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Expectation value of spin 1/2 particles along different axes

  1. Nov 23, 2015 #1
    1. The problem statement, all variables and given/known data
    Show that for a two spin 1/2 particle system, the expectation value is [itex]\langle S_{z1} S_{n2} \rangle = -\frac{\hbar^2}{4}\cos \theta[/itex] when the system is prepared to be in the singlet state [itex]|00\rangle=\frac{1}{\sqrt{2}}\left(|\uparrow\rangle|\downarrow\rangle-|\downarrow\rangle|\uparrow\rangle\right)[/itex].

    Matrix [itex]S_{z1}[/itex] returns the z component of the spin angular momentum for particle 1. For particle 2 the matrix [itex]S_{n2}[/itex] gives the component of spin angular momentum along an axis denoted by the unit vector [itex]\hat{n}[/itex]. where [itex]\theta[/itex] is the angle between the z-axis and [itex]\hat{n}[/itex].

    2. Relevant equations

    My first guess was there must be some relation between [itex]S_{z1}[/itex] and [itex]S_{n2}[/itex] to get the required relation. I got the right answer but after thinking about it I think I got the right answer with entirely erroneous assumptions. See below.


    3. The attempt at a solution
    My first attempt was to state that a measurement of both particles' z component would return [itex]\frac{\hbar}{2}[/itex] or [itex]-\frac{\hbar}{2}[/itex]. For particle 1, I stated.

    [tex]
    S_{z1}=\frac{\hbar}{2}\begin{pmatrix}1& 0 \\0&-1\end{pmatrix}
    [/tex]

    Then I basically argued that for particle 2, the component along [itex]\hat{n}[/itex] would just be the projection of particle 2's z-component projected onto [itex]\hat{n}[/itex] and thus
    [tex]
    S_{n2}=\frac{\hbar}{2}\cos \theta \begin{pmatrix}1& 0 \\0&-1\end{pmatrix}
    [/tex]

    It gives the right answer when I do the expectation value
    [tex]
    \langle S_{z1}S_{n2} \rangle =\langle 0 0|S_{z1}S_{n2}|0 0\rangle\\
    =-\frac{\hbar^2}{4} \cos \theta\\
    [/tex]

    But I got to thinking, when you make a measurement along any axis, shouldn't it always return either [itex]\frac{\hbar}{2}[/itex] or [itex]-\frac{\hbar}{2}[/itex] and thus the matrix [itex]S_{n2}[/itex] should be the same eigenvalues as [itex]S_{z1}[/itex]? Thus
    [tex]
    S_{n2}=\frac{\hbar}{2}\begin{pmatrix}1& 0 \\0&-1\end{pmatrix}
    [/tex]

    So now I am at a total loss on how to find the correct relation. Would it instead be plausible to argue that the state vectors for particle 2, as expressed in the singlet state, have to be expanded in the eigenbasis for [itex]S_{n2}[/itex]? Then you do the do the expectation value calculation?
     
  2. jcsd
  3. Nov 23, 2015 #2

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Have you tried taking ## \vec{S}_{2n} = \hat{n} \cdot \vec{S}_2 ##?
     
  4. Nov 23, 2015 #3

    DrClaude

    User Avatar

    Staff: Mentor

    What if [itex]\hat{n} = \hat{x}[/itex]? Does that equation give you back ##S_x##?
     
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