}{2}[/itex]:
    1) Don't you have [itex]\frac{3}{4} \hbar^{2} = \frac{1}{2} ( \frac{1}{2} + 1 ) \hbar^{2} , \ \ \Rightarrow s = \frac{1}{2}[/itex] ?
    2) For what [itex]s > 0[/itex], do you have [itex]s^{2} + s - \frac{3}{4} = ( s - \frac{1}{2} ) ( s + \frac{3}{2} ) = 0[/itex] ?
     
  11. Feb 17, 2015 #10
    I see, thank you very much guys!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Quantum Mechanics: Three Spin Observables
  1. Spin observables (Replies: 1)

Loading...