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Finding the eigenvalues of the spin operator

  1. Nov 7, 2013 #1
    1. What are the possible eigenvalues of the spin operator [itex] \vec{S} [/itex] for a spin 1/2 particle?

    2. Relevant equations
    I think these are correct:
    [tex] \vec{S} = \frac{\hbar}{2} ( \sigma_x + \sigma_y + \sigma_z ) [/tex]
    [tex] \sigma_x = \left(\begin{array}{cc}0 & 1\\1 & 0\end{array}\right),\quad
    \sigma_y = \left(\begin{array}{cc}0 & -i\\i & 0\end{array}\right),\quad
    \sigma_z = \left(\begin{array}{cc}1 & 0\\0 & -1\end{array}\right),\quad
    [/tex]

    3. The attempt at a solution
    [tex] \text{Define } {\bf \sigma} = \sigma_x + \sigma_y + \sigma_z = \left(\begin{array}{cc}1 & 1-i\\1+i & -1\end{array}\right) [/tex]
    To find the eigenvalues, solve the characteristic polynomial:
    [tex] \det (\sigma - \lambda {\bf I}) = 0 [/tex]
    [tex] \Rightarrow \lambda = \pm \sqrt{3} [/tex]
    So that the eigenvalues of the original operator, [itex] \vec{S} [/itex] are [itex] \pm \frac{\hbar}{2} \sqrt{3} [/itex]?

    I'm not sure if I can just add the pauli matrices like that
     
  2. jcsd
  3. Nov 8, 2013 #2

    vanhees71

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    The spin operator is a vector operator. Thus for spin 1/2 you have
    [tex]\vec{S}=\frac{\hbar}{2} (\sigma_x \vec{e}_x+\sigma_{y} \vec{e}_y + \sigma_z \vec{e}_z),[/tex]
    where the [itex]\vec{e}_j[/itex] are a Cartesian basis system.

    Your should always have a "syntax checker" switched on in your mind. An equation, where you have on one side a vector and on the other side a scalar-like quantity (your expression isn't even a scalar under rotations!) that expression for sure must be a wrong statement. It doesn't even make sense!
     
  4. Nov 8, 2013 #3

    BruceW

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    I've usually just seen the spin operator as a matrix. But still, the OP's equation is not correct I think. If you want to measure the spin in the direction ##(u_x,u_y,u_z)## then you will have spin operator:
    [tex]S = \frac{\hbar}{2} ( u_x \sigma_x + u_y \sigma_y + u_z \sigma_z ) [/tex]
    And sbryant seems to have ##u_x=u_y=u_z=1## this is weird for 2 reasons: 1) this is not allowed, because it must be a unit vector. (This is why you get the incorrect square root of 3 in your answer). 2) why choose that direction? surely it would be simpler to choose something like ##(1,0,0)## as the direction to measure the spin along?
     
  5. Nov 8, 2013 #4

    vanhees71

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    Yes, and [itex]\vec{S}[/itex] is a vector with matrices (operators) as "components".

    Of course, you cannot diagonalize all vector components at the same time, because they do not commute, because they fulfill the Lie algebra of the rotation group
    [tex][s_j,s_k]=\mathrm{i} \hbar \epsilon_{jkl} s_l.[/tex]
    You can only diagonalize one component in an arbitrary direction, i.e.,
    [tex]S_u=\vec{u} \cdot \vec{S},[/tex]
    where [itex]\vec{u}[/itex] is a usual unit vector in [itex]\mathbb{R}^3[/itex].
     
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